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https://mathoverflow.net/questions/135946/classes-of-non-continuous-functions-with-the-fixed-point-property	Classes of (non-continuous) functions with the fixed point property Let $K$ be a convex body in $R^d$. (Say, a ball, say a cube...) For which classes $\cal C$ of functions, every function $f \in {\cal C}$ which takes $K$ into itself admits a fixed point in $K$. motivation: Of course this holds for continuous functions so we want more general classes of functions. For $d=1$ we ask for extensions of the intermediate value function, and the first such extension known to me is a theorem of Darboux which gives the intermediate value function for differentials of continuous functions. I asked about extensions for Darboux theorem in this post, and Marton Elekes proved it for high dimensional cubes.I recalled the question again after hearing a lecture by Haim Brezis, where he described three classes of functions that satisfy the intermediate valye theorem for $d=1$. He has other results about when you can "hear" the degree of a (non-contiuous) map which seems very relevant. (See Haim's publication list for some relevant papers with Nirenberg, Bourgain, Mironescu, Nguyen, and others.) Warning: The answer may depend on $K$. Related question: Fixed point theorems - Another class comes from defining a complete lattice on K and taking a function that is monotone w/r/t this lattice. This has a fixed point as shown by the Knaster–Tarski theorem. Many functions can be obtained this way, but e.g. rotations cannot be. – domotorp Nov 10 '13 at 22:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9138495326042175, "perplexity": 477.45771175038044}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00069-ip-10-236-182-209.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/644342/why-do-we-treat-the-whole-capacitor-as-if-it-would-be-a-single-conductor	# Why do we treat the whole capacitor as if it would be a single conductor? I am a high school and I am very confused about redistribution of charges when we connect 2 capacitors, my problems are: 1. why are we treating the whole capacitor as if it would be a single conductor and say that charge will distribute equally in both plates (which doesn't make sense to me) unless the potential/better to say a potential difference of both plates becomes equal so, if that's the case let's take the scenario of what I have shown in the image. why the charge redistribution doesn't occur here? there is a potential difference between both the plates connected, so the charges should flow? 1. why is always the charge redistribution occurs symmetrically on both plates? say the potential of the right plate of below capacitor be "V/2" instead of 0 then the potential difference between both plates connected is different so the charge redistribution should be asymmetrical I,e both plates doesn't necessarily have equal and opposite charges? 2. In my textbook, for series combination it is written that the capacitors must have equal and opposite charges on both plates because if they wouldn't then there will be an electric field inside the conductor connecting both capacitors and that would redistribute the charges until they acquire equal and opposite charge because as the plates as very close the electric field due to positive plate and negative plate are almost equal and opposite at any point inside the conductor, which seems to be logical but there is a problem in it. if we go by this explanation then when two capacitors having equal and opposite charges on both plates are connected then the charge redistribution should never occur because the net electric field inside the conductors connecting the two is 0 always, isn't it? EDIT-since my people are confused with what I am thinking I am adding one more image with what I meant by 0 electric field due to equal and opposite charges on the plates, • Your question as it stands requires a heavy amount of editing as it isn't clear. Try to be concise. Label the plates and capacitors in your diagrams. Think about how a stranger reading your description would infer its meaning. This lack of clarity may be discouraging potential 'answerers'. Jun 13 '21 at 15:11 • I think ill add that there is an insulator between the two plates that prevent the flow of electrons from one plate to another. This should answer your first question(maybe?), Could you also elaborate on what you mean by why are we treating the whole capacitor as if it would be a single conductor? which property are you talking about? Jun 13 '21 at 18:01 • This question should have been closed as "needs more focus" for asking so many questions. Jun 14 '21 at 3:39 In my second statement I am asking why do always the plates of capacitor gets an equal and opposite charge no matter how we connect it in any circuit Allow me to show that this is in general (and in general I mean not in a circuit) not the case. ### The isolated capacitor The plates of an isolated capacitor can host different amounts of charge, if the net charge of the system is not zero to begin with. Using parallel plate capacitors makes it easy to see that what is equal (and opposite in sign) is the charge on the facing sides of each plate. This should actually represent portions of infinite facing planes, to the electric field lines should all be vertical They must have taught you the phenomenon of electrostatic induction. So try to think what happens when - in the distant voids of sidereal space, far from any other conductors - you bring a charged conducting plate with positive net charge Q near another identical, but neutral, plate. Let's say we just materialize it there, at a distance d, like so: The positive charge on the first plate will attract the negative charge on the nearest face of the neutral plate, leaving the opposite face positively charged. At equilibrium (we need a little bit of time for the electric field to propagate - say d/c - and the charge to rearrange - say a multiple of the relaxation time of the conductor's material), only half of the total charge Q will be present (with opposite sign) on the inner sides of the newly formed capacitor. If I understand you right, you have a problem in accepting the fact that the 'inner' charges are equal (and opposite). Well, mathematically this results from solving a system of equations that relate voltage and charge in a multi-conductor system. This might be above the level taught in high school (but you can find a lucid description in Pollack and Stump's "Electromagnetism" textbook, if you wish), so let's try to see it in a more intuitive way. Faraday Lines and Tubes of Flux Are you familiar with the concept of Faraday's lines? They should be taught in a high school course. Basically, they are a way to represent the electric field orientation and strength in space. Electric field lines emerge from positive charges and sink into negative charges. The more lines are present in a region of space, the stronger is the electric field. A better way to see this is through the concept of tubes of flux. If you have been introduced to such a concept, then you should know that 1. The charges Q1 and Q2 on the conductor's areas that are at the extremes of a tube of flux are equal in magnitude and opposite in sign. $$Q1= Q, Q2 = -Q$$ 1. The electric flux through an arbitrary cross section of the tube of flux is given by $$flux = Q/{\epsilon}_0$$ The relevant consequences are summarized in this quote from Branko Popovic's "Introductory Engineering Electromagnetism" textbook (p. 49): "The entire region in which the electric flux exists can be divided into tubes of equal flux. Each tube can then be represented by a single line of force (say, the line of force along its axis). Since the tube are supposed to be of equal flux, the charges on which they end are also equal. We can represent these charges by a single plus or minus sign. If these conventions are adopted, the magnitude of the electric field intensity is proportional to the density of the lines of force at a point, and the charge density is proportional to the density of the plus and minus signs." Now, try to imagine the field lines leaving the inner side of the first plate originating from a given fraction of charge there, they will have to end on the the inner side of the second plate on the same amount of charge there. If it can help you, imagine that each charge represented by a single plus or minus sign can only shoot out (if positive) or sink (if negative) one field line, do you see now why you need to have the same amount of charge facing in the interior of the capacitor? You can't have dangling electric field lines starting from a charge and not ending in another charge. Likewise, the field lines reaching a negative charge cannot come from an empty point in space. ### The connected capacitor Now, when you connect one plate to Earth - i.e. a reservoir of charge that can supply and balance any charge you need without changing its potential, you end up losing the extra charge on the outer sides of the plates and all you are left with is the equal and opposing charges on the inner faces. The rationale behind this is that the Earth is so big (i.e. it has a humongous self-capacitance) that whatever charge resides on its surface (if you are curios follow this link) ends up so diluted that it appears to have none on the small partial surfaces offered by everyday objects and electronic components connected to it. It just appears that earthing a conducting object drives away any excess charge on it, leaving only the charge that is electrostatically induced by the nearby non-grounded objects. When we connect a capacitor in a circuit, even if it is not grounded, there is another mechanism that ensures that there won't be excess charge on the outer sides of the plates. Batteries are inherently neutral, so when they offer a charge +Q at the plus terminal, they will have a charge -Q at the minus terminal. If the capacitor you connect is neutral to begin with, you will necessarily reach an equilibrium where positive and negative plates of the capacitor will have identical but opposite charge - charge that once equilibrium is established will end up on the internal facing plates. The plates ends up as being an extension of the battery contacts. That's why in a circuit context we say that capacitors do NOT store charge, but instead they displace it. In the following I will only consider neutral capacitors, meaning that each capacitor is 'charged' with equal and opposite charges on its facing plates and no residual charge is to be found (in the approximation of negligible fringe effects or of infinite parallel plated capacitor) on the exterior surfaces of the plates. ### Multiple connected capacitors I believe part of your confusion stems from the fact that much depends on how the connection happens: are the capacitors isolated and already charged? Are they connected in a (possibly but not necessarily earthed) circuit and then 'charged'? Let's see what happens in the parallel and series configuration when isolated, pre-charged capacitors are connected and then when the same 'uncharged' configuration is connected to a battery (with a convenient internal resistance to avoid un-physical behavior). Let's start with two isolated capacitors, each charged independently with charges Q1 for the first one, and Q2>Q1 for the second one. The caps are overall neutral, so the charge will be on the opposing internal faces of the plates. For the intensity of the electric field I am using the number of lines per unit area (the caps are flat and since we neglect fringe effects the field lines are perpendicuar to the plates' surface, ie the flux is the product of the field strenght with the area - if you want to know the voltage, integrate the field or just multiply by the distance, with the correct sign, I do not want to be bothered by these details) Parallel capacitors Now, let's see what happens when we connect them in parallel with the same polarity. The left plate of the first cap, which carried charge +Q1, and the left plate of the second cap, which carried charge +Q2 are now basically a single plate with total charge +(Q1+Q2). The charge cannot go anywhere, and all you have by connecting the plate with a piece of conductor is another conductor. So, in this sense you consider each couple of plates as a single conductor. The charge will in general redistribute in order to give a uniform electric field between the newly formed plates and zero field inside the conductor. If the caps have the same area, and we do not mess with the distance d between plates, we end up with a two-plates structure with charge Q1+Q2 on a plate of double area on one side, and the opposite charge on the other side. What would the field be in our units of lines per unit area? Correct, the average of the field of the two isolated caps. I have added two more pictures where the area of the capacitors are different (with the same overall amount of total charge Q1+Q2) to show that the field inside is the same, and so is the potential difference across the distance d. That's what you would expect from a parallel connection: both devices are subject to the same voltage which, for identical capacitors, happens to be the average of the voltages of the separated charged caps. If we connect the isolated caps with reverse polarity, we get a partial cancellation of charge on each plate (the total charge on each newly formed plate would be Q2-Q1 (if we invert the first one wrt to the previous configuration) and the field inside the newly formed cap will be considerably weakened (zero, if the separated charges were identical). In this case the total field witl be only 2 lines per unit area and the resulting voltage will be half the difference of the voltages of the separated caps. Now, when you connect the parallel to a battery you won't see anything particularly different because the bottom line of the situation is that of having identical opposite charge on the two plates of the parallel configuration. Of course, now it's the battery that imposes the voltage, so the charge will follow from that. Things can be different though when we consider a series configuration, as your instinct told you (so, here is my +1 to your question) Series capacitors Again, let's start with two isolated charged capacitor, one with charge Q1 (meaning +Q1 on one plate and -Q1 on the other) and the other with charge Q2 > Q1. We bring them together putting one after another, but still as an isolated system. When we put them in series with the same polarity we get something like this (I added vectors that represent the field generated by the sheets of charge) Note that we have different charges on the end plates: its Q1 on the leftmost plate and -Q2 on the rightmost plate. It has to be this way because the plates are isolated and they cannot change their charge (this is an ideal system without any leakages). The middle section, composed of two joined plates now is a single piece of conductor with charge Q2-Q1. In the case of Q1=Q2 the net charge on this section would be zero, but charge will still be separated due to the electrostatic induction effect of the charged exterior plates. So, in this case I would say that there is no charge redistribution, but you could see charge separation if you assembled this capacitor by bringing close to each other the two outermost charged plates and the neutral middle section. Whether or not charge will be distributed will depend on how you assemble the final configuration. Now, something different happens when we have a series capacitor connected to a battery. In this case the charge on the outermost plates is imposed by the battery connection and we can no longer have different charges on the outer plates. We start from +Q on the leftmost plate and -Q on the rightmost one, and the inner section responds by polarizing itself via electrostatic induction. fig series of two caps - connected to battery At equilibrium, the neutral section will see its charge displaced so that -Q faces the +Q on the left, while a charge +Q faces the -Q charge on the right. This will also happen with any series of capacitors: the charge is the same (sign apart) on all the plates: what changes when the capacity is different, is the electric field between plates, hence the voltage drop across each capacitor. One more thing: What happens in the isolated case when we place the series capacitors in opposition? If we place the caps in series with opposite polarity, we will have a new structure with +Q1 on the first plate, -Q1+Q2 on the central section, and +Q2 on the rightmost plate. If Q2>Q1, we have a central section that has a negative overall charge, while both exterior plates carry a positive charge. The system is overall neutral, as were the two separated capacitors before connection, but the new isolated structure will show a field like this fig isolated series of two caps - opposite polarity As you can see, the final result depends on how the constituent parts are put together, and on the connections - or more generally, interactions - with the rest of the world. If we bring a non-neutral isolated capacitor (or even a neutral real capacitor with fringe effects) near a bigger conducting body connected to the Earth - something we could call a ground plane - even if we do not connect any part of it to the body, electrostatic induction will displace charges in the bigger body producing a different configuration in the distribution of charge and the values of the potentials. When you have several conductors, it is best to approach the problem by writing a system of equations in the coefficients of potential or in the coefficients of electrostatic induction (or, as some call them, coefficients of capacity). Trying to solve such problems with intuition could easily lead to the wrong solution (and this post is no exception!). • but the statement, hat charge should be conserved in a circuit, can be fulfilled even if all the plates are not charges equally and opposite, but the sum overall charge on all capacitors is 0 Jun 16 '21 at 7:22 • also please see my added image If u wanna know Why I am saying that there is a contradiction between the 3rd statement and 1st statement Jun 16 '21 at 7:39 • @ArunBhardwaj I came here yesterday to complete my answer, but then I saw it could be closed so I gave up. But maybe you can answer this before it gets closed: do you know how to compute the electrostatic field of an infinite sheet of charge? Do you know that the field is the same in every point of each halfspace? Do you know how to apply Gauss' Law (this is just a bonus question) - ALSO: do you know why the electrostatic charge on a conductor can only be on its surface? (now, think at the first charged plate - how can the charge distribute on the two opposite faces?) Jun 16 '21 at 7:40 • yes I know , how to use gauss law and what is the electric field at any point due to infinte sheet, Jun 16 '21 at 7:42 • Ok, then tomorrow, if the question is still open I will complete my answer. For the time being I suggest you draw the electric field due to each sheet of charge in the three capacitors in my first picture and convince yourself that you need to have the charge in that way if you want to have zero field inside the conductors (just apply superposition). Jun 16 '21 at 7:44 Excellent questions for a student at your level. Here's an attempt at answering them. 1. The situation you've drawn can't occur. The potential difference is determined by the electric field, which in turn is determined by the charge distribution. In the absence of external charges, you're not free to set the potential any way you want, and then also set the charge distribution independently. In addition, if $$Q_1 \neq Q_2$$, then you have a net charge on each disconnected component of your circuit, so you'd expect that in the steady state there would be a net potential difference between these two components. 2. I'm not sure I understand this question. "the potential difference between both plates connected is different"; the potential difference between which plates and which plates? From your wording, it sounds like you're associating a potential difference to each plate, but a potential difference, as the name suggests, is a property defined between two points. 3. I assume you mean when you connect the two capacitor plates with a wire going around the outside of the capacitor. In that case, we're no longer at a steady state: the wire is able to propagate an electric field from one plate to the other. There will be a field inside the conducting plates. • In my 1st statement , I am asking that say that we have charged any capacitor with a battery of potential difference say potential at its positive terminal is 3V(reference w.r.t infinite) and potential at negative terminal is 2V and after charging we disconnect it and say we charge an another identical capacitor with a battery of potential difference V (say V at its positive terminal and 0 at its negative terminal) then we disconnect it and connect it with the capacitor we have charged before, then will the charge redistribution will occur or not? if not why not? Jun 12 '21 at 6:11 • it should occur because there is potential difference between both the plates connected via a conducting wire Jun 12 '21 at 6:12 • In my second statement I am asking why do always the plates of capacitor gets an equal and opposite charge no matter how we connect it in any circuit and also when we connect it with another capacitor like I have shown above? I am saying that if the potential of the plate was" v/2" instead of 0( reference is w.r.t infinite) and we then connect it like I have shown above then the potential difference between positive plates of both capacitor is not same as the potential difference between negative plates of both capacitors connected then why still the charge redistribution occurs symmetrically? Jun 12 '21 at 6:18 • In my third statement I am saying that in my textbook and my logic it is saying that if capacitors are connected in series charge will pass from one capacitor to the other by induction/polarization and it will occur until the net charge on each capacitor becomes 0 because then electric field at any point inside the conducting wire connecting them will be 0 because as both are very near electric field due to one plate at any point is equal and opposite to the other, ,,it seems logical and true....but there lies a contradiction in it,, Jun 12 '21 at 6:23 • continuing: if that's the case then if any two capacitors are connected then charge redistribution should never occur between them because both plates of both capacitor have equal and opposite charge in starting the field inside the conductor connecting them will be 0 then why would the charges flow? Jun 12 '21 at 6:25 Question 3: If two caps, each one's plates possessing equal and opposite charge, are connected in series, then this supposed redistribution of charges should never occur. This follows from the last point above, doesn't it? First and foremost is the answer to the third question. Your understanding that unbalanced charges within the capacitors lead to an electric field in the connecting conductor, causing charges to flow till balancing occurs - is correct and in line with your textbook's. The source of your confusion is that you have taken the explanation provided in the book about what would have happened if the charges within had been unbalanced and concluded that this is what always happens. Not at all. If you connect capacitors whose charges are already balanced within they, will stay so. The books argues in the manner of a counterfactual. In fact, it is impossible to get an ideal capacitor charged in such a way that its plates have different charges. That violates charge conservation as the total charge before, during and at the end of charging must remain same. If a cap is uncharged initially, it stays uncharged on the whole.$$^1$$ (Also see answer to Question 2.2 below) Question 1.1 Why do we treat a whole capacitor as if it is a single conductor and say that charge will distribute equally on both plates? This doesn't makes sense to me. The fact that charge redistributes equally on both plates of a parallel plate cap. doen't imply that we are treating the whole cap. as a single conductor. What makes you think that? The charges (be careful here, counterfactual alert) if were different on the two plates won't need the cap. to be a single conductor to redistribute. They would instead utilise the external circuit in which they are connected to physically redistribute themselves. Question 1.2 Unless the potential, or should I say potential difference, of both plates becomes equal. Does it? It isn't clear what you are trying to say here. First of all it would be either "potential of both plates"(relative to what?) or "potential difference between the plates" but not both in same sentence. Refrain from such careless language. Are you trying to say that charges will redistribute till the potential of the two plates of a cap becomes (their potential difference becomes $$0$$)? If so, that is not the case. After balancing of charges within a cap., it has a potential given by its defining property $$CV$$ Question 2.1 Why doesn't charge redistribution occur here? There is a potential difference between both the plates connected, so the charges should flow, shouldn't they? The labeling of your circuit diagram without any explanation is hard to interpret. Is there some external field keeping the plates at the potentials they are at? Even so how are to two conductors connected by a wire but aren't equipotential? Refering to your comment (even though you say its regarding your first statement) In my 1st statement , I am asking that say that we have charged any capacitor with a battery of potential difference say potential at its positive terminal is 3V(reference w.r.t infinite) and potential at negative terminal is 2V and after charging we disconnect it and say we charge an another identical capacitor with a battery of potential difference V (say V at its positive terminal and 0 at its negative terminal) then we disconnect it and connect it with the capacitor we have charged before, then will the charge redistribution will occur or not? if not why not? I think you simply meant two caps both charged to potential V. There shall be no charge flow in this case. (By using explicit potential values without any explanation, you made it quite difficult understand your question.) Keep in mind that after disconnecting the caps from their charger batteries, their ends have no memory of what the absolute potentials were wrt. infinity. All that matters is the difference be $$V$$. This is because potential everywhere is only defined upto addition of a constant. And so There is a potential difference between both the plates connected... isn't true as there is no potential difference between the plates on the left (or right). Question 2.2: Why does the charge redistribution occur symmetrically on both the plates of a capacitor in the circuit above? Say the potential of the plate with potential $$0$$ be changed to $$V/2$$. Now the potential difference between the bottom capacitor's plates is different than the top one's, so why is the charge redistribution not asymmetric within each cap i.e. why are each cap's plates still exhibiting equal and opposite charge? I take it you meant In my second statement I am asking why do always the plates of capacitor gets an equal and opposite charge no matter how we connect it in any circuit and also when we connect it with another capacitor like I have shown above? I am saying that if the potential of the plate was" v/2" instead of 0( reference is w.r.t infinite) and we then connect it like I have shown above then the potential difference between positive plates of both capacitor is not same as the potential difference between negative plates of both capacitors connected then why still the charge redistribution occurs symmetrically? First of all, since one cap. is charged to $$V$$ while the other to $$V/2$$ charge redistribution will occur as there is now relative potential difference between the plates on the left (or right). While you may understand why the capacitors still stay neutral on the whole by using the answer above to question 3, there is another way to look at it. There is charging/discharging current created during charge redistribution. This current is same throughout the circuit. This current deposits a charge $$Idt$$ on one plate of the capacitor and extracts the same from the other. Thus both increasing the charge while keeping net charge zero. The reverse happens at the other cap. In this way though chrges are moving across the caps. within each net neutrality is maintained. $$^1$$ Note that this argument doesn't depend on the geometry of the capacitor. Though you mayn't realize this know, the electrical properties of a capacitor that you study may study like its $$IV$$ relation, its energy content, its $$AC$$ response, its behaviour in circuits - its all independent of the geometry of the circuit. • please read to my comments below, the edit was not upto the point of what I am asking,, I am very confused in this and trying to understand just from logical way, like what is happening physically.. Jun 16 '21 at 7:20 This configuration is an example of a more general case of the behaviour of a material under an electric field. If between the plate of the left side of C1 and the plate of the right side of C2 in the fig 2.26, there was an isolant, the result would be an electric field in the region. Because it is a conductor, any electric field results in a displacement of charges. So the electric field in the region is zero. This effect is produced by the displaced charges. For E be zero in the region, the charges of the right plate of C1 and left plate of C2 must be such to generate an E-field that exactly opposes the original. So, the charges are respectively -Q and Q. The charge equalization does occur in your first picture. You've drawn an unbalanced situation that can't exist at steady-state (i.e. equilibrium). The charges between the plates will redistribute so that the net electric field between the plates is zero. This occurs even when the voltage supply is 0V. As long as there is a net electric field between the plates, the forces are unbalanced and will work to redistribute the charges until the electric field is zero. It doesn't matter whether this field is being produced by the charges themselves or by the voltage supply. So even if the voltage supply was 0V, if you had an initial charge distribution producing 3V and 2V on the plates, the charges (either positive or negative depend on which you want to look at) would still redistribute because the charges themselves are producing a net electric field between the plates that repels them away from each other. They'll keep repelling until an equal number of charges ends up on both plates so that the net electric field is zero at which points there is no more net force and the charges stop moving. If the voltage supply is not 0V, everything still behaves the same except now there is also a second electric field being applied across the plates by the voltage source which also acts on the charges. The charges still have to redistribute until the net electric field between the plates is zero, because as long as it is non-zero there is a force acting to move the charges. It doesn't matter whether this field is externally applied or whether it is from the unbalanced charges themselves. When the supply was 0V the only fields being produced anywhere in the circuit were from the charges themselves, so in order to produce a net electric field of zero between the plates the charges just had to cancel their own fields out which means an equal number of charges on both plates. When the voltage supply is non-zero, a second electric field is introduced in the circuit and produces an "offset" or asymmetry. So in order to cancel out the electric field between the plates, the number of charges now has to be different on each plate in order to compensate for the voltage supply's electric field. Q1 To begin, we are not treating "the whole capacitor as if it would be a single conductor". A capacitor consists of two separate conductors (plates) separated by an insulating medium. The upper and lower capacitors are connected in parallel. Therefore the voltage across each capacitor has to be the same (3-2=1v), not different as you have shown, and the potential on the plates connected by the wires must also be the same. If you assign a potential of zero on the lower right plate (which is theoretically a purely arbitrary decision) the potential of the upper right plate also has to be zero and the potential everywhere else is measured with respect to the assigned point of zero potential. Then the voltage on the left plate of each capacitor would have to be 1 volt for a potential difference of 1 volt. Since the charge is different on the pairs of plates, their capacitances also have to be different because $$C=\frac{Q}{V}$$ Then, if $$C_1$$ and $$C_2$$ are the capacitances of the upper and lower pairs of plates, respectively, $$C_{1}=\frac{Q_1}{V}$$ $$C_{2}=\frac{Q_2}{V}$$ Where, in this case, $$V$$ = 1 volt. The charge will not redistribute because there is no potential difference between the plates connected by the wires and the electric field of each capacitor will hold the charge in place. Q2 why is always the charge redistribution occurs symmetrically on both the plates? To understand why the charge has to be symmetric on both plates of a capacitor you need to understand how a capacitor gets "charged" in the first place. Consider the charging of a capacitor by a battery in series with a resistor (to limit the current). Assume the capacitor is not initially charged. Then before it is connected to the battery each metal plate has an equal amount of protons (positive charge) and highly mobile electrons (negative charge) so that each plate is electrically neutral and there is no voltage (potential difference) between the plates. When the capacitor is connected to a battery, the positive terminal of the battery attracts electrons off of the plate connected to it moving them to the positive terminal of the battery. This leaves a deficit of electrons on that plate making it positively charged. Simultaneously, the negative terminal of the battery supplies an equal amount of electrons to the plate connected to it giving it a surplus of electrons making the plate negatively charged. In effect, there is no change in the total charge of the capacitor. Charge has simply been moved from one plate to another. Now let's say the capacitor we just charged is your upper capacitor, but it is not yet connected in parallel to the lower capacitor. What will happen if we connect them in parallel? Well, that will depend on the capacitance of the two capacitors and whether or not the lower capacitor has been charged. Let's assume the lower capacitor is initially uncharged. Here's what we know will happen: 1. The upper capacitor will charge the lower capacitor until the potential difference for each capacitor is the same. (Generally the charging is through a resistor to limit potentially damaging high currents). 2. Because total charge has to be conserved, the sum of the charges on the two capacitors will have to equal $$Q_1$$, the initial charge of the upper capacitor. So, if $$Q_U$$ and $$Q_L$$ are the final charge on the upper and lower capacitors, respectively, $$Q_{U}+Q_{L}=Q_1$$ 3. Since the final voltage on the two capacitors has to be the same, lets call it $$V_f$$, we have the final conditions of $$Q_{U}=C_{U}V_f$$ $$Q_{L}=C_{L}V_f$$ $$Q_{U}+Q_{L}=Q1$$ And $$Q_{1}=C_{U}V$$ Where $$V=$$ 1 volt and $$C_U$$ is the capacitance of your upper capacitor. Q3 Your textbook is correct that series capacitors must always have the same charge. But a simpler explanation is as follows: Capacitors in series all have the same total current flowing through them, or $$I_{T}=I_{1}+I_{2}=I_{3}=$$etc. Therefore each capacitor will accumulate the same amount of electrical charge, $$Q$$, on its plates regardless of its capacitance. This is because the charge accumulated by a plate of any one capacitor must have come from the plate of its adjacent capacitor. Consequently, $$Q_{1}=Q_{2}=Q_{3}=$$ etc.. ..if we go by this explanation then when two capacitors having equal and opposite charges on both plates are connected together then the charge redistribution should never occur because the net electric field inside the conductors connecting the two is 0 always ,isn't it? When you say "connected together" I assume you mean taking two of those series capacitors and connecting them in parallel. If that's the case, if the two capacitors have different capacitance but the same charge, then their voltages will be different before being connected together because $$V=\frac{Q}{C}$$ In which case the capacitor having the higher voltage will charge the capacitor of lower voltage, redistributing the charge until the voltages become the same. Hope this helps. • I am not very clear with what u are saying, please read to my comments above to understand what I am actually thinking, Jun 16 '21 at 7:29 Key concepts when dealing with capacitors in a DC circuit: 1. Charge is conserved, so the battery does NOT create or destroy any electrons. 2. The battery acts as a "pump" for electrons, and moves electrons around in the circuit. This means that for one capacitor in the circuit, the battery removes electrons from the positive plate and pushes them onto the negative plate, until the potential difference across the capacitor equals the emf of the battery. It also means that the net charge on the capacitor is zero. For capacitors in series, this means that the total number of electrons pushed onto the negative plate of each capacitor (aka the charge) must be equal. 3. For capacitors connected in parallel with no battery in the circuit (your drawing in question 1.2) charge will redistribute until the the negative plates are at the same voltage (usually arbitrarily assigned a value of zero volts) and the positive plates are at the same voltage. The amount of charge on each of the parallel plates will depend on plate area, spacing between plates, the dielectric between the plates, etc.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.813612699508667, "perplexity": 326.6593975355196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00287.warc.gz"}
https://web2.0calc.com/questions/helpp-please-i-need-this-done-by-today-please	+0 0 64 1 What is the value of x? ?cm http://prntscr.com/i1bhk7 (I just added the pic - Melody) Guest Jan 16, 2018 edited by Guest Jan 16, 2018 edited by Melody Jan 16, 2018 #1 +1720 +1 $$\triangle MNP\sim\triangle MAB$$ because a segment located in the interior of the triangle is parallel to a side. You could also prove similarity by Angle-Angle Similarity Theorem. Using the above similarity statement, one can create a proportion because each side is proportional. The one I will use is $$\frac{MA}{MB}=\frac{MN}{MP}$$. This comes from the similarity statement. Although we do not know the length of $$\overline{MA}$$ directly, we can find it by subtracting the length of $$\overline{AN}$$ from the length of $$\overline{MN}$$. When we plug in these numbers, we can then solve for the unknown side length. $$\frac{MA}{MB}=\frac{MN}{MP}$$ Plug in the known information and solve for the unknown. $$\frac{46.2-14}{x}=\frac{46.2}{72.6}$$ Let's simplify the numerator of the left hand side of the equation first. $$\frac{32.2}{x}=\frac{46.2}{72.6}$$ Before proceeding, it may be wise to multiply the fractions by 10/10 so that the numbers are whole numbers. $$\frac{322}{10x}=\frac{462}{726}$$ Some simplification can occur here. The numerator and denominator of the right hand side happen to have a greatest common factor of 66. That's something you don't see every day! $$\frac{161}{5x}=\frac{7}{11}$$ Now, let's do the cross multiplication with simplified numbers. $$1771=35x$$ Divide by 35 from both sides. $$x=\frac{1771}{35}=\frac{253}{5}=50.6$$ TheXSquaredFactor Jan 17, 2018 Sort: #1 +1720 +1 $$\triangle MNP\sim\triangle MAB$$ because a segment located in the interior of the triangle is parallel to a side. You could also prove similarity by Angle-Angle Similarity Theorem. Using the above similarity statement, one can create a proportion because each side is proportional. The one I will use is $$\frac{MA}{MB}=\frac{MN}{MP}$$. This comes from the similarity statement. Although we do not know the length of $$\overline{MA}$$ directly, we can find it by subtracting the length of $$\overline{AN}$$ from the length of $$\overline{MN}$$. When we plug in these numbers, we can then solve for the unknown side length. $$\frac{MA}{MB}=\frac{MN}{MP}$$ Plug in the known information and solve for the unknown. $$\frac{46.2-14}{x}=\frac{46.2}{72.6}$$ Let's simplify the numerator of the left hand side of the equation first. $$\frac{32.2}{x}=\frac{46.2}{72.6}$$ Before proceeding, it may be wise to multiply the fractions by 10/10 so that the numbers are whole numbers. $$\frac{322}{10x}=\frac{462}{726}$$ Some simplification can occur here. The numerator and denominator of the right hand side happen to have a greatest common factor of 66. That's something you don't see every day! $$\frac{161}{5x}=\frac{7}{11}$$ Now, let's do the cross multiplication with simplified numbers. $$1771=35x$$ Divide by 35 from both sides. $$x=\frac{1771}{35}=\frac{253}{5}=50.6$$ TheXSquaredFactor Jan 17, 2018 ### 5 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9049229621887207, "perplexity": 406.1100063121692}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00521.warc.gz"}
https://hal.archives-ouvertes.fr/hal-01287143	# The comb representation of compact ultrametric spaces Abstract : We call a \emph{comb} a map $f:I\to [0,\infty)$, where $I$ is a compact interval, such that $\{f\ge \varepsilon\}$ is finite for any $\varepsilon$. A comb induces a (pseudo)-distance $d_f$ on $\{f=0\}$ defined by $d_f(s,t) = \max_{(s\wedge t, s\vee t)} f$. We describe the completion $\bar I$ of $\{f=0\}$ for this metric, which is a compact ultrametric space called \emph{comb metric space}. Conversely, we prove that any compact, ultrametric space $(U,d)$ without isolated points is isometric to a comb metric space. We show various examples of the comb representation of well-known ultrametric spaces: the Kingman coalescent, infinite sequences of a finite alphabet, the $p$-adic field and spheres of locally compact real trees. In particular, for a rooted, locally compact real tree defined from its contour process $h$, the comb isometric to the sphere of radius $T$ centered at the root can be extracted from $h$ as the depths of its excursions away from $T$. Keywords : Type de document : Pré-publication, Document de travail 2016 https://hal.archives-ouvertes.fr/hal-01287143 Contributeur : Amaury Lambert <> Soumis le : vendredi 11 mars 2016 - 18:38:34 Dernière modification le : mercredi 21 mars 2018 - 18:56:49 ### Identifiants • HAL Id : hal-01287143, version 1 • ARXIV : 1602.08246 ### Citation Amaury Lambert, Geronimo Uribe Bravo. The comb representation of compact ultrametric spaces. 2016. 〈hal-01287143〉 ### Métriques Consultations de la notice	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.931257426738739, "perplexity": 1512.922142692759}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647883.57/warc/CC-MAIN-20180322112241-20180322132241-00261.warc.gz"}
http://www.reddit.com/r/math/comments/2a5r5h/linear_order_for_sets_of_arbitary_size/	[–] 4 points5 points (0 children) It turns out that "Every set has a linear order" is independent from ZF, but strictly weaker than the Axiom of Choice. http://mathoverflow.net/questions/37272/are-all-sets-totally-ordered [–]Logic 0 points1 point (0 children) It seems that ZF+"every set can be linearly ordered" is strictly between ZF and ZFC in terms of strength. See: http://cantorsattic.info/Dedekind_finite By adding a Dedekind finite set of real numbers it is possible to show that the ultrafilter lemma holds; that every set can be linearly ordered; every set can be mapped onto ω; however the axiom of choice fails (for countable families). I can't say why choice is strictly stronger, but for the linear ordering axiom being independent of ZF: An amorphous set is an infinite set which cannot be split into two disjoint infinite pieces. Amorphous sets cannot be linearly ordered, but it is consistent with ZF that amorphous sets exist.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9816741943359375, "perplexity": 473.0452518081651}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137841.57/warc/CC-MAIN-20140914011217-00187-ip-10-234-18-248.ec2.internal.warc.gz"}
http://www.theinfolist.com/php/SummaryGet.php?FindGo=Generalizations_of_the_derivative	Generalizations of the derivative TheInfoList In mathematics, the derivative is a fundamental construction of differential calculus and admits many possible generalizations within the fields of mathematical analysis, combinatorics, algebra, and geometry. # Derivatives in analysis In real, complex, and functional analysis, derivatives are generalized to functions of several real or complex variables and functions between topological vector spaces. An important case is the functional derivative, variational derivative in the calculus of variations. Repeated application of differentiation leads to derivatives of higher order and differential operators. ## Multivariable calculus The derivative is often met for the first time as an operation on a single real function of a single real variable. One of the simplest settings for generalizations is to vector valued functions of several variables (most often the domain forms a vector space as well). This is the field of multivariable calculus. In one-variable calculus, we say that a function $f: \R \to \R$ is differentiable at a point ''x'' if the limit :$\lim_\frac$ exists. Its value is then the derivative ƒ'(''x''). A function is differentiable on an Interval (mathematics), interval if it is differentiable at every point within the interval. Since the line $L\left(z\right) = f\text{'}\left(x\right)\left(z - x\right) + f\left(x\right)$ is tangent to the original function at the point $\left(x, f\left(x\right)\right),$ the derivative can be seen as a way to find the ''best linear approximation'' of a function. If one ignores the constant term, setting $L\left(z\right) = f\text{'}\left(x\right)z$, ''L''(''z'') becomes an actual linear operator on R considered as a vector space over itself. This motivates the following generalization to functions mapping $\R^m$ to $\R^n$: ƒ is differentiable at ''x'' if there exists a linear operator ''A''(''x'') (depending on ''x'') such that :$\lim_\frac = 0.$ Although this definition is perhaps not as explicit as the above, if such an operator exists, then it is unique, and in the one-dimensional case coincides with the original definition. (In this case the derivative is represented by a 1-by-1 matrix consisting of the sole entry ''f(''x'').) Note that, in general, we concern ourselves mostly with functions being differentiable in some open neighbourhood (mathematics), neighbourhood of $x$ rather than at individual points, as not doing so tends to lead to many Pathological (mathematics), pathological counterexamples. An ''n'' by ''m'' matrix (mathematics), matrix, of the linear operator ''A''(''x'') is known as Jacobian matrix and determinant, Jacobian matrix J''x''(ƒ) of the mapping ƒ at point ''x''. Each entry of this matrix represents a partial derivative, specifying the rate of change of one range coordinate with respect to a change in a domain coordinate. Of course, the Jacobian matrix of the composition ''g°f'' is a product of corresponding Jacobian matrices: J''x''(''g°f'') =Jƒ(''x'')(''g'')J''x''(ƒ). This is a higher-dimensional statement of the chain rule. For real valued functions from R''n'' to R (scalar fields), the total derivative can be interpreted as a vector field called the gradient. An intuitive interpretation of the gradient is that it points "up": in other words, it points in the direction of fastest increase of the function. It can be used to calculate directional derivatives of Scalar (mathematics), scalar functions or normal directions. Several linear combinations of partial derivatives are especially useful in the context of differential equations defined by a vector valued function R''n'' to R''n''. The divergence gives a measure of how much "source" or "sink" near a point there is. It can be used to calculate flux by divergence theorem. The curl (mathematics), curl measures how much "rotation" a vector field has near a point. For vector-valued functions from R to R''n'' (i.e., parametric curves), one can take the derivative of each component separately. The resulting derivative is another vector valued function. This is useful, for example, if the vector-valued function is the position vector of a particle through time, then the derivative is the velocity vector of the particle through time. The convective derivative takes into account changes due to time dependence and motion through space along vector field. ## Convex analysis The subderivative and subgradient are generalizations of the derivative to convex functions. ## Higher-order derivatives and differential operators One can iterate the differentiation process, that is, apply derivatives more than once, obtaining derivatives of second and higher order. A more sophisticated idea is to combine several derivatives, possibly of different orders, in one algebraic expression, a differential operator. This is especially useful in considering ordinary linear differential equations with constant coefficients. For example, if ''f''(''x'') is a twice differentiable function of one variable, the differential equation $f'' + 2f' - 3f = 4x - 1$ may be rewritten in the form : $L\left(f\right)=4x-1,$ where $L=\frac+2\frac-3$ is a ''second order linear constant coefficient differential operator'' acting on functions of ''x''. The key idea here is that we consider a particular linear combination of zeroth, first and second order derivatives "all at once". This allows us to think of the set of solutions of this differential equation as a "generalized antiderivative" of its right hand side 4''x'' − 1, by analogy with ordinary Integral, integration, and formally write $f(x)=L^(4x-1).$ Higher derivatives can also be defined for functions of several variables, studied in multivariable calculus. In this case, instead of repeatedly applying the derivative, one repeatedly applies partial derivatives with respect to different variables. For example, the second order partial derivatives of a scalar function of ''n'' variables can be organized into an ''n'' by ''n'' matrix, the Hessian matrix. One of the subtle points is that the higher derivatives are not intrinsically defined, and depend on the choice of the coordinates in a complicated fashion (in particular, the Hessian matrix of a function is not a tensor). Nevertheless, higher derivatives have important applications to analysis of maxima and minima, local extrema of a function at its critical point (mathematics), critical points. For an advanced application of this analysis to topology of manifolds, see Morse theory. As in the case of functions of one variable, we can combine first and higher order partial derivatives to arrive at a notion of a partial differential operator. Some of these operators are so important that they have their own names: * The Laplace operator or Laplacian on R3 is a second-order partial differential operator given by the divergence of the gradient of a scalar function of three variables, or explicitly as $\Delta = \frac + \frac + \frac.$ Analogous operators can be defined for functions of any number of variables. * The d'Alembertian or wave operator is similar to the Laplacian, but acts on functions of four variables. Its definition uses the indefinite metric tensor of Minkowski space, instead of the Euclidean space, Euclidean dot product of R3: $\square = \frac + \frac + \frac - \frac\frac.$ ## Weak derivatives Given a function $u:\R^n\to\R$ which is Locally integrable function, locally integrable, but not necessarily classically differentiable, a weak derivative may be defined by means of integration by parts. First define test functions, which are infinitely differentiable and compactly supported functions $\varphi \in C^_c\left\left(\R^n\right\right)$, and Multi-index notation, multi-indices, which are length $n$ lists of integers $\alpha = \left(\alpha_1, \dots, \alpha_n\right)$ with $, \alpha, := \sum_1^n \alpha_i$. Applied to test functions, $D^\alpha \varphi := \frac$. Then the $\alpha^$ weak derivative of $u$ exists if there is a function $v:\R^n\to\R$ such that for ''all'' test functions $\varphi$, we have : $\int_ u\ D^ \!\varphi\ dx = \left(-1\right)^\int_ v\ \varphi\ dx$ If such a function exists, then $D^ u := v$, which is unique almost everywhere. This definition coincides with the classical derivative for functions $u \in C^\left\left(\R^n\right\right)$, and can be extended to a type of generalized functions called Distribution (mathematics), distributions, the dual space of test functions. Weak derivatives are particularly useful in the study of partial differential equations, and within parts of functional analysis. ## Analysis on fractals Laplacians and differential equations can be defined on analysis on fractals, fractals. ## Fractional derivatives In addition to ''n'' th derivatives for any natural number ''n'', there are various ways to define derivatives of fractional or negative orders, which are studied in fractional calculus. The −1 order derivative corresponds to the integral, whence the term differintegral. ## Complex analysis In complex analysis, the central objects of study are holomorphic functions, which are complex-valued functions on the complex numbers satisfying a Fréchet derivative, suitably extended definition of differentiability. The Schwarzian derivative describes how a complex function is approximated by a fractional-linear map, in much the same way that a normal derivative describes how a function is approximated by a linear map. The Wirtinger derivatives are a set of differential operators that permit the construction of a differential calculus for complex functions that is entirely analogous to the ordinary differential calculus for functions of real variables. ## Quaternionic analysis In quaternionic analysis, derivatives can be defined in a similar way to real and complex functions. Since the Quaternion, quaternions $\mathbb$ are not commutative, the limit of the difference quotient yields two different derivatives: A left derivative :$\lim_\left\left[ h^ \,\left\left( f\left\left( a + h \right\right) - f\left\left( a \right\right) \right\right) \right\right]$ and a right derivative :$\lim_\left\left[ \left\left( f\left\left( a + h \right\right) - f\left\left( a \right\right) \right\right) \, h^ \right\right]~.$ The existence of these limits are very restrictive conditions. For example, if $f:\mathbb \to \mathbb$ has left-derivatives at every point on an open connected set $U \subset \mathbb$, then $f\left(q\right) = a + q\,b$ for $a,\,b \in \mathbb$. ## Functional analysis In functional analysis, the functional derivative defines the derivative with respect to a function of a functional on a space of functions. This is an extension of the directional derivative to an infinite dimensional vector space. The Fréchet derivative allows the extension of the directional derivative to a general Banach space. The Gateaux derivative extends the concept to locally convex topological vector spaces. Fréchet differentiability is a strictly stronger condition than Gateaux differentiability, even in finite dimensions. Between the two extremes is the quasi-derivative. In measure theory, the Radon–Nikodym derivative generalizes the Jacobian matrix and determinant, Jacobian, used for changing variables, to measures. It expresses one measure μ in terms of another measure ν (under certain conditions). In the theory of abstract Wiener spaces, the H-derivative, ''H''-derivative defines a derivative in certain directions corresponding to the Cameron-Martin Hilbert space. On a function space, the linear operator which assigns to each function its derivative is an example of a differential operator. General differential operators include higher order derivatives. By means of the Fourier transform, pseudo-differential operators can be defined which allow for fractional calculus. ## Analogues of derivatives in fields of positive characteristic The Carlitz derivative is an operation similar to usual differentiation have been devised with the usual context of real or complex numbers changed to local fields of positive Characteristic_(algebra), characteristic in the form of formal Laurent series with coefficients in some finite field F''q'' (it is known that any local field of positive characteristic is isomorphic to a Laurent series field). Along with suitably defined analogs to the exponential function, logarithms and others the derivative can be used to develop notions of smoothness, analycity, integration, Taylor series as well as a theory of differential equations. # Difference operator, q-analogues and time scales * The q-derivative of a function is defined by the formula $D_q f(x)=\frac.$ For ''x'' nonzero, if ''f'' is a differentiable function of ''x'' then in the limit as we obtain the ordinary derivative, thus the ''q''-derivative may be viewed as its q-deformation. A large body of results from ordinary differential calculus, such as binomial formula and Taylor expansion, have natural ''q''-analogues that were discovered in the 19th century, but remained relatively obscure for a big part of the 20th century, outside of the theory of special functions. The progress of combinatorics and the discovery of quantum groups have changed the situation dramatically, and the popularity of ''q''-analogues is on the rise. * The difference operator of difference equations is another discrete analog of the standard derivative. $\Delta f(x)=f(x+1)-f(x)$ * The q-derivative, the difference operator and the standard derivative can all be viewed as the same thing on different time scale calculus, time scales. For example, taking $\varepsilon = \left(q-1\right)x$, we may have $\frac = \frac.$ The q-derivative is a special case of the Wolfgang Hahn, Hahn difference, $\frac.$The Hahn difference is not only a generalization of the q-derivative but also an extension of the forward difference. * Also note that the q-derivative is nothing but a special case of the familiar derivative. Take $z = qx$. Then we have, $\lim_\frac = \lim_\frac = \lim_\frac.$ # Derivatives in algebra In algebra, generalizations of the derivative can be obtained by imposing the product rule, Leibniz rule of differentiation in an algebraic structure, such as a ring (mathematics), ring or a Lie algebra. ## Derivations A derivation (abstract algebra), derivation is a linear map on a ring or algebra over a field, algebra which satisfies the Leibniz law (the product rule). Higher derivatives and algebraic differential equation, algebraic differential operators can also be defined. They are studied in a purely algebraic setting in differential Galois theory and the theory of D-modules, but also turn up in many other areas, where they often agree with less algebraic definitions of derivatives. For example, the differential algebra, formal derivative of a polynomial over a commutative ring ''R'' is defined by :$\left\left(a_d x^d + a_x^ + \cdots + a_1x + a_0\right\right)\text{'} = da_d x^ + \left(d - 1\right)a_x^ + \cdots + a_1.$ The mapping $f\mapsto f\text{'}$ is then a derivation on the polynomial ring ''R''[''X'']. This definition can be extended to rational functions as well. The notion of derivation applies to noncommutative as well as commutative rings, and even to non-associative algebraic structures, such as Lie algebras. See also Pincherle derivative and Arithmetic derivative. ## Commutative algebra In commutative algebra, Kähler differentials are universal derivations of a commutative ring or module (algebra), module. They can be used to define an analogue of exterior derivative from differential geometry that applies to arbitrary algebraic varieties, instead of just smooth manifolds. ## Number theory In p-adic analysis, the usual definition of derivative is not quite strong enough, and one requires strictly differentiable, strict differentiability instead. Also see arithmetic derivative and Hasse derivative. ## Type theory Many abstract data types in mathematics and computer science can be described as the universal algebra, algebra generated by a transformation that maps structures based on the type back into the type. For example, the type T of binary trees containing values of type A can be represented as the algebra generated by the transformation 1+A×T2→T. The "1" represents the construction of an empty tree, and the second term represents the construction of a tree from a value and two subtrees. The "+" indicates that a tree can be constructed either way. The derivative of such a type is the type that describes the context of a particular substructure with respect to its next outer containing structure. Put another way, it is the type representing the "difference" between the two. In the tree example, the derivative is a type that describes the information needed, given a particular subtree, to construct its parent tree. This information is a tuple that contains a binary indicator of whether the child is on the left or right, the value at the parent, and the sibling subtree. This type can be represented as 2×A×T, which looks very much like the derivative of the transformation that generated the tree type. This concept of a derivative of a type has practical applications, such as the zipper (data structure), zipper technique used in functional programming languages. # Derivatives in geometry Main types of derivatives in geometry is Lie derivatives along a vector field, exterior differential, and covariant derivatives. ## Differential topology In differential topology, a vector field may be defined as a derivation on the ring of smooth functions on a manifold, and a tangent vector may be defined as a derivation at a point. This allows the abstraction of the notion of a directional derivative of a scalar function to general manifolds. For manifolds that are subsets of R''n'', this tangent vector will agree with the directional derivative defined above. The pushforward (differential), differential or pushforward of a map between manifolds is the induced map between tangent spaces of those maps. It abstracts the Jacobian matrix. On the exterior algebra of differential forms over a smooth manifold, the exterior derivative is the unique linear map which satisfies a Graded Leibniz rule, graded version of the Leibniz law and squares to zero. It is a grade 1 derivation on the exterior algebra. The Lie derivative is the rate of change of a vector or tensor field along the flow of another vector field. On vector fields, it is an example of a Lie bracket (vector fields form the Lie algebra of the diffeomorphism group of the manifold). It is a grade 0 derivation on the algebra. Together with the interior product (a degree -1 derivation on the exterior algebra defined by contraction with a vector field), the exterior derivative and the Lie derivative form a Lie superalgebra. ## Differential geometry In differential geometry, the covariant derivative makes a choice for taking directional derivatives of vector fields along curves. This extends the directional derivative of scalar functions to sections of vector bundles or principal bundles. In Riemannian geometry, the existence of a metric chooses a unique preferred Torsion tensor, torsion-free covariant derivative, known as the Levi-Civita connection. See also gauge covariant derivative for a treatment oriented to physics. The exterior covariant derivative extends the exterior derivative to vector valued forms. ## Geometric calculus In geometric calculus, the Geometric_calculus#Differentiation, geometric derivative satisfies a weaker form of the Leibniz rule. It specializes the Frechet derivative to the objects of geometric algebra. Geometric calculus is a powerful formalism that has been shown to encompass the similar frameworks of differential forms and differential geometry.David Hestenes, Garrett Sobczyk: Clifford Algebra to Geometric Calculus, a Unified Language for mathematics and Physics (Dordrecht/Boston:G.Reidel Publ.Co., 1984, # Other generalizations It may be possible to combine two or more of the above different notions of extension or abstraction of the original derivative. For example, in Finsler geometry, one studies spaces which look locally like Banach spaces. Thus one might want a derivative with some of the features of a functional derivative and the covariant derivative. The study of stochastic processes requires a form of calculus known as the Malliavin calculus. One notion of derivative in this setting is the H-derivative, ''H''-derivative of a function on an abstract Wiener space. Multiplicative calculus replaces addition with multiplication, and hence rather than dealing with the limit of a ratio of differences, it deals with the limit of an exponentiation of ratios. This allows the development of the geometric derivative and bigeometric derivative. Moreover, just like the classical differential operator has a discrete analog, the difference operator, there are also List of derivatives and integrals in alternative calculi, discrete analogs of these multiplicative derivatives.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 44, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9566925764083862, "perplexity": 251.98608283212096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103337962.22/warc/CC-MAIN-20220627164834-20220627194834-00768.warc.gz"}
http://math-mprf.org/journal/articles/id857/	Exact Asymptotics for a Large Deviations Problem for the GI/G/1 Queue #### S. Asmussen, J.F. Collamore 1999, v.5, №4, 451-476 ABSTRACT Let $V$ be the steady-state workload and $Q$ the steady-state queue length in the GI/G/1 queue. We obtain the exact asymptotics for probabilities of the form $\P\{V\ge a(t),\, Q\ge b(t)\}$ as $t\to\infty$. In the light-tailed case, there are three regimes according to the limiting value of $a(t)/b(t)$. Our analysis here extends and simplifies recent work of Aspandiiarov and Pechersky [S. Aspandiiarov and E.A. Pechersky, A large deviations problem for compound Poisson processes in queueing theory, Markov Processes Relat. Fields, 1997, v.3, pp. 333-366]. In the heavy-tailed subexponential case, a lower asymptotic bound is derived and shown to be the exact asymptotics in a regime where $a(t)$, $b(t)$ vary in a certain way determined by the service time distribution.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9558724761009216, "perplexity": 790.3656792101665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119637.34/warc/CC-MAIN-20170423031159-00347-ip-10-145-167-34.ec2.internal.warc.gz"}
https://www.examfriend.in/Forum/Arithmetic-aptitude/Heights-and-Distances/361	1) A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 metres from the tower. After 5'seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water ? A) 32 kraph B) 36 kraph C) 38 kraph D) 40 kraph E) 42 kraph	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.820765495300293, "perplexity": 1395.546341953479}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585516.51/warc/CC-MAIN-20211022145907-20211022175907-00594.warc.gz"}
https://www.arxiv-vanity.com/papers/q-bio/0510051/	arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org. \section Main discussion In nonlinear contraction theory, the analysis of dynamical systems is greatly simplified by studying stability and nominal motion separately. We propose a similar point of view for analyzing synchronization in networks of dynamical systems. In section \refsec:analysis, we study specific conditions on the coupling structure which guarantee exponential convergence to a linear subspace. In section \refsec:syminv, we examine how symmetries and/or diffusion-like couplings can give rise to specific flow-invariant subspaces corresponding to concurrent synchronized states. \subsection Some coupling structures and conditions for exponential synchronization \label sec:analysis \subsubsection Balanced diffusive networks \label sec:balanced A balanced network \citeOlfMur is a directed diffusive network which verifies the following equality for each node (see figure \reffig:balanced for an example) $∑_j ≠i \bfK_ij= ∑_j ≠i \bfK_ji$ Because of this property, the symmetric part of the Laplacian matrix of the network is itself the Laplacian matrix of the underlying undirected graph to the network In fact, it is easy to see that the symmetric part of the Laplacian matrix of a directed graph is the Laplacian matrix of some undirected graph \emphif and only if the directed graph is balanced. . Thus, the positive definiteness of for a balanced network is equivalent to the connectedness of some well-defined undirected graph. For general directed diffusive networks, finding a simple condition implying the positive definiteness of (such as the connectivity condition in the case of undirected networks) still remains an open problem. However, given a particular example, one can compute and determine directly whether it is positive definite. ### \thesubsubsection Extension of diffusive connections In some applications [WangSlo2], one might encounter the following dynamics {˙\bfx1=\bff1(\bfx1,t)+k\bfA⊤(\bfB\bfx2−\bfA\bfx1)˙\bfx2=\bff2(\bfx2,t)+k\bfB⊤(\bfA\bfx1−\bfB\bfx2) Here and can be of different dimensions, say and . and are constant matrices of appropriate dimensions. The Jacobian matrix of the overall system is \bfJ=⎛⎜ ⎜⎝∂\bff1∂\bfx1∂\bff2∂\bfx2⎞⎟ ⎟⎠−k\bfL,where \bfL=(\bfA⊤\bfA−\bfA⊤\bfB−\bfB⊤\bfA\bfB⊤\bfB) Note that is symmetric positive semi-definite. Indeed, one immediately verifies that ∀\bfx1,\bfx2:(\bfx1\bfx2)\bfL(\bfx1\bfx2)=(\bfA\bfx1−\bfB\bfx2)⊤(\bfA\bfx1−\bfB\bfx2)≥0 Consider now the linear subspace of defined by \sM={(\bfx1\bfx2)∈Rd1×Rd2:\bfA\bfx1−\bfB\bfx2=\zeros} and use as before the orthonormal projection on , so that is positive definite. Assume furthermore that is flow-invariant, i.e. ∀(\bfx1,\bfx2)∈Rd1×Rd2,[\bfA\bfx1=\bfB\bfx2]⇒[\bfA\bff1(\bfx1)=\bfB\bff2(\bfx2)] and that the Jacobian matrices of the individual dynamics are upper-bounded. Then large enough , i.e. for example kλmin(\bfV\bfL\bfV⊤)>maxi=1,2(sup\bfai,tλmax∂\bffi∂\bfxi(\bfai,t)) ensures exponential convergence to the subspace . The state corresponding to can be viewed as an extension of synchronization states to systems of different dimensions. Indeed, in the case where and have the same dimension and where are non singular, we are in the presence of classical diffusive connections, which leads us back to the discussion of section LABEL:sec:global. As in the case of diffusive connections, one can consider networks of so-connected elements, for example : ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩˙\bfx1=\bff1(\bfx1,t)+\bfA⊤B(\bfBA\bfx2−\bfAB\bfx1)+\bfA⊤C(\bfCA\bfx3−\bfAC\bfx1)˙\bfx2=\bff2(\bfx2,t)+\bfB⊤C(\bfCB\bfx3−\bfBC\bfx2)+\bfB⊤A(\bfAB\bfx1−\bfBA\bfx2)˙\bfx3=\bff3(\bfx2,t)+\bfC⊤A(\bfAC\bfx1−\bfCA\bfx3)+\bfC⊤B(\bfBC\bfx2−\bfCB\bfx3) leads to a positive semi-definite Laplacian matrix ⎛⎜ ⎜⎝\bfA⊤B\bfAB−\bfA⊤B\bfBA\zeros−\bfB⊤A\bfAB\bfB⊤A\bfBA\zeros\zeros\zeros\zeros⎞⎟ ⎟⎠+⎛⎜ ⎜⎝\zeros\zeros\zeros\zeros\bfB⊤C\bfBC−\bfB⊤C\bfCB\zeros−\bfC⊤B\bfBC\bfC⊤B\bfCB⎞⎟ ⎟⎠+⎛⎜ ⎜⎝\bfA⊤C\bfAC\zeros−\bfA⊤C\bfCA\zeros\zeros\zeros−\bfC⊤A\bfAC\zeros\bfC⊤A\bfCA⎞⎟ ⎟⎠ and potentially a flow-invariant subspace \sM={\bfAB\bfx1=\bfBA\bfx2}∩{\bfBC\bfx2=\bfCB\bfx3}∩{\bfCA\bfx3=\bfAC\bfx1} The above coupling structures can be implemented in nonlinear versions of the predictive hierarchies used in image processing (e.g. [LueWil, DayHin, RaoBal, Korner, GeoHaw, Rao]). ### \thesubsubsection Excitatory-only networks One can also address the case of networks with excitatory-only connections. Consider for instance the following system and its Jacobian matrix 111For the sake of clarity, the elements are assumed to be 1-dimensional. However, the same reasoning applies for the multidimensional case as well: instead of , one considers as in section LABEL:sec:global. Clearly, is flow-invariant. Applying the methodology described above, we choose , so that the projected Jacobian matrix is . Thus, for , the two elements synchronize exponentially. In the case of diffusive connections, once the elements are synchronized, the coupling terms disappear, so that each individual element exhibits its natural, uncoupled behavior. This is not the case with excitatory-only connections. This is illustrated in figure Document using FitzHugh-Nagumo oscillator models (see appendix LABEL:sec:oscillators for the contraction analysis of coupled FitzHugh-Nagumo oscillators). ### \thesubsubsection Rate models for neuronal populations In computational neuroscience, one often uses the following simplified equations to model the dynamics of neuronal populations τ˙\bfxi=−\bfxi+Φ⎛⎝∑j≠ikij\bfxj(t)⎞⎠+\bfui(t) Assume that the external inputs are all equal, and that the synaptic connections verify (i.e., that they induce input-equivalence, see section Document). Then the synchronization subspace is flow-invariant. Furthermore, since each element, taken in isolation, is contracting with contraction rate , synchronization should occur when the coupling is not too strong (see remark (ii) in section LABEL:sec:global). Specifically, consider first the case where is a linear function : . The Jacobian matrix of the global system is then , where is the matrix of . Using the result of remark (ii) in section LABEL:sec:global, a sufficient condition for the system to be contracting (and thus synchronizing) is that the couplings are weak enough (or more precisely, such that ). The same condition is obtained if is now e.g. a multidimentional sigmoid of maximum slope (see remark (iii) in section LABEL:sec:global). Besides the synchronization behavior of these models, their natural contraction property for weak enough couplings of any sign is interesting in its own right. Indeed, given a set of (not necessarily equal) external inputs , all trajectories of the global system will converge to a unique trajectory, independently of initial conditions. ## \thesubsection Symmetries, diffusion-like couplings, flow-invariant subspaces and concurrent synchronization Synchronized states can be created in at least two ways : by architectural and internal222Internal symmetries can easily be analyzed within our framework as leading to flow-invariant subspaces, and we shall use this property in section LABEL:sec:cpg for building central pattern generators. However, they will not be discussed in detail in this article. The interested reader can consult [DioGolSte]. symmetries [GolSte, GolSteTor, DioGolSte, PogSanNij] or by diffusion-like couplings [WangSlo, JadMotBar, OlfMur, LinBroFra, Belykh1]. Actually, we shall see that both, together or separately, can create flow-invariant subspaces corresponding to concurrently synchronized states. ### \thesubsubsection Symmetries and input-equivalence In section LABEL:sec:global, we argued that, in the case of coupled identical elements, the global synchronization subspace represents a flow-invariant linear subspace of the global state space. However, several previous works have pointed out that larger (less restrictive) flow-invariant subspaces may exist if the network exhibits symmetries [Zhang, Belykh1, PogSanNij], even when the systems are not identical [GolSte]. The main idea behind these works can be summarized as follows. Assume that the network is divided into aspiring synchronized groups 333Some groups may contain a single element, see section LABEL:sec:single.. The flow-invariant subspace corresponding to this regime (in the sequel, we shall call such a subspace a concurrent synchronization subspace), namely {(\bfx1;…;\bfxn):∀1≤m≤k,∀i,j∈Sm:\bfxi=\bfxj} is flow-invariant if, for each , the following conditions are true : 1. if , then they have a same individual (uncoupled) dynamics 2. if , and if they receive their input from elements and respectively, then and must be in a same group , and the coupling functions (the synapses) and must be identical. If and have more than one input, they must have the same number of inputs, and the above conditions must be true for each input. In this case, we say that and are input-symmetric, or more precisely, input-equivalent (since formally “symmetry” implies the action of a group). One can see here that symmetry, or more generally input-equivalence, plays a key role in concurrent synchronization. For a more detailed discussion, the reader is referred to [GolSte, GolSteTor]. Remark : One can thus turn on/off a specific symmetry by turning on/off a single connection. This has similarities to the fact that a single inhibitory connection can turn on/off an entire network of synchronized identical oscillators [WangSlo]. ### \thesubsubsection Diffusion-like couplings The condition of input-equivalence can be relaxed when some connections within a group are null when the connected elements are in the same state. Such connections are pervasive in the literature : diffusive connections (in a neuronal context, they correspond to electrical synapses mediated by gap junctions [SheRin, fukuda06], in an automatic control context, they correspond to poursuit or velocity matching strategies [OlfMur, LinBroFra], …), connections in the Kuramoto model [IzhiKura, JadMotBar, Strogatz] (i.e. in the form ), etc. Indeed, consider for instance diffusive connections and assume that • has the form • has the form with possibly Here, and are not input-equivalent in the sense we defined above, but the subspace is still flow-invariant. Indeed, once the system is on this synchronization subspace, we have , , so that the diffusive couplings and vanish. One can also view the network as a directed graph , where the elements are represented by nodes, and connections by directed arcs . Then, the above remark can be reformulated as 1 : for all , color the nodes of with a color , 2 : for all , erase the arcs representing diffusion-like connections and joining two nodes in , 3 : check whether the initial coloring is balanced (in the sense of [GolSte]) with respect to the so-obtained graph. It should be clear by now that our framework is particularly suited to analyze concurrent synchronization. Indeed, a general methodology to show global exponential convergence to a concurrent synchronization regime consists in the following two steps • First, find an flow-invariant linear subspace by taking advantage of potential symmetries in the network and/or diffusion-like connections. • Second, compute the projected Jacobian matrix on the orthogonal subspace and show that it is uniformly negative definite (by explicitly computing its eigenvalues or by using results regarding the form of the network, e.g. remark (i) in section LABEL:sec:global or section LABEL:sec:analysis). ## \thesubsection Illustrative examples 1. The first network has three non-trivial flow-invariant subspaces other than the global sync subspace, namely , , and . Any of these subspaces is a strict superset of the global sync subspace, and therefore one should expect that the convergence to any of the concurrent sync state is “easier” than the convergence to the global sync state [Zhang, Belykh1, PogSanNij]. This can be quantified from (LABEL:equ:synccond), by noticing that \sMA⊃\sMB⇒\sM⊥A⊂\sM⊥B ⇒λmin(\bfVA\bfL\bfV⊤A)≥λmin(\bfVB\bfL\bfV⊤B) (\theequation) While in the case of identical systems and relatively uniform topologies, this “percolation” effect may often be too fast to observe, (\theequation) applies to the general concurrent synchronization case and quantifies the associated and possibly very distinct time-scales. 2. The second network has only one non-trivial flow-invariant subspace . 3. If the dashed blue arrows represent diffusive connections then the third network will have one non-trivial flow-invariant subspace , even if these extra diffusive connections obviously break the symmetry. Let’s study in more detail this third network, in which the connections between the round element and the square ones are modelled by trigonometric functions (we shall see in section LABEL:sec:single that their exact form has no actual influence on the convergence rate). ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩˙v1=f(v1)+a1cos(v2)+a2sin(v3)˙v2=g(v2)+a4sin(v1)+c1v6˙v3=g(v3)+a4sin(v1)+b1(v2−v3)+b2(v4−v3)+c1v5˙v4=g(v4)+a4sin(v1)+b3(v3−v4)+c1v7˙v5=h(v5)+c2v2+(d2v7−d1v5)˙v6=h(v6)+c2v3+(d2v5−d1v6)˙v7=h(v7)+c2v4+(d2v6−d1v7) The Jacobian matrix of the couplings is \bfL=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝0a1˙v2sin(v2)−a2˙v3cos(v3)0000−a4˙v1cos(v1)0000−c10−a4˙v1cos(v1)−b1b1+b2−b2−c100−a4˙v1cos(v1)0−b3b300−c10−c200d10−d200−c20−d2d10000−c20−d2d1⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ As we remarked previously, the concurrent synchronization regime is possible. Bases of the linear subspaces and corresponding to this regime are ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1000000⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝0111000⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝0000111⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠for \sM, and⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝[]r0√63−√66−√66000⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝00−√22√22000⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝0000√63−√66−√66⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝00000−√22√22⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠for \sM⊥. Group together the vectors of the basis of into a matrix and compute As a numerical example, let , , , , , , and evaluate the eigenvalues of . We obtain approximately for the smallest eigenvalue. Using again FitzHugh-Nagumo oscillators and based on their contraction analysis in appendix LABEL:sec:oscillators, concurrent synchronization should occur for . A simulation is shown in figure Document. One can see clearly that, after a transient period, oscillators 2, 3, 4 are in perfect sync, as well as oscillators 5, 6, 7, but that the two groups are not in sync with each other. ## \thesubsection Robustness of synchronization So far, we have been considering exact synchronization of identical elements. However this assumption may seem unrealistic, since real systems are never absolutely identical. We use here the robustness result for contracting systems (see theorem LABEL:theorem:robust) to guarantee approximate synchronization even when the elements are not identical. Consider, as in section LABEL:sec:global, a network of dynamical elements ˙\bfxi=\bffi(\bfxi,t)+∑j≠i\bfKij(\bfxj−\bfxi)i=1,…,n (\theequation) with now possibly for . This can be rewritten as ⎛⎜ ⎜ ⎜⎝˙\bfx1⋮˙\bfxn⎞⎟ ⎟ ⎟⎠=⎛⎜ ⎜⎝\bfc(\bfx1,t)⋮\bfc(\bfxn,t)⎞⎟ ⎟⎠−\bfL⎛⎜ ⎜⎝\bfx1⋮\bfxn⎞⎟ ⎟⎠+⎛⎜ ⎜⎝\bff1(\bfx1,t)−\bfc(\bfx1,t)⋮\bffn(\bfxn,t)−\bfc(\bfxn,t)⎞⎟ ⎟⎠ (\theequation) where is some function to be defined later. Keeping the notations introduced in section LABEL:sec:global, one has ˙\bfxg=\bfcg(\bfxg,t)−\bfL\bfxg+\bfd(\bfxg,t) where stands for the last term of equation (\theequation). Consider now the projected auxiliary system on ˙\bfy=\bfV\bfcg(\bfV⊤\bfy+\bfU\bfU⊤\bfxg,t)−\bfV\bfL\bfV⊤\bfy+\bfV\bfd(\bfV⊤\bfy+\bfU\bfU⊤\bfxg,t) (\theequation) Assume that the connections represented by are strong enough (in the sense of equation (LABEL:equ:synccond)), so that the undisturbed version of (\theequation) is contracting with rate . Let , where can be viewed as a measure of the dissimilarity of the elements. Since is a particular solution of the undisturbed system, theorem LABEL:theorem:robust implies that the distance between any trajectory of (\theequation) and verifies, after a transient period, . In the -space, it means that any trajectory will eventually be contained in a boundary layer of thickness around the synchronization subspace . The choice of can now be specified so as to minimize . Neglecting for simplicity the variation of , a possible choice for is then the center of the ball of smallest radius containing , with being the radius of that ball. Consider for instance, the following system (similar to the model used for coincidence detection in [WangSlo] and section LABEL:sec:manies) ˙xi=f(xi)+Ii+k(x0−xi)where Imin≤Ii≤Imax, ∀i In this case, choosing , one can achieve the bound , where is the contraction rate of and . Remark : Assume that two spiking neurons are approximately synchronized, as just discussed. Then, since spiking induces large abrupt variations, the neurons must spike approximately at the same time. More specifically, if the bound on their trajectory discrepancy guaranteed by the above robustness result is significantly smaller than spike size, then this bound will automatically imply that the two neurons spike approximately at the same time.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9632886648178101, "perplexity": 2240.494246899733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00089.warc.gz"}
https://math.stackexchange.com/questions/734963/how-much-close-should-two-spaces-be-in-the-gromov-hausdorff-distance-to-be-homeo	# How much close should two spaces be in the Gromov-Hausdorff distance to be homeomorphic? Here I am considering the Gromov-Hausdorff convergence for metric spaces. I know two compact metric spaces are isometric if and only if $d_{GH}(X,Y)=0$, where $d_{GH}$ denotes the Gromov-Hausdorff (pseudo)distance. I was wondering if a weaker assumption (e.g. $d_{GH}(X,Y)$ small enough) could imply homeomorphism between $X$ and $Y$. To be more concrete, I am especially interested in the following problem: let $(X,d)$ be a metric space and let $d_n$ a sequence of metrics on $X$ such that $(X,d_n)$ converges to $(X,d)$ in the Gromov-Hausdorff convergence. Can I say that the topology induced by $d_n$ is equivalent to the native topology (induced by $d$) of $X$, at least for $n$ large enough? Consider the sequence of annuli $A=\{1\le \|z\|\le 1+1/n\}$ in the complex plane. Their limit in the GH-sense is the unit circle. Since the annuli and the circle have the same cardinality, you can assume (although it is very unnatural, GH-convergence is set up to avoid this) that the corresponding metrics are defined on the same set. If you want Riemannian manifolds as your examples, think of a sequence of tori converging to the unit circle. On the positive side, Perelman's stability theorem (used in the proof of his more famous theorem) states that under certain conditions ("Alexandrov", "curvature bounds", "noncollapsing"), if $X_n$'s GH-converge to $X$ then for large $n$ indeed $X$ is homeomorphic to $X_n$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9899670481681824, "perplexity": 61.45314174412135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00091.warc.gz"}
https://mathoverflow.net/questions/80163/are-g-spectra-the-same-as-modules-over-a-group-ring-spectrum	# Are $G$-spectra the same as modules over a “group ring spectrum”? Let $G$ be a finite group (maybe this will also work when $G$ is compact, or something, but to be safe we'll let it be finite). I imagine it's quite natural to ask: is the category of $G$-spectra equivalent to the category of module spectra over some ring spectrum, probably denoted by $SG$? For definiteness, we can take "the" category of $G$-spectra to be the symmetric-spectra model (or orthogonal if we want to try and deal with compact groups), and similarly for modules over a ring spectrum. I imagine if such a thing existed it would have the property that $\pi_0SG$ should be the actual group ring, $SG$ should be an $A_\infty$-ring, and if $G$ is abelian it should be an $E_\infty$-ring. It seems like such a thing should exist since the ($\infty$-)category of $G$-spectra (at least coming from, say, the orthogonal model) looks presentable and generated by the equivariant sphere... so by some theorem in Lurie it should be equivalent to an ($\infty$ -)category of modules over some ring spectrum. A brief search online and glance at May's book on the subject revealed nothing, but I could have easily missed it. Any pointers to the literature or brief epositions of a construction of such a thing would be much appreciated! (I would say "take the free spectrum on the objects of $G$ and mod out by some relations" and maybe this is how it's done, but I wonder how to make this precise in the "brave new" world, also such a construction probably would not generalize to the compact group case, so maybe there's some better way of doing it.) - *I guess, specifically, it the ring spectrum should be the spectrum $End(S,S)$ of endomorphisms of the equivariant sphere in the category of $G$-spectra... but there has to be a description that doesn't presume a construction of the category of $G$-spectra! – Dylan Wilson Nov 5 '11 at 22:45 Eric wrote a really nice response telling that your initial hope is incorrect and why. I'd just like to write some positive results that you can find. Disclaimer: I understand little to nothing about the case of a compact Lie group. Schwede and Shipley have a paper entitled "Stable model categories are categories of modules" from 2003. In particular, G-spectra form a stable model category to which their results apply. Schwede-Shipley show that if you pick a set of "generators", then you'll get an category $I$ enriched in spectra, with $Hom_I(i,j)$ being a spectrum together with units $\mathbb S \to Hom_I(i,i)$ and composition maps $$Hom_I(j,k) \wedge Hom_I(i,j) \to Hom_I(i,k)$$ which are unital and associative. (This is the spectrum version of a DG-category, if you like). Then there is an equivalence between $G$-spectra and enriched functors from $I$ to the category of spectra. In $G$-spectra, we can pick generators given by the spectra $\Sigma^\infty G/H_+$, which are representing objects for the standard "fixed point" functors. So a $G$-spectrum is equivalent to the data of a collection of spectra $Y^H$ as $H$ ranges over the subgroups of $G$, together with "action maps" $$F(\Sigma^\infty G/H_+, \Sigma^\infty G/K_+) \wedge Y^H \to Y^K$$ that are unital and associative. If you're feeling like it, you could instead replace several generators with $\bigvee_H G/H_+$, and establish $G$-spectra as equivalent to modules over one ring spectrum which is a big "matrix algebra" containing a bunch of commuting idempotents. It's not clear to me whether this is generally profitable. (It certainly doesn't make looking at the symmetric monoidal structure on $G$-spectra easier.) To go further, we need to specify a little about which category of $G$-spectra you're interested in. This is often phrased in terms of a choice of universe. At one end, you have $G$-spectra indexed on the trivial universe, which are formed by taking the category of $G$-spaces and inverting the suspension functor. There's a "coalescing" result of Elmendorf (his "Systems of fixed point sets") that essentially shows that the homotopy category of $G$-spaces is equivalent to the homotopy category of functors from the orbit category of $G$ to spaces; $G$-spectra indexed on the trivial universe satisfy a similar result. At the other end, you have $G$-spectra indexed on a complete universe, where all the spheres based on orthogonal representations of $G$ become invertible. These are more complicated, because they're generated by more than just actions $g: Y^H \to Y^{gHg^{-1}}$ and restrictions $Y^K \to Y^H$ for $H < K$. They also have transfer maps. If you've done any looking into $G$-spectra, you've probably encountered the notion of a Mackey functor, which is a collection of abelian groups with restriction, transfer, and conjugation maps. One compact way to phrase this is that Mackey functors are additive functors from the "Burnside category" to the category of abelian groups. $G$-spectra indexed on a complete universe satisfy a similar result: they are enriched functors from a topological Burnside category to the category of spectra. In particular, every $G$-spectrum produces a Mackey functor to the stable homotopy category. (There are several places I could insert some more or less gratuitous $\infty$-category theory here.) I know that Clark Barwick has given several talks on this, and is likely in the process of writing it up. Whether Mackey functors make you happy might depend on whether you're in that pleasant zone between understanding their definitions and trying to do serious homological algebra with them. While I'm writing pithy asides, it's kind of depressing that the Burnside category doesn't have entries on Wikipedia or the nLab for me to link to, and Mackey functors only have this. Many of the presentations in the literature are worth looking at. There are several categories of $G$-spectra in between, and I don't know much about the general properties of those. - This was very helpful! Thank you for the wonderful references! I thought that paper by Schwede and Shipley would be relevant, but for some reason I wasn't thinking that we would need to consider diagrams of ring spectra. If you allow for the Hom-sets to be considered as Mackey functors, then I think the category of G-spectra is monogenic... so maybe Clark Barwick sets up a framework for equivariant infty-categories where something like this happens naturally? I'd love to see it! – Dylan Wilson Nov 7 '11 at 16:01 Yes, if you consider some kind of enrichment then the sphere spectrum could become monogenic. – Tyler Lawson Nov 7 '11 at 16:56 I should note that there is work in progress (which seems to be close to completion) of Guillou and May that addresses some of the constructive aspects of this description. – Tyler Lawson Feb 21 '12 at 2:57 There is now at least a useful pointer on the nLab page on Mackey functors. Maybe somebody has the energy to add more. – Urs Schreiber Apr 12 '14 at 21:26 The "group ring spectrum" $SG$ you ask for does indeed exist, but modules over it are not the same as $G$-spectra. Indeed, $SG$ is just the suspension spectrum $\Sigma^\infty_+ G$ of $G$ as a discrete space. The suspension spectrum of any $A_\infty$ space is naturally an $A_\infty$ ring spectrum, so the group structure on $G$ gives an $A_\infty$ structure on $SG$. An $SG$-module can be shown to be the same thing as a spectrum with a (coherent) action of $G$. Any $G$-spectrum has an "underlying" non-equivariant spectrum which has the natural structure of an $SG$-module. However, this is not an equivalence of categories, and the basic reason is that they correspond to different notions of "weak equivalence" of equivariant objects. For simplicity, I'll describe this in the unstable setting of $G$-spaces. If $X$ and $Y$ are two spaces with an action of a group $G$ and $f:X\to Y$ is an equivariant map, there are two things we might mean when we say $f$ is an "equivariant homotopy equivalence". The first is that $f$ is an ordinary homotopy equivalence of the spaces $X$ and $Y$ which happens to also be an equivariant map. The second is that $f$ has a homotopy inverse internal to the category of $G$-spaces: there exists another equivariant map $g:Y\to X$ such that the compositions $fg$ and $gf$ are homotopy to the identity through equivariant maps. This notion is much stronger. For example, the map $EG\to $ is a homotopy equivalence in the weaker sense, but not in this stronger sense, because $EG$ has no fixed points so there are no equivariant maps $*\to EG$. If we restrict to $G$-CW complexes (a natural equivariant generalization of CW-complexes), it turns out that a map $X\to Y$ is an equivariant equivalence in this stronger sense iff for every subgroup $H\subseteq G$, the induced map $X^H\to Y^H$ on fixed points is a homotopy equivalence. The category of $SG$-modules is a stable version of $G$-spaces under the first, weaker notion of equivalence. Indeed, as with any ring spectrum, a weak equivalence of $SG$-modules is just a map of $SG$-modules which happens to be a weak equivalence of underlying spectra (though an inverse which is actually an $SG$-module map can be found if we're willing to take cofibrant and fibrant replacements of our modules, which corresponds to replacing a $G$-space $X$ with $EG \times X$). On the other hand, the category of $G$-spectra is a stable version of $G$-spaces under the second, stronger notion of equivalence. - I also quite like this answer (especially because it answered my actual question...). But I'm not sure if I understand correctly: are you saying that the objects in $SG$-mod are basically the same as in $G$-spectra, but the morphisms are different? Or is the distinction more drastic? Also- is there any use for this weak approximation to the category of $G$-spectra? – Dylan Wilson Nov 7 '11 at 16:03 There is a slight inaccuracy in the first paragraph of Eric Wofsey's otherwise excellent answer. It depends on which category of spectra is used. Using suspension spectra (the original definition) then everything Eric says is correct. But With symmetric spectra or orthogonal spectra there is a non-obvious way of defining equivalences so that the corresponding homotopy theory is $G$-equivariant spectra. So one category with two notions of equivalence. With $G$-spaces, using the weak approximation means you are studying only the free actions. – Jeff Smith Dec 6 '11 at 8:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.877286970615387, "perplexity": 298.04651674617253}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824037.46/warc/CC-MAIN-20160723071024-00028-ip-10-185-27-174.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/266461/convergence-of-bessel-sturm-liouville-expansions-at-the-end-points	# Convergence of Bessel (Sturm-Liouville) Expansions at the End Points I have asked this question before on MSE but received no answer at all. So I assume that it is proper to ask it here. I am not a mathematician so my language may not be too precise, please correct me wherever it is necessary. Suppose that a function $f:[0,a]\to\Bbb{R}$ is given. We are interested to find its Bessel-Expansion and study the convergence of this expansion specially at the end points $r=0$ and $r=a$. Consider the following singular Sturm-Liouville systems for zero and first order Bessel functions \begin{align} A.\,&\frac{d}{dr}\left[r\frac{dR}{dr}\right]+\lambda r R=0 & B.\,&\frac{d}{dr}\left[r\frac{dR}{dr}\right]+\left[\lambda r + \frac{1}{r}\right]R=0\\ &R(0)\lt\infty & &R(0)\lt\infty\\ &\frac{dR}{dr}(a)=0 & &R(a)=0 \tag{1} \end{align} I know that the eigen-functions of systems $A$ and $B$ can form a basis for all peice-wise continuous functions with peice-wise continuous derivatives (which I denote by $C^1[0,a]$). Then we can construct the Bessel expansions corresponding to systems $A$ and $B$ as \begin{align} S(r):=\sum_{i=1}^{\infty}C_iR(\lambda_i,r) \tag{2} \end{align} I know that $S(r)$ converges to $\frac{f(r^+)+f(r^-)}{2}$ at the points in the interval $(0,a)$. Consequently, $S$ will converge to $f$ at the points of coninuity of $f$ inside the aforementioned interval but I don't know what happens at the end points. Here is my question 1. What are the necessary, sufficient, necessary and sufficient conditions that $S$ converges to $f$ at $r=0$ and $r=a$? 2. Suppose we answered question $1$. Can this be generalized to any Sturm-Liouville expansions like Fourier, Chebyshev, Hermit, etc? This animation shows the convergence of the eigen-function expansion of system $A$ to the modified Bessel function of order zero $f(r)=I_0(r)$. In this case convergence at the end points is achieved although $f(r)$ does not satisfy the second BC mentioned in Sturm-Liouville system A. and this one shows the convergence of the eigen-function expansion of system $B$ to $f(r)=I_0(r)$. It seems that the series does not converge to the function at the end points in this case and $f(r)$ is not satistying the second BC of Sturm-Liouville system B. Here is the mathematica code for making the animations. ClearAll["Global"] f[r_] := BesselI[0, r] a = 1; Subscript[N, max] = 40; A[n_] = Simplify[\!$$\SubsuperscriptBox[$\[Integral]$$, $$0$$, $$a$$]$$rf[r] BesselJ[0, \[Alpha][n]r] \[DifferentialD]r$$$/\!$$\SubsuperscriptBox[$\[Integral]$$, $$0$$, $$a$$]$$r* \SuperscriptBox[\(BesselJ[ 0, \[Alpha][n]r]$$, $$2$$] \[DifferentialD]r$\), Assumptions -> {BesselJ[1, \[Alpha][n]a] == 0}] (2 BesselI[1, 1])/(BesselJ[0, \[Alpha][n]] (1 + \[Alpha][n]^2)) Eig = Table[{i, N[BesselJZero[1, i]/a]}, {i, 1, Subscript[N, max]}]; Eig = Prepend[Eig, {0, 0}]; (\[Alpha][#] = #2) & @@@ Eig; BesselSeries[r_, n_] := \!$$\SubsuperscriptBox[$\[Sum]$$, $$i = 0$$, $$n$$]$$A[i]* BesselJ[0, \[Alpha][i]r]$$$ ConvAnim = Table[Plot[Evaluate[{f[r], BesselSeries[r, n]}], {r, 0, a}, PlotRange -> {0.9f[0], 1.1f[a]}, ImageSize -> Large, AspectRatio -> Automatic], {n, 1, Subscript[N, max]}]; (Export["Convergence.gif",ConvAnim,"DisplayDurations"\[Rule]{0.25}]*) `	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9343022108078003, "perplexity": 852.398727431183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00465.warc.gz"}
http://www.theufochronicles.com/2012/11/the-strange-planets-of-fomalhaut.html	## Saturday, November 03, 2012 ### The Strange Planets of 'Fomalhaut' --A Spectacular Alien Star System Tweet By The Daily Galaxy 10-26-12 Astronomers using the Atacama Large Millimeter/submillimeter Array (ALMA) in Chile discovered that planets orbiting the star Fomalhaut must be much smaller than originally thought. The discovery was made possible by exceptionally sharp ALMA images of a disc, or ring, of dust orbiting Fomalhaut, which lies about 25 light-years from Earth. It helps resolve a controversy among earlier observers of the system. The ALMA images showed that both the inner and outer edges of the thin, dusty disc have very sharp edges. That fact, combined with computer simulations, led the scientists to conclude that the dust particles in the disc are kept within the disc by the gravitational effect of two planets — one closer to the star than the disc and one more distant. Their calculations also indicated the probable size of the planets — larger than Mars but no larger than a few times the size of the Earth. This is much smaller than astronomers had previously thought. In 2008, a NASA/ESA Hubble Space Telescope image had revealed the inner planet, then thought to be larger than Saturn, the second largest planet in our Solar System. However, later observations with infrared telescopes failed to detect the planet. "Combining ALMA observations of the ring's shape with computer models, we can place very tight limits on the mass and orbit of any planet near the ring," said Aaron Boley (a Sagan Fellow at the University of Florida, USA) who was leader of the study. "The masses of these planets must be small; otherwise the planets would destroy the ring," he added. The small sizes of the planets explain why the earlier infrared observations failed to detect them, the scientists said. The original ALMA research shows that the ring's width is about 16 times the distance from the Sun to the Earth, and is only one-seventh as thick as it is wide. "The ring is even more narrow and thinner than previously thought," said Matthew Payne, also of the University of Florida. . . . ~~BOOK SALE~~	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127442240715027, "perplexity": 1611.0120662229244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430456358323.20/warc/CC-MAIN-20150501045918-00015-ip-10-235-10-82.ec2.internal.warc.gz"}
http://web.math.rochester.edu/news-events/events/single/1843	# Probability, Ergodic Theory, Mathematical Physics Seminar ## Stationary Coalescing Walks: Entropy Arjun Krishnan, University of Rochester Friday, February 15th, 2019 3:00 PM - 4:00 PM Hylan 1106A This talk is a sequel to the one I gave last year, but I will remind you about the original setting and results. The model is simple: consider a translation invariant measure on arrow configurations on the usual nearest-neighbor lattice in d dimensions. There is an arrow on every point on Z^d, and following arrows produces a semi-infinite trajectory. Last time, I showed you an striking dichotomy theorem: either walks from every pair of points coalesce with probability one, or they form bi-infinite trajectories. We believe that when trajectories are random enough, bi-infinite trajectories do not exist. One such measure of randomness is entropy. So we consider various entropic properties of these systems. We show that in systems with completely positive entropy, bi-infinite trajectories must carry entropy. In some specialized situations, we show that positive entropy guarantees that bi-infinite trajectories do not exist. Many classical models fall into our simple framework: we construct a stationary discrete-time symmetric exclusion process whose particle trajectories form bi-infinite trajectories carrying entropy. Event contact: arjun dot krishnan at rochester dot edu	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9630838632583618, "perplexity": 2222.2888425574924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578760477.95/warc/CC-MAIN-20190426053538-20190426075538-00052.warc.gz"}
https://www.physicsforums.com/threads/the-state-of-a-particle-changes-when-viewed.263268/	# The state of a particle changes when viewed 1. ### WhoShot 1 Only if particles are viewed, do they conform to a non-quantum state. And when viewed, the light governs the state of the particle, as if the particle was a class, with a boolean property, normal state = yes, or normal-state = no (quantum). So the primary principle which governs particles is light, but how can light be fast enough to change the state of two particles existing in seperate spaces, trillions of miles apart? I therefore expand that these two points (in non-normal state) must exist as one, and space as a whole is like one MASSIVE container. So how does matter behave outside of this container. If matter past through the boundaries of space fabric, would it still exist?, could it exist? and is matter limited to the boundaries of space? 2. ### Xelthen 4 Sounds like you're talking about uncertainty principle. 3. ### Domenicaccio 86 Or perhaps in this case the container is always and only where matter (or energy) is or has been, and therefore matter can never cross its boundary, because the boundary is defined by matter. Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook Have something to add? Similar discussions for: The state of a particle changes when viewed	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8429954648017883, "perplexity": 1506.795733249925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988065.26/warc/CC-MAIN-20150728002308-00257-ip-10-236-191-2.ec2.internal.warc.gz"}
https://ai.stackexchange.com/tags/problem-solving/info	The Stack Overflow podcast is back! Listen to an interview with our new CEO. Tag Info For questions about AI problem solving in terms of approaches, theory, logic, and other aspects where the problem is well defined and the objective is to find a solution to the problem. Problem solving in artificial intelligence is the study of how an AI can solve a given problem. The usual approach to problem solving is state search. The problem was described as an initial state, conditions for a final state, and a set of transition rules. A transition rule changes takes a state as input and outputs a new state. The solution of the problem then consists of applying the right transitions, until a state is reached where that satisfies the condition for a final state. As a concrete example, there is the problem of the Farmer, the Goat, the Cabbage and the Wolf. The farmer must row each of these to the other side of a river, but his boat is only big enough that he can transport only one of them at a time. If he leaves the goat with the cabbage, the goat will eat the cabbage; if he leaves the wolf with the goat, the wolf will eat the goat. The initial state has the farmer, cabbage, goat and wolf on one side of the river. A final state has them all on the other side. The transition rules are all "row the cabbage OR the goat OR the wolf from the current side to the other". There are several state search algorithms, where the purpose is to arrive at a final state in an efficient manner.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8577144145965576, "perplexity": 486.22010004281026}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986710773.68/warc/CC-MAIN-20191020132840-20191020160340-00382.warc.gz"}
http://dlmf.nist.gov/21.2	§21.2 Definitions §21.2(i) Riemann Theta Functions 21.2.1 This -tuple Fourier series converges absolutely and uniformly on compact sets of the and spaces; hence is an analytic function of (each element of) and (each element of) . is also referred to as a theta function with components, a -dimensional theta function or as a genus theta function. For numerical purposes we use the scaled Riemann theta function , defined by (Deconinck et al. (2004)), 21.2.2 is a bounded nonanalytic function of . Many applications involve quotients of Riemann theta functions: the exponential factor then disappears. See also §21.10(i). §21.2(ii) Riemann Theta Functions with Characteristics Let . Define This function is referred to as a Riemann theta function with characteristics . It is a translation of the Riemann theta function (21.2.1), multiplied by an exponential factor: and Characteristics whose elements are either 0 or are called half-period characteristics. For given , there are -dimensional Riemann theta functions with half-period characteristics.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9894537925720215, "perplexity": 1066.855625159066}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164038538/warc/CC-MAIN-20131204133358-00022-ip-10-33-133-15.ec2.internal.warc.gz"}
https://cleantechlaw.com/2021/01/rhode-island-report-100-renewable-energy-by-2030-possible/	# Blog ### Rhode Island Report: 100% Renewable Energy by 2030 Possible Achieving Rhode Island’s goal of using 100% renewable energy by 2030 is possible, a state report said, but it will require the ongoing construction of renewable energy projects as transportation and heating transition to electric power. The report released Wednesday from the state’s Office of Energy Resources was produced by The Brattle Group. The consulting firm described the production capacity needed and the estimated costs required to reach the ambitious clean energy goal. Gov. Gina Raimondo signed an executive order last year making 2030 the target date for the state to completely transition to renewable energy sources. It directed the state to study and develop ways to achieve that goal, the state’s Office of Energy Resources said in a statement. The state’s Renewable Energy Standard actually sets standards for the percentage of renewable energy supplied in the state, the Providence Journal reported. An offshore wind farm that is waiting for federal approval would provide a large portion of the required clean energy, and another proposed offshore windfarm would make up another major portion, the newspaper reported. The remaining clean energy production would come from various solar installations and the purchase of renewable energy certificates. Source	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.883626401424408, "perplexity": 2561.609224033352}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00072.warc.gz"}
https://chemistry.stackexchange.com/questions/108838/reaction-between-naoh-and-carbonic-acid	# Reaction between NaOH and Carbonic acid I was presented with the following reaction when we were just learning about acids and bases: $$\ce{2NaOH + H2CO3 -> H2O + Na2CO3}$$ however I had thought it might've been something like $$\ce{NaOH + H2CO3 -> H2O + NaHCO3}$$ They both seem to fit the notion of a neutralization reaction creating a salt so how do I know that the first one is correct? The second reaction looks fine to me (is it that bicarbonate is too reactive?) and more intuitive as the carbonic acid just gives a proton away ending up as an ion to bond with $$Na$$, so how do I distinguish between these kind of cases? You're right, both of these reactions take place. In aqueous solution, carbonate, bicarbonate, carbon dioxide, and carbonic acid exist together in a dynamic equilibrium. Now, the answer to the question depends on context. If you're asked to do this reaction in a mildly acidic/weakly basic medium, then the answer would be the bicarbonate, while in a strongly basic medium, the answer would be the carbonate. This $$\ce{pH}$$ dependent equilibrium between carbonate and bicarbonate is one of the ways in which the blood $$\ce{pH}$$, and the $$\ce{pH}$$ of the mouth is maintained. It also helps aquatic life to survive in ocean waters. (Also see urea cycle and carbon cycle) • @ AbhigyanC What about $\pu{pH}$ at the second equivalence point, more or less than 7? – Adnan AL-Amleh Jan 31 '19 at 15:35 • @AdnanAL-Amleh $\ce{pK_{a1}}$ and $\ce{pK_{a2}}$ of carbonic acid are $4.5\times10^{-7}$ and $4.7\times10^{-11}$, respectively. So it happens at $\ce{pH}=8.34$, using the formula for pH of a amphiprotic anion is the average of the two pK values. See chembuddy.com/… – Abhigyan Chattopadhyay Jan 31 '19 at 15:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8685757517814636, "perplexity": 894.1611374901646}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250616186.38/warc/CC-MAIN-20200124070934-20200124095934-00007.warc.gz"}
http://www.zora.uzh.ch/id/eprint/108057/	# Identification techniques for highly boosted W bosons that decay into hadrons CMS Collaboration; Canelli, M F; Chiochia, V; Kilminster, B; Robmann, P; et al (2014). Identification techniques for highly boosted W bosons that decay into hadrons. Journal of High Energy Physics:017. ## Abstract In searches for new physics in the energy regime of the LHC, it is becoming increasingly important to distinguish single-jet objects that originate from the merging of the decay products of W bosons produced with high transverse momenta from jets initiated by single partons. Algorithms are defined to identify such W jets for different signals of interest, using techniques that are also applicable to other decays of bosons to hadrons that result in a single jet, such as those from highly boosted Z and Higgs bosons. The efficiency for tagging W jets is measured in data collected with the CMS detector at a center-of-mass energy of 8 TeV, corresponding to an integrated luminosity of 19.7 fb−1. The performance of W tagging in data is compared with predictions from several Monte Carlo simulators. ## Abstract In searches for new physics in the energy regime of the LHC, it is becoming increasingly important to distinguish single-jet objects that originate from the merging of the decay products of W bosons produced with high transverse momenta from jets initiated by single partons. Algorithms are defined to identify such W jets for different signals of interest, using techniques that are also applicable to other decays of bosons to hadrons that result in a single jet, such as those from highly boosted Z and Higgs bosons. The efficiency for tagging W jets is measured in data collected with the CMS detector at a center-of-mass energy of 8 TeV, corresponding to an integrated luminosity of 19.7 fb−1. The performance of W tagging in data is compared with predictions from several Monte Carlo simulators. ## Statistics ### Citations 8 citations in Web of Science® 12 citations in Scopus® ### Altmetrics Detailed statistics Item Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics English 2014 24 Feb 2015 11:42 08 Dec 2017 11:45 Springer 1029-8479 Publisher DOI. An embargo period may apply. https://doi.org/10.1007/JHEP12(2014)017	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9234069585800171, "perplexity": 1646.3449264387896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00070.warc.gz"}
https://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/	# Almost Sure ## 1 April 10 ### Continuous Local Martingales Continuous local martingales are a particularly well behaved subset of the class of all local martingales, and the results of the previous two posts become much simpler in this case. First, the continuous local martingale property is always preserved by stochastic integration. Theorem 1 If X is a continuous local martingale and ${\xi}$ is X-integrable, then ${\int\xi\,dX}$ is a continuous local martingale. Proof: As X is continuous, ${Y\equiv\int\xi\,dX}$ will also be continuous and, therefore, locally bounded. Then, by preservation of the local martingale property, Y is a local martingale. ⬜ Next, the quadratic variation of a continuous local martingale X provides us with a necessary and sufficient condition for X-integrability. Theorem 2 Let X be a continuous local martingale. Then, a predictable process ${\xi}$ is X-integrable if and only if $\displaystyle \int_0^t\xi^2\,d[X]<\infty$ for all ${t>0}$. Proof: If ${\xi}$ is X-integrable then the quadratic variation ${V_t\equiv\int_0^t\xi^2\,d[X]}$ is finite. Conversely, suppose that V is finite at all times. As X and, therefore, [X] are continuous, V will be continuous. So, it is locally bounded and as previously shown, ${\xi}$ is X-integrable. ⬜ In particular, for a Brownian motion B, a predictable process ${\xi}$ is B-integrable if and only if, almost surely, $\displaystyle \int_0^t\xi^2_s\,ds<\infty$ for all ${t>0}$. Then, ${\int\xi\,dB}$ is a continuous local martingale. Quadratic variations also provide us with information about the sample paths of continuous local martingales. Theorem 3 Let X be a continuous local martingale. Then, • X is constant on the same intervals for which [X] is constant. • X has infinite variation over all intervals on which [X] is non-constant. Proof: Consider a bounded interval (s,t) for any ${s, and set ${t^n_k=s+(t-s)k/n}$ for k=0,1,…,n. By the definition of quadratic variation, using convergence in probability, $\displaystyle [X]_t-[X]_s=\lim_{n\rightarrow\infty}\sum_{k=1}^n(X_{t^n_k}-X_{t^n_{k-1}})^2 \le \lim_{n\rightarrow\infty}V\max _{k=1,...,n}\vert X_{t^n_k}-X_{t^n_{k-1}}\vert$ where V is the variation of X over the interval (s,t). By continuity, ${\vert X_{t^n_k}-X_{t^n_{k-1}}\vert}$ tends uniformly to zero as n goes to infinity, so ${[X]_t=[X]_s}$ and [X] is constant over (s,t) whenever the variation V is finite. This proves the second statement of the theorem, which also implies that [X] is constant on all intervals for which X is constant. It only remains to show that ${X_t=X_s}$ whenever ${[X]_t=[X]_s}$. Applying this also to the countable set of rational times u in (s,t) will then show that X is constant on this interval whenever [X] is. The process ${Y_u\equiv X_u-X_{u\wedge s}}$ is a local martingale constant up until s, with quadratic variation ${[Y]_u=[X]_u-[X]_s}$ for ${u\ge s}$. Then ${\tau=\inf\{u\colon [ Y]>0\}}$ is a stopping time with respect to the right-continuous filtration ${\mathcal{F}_{\cdot+}}$ and, by stopping, ${Y^{\tau}}$ is a local martingale with zero quadratic variation ${[Y^\tau]=[Y]^\tau=0}$. Then, as previously shown, ${Y^2=Y^2-[Y]}$ is a martingale and, therefore, ${{\mathbb E}[Y_t^2]=0}$. This shows that ${X_{t\wedge\tau}=X_s}$ almost surely. Finally, on the set ${[X]_t=[X]_s}$, we have ${\tau\ge t}$ and, hence, ${X_t=X_{t\wedge\tau}=X_s}$. ⬜ Theorem 3 has the following immediate consequence. Corollary 4 Any continuous FV local martingale is constant. Proof: By the second statement of Theorem 3, the quadratic variation [X] is constant. Then, by the first statement, X is constant. ⬜ The quadratic covariation also tells us exactly when X converges at infinity. Theorem 5 Let X be a continuous local martingale. Then, with probability one, the following both hold. • ${X_\infty=\lim_{t\rightarrow\infty}X_t}$ exists and is finite whenever ${[X]_\infty<\infty}$. • ${\limsup_{t\rightarrow\infty}X_t=\infty}$ and ${\liminf_{t\rightarrow\infty}X_t=-\infty}$ whenever ${[X]_\infty=\infty}$. Proof: By martingale convergence, with probability one either ${X_\infty}$ exists and is finite or ${\limsup_{t\rightarrow\infty}X_t}$ and ${\limsup_{t\rightarrow\infty}(-X_t)}$ are both infinite. It just remains to be shown that, with probability one, ${X_\infty}$ exists if and only if ${[X]_\infty}$ is finite.. Let ${\tau_n=\inf\{t\colon [X]_t\ge n\}}$. Then, ${X^{\tau_n}}$ is a local martingale with quadratic variation ${[X^{\tau_n}]=[X]^{\tau_n}}$ bounded by n. So, ${{\mathbb E}[(X^{\tau_n}_t)^2]\le n}$ and ${X^{\tau_n}}$ is an ${L^2}$-bounded martingale which, therefore, almost surely converges at infinity. In particular, on the set $\displaystyle \left\{[X]_\infty we have ${X_\infty=\lim_{t\rightarrow\infty}X_t=\lim_{t\rightarrow\infty}X^{\tau_n}_t}$ outside of a set of zero probability. Therefore, ${X_\infty}$ almost surely exists on $\displaystyle \left\{[X]_\infty<\infty\right\}=\bigcup_{n=1}^\infty\left\{[X]_\infty For the converse statement, set ${\tau_n=\inf\{t\colon\vert X_t\vert\ge n\}}$. Then, ${X^{\tau_n}}$ is a local martingale bounded by n and ${{\mathbb E}[[X]_{\tau_n}]={\mathbb E}[X_{\tau_n}^2]\le n^2}$. Hence, ${[X]_{\tau_n}}$ is almost surely finite and ${[X]_\infty}$ is finite on the set $\displaystyle \left\{\sup_t\vert X_t\vert outside of a set of zero probability. Therefore, ${[X]_\infty}$ is almost surely finite on the set $\displaystyle \left\{X_\infty{\rm\ exists}\right\}\subseteq\left\{\sup_t\vert X_t\vert<\infty\right\}=\bigcup_n\left\{\sup_t\vert X_t\vert Theorems 3 and 5 are easily understood once it is known that all local martingales are random time-changes of standard Brownian motion, as will be covered in a later post. The topology of uniform convergence on compacts in probability (ucp convergence) was introduced in a previous post, along with the stronger semimartingale topology. On the space of continuous local martingales, these two topologies are actually equivalent, and can be expressed in terms of the quadratic variation. Recalling that semimartingale convergence implies ucp convergence and that quadratic variation is a continuous map under the semimartingale topology, it is immediate that the first and third statements below follow from the second. However, the other implications are specific to continuous local martingales. Lemma 6 Let ${\{M^n\}_{n\in{\mathbb N}}}$ and M be continuous local martingales. Then, as n goes to infinity, the following are equivalent. 1. ${M^n}$ converges ucp to M. 2. ${M^n}$ converges to M in the semimartingale topology. 3. ${(M^n_0-M_0)^2+[M^n-M]_t\rightarrow0}$ in probability, for each ${t\ge0}$. Proof: As semimartingale convergence implies ucp convergence, the first statement follows immediately from the second. So, suppose that ${M^n\xrightarrow{\rm ucp}M}$. Write ${N^n\equiv M^n-M}$ and let ${\tau_n}$ be the first time at which ${\vert N^n-N^n_0\vert\ge1}$. Ucp convergence implies that ${\tau_n}$ tends to infinity in probability, so to prove the third statement it is enough to show that ${[N^n]_{\tau_n\wedge t}}$ tends to zero in probability. By continuity, the stopped process ${(N^n-N^n_0)^{\tau_n}}$ is uniformly bounded by 1, so is a square integrable martingale, and Ito’s isometry gives $\displaystyle \displaystyle{\mathbb E}\left[[N^n]_{\tau_n\wedge t}\right]={\mathbb E}\left[(N^n_{\tau_n\wedge t}-N^n_0)^2\right]\rightarrow0$ as n goes to infinity. The limit here follows from the fact that ${N^n_{\tau_n\wedge t}-N^n_0}$ is bounded by 1 and tends to zero in probability. So, we have shown that ${N^n_{\tau_n\wedge t}}$ tends to zero in the ${L^2}$ norm and, hence, in probability. Now suppose that the third statement holds. This immmediately gives ${N^n_0\rightarrow0}$ in probability. Letting ${\tau_n}$ be the first time at which ${[N^n]\ge1}$ and ${\vert\xi^n\vert\le1}$ be elementary predictable processes, Ito’s isometry gives $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[\left(\int_0^{\tau_n\wedge t}\xi^n\,dN^n\right)^2\right] &\displaystyle\le{\mathbb E}\left[\int_0^{\tau_n\wedge t}(\xi^n)^2\,d[N^n]\right]\smallskip\\ &\displaystyle\le{\mathbb E}\left[[N^n]_{\tau_n\wedge t}\right]\smallskip\\ &\displaystyle\le{\mathbb E}\left[[N^n]_t\wedge1\right] \rightarrow0. \end{array}$ So, in particular, ${\int_0^{\tau_n\wedge t}\xi^n\,dN^n\rightarrow0}$ in probability. Finally, as ${\tau_n > t}$ whenever ${[N^n]_t < 1}$, which has probability one in the limit ${n\rightarrow\infty}$, this shows that ${\int_0^t\xi^n\,dN^n}$ tends to zero in probability and ${N^n}$ tends to zero in the semimartingale topology. ⬜ Applying the previous result to stochastic integrals with respect to a continuous local martingale gives a particularly strong extension of the dominated convergence theorem in this case. Note that this reduces convergence of the stochastic integral to convergence in probability of Lebesgue-Stieltjes integrals with respect to ${[X]}$. Theorem 7 Let X be a continuous local martingale and ${\{\xi^n\}_{n\in{\mathbb N}}}$, ${\xi}$ be X-integrable processes. Then, the following are equivalent. 1. ${\int\xi^n\,dX}$ converges ucp to ${\int\xi\,dX}$. 2. ${\int\xi^n\,dX}$ converges to ${\int\xi\,dX}$ in the semimartingale topology. 3. ${\int_0^t(\xi^n-\xi)^2\,d[X]\rightarrow0}$ in probability, for each ${t\ge0}$. Proof: This follows from applying Lemma 6 to the continuous local martingales ${M^n=\int\xi^n\,dX}$ and ${M=\int\xi\,dX}$. ⬜ Theorem 7 also provides an alternative route to constructing the stochastic integral with respect to continuous local martingales. Although, in these notes, we first proved that continuous local martingales are semimartingales and used this to imply the existence of the quadratic variation, it is possible to construct the quadratic variation more directly. Once this is done, the space ${L^1(X)}$ of X-integrable processes can be defined to be the predictable processes ${\xi}$ such that ${\int_0^t\xi^2\,d[X]}$ is almost surely finite for all times t. Define the topology on ${L^1(X)}$ so that ${\xi^n\rightarrow\xi}$ if and only if ${\int_0^t(\xi^n-\xi)^2\,d[X]\rightarrow0}$ in probability as ${n\rightarrow\infty}$ for each t, and use ucp convergence for the topology on the integrals ${\int\xi\,dX}$. Then, Theorem 7 says that ${\xi\mapsto\int\xi\,dX}$ is the unique continuous extension from the elementary integrands to all of ${L^1(X)}$. 1. Hi, nice blog! In Theorem 5, what do you mean by [X]_{\infty} < \infty. That the quadratic variation is uniformly bounded by a constant C? Otherwise your arguments would probably not work? If [X] would only be a.s. finite, then an "n" such that [X] \leq n a.s. would not exist (example: normal random variable takes on only finite values and is therefore a.s. finite but there is no upper bound to the values it takes)… Is that the way to understand your statement? Cheers Chris Comment by Chris — 12 April 10 @ 8:54 PM • No, all that matters is that [X]_\infty < n for some n. n is not fixed, so uniform boundedness is not needed. You don't even need [X]_\infty to be almost surely finite. Even if it is only finite on a set of probability p < 1, X will converge on that set (up to a zero probability set). Comment by George Lowther — 12 April 10 @ 9:38 PM 2. Ok, still not sure whether I can follow you. When [X]_\infty < n for some n and this holds for all \omega in the sample space then it is uniformly bounded by a constant. When you mean [X]_\infty(\omega) < n(\omega), that is for every element \omega in the sample space you find an n such that [X]_\infty is bounded by n on that given \omega only then the subsequent estimate E[(X^{\tau_n}_t)^2] \leq n does not hold anymore. That estimate would only hold if you have on the right of the estimate something like the ess sup n(\omega), so basically the essential supremum of n over all elements in the sample space. However nothing gurantees you, that this supremum even exists…. Comment by Chris — 13 April 10 @ 12:01 PM 3. I think you’re still misunderstanding my argument. There is no need to be thinking about essential supremums. I’ll try modifying the argument and proof to make it clearer. I’m not able to do this right now as I’m away from my computer (just replying by mobile). Hopefully will have time later tonight. For now, the theorem could be stated more precisely as follows. $\displaystyle A=\{\omega\in\Omega: [X]_\infty(\omega) < \infty \}$ then, outside of a set of probability zero, $X_\infty =lim_{t\rightarrow\infty} X_t$ exists and is finite on A, $\limsup_{t\rightarrow\infty} X_t=\infty$ and $\liminf_{t\to\infty} X_t=-\infty$ outside of A. Comment by George Lowther — 13 April 10 @ 2:10 PM 4. Hey, a lot clearer now thanks…. Comment by Chris — 17 April 10 @ 11:12 AM 5. Update: I have added a couple of extra results to this post. Lemma 6 shows that ucp convergence, semimartingale convergence and convergence in probability of quadratic variations all coincide. Theorem 7 uses this to give a much stronger version of the dominated convergence theorem for continuous local martingales. Comment by George Lowther — 13 May 11 @ 11:39 PM 6. Hi George, I have been looking for an answer to the following question on continuous local martingales: Can one construct a non-zero (in sense of having P>0 of being nonzero) continuous local martingale which is identically equal to $0$ P-a.e. at a fixed time $T$? It has to do with an option hedging problem I am working on. Tigran Comment by Anonymous — 28 May 12 @ 9:27 AM 7. […] with respect to $M$, iff $mathbb{E}left[int_0^t frac{1}{X^2_s}d[M]_sright]0$ (see e.g. Thm 2 in https://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/). Taking e.g. $M_t$ as Brownian motion we have $[M]_t = t$. Now it is known that the second inverse […] Pingback by Integrating the inverse of a squared bessel process - integrability - MathHub — 4 April 16 @ 1:45 PM 8. Why is the quadratic variation of the stopped process Y^\tau zero? Comment by Anonymous — 6 October 16 @ 8:54 PM • in the proof of theorem 3 Comment by Anonymous — 6 October 16 @ 8:54 PM • At least, there is an index missing in the definition of \tau. Comment by Anonymous — 8 October 16 @ 4:21 PM • Right. It should be $[Y]_u$. I’ll correct the the post in a moment. Thanks Comment by George Lowther — 8 October 16 @ 4:29 PM Blog at WordPress.com.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 114, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9007634520530701, "perplexity": 2587.1582385236734}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00530-ip-10-145-167-34.ec2.internal.warc.gz"}
http://jgaa.info/getPaper?id=457	Special Issue on Graph Drawing Beyond Planarity Parameterized Complexity of 1-Planarity Vol. 22, no. 1, pp. 23-49, 2018. Regular paper. Abstract We consider the problem of drawing graphs with at most one crossing per edge. These drawings, and the graphs that can be drawn in this way, are called $1$-planar. Finding $1$-planar drawings is known to be ${\mathsf{NP}}$-hard, but we prove that it is fixed-parameter tractable with respect to the vertex cover number, tree-depth, and cyclomatic number. Special cases of these algorithms provide polynomial-time recognition algorithms for $1$-planar split graphs and $1$-planar cographs. However, recognizing $1$-planar graphs remains ${\mathsf{NP}}$-complete for graphs of bounded bandwidth, pathwidth, or treewidth. Submitted: May 2017. Reviewed: September 2017. Revised: October 2017. Accepted: October 2017. Final: October 2017. Published: January 2018. Communicated by Michael A. Bekos, Michael Kaufmann, and Fabrizio Montecchiani article (PDF) BibTeX	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8254417181015015, "perplexity": 1808.486003409166}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999838.27/warc/CC-MAIN-20190625132645-20190625154645-00028.warc.gz"}
http://physics.stackexchange.com/questions/28891/gravimagnetic-monopole-and-general-relativity	Gravimagnetic monopole and General relativity Review and hystorical background: Gravitomagnetism (GM), refers to a set of formal analogies between Maxwell's field equations and an approximation, valid under certain conditions, to the Einstein field equations for general relativity. The most common version of GM is valid only far from isolated sources, and for slowly moving test particles. The GM equations coincide with equations which were first published in 1893, before general relativity, by Oliver Heaviside as a separate theory expanding Newton's law: $\nabla \cdot \vec G = -4\pi\gamma\rho$ $\nabla \cdot \vec \Omega = 0$ $\nabla \times \vec G = - \dfrac{\partial \vec \Omega}{\partial t}$ $\nabla \times \vec \Omega = -\dfrac{4\pi\gamma}{c^2} \vec J + \dfrac{1}{c^2} \dfrac{\partial \vec G}{\partial t}$ $\vec G$ is gravitational field strength or gravitational acceleration, also called gravielectric for the sake of analogy; $\vec\Omega$ is intensity of torsion field or simply torsion, also called gravitomagnetic field; $\vec J$ is mass current density; $\gamma$ is gravitational constant. Magnetic monopole and Maxwell's field equations: It is known that the Maxwell's field equations have some asymmetry, in the absence of a magnetic monopole, although formally we can say that the problem can be solved theoretically (PAM Dirac and other works). $\nabla \cdot \vec E = \dfrac{1}{\epsilon_0}\rho_e$ $\nabla \cdot \vec B = \mu_0 c \cdot g_m$, $g_m$ - magnetic monopole charge dencity. $\nabla \times \vec E = \mu_0 J_{mag} - \dfrac{\partial \vec B}{\partial t}$, $J_{mag}$ - magnetic charge current $\nabla \times \vec B = -\dfrac{1}{c^2 \epsilon_0} \vec J_{el} + \dfrac{1}{c^2} \dfrac{\partial \vec E}{\partial t}$, $J_{el}$ - electric charge current General relativity and gravitomagnetic monopole: Formally, a massive body in the linearized general relativity, is the gravielectric charge. Now there is another interesting issue associated with the hypothesis of the existence gravimagnetic charge. If we suppose its existence, what changes should be made to the equations of general relativity $G_{ik}= \kappa T_{ik}$ ($G_{ik}$ - Einstein tensor, $T_{ik}$ stress-energy tensor)? And what are the properties of such a charge? - – Qmechanic May 24 '12 at 10:36 It looks like an exact duplicate to me. – Ron Maimon May 24 '12 at 18:53 There isn't any precise analogy between electromagnetism and gravity. In Newton's theory, only the gravitational acceleration (the gradient of the gravitational potential) exists at each point of space as an independent field; there is no independent extra "gravimagnetic" field. In GR, the gravitational acceleration is given by $\Gamma^a_{bc}$, the Christoffel symbol, but it isn't antisymmetric in the same sense as $F_{\mu\nu}$ so one can't really Hodge-dualize it to get the dual magnetic field. One may speculate about the torsion, extra fields added to GR, but the observations exclude their existence at long distances at any significant couplings. "Gravimagnetic" (literally) isn't an adjective used anywhere in serious physics literature. Instead, "gravitomagnetic" is sometimes used. But it refers to any effects - in GR or whatever right theory we consider - in which masses are moving and their impact is proportional to the velocity, much like Lorentz's force $v\times B$ in magnetism. See e.g. http://arxiv.org/abs/gr-qc/0207065 for a review. No full analogy with electromagnetism, of course. Kaluza-Klein monopoles The most interesting insight about "magnetic monopoles constructed purely from gravitational degrees of freedom" that one may discuss in serious physics are the so-called Kaluza-Klein monopoles. They appear in the Kaluza-Klein theory where the electromagnetic $U(1)$ gauge symmetry is geometrized as the group of rotations of an extra, circular coordinate of spacetime. In this setup, one may find the solutions of higher-dimensional Einstein's equations that looks like the Dirac magnetic monopole if the $g_{\mu 5}$ components of the metric are related to the electromagnetic potential $A_\mu$ according to the usual Kaluza-Klein dictionary. In 4+1D gravity compactified on circle, to imitate 3+1D gravity coupled to electromagnetism, Kaluza-Klein monopoles are point-like objects in 3+1D. The KK monopoles play an important role in string/M-theory. In particular, D6-branes in type IIA string theory become KK monopoles (with 6 extra spatial dimensions in which the solution is extended/constant) if the coupling of type IIA string theory is sent to infinity to get M-theory in 11 dimensions. The M-theory (or 11D supergravity) solution for the KK monopole that becomes a D6-brane is actually completely non-singular and smooth. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9032033681869507, "perplexity": 501.24225300890544}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860121737.31/warc/CC-MAIN-20160428161521-00094-ip-10-239-7-51.ec2.internal.warc.gz"}
https://brilliant.org/problems/nimo-2014-reciprocal-of-interval/	# (NIMO 2014) Reciprocal of Interval Number Theory Level pending For any interval $$\mathcal{A}$$ in the real number line not containing zero, define its $$\textit{reciprocal}$$ to be the set of numbers of the form $$\frac1x$$ where $$x$$ is an element in $$\mathcal{A}$$. Compute the number of ordered pairs of positive integers $$(m,n)$$ with $$m<n$$ such that the length of the interval $$[m,n]$$ is $$10^{10}$$ times the length of its reciprocal. ×	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9162587523460388, "perplexity": 84.49717407258852}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424721.74/warc/CC-MAIN-20170724042246-20170724062246-00194.warc.gz"}
http://www.milefoot.com/math/stat/pdfc-normaldisc.htm	## Approximately Normal Distributions with Discrete Data If a random variable is actually discrete, but is being approximated by a continuous distribution, a continuity correction is needed. This is to more closely match the areas of bars in a discrete distribution with the areas under the curve of a continuous distribution. ### Continuity Corrections Discrete data take on only integer values, while continuous data can have any real number value. Since areas under a continuous distribution give probabilities, and not the height of the curve, we need to adjust our approach to finding probabilities when discrete data are involved. So we think of the half-way points between consecutive discrete values, and these become the boundaries of the bars that represent the area under the curve for that discrete value. As an example, if the discrete values include ${..., 75, 76, 77, ...}$, then the bar for 76 would run from 75.5 to 76.5, and we would find $P(X = 76)$ by computing $P(75.5 < X < 76.5)$ instead. Example: The number of bags lost by a small airline per week is approximately normally distributed with a mean of 427 bags and a standard deviation of 35 bags. What is the probability that they lose exactly 430 bags next week? Since the airline cannot lose a fraction of a bag, the random variable is discrete. The bar for 430 in the histogram of the probability distribution would have its left and right endpoints at 429.5 and 430.5. Therefore, we find z-scores for these two x-values. They are $z = \dfrac{429.5-427}{35} \approx 0.0714$ and $z = \dfrac{430.5-427}{35} = 0.10$. Then \begin{align} P(x = 430) &= P(429.5 < x < 430.5) = P(0.0714 < z < 0.10) \\ &= \Phi(0.10) - \Phi(0.0714) = \operatorname{normalcdf}(0.0714,0.10) \approx 0.0114 \end{align} In other words, there is a 1.14% probability that the airline would lose exactly 430 bags next week. Example: For the same airline, what is the probability that they lose at most 420 bags next week? The discrete event "at most 420" is approximated with the continuous event "420.5 or less", as both will include the random variable value 420, but not 421. Therefore, $z = \dfrac{420.5-427}{35} = -0.1857$. Then we have \begin{align} P(x \le 420) &= P(X < 420.5) = P(z < -0.1857) \\ &= \Phi(-0.1857) = \operatorname{normalcdf}(-\infty,-0.1857) = 0.4263 \end{align} In other words, there is a 42.63% probability that they will have at most 420 lost bags. ### Using the Normal Distribution to Approximate a Binomial Distribution Binomial distributions are considered approximately normal when the expected number of successes and the expected number of failures are both at least 5. Formulaically, this translates into the requirement that $np \ge 5$ and $n(1-p) \ge 5$. Example: On every one of the 100 questions on a multiple choice test, Tom randomly chooses one of the four answer options. What can Tom expect for the number of correct answers? To answer this, we use the expected value formula for a binomial distribution. Therefore, $E(X) = np = 100 \left( \dfrac14 \right) = 25$ . He can expect 25 correct answers on average. Example: What is the standard deviation for the number of correct answers on Tom's test? From the variance formula for a binomial distribution, we have $Var(X) = np(1-p) = 100 \left( \dfrac14 \right) \left(1 - \dfrac14 \right) = \dfrac{75}{4}$. Therefore, the standard deviation is $\sigma = \sqrt{\dfrac{75}{4}} \approx 4.33$ bags. Example: What is the probability that Tom would obtain at least 35 correct answers? We first note that Tom is expecting 25 correct answers, and 75 incorrect answers. Both of these values are greater than 5, and therefore the normal distribution will provide a reasonable approximation to this binomial problem. Since the number of correct answers is discrete, a continuity correction is needed, and we use $x = 34.5$. Thus $z = \dfrac{34.5 - 25}{4.33} = 2.19$, and \begin{align} P(x \ge 35) &= P(x > 34.5) = P(z > 2.19) \\ &= 1 - \Phi(2.19) = \operatorname{normalcdf}(2.19,\infty) \approx 0.0143 \end{align} Tom has a 1.43% chance of obtaining at least 35 correct answers on his test. Example: What is the probability that Tom would obtain a passing score of at least 60 correct answers? This is very similar to the previous question. We use $x = 59.5$, which gives $z = 7.97$. We compute \begin{align} P(x \ge 60) &= P(x > 59.5) = P(x > 7.97) \\ &= 1 - \Phi(7.97) = \operatorname{normalcdf}(7.97,\infty) \approx 8 \times 10^{-16} \end{align} With a probability of less than $10^{-15}$, Tom has almost no chance of passing the test. ### Using the Normal Distribution to Approximate a Poisson Distribution Poisson distributions are considered to be approximately normal when the expected rate of successes per unit time is at least 10. Since the expected value formula for a Poisson distribution is $E(X) = \lambda$, the requirement is that $\lambda \ge 10$. Example: Cars arrive at the drive-up window of a local fast food restaurant at the rate of 40 per hour. Assume the arrivals are independent of one another, so that the Poisson distribution applies. What are the mean and standard deviation of the number of arrivals per hour? Using the Poisson distribution formulas, we have $E(X) = \lambda = 40$ and $\sigma = \sqrt{\lambda} \approx 6.32$. We can expect 40 cars per hour to arrive on average, with a standard deviation of 6.32 cars. Example: At the same drive-up window, What is the probability that exactly 25 cars will arrive in the next hour? Since $\lambda \ge 10$, we can use the normal distribution as an approximation. The required continuity correction gives x-values of 24.5 and 25.5, which produces z-scores of $z = \dfrac{24.5-40}{6.32} = -2.45$ and $z = \dfrac{25.5-40}{6.32} = -2.29$. Therefore \begin{align} P(x = 25) &= P(24.5 < x < 25.5) = P(-2.45 < z < -2.29) \\ &= \Phi(-2.29) - \Phi(-2.45) = \operatorname{normalcdf}(-2.45,-2.29) \approx 0.0039 \end{align} There is a 0.39% probability that exactly 25 cars would arrive in the next hour.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9951696991920471, "perplexity": 455.702799014897}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827097.43/warc/CC-MAIN-20181215200626-20181215222626-00038.warc.gz"}
http://math.stackexchange.com/questions/272242/max-min-of-reals-as-cauchy-sequences/273130	# Max/min of reals as Cauchy sequences Apologies if this seems overly simplified, I'm just getting to grips with this. This is related to the computability of the min() and max() functions of real numbers reconstructed using Cauchy sequences of rational numbers. To check my understanding first: Cauchy sequences are a sequence of rational numbers converging to a real value with some accuracy. They might look like: [2, 1.7, 1.24, 0.8] etc. The addition of two reals represented like this is done by: x + y = [x0 + y0, ...xn + yn] and similar for subtraction and multiplication. In my lecture notes however, the max() and min() functions are show as: max(x, y) = [max(x0, y0), ...max(xn, yn)] min(x, y) = [min(x0, y0), ...min(xn, yn)] which would surely have the some weird effect of returning some value nering x + y? Could someone please explain how these functions should work? Thanks - Why do you think that $\max\{x_n,y_n\}$ approaches $x+y$ as $n\to\infty$? In fact it approaches $\max\{x,y\}$. – Brian M. Scott Jan 7 '13 at 16:39 Given max([2, 1.7, 1.24, 0.8], [5, 1.5, 1.04, 0.6]), wouldn't the answer (by the method I gave above) be [5, 1.7, 1.24, 0.8]? A number larger than either individual number. Not approaching x+y I agree, but still... – Karl Barker Jan 7 '13 at 16:47 But Cauchy sequences are infinite sequences; what happens in the first four terms really tells you nothing. – Brian M. Scott Jan 7 '13 at 16:50 Okay, so is there no property that elements in a Cauchy sequence are in decreasing order, as in my examples. Just realised that. Also, since I'm talking about computable reals, these Cauchy sequences are either finite, or only considered to some finite precision? – Karl Barker Jan 7 '13 at 16:55 There is no restriction on how the terms of a Cauchy sequence bounce around. I strongly suspect that you’ve misunderstood. So far as I know, you’re still looking at infinite sequences; they just have a computable modulus of convergence. – Brian M. Scott Jan 7 '13 at 17:03 A CS major can easily check the claims about sequences against numerical experiments (keeping in mind that an experiment does not prove the theory, but may disprove it with a counterexample). For example, I used RAND() to generate two sequences: $x_n\to 3$ and $y_n\to 2$, and calculated (in OpenOffice) their maximum $\max(x_n,y_n)$ and minimum $\min(x_n,y_n)$. Do you see what numbers $\max(x_n,y_n)$ and $\min(x_n,y_n)$ are approaching? Aside: a lot of interesting sequences can be quickly generated in a spreadsheets. For example, $x_{n+1}=\frac12 (x_n+\frac{a}{x_n})$ is a nice sequence for approximation of $\sqrt{a}$. If $a$ is rational and the initial value $x_1$ (which is ours to choose) is also rational, then the sequence consists of rational numbers quickly converging to $\sqrt{a}$. - That's a great demonstration, thanks. As you can probably tell, I'm really not versed in Cauchy sequences. How did you generate those sequences? I'm having trouble understanding this wiki page on constructing sequences for reals. – Karl Barker Jan 8 '13 at 10:42 @Karl For example =3+(RAND()-0.5)*500/ROW()^3 which adds random numbers that tend to zero because of the division by ROW(). Particular numbers there were picked for no particular reason. – user53153 Jan 8 '13 at 13:12 Since you have expressed a lack of "rigorous" mathematical background, I'll try to stay simple. Cauchy sequences are infinite combinations of numbers that get closer together as you go on. It really means that for any distance $\epsilon$>0, we can find a value in the sequence such that all of the remaining values are within $\epsilon$ of any other value. More technically, there exists an $N$, such that for all $n, m \ge N, \|x_n-x_m\|<\epsilon$. The absolute value bars being the distance from $x_m$ to $x_ n$, all the later terms fit in an interval around $x_N$ of length $2\epsilon$, and since we can make $\epsilon$ really small, they all bunch up around the limit. The limit doesn't need to exist, it is just the "value" for which they all clump around. Now to the problem at hand. We now see that the max and min functions, as we have defined them, work. Let's look at $max(x_n, y_n)$, Then since $x_n \to x$ and $y_n \to y$, so $max(x_n, y_n)\to max(x, y)$. Since large values of our "max sequence" have arguments really close to the limits, the value is close to the limit. To see this, take $\epsilon$>0, and take $N=max(N', N'')$, where $N'$ is the value for $x$ and $N''$ for $y$ for $\epsilon$. Then the max has bunched itself in a little interval for values greater then $N$. Since $\epsilon$ can be as small as we like, we have shown that it clumps up around the limit. Convergence and the idea of getting arbitrary close are closely tied concepts mathematics. Your definition works for any converging sequence, which is implied by the limits. A good book if you don't have much experience in mathematics is Rosenlicht's, "Introduction to Analysis". It's not really in your field, but it's short and well written. It will give you some fundamental mathematics, some convergence and topology, and then rigorous limits, continuity, and calculus. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9205862283706665, "perplexity": 535.447910621762}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500835872.63/warc/CC-MAIN-20140820021355-00386-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/deviation-in-the-direction-of-current.394513/	Deviation in the direction of current . 1. Apr 12, 2010 ankur.schwing Deviation in the direction of current..... Is there any law of refraction of direct current lines at the boundry between two conducting media relating with conductivities and the angles between the current lines and the normal of the boundry surface..? 2. Apr 17, 2010 ankur.schwing Re: Deviation in the direction of current..... Similar Discussions: Deviation in the direction of current .	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8077102899551392, "perplexity": 1318.4243436935726}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00337.warc.gz"}
https://mathoverflow.net/questions/181284/compositional-inversion-and-generating-functions-in-algebraic-geometry	# Compositional inversion and generating functions in algebraic geometry The exponential generating function of the graded dimension of the cohomology ring of the moduli space of n-pointed curves of genus zero satisfying the associativity equations of physics (the WDVV equations) (cf. OEIS-A074060) is the compositional inverse of the generating function for the Betti numbers for $M_{0,n}$ (cf. OEIS-A049444 and A143491). And, Brown and Bergstrom in "Inversion of series and the cohomology of the moduli spaces of $M_{0,n}^\delta$" state in the abstract: For $n \geq 3$, let $M_{0,n}$ denote the moduli space of genus $0$ curves with $n$ marked points, and $\overline{M}_{0,n}$ its smooth compactification. A theorem due to Ginzburg, Kapranov and Getzler states that the inverse of the exponential generating series for the Poincare polynomial of $H^\bullet(M_{0,n})$ is given by the corresponding series for $H^\bullet(\overline{M}_{0,n})$. In this paper, we prove that the inverse of the ordinary generating series for the Poincare polynomial of $H^\bullet(M_{0,n})$ is given by the corresponding series for $H^\bullet(M^{\delta}_{0,n})$, where $\overline{M}_{0,n}\subset M^{\delta}_{0,n} \subset \overline{M}_{0,n}$ is a certain smooth affine scheme. Is there an intuitive principle underlying this relationship of compositional inversion between the generating functions? (Getzler makes use of a generalized Legendre transform-compositional inversion in disguise-to relate the generating functions.) • There is an intriguing tale told by He and Jejalla in "Modular matrix models" interweaving the compositional inversion of generating series as in free probability theory, matrix models, gauge field theory, Calabi-Yau geometry, the Klein modular invariant j-function, and some monstrous moonshine. – Tom Copeland Jan 31 '15 at 10:10 • To the degree that the inverse pair can be represented by certain types of trees, Drakes' thesis "An inversion theorem for labelled trees ..." people.brandeis.edu/~gessel/homepage/students/drakethesis.pdf has some insights. – Tom Copeland Dec 8 '15 at 6:56 • See also "Brown's moduli spaces of curves and the gravity operad" by Dupont and Vallette. arxiv.org/abs/1509.08840 – Tom Copeland Apr 14 '16 at 20:49 I think you would enjoy reading Curt McMullen's paper "Moduli spaces in genus zero and inversion of power series". In some sense there is nothing there that isn't already in Getzler's paper, but everything is stated in a down-to-earth and combinatorial fashion. Let me summarize the story, first for the spaces $\overline M_{0,n}$ and $M_{0,n}$. The space $\overline M_{0,n}$ has a stratification where a stratum corresponds to a tree with no vertices of valence two. The stratum itself is isomorphic to $\prod_v M_{0,\mathrm{val}(v)}$ where $v$ runs over interior vertices of the tree and $\mathrm{val}(v)$ denotes the number of incident edges. Since the virtual Poincaré polynomial is additive over stratifications, this shows that the virtual Poincaré polynomial of $\overline M_{0,n}$ is given by a sum over trees involving the virtual Poincaré polynomials of $M_{0,n'}$ for $n' \leq n$. Now using the relationship between compositional inversion and summing over trees, well-known to combinatorists, one can thus show that the exponential generating series of virtual Poincaré polynomials of $\overline M_{0,n}$ and $M_{0,n}$ are compositional inverses of each other. (If you don't know virtual Poincaré polynomials, think about any other invariant additive under stratification, e.g. Euler characteristic.) Finally, both the spaces $\overline M_{0,n}$ and $M_{0,n}$ have pure cohomology in every degree: $H^k (\overline M_{0,n})$ is pure of weight $k$, and $H^k(M_{0,n})$ us pure of weight $2k$. Thus in both cases, the virtual Poincaré polynomial concides with the usual Poincaré polynomial (in the latter case up to a substitution $t \mapsto t^2$). This explains the second sentence in Bergström-Brown's abstract. The story for $M_{0,n}^\delta$ and $M_{0,n}$ is completely similar, the only difference being that $M_{0,n}^\delta$ has a stratification indexed by trees without vertices of valence two and with a cyclic ordering of the edges incident to each vertex. In the same way as compositional inversion of exponential generating functions corresponds to sums over trees, compositional inversion of ordinary generating functions corresponds to sums over trees with such cyclic structure. McMullen touches upon something very similar at the very end of his paper. He doesn't consider $M_{0,n}$ and $M_{0,n}^\delta$, but instead considers the choice of a connected component of $M_{0,n}(\mathbf R)$ and its closure. Combinatorially this amounts to exactly the same thing: $M_{0,n}^\delta$ is defined by choosing a connected component of $M_{0,n}(\mathbf R)$ and taking the union of all strata meeting the closure of this component. A final remark is that the duality between $H^\bullet(\overline M_{0,n})$ and $H^\bullet(M_{0,n})$ can be upgraded to a Koszul duality of two cyclic operads, the "Hypercommutative" and "Gravity" operads. This is a much stronger result than just that their generating series are compositional inverses, and this is what Getzler proves. On the other hand the cohomologies of $M_{0,n}^\delta$ and $M_{0,n}$ give rise to nonsymmetric cyclic operads (this notion is not defined in the literature, but it's not hard to give the definition). However, it turns out that they are not in any natural sense Koszul dual of each other, but it is still true that they are interchanged with each other under bar-cobar-duality, up to homotopy. (But first one needs to define a bar transform of nonsymmetric cyclic operads...) This is an operad-theoretic statement that improves on what Bergström-Brown proved. I worked this out with Johan Alm at one point but we never wrote it down properly. • Can you give a bit more detail about how one goes from a Koszul duality between two operads to an observation about two generating functions being compositional inverses? – Qiaochu Yuan Sep 19 '14 at 21:03 • Thanks Dan for the great overview. Hadn't seen McMullen's paper. (Murri gives nice diagrams of ribbons and marked Riemann surfaces. For the interested, see the OEIS refs.). The relations of diff. ops to inversion inspired me to climb the Cayley trees for a clear view of the landscape years ago. I guess I must put my nose to the grindstone to familiarize myself with stratified spaces so that I can feel it's intuitive. – Tom Copeland Sep 20 '14 at 3:19 • Brown and Bergstrom are inverting o.g.f.s, so the associahedra face vectors occur oeis.org/A133437, and McMullen is inverting e.g.f.s oeis.org/A134685. – Tom Copeland Sep 20 '14 at 5:58 • Alm and Petersen, "Brown's dihedral moduli space and freedom of the gravity operad" arxiv.org/abs/1509.09274 – Tom Copeland Apr 14 '16 at 20:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8628173470497131, "perplexity": 321.475076848219}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00307.warc.gz"}
https://www.physicsforums.com/threads/black-scholes-equation-a-type-of-diffusion-equation.598841/	# Homework Help: Black-Scholes equation (a type of diffusion equation) 1. Apr 21, 2012 ### tjackson3 1. The problem statement, all variables and given/known data The equation for the probability distribution of the price of a call option is $$\frac{\partial P}{\partial t} = \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 P}{\partial S^2} + rS\frac{\partial P}{\partial S} - rP$$ with the conditions $P(0,t) = 0, P(S,0) = \max(S-K,0)$, and the goal is to find the current price of the option, which is given by $$\int_K^{\infty}\ (S-K)P(S,T)\ dS$$ 2. Relevant equations 3. The attempt at a solution The obvious thing to try is separation of variables: $P(S,t) = u(S)v(t)$. The time ODE is not homogeneous, so the eigenvalue problem is the spatial problem: $\frac{\sigma^2}{2}S^2 u'' + rSu' - (r-\lambda)u = 0, u(0) = 0$ (with the understanding that u is finite as S goes to infinity) This is an equidimensional equation, so its solutions are of the form $S^n$. Unfortunately, the form of n is not so nice: $$n = \frac{(1/2)\sigma^2-r \pm \sqrt{r^2+\sigma^2 r + (1/4)\sigma^4 - 2\sigma^2\lambda}}{\sigma^2}$$ Without knowing anything about the constants σ and r, I don't see how to proceed from here. I don't think you can tell a priori what you'll get: two positive exponents, one of each sign (though with the boundary conditions, you would only get the trivial solution in these cases), two negative exponents, complex exponents... it's hard to say. The other thing that occurs to me to try is a Laplace transform. Doing that doesn't get me any farther, since it results in the same equation, except instead of the equation being homogeneous, it has that weird boundary condition on the right hand side. Since we have a semi-infinite interval with Dirichlet boundary conditions, you would expect to use a Fourier sine transform somewhere in here, but it's not directly applicable. I've seen in different places that you can make a change of variables to condense this into a typical diffusion equation. However, I have no idea what motivates the particular change of variables they use, so I wouldn't feel right approaching it that way unless I could figure out why people make that transformation (no resource I've found has adequately explained this. For example, they change the independent variables to $S = Ke^x,t = T-\tau/(\sigma^2/2)$. I can sort of understand the S change, since that is how you solve an equidimensional equation. I think the t change is an artifact from this equation normally taking place in finite time) Any thoughts? Thank you so much! :) 2. Apr 22, 2012 ### tjackson3 I'm noticing that the solution doesn't even look separable	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9550610780715942, "perplexity": 265.9919846584723}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00572.warc.gz"}
https://www.coursehero.com/file/47000882/Newtons-Third-Law/	Newton's Third Law - Section 3 Newton\u2019s Third Law \u25cf Newton\u2019s third law describes something else that happens when one object exerts a force on # Newton's Third Law - Section 3 Newtonu2019s Third Law... • 1 This preview shows page 1 out of 1 page. Section 3: Newton’s Third LawNewton’s third law describes something else that happens when one object exerts a force on another object. According to Newton’s third law of motion, forces always act in equal but opposite pairs. In simple terms, for every action, there is an equal and opposite reaction. This means when you push on a wall, the wall pushes back on your with a force equal in strength to the force you exerted. The forces exerted by two objects on each other can be considered the action force or the reaction force. Either force can be considered the action force or the reaction force. Action and reaction pairs don’t cancel because they act on different object. Force can only cancel if they act on the same object. Motion depends on mass and the effects of the forces in an action-reaction pair may seemnoticeable. The launching of a space shuttle is a spectacular example of Newton’s third law. Three rocket engines supply the force, called thrust, that lifts the rocket. The gas molecules	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8453118205070496, "perplexity": 367.5502322325095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00503.warc.gz"}
http://math.stackexchange.com/questions/119936/equality-in-the-schwarz-pick-theorem-implies-function-is-a-linear-fractional/119939	# Equality in the Schwarz-Pick theorem implies function is a linear fractional? Part of the Schwarz-Pick Theorem states that for an analytic automorphism of the unit disk, then $$\frac{\|f'(z)\|}{1+\|f(z)\|^2}\leq\frac{1}{1-\|z\|^2}.$$ In the wikipedia article of the Schwarz-Pick theorem, it is mentioned that if equality holds, then $f$ is a Moebius transformation on the unit disk without proof. Is there a proof of this detail? Thank you. - $f$ is assumed to be a holomorphic function that maps the disk to itself, not necessarily an automorphism. Let $a$ be in the unit disk, and let $b=f(a)$. Let $\phi_a(z)=\frac{a-z}{1-\overline az}$ be the holomorphic automorphism of the disk that swaps $a$ and $0$, and similarly $\phi_b(z)=\frac{b-z}{1-\overline b z}$. If $g = \phi_b\circ f\circ \phi_a$, then $g$ is a holomorphic function that maps the disk to itself, and $g(0)=0$, so by Schwarz (or Cauchy's estimate) $\|g'(0)\|\leq 1$. Since $\phi_a'(0)=\|a\|^2-1$ and $\phi_b'(b)=\frac{1}{\|b\|^2-1}$, this yields by the chain rule $$\frac{1}{1-\|b\|^2}\|f'(a)\|(1-\|a\|^2)\leq 1,$$ or $$\frac{\|f'(a)\|}{1-\|b\|^2}\leq \frac{1}{1-\|a\|^2}.$$ Since $a$ was arbitrary and $b=f(a)$, this means that $$\frac{\|f'(z)\|}{1-\|f(z)\|^2}\leq \frac{1}{1-\|z\|^2}$$ for all $z$ in the disk. If for some $a$ the inequality had been equality, then by Schwarz we would have $g(z)=cz$ for some $c$ with $\|c\|=1$. This means that $f=\phi_b\circ g\circ \phi_a$ is a composition of three Möbius transformations and automorphisms of the unit disk, hence is one itself. Yes, it implies that $\|g'(0)\|=1$. The inequality is a rearrangment of (and equivalent to) $\|g'(0)\|\leq 1$. – Jonas Meyer Mar 14 '12 at 5:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9984667301177979, "perplexity": 83.13673453517508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00101-ip-10-185-27-174.ec2.internal.warc.gz"}
https://socratic.org/questions/68-ml-of-a-0-28-m-cacl2-solution-is-added-to-92-ml-of-a-0-46-m-cacl2-solution-de	Chemistry Topics # 68 mL of a 0.28 M CaCl2 solution is added to 92 mL of a 0.46 M CaCl2 solution. Determine the concentration of the combined solution. Show all work with units. How would you solve this? May 8, 2017 I hope I would solve it correctly. ......I finally get a concentration with respect to $C a C {l}_{2}$ of $\cong 0.4 \cdot m o l \cdot {L}^{-} 1$ #### Explanation: The basic definition of concentration is as amount of solute per unit volume. That is.......... $\text{Concentration"="Moles of solute"/"Volume of solution}$. Most of the time we want to assess $\text{moles of solute}$, and this is simply the product: $\text{Moles of solute"="Volume"xx"concentration}$. I would get familiar with this expression, because you will use it a lot. .........And in problems like these we must assume (REASONABLY!) that the volumes are additive. And so to address your problem (finally!), we solve the quotient: $\frac{0.092 \cdot L \times 0.46 \cdot m o l \cdot {L}^{-} 1 + 0.068 \cdot L \times 0.28 \cdot m o l \cdot {L}^{-} 1}{\left(92 + 68\right) \times {10}^{-} 3 \cdot L}$ $\cong 0.4 \cdot m o l \cdot {L}^{-} 1$ Do the units in the quotient cancel to give an answer in $m o l \cdot {L}^{-} 1$? It is your problem not mine. May 8, 2017 The concentration of the combined solution is 0.38 mol/L. #### Explanation: 1. Calculate the number of moles in each solution. 2. Use the total moles and the total volume to calculate the molarity of the combined solution. 1. Number of moles in each solution (a) Solution 1 ${\text{Moles of CaCl"_2 = 0.068 color(red)(cancel(color(black)("L solution"))) × ("0.28 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0190 mol CaCl}}_{2}$ (b) Solution2 ${\text{Moles of CaCl"_2 = 0.092 color(red)(cancel(color(black)("L solution"))) × ("0.46 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0423 mol CaCl}}_{2}$ Molarity of combined solutions $\text{Total moles" = "(0.0190 + 0.0423) mol" = "0.0613 mol}$ $\text{Total volume" = "(68 + 92) mL" = "160 mL" = "0.160 L}$ $\text{Molarity" = "moles"/"litres" = "0.0613 mol"/"0.160 L" = "0.38 mol/L}$ ##### Impact of this question 1678 views around the world	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8408674001693726, "perplexity": 1206.2391459320577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00279.warc.gz"}
https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_1&diff=prev&oldid=124249	# Difference between revisions of "2010 USAJMO Problems/Problem 1" ## Problem A permutation of the set of positive integers is a sequence such that each element of appears precisely one time as a term of the sequence. For example, is a permutation of . Let be the number of permutations of for which is a perfect square for all . Find with proof the smallest such that is a multiple of . ## Solutions We claim that the smallest is . ### Solution 1 Let be the set of positive perfect squares. We claim that the relation is an equivalence relation on . • It is reflexive because for all . • It is symmetric because . • It is transitive because if and , then , since is closed under multiplication and a non-square times a square is always a non-square. We are restricted to permutations for which , in other words to permutations that send each element of into its equivalence class. Suppose there are equivalence classes: . Let be the number of elements of , then . Now . In order that , we must have for the class with the most elements. This means , since no smaller factorial will have as a factor. This condition is sufficient, since will be divisible by for , and even more so . The smallest element of the equivalence class is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of . Also, each prime that divides divides all the other elements of , since and thus . Therefore for all . The primes that are not in occur an even number of times in each . Thus the equivalence class . With , we get the largest possible . This is just the set of squares in , of which we need at least , so . This condition is necessary and sufficient. ### Solution 2 This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation": It is possible to write all positive integers in the form , where is the largest perfect square dividing , so is not divisible by the square of any prime. Obviously, one working permutation of is simply ; this is acceptable, as is always in this sequence. Lemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities . Proof. Let and be the values of and , respectively, for a given as defined above, such that is not divisible by the square of any prime. We can obviously permute two numbers which have the same , since if where and are 2 values of , then , which is a perfect square. This proves that we can permute any numbers with the same value of . End Lemma Lemma 2. We will prove the converse of Lemma 1: Let one number have a value of and another, . and are both perfect squares. Proof. and are both perfect squares, so for to be a perfect square, if is greater than or equal to , must be a perfect square, too. Thus is times a square, but cannot divide any squares besides , so ; . Similarly, if , then for our rules to keep working. End Lemma We can permute numbers with the same in ways. We must have at least 67 numbers with a certain so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as , in general, we need numbers all the way up to , so obviously, is the smallest such number such that we can get a term; here 67 terms are 1. Thus we need the integers , so , or , is the answer. ## Solution Number Sense We have to somehow calculate the number of permutations for a given . How in the world do we do this? Because we want squares, why not call a number , where is the largest square that allows to be non-square? is the only square can be, which only happens if is a perfect square. For example, , therefore in this case . I will call a permutation of the numbers , while the original I will call . Note that essentially we are looking at "pairing up" elements between and such that the product of and is a perfect square. How do we do this? Using the representation above. Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes and if there are any odd numbers, those are the ones we have to match- in effect, they are the numbers mentioned at the beginning. By listing the values, in my search for "dumb" or "obvious" ideas I am pretty confident that only values with identical s can be matched together. With such a solid idea let me prove it. If we were to "pair up" numbers with different s, take for example with an of and with an of , note that their product gives a supposed of because the values cancel out. But then, what happens to the extra left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a "pair" has different values, and the smaller one is , in order for the product to leave a square, the larger value has to have not just but another square inside it, which is absurd because we stipulated at the beginning that was square-free except for the trivial multiplication identity, 1. Now, how many ways are there to do this? If there are numbers with , there are clearly ways of sorting them. The same goes for by this logic. Note that the as stated by the problem requires a thrown in there because , so there has to be a with 67 elements with the same . It is evident that the smallest will occur when , because if is bigger we would have to expand to get the same number of values. Finally, realize that the only numbers with are square numbers! So our smallest , and we are done. I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers! -expiLnCalc ## Solution Easy Consider the set of numbers such that is not divisible by any squares other than 1 and By changing we can encompass all numbers less than or equal to . Now notice that for a working arrangement these numbers can be permutated in any way to create a new one; for instance, the numbers have and they can be arranged in ways. Thus, since we need a permutation to have at least 67 elements (since 67 is prime). To minimize , we let and we have and we stop at to get . ~Leonard_my_dude~ ## Solution 5 It's well known that there exists and s.t. , s.t. no square divides other than 1, and is a perfect square. Lemma: is a perfect square if and only if We prove the backwards direction first. If , , which is a perfect square. We will now prove the forwards direction. We will prove the contrapositive: If , is not a perfect square. Note that if , There exists a prime p, s.t. . Also, . Thus, , making not a square. Thus, we can only match k with a_k if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is: For all , doesn't have a factor of 67. However, if , the first term will be a multiple of 2010, and thus the answer is	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9401416778564453, "perplexity": 255.78703446826668}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00532.warc.gz"}
http://www.cs.nyu.edu/pipermail/fom/2007-July/011711.html	# [FOM] 308:Large large cardinals Robert M. Solovay solovay at Math.Berkeley.EDU Sat Jul 7 03:34:14 EDT 2007 In a recent posting, Joe Shipman asks: How do you express "the existence of a nontrivial elementary embedding j:V into M, where V(lambda) containedin M." as a statement or scheme in ZFC, that refers only to sets and not classes? ********************************************************************** A good reference for this is Theorem 1.3 of the paper by Rich Laver "Implications between strong large cardinal axioms" (Annals of Pure and Applied Logic v. 90 (1997) 79-90.) [The cited theorem appears on page 83.] of them are assertions which are sentences of ZFC, The general context is that j is an elementary embedding of V_lambda into itself which is not the identity; lambda is a limit ordinal [necessarily of cofinality omega]. Let kappa_0 be the critical point of j and for n > 0 define kappa_n inductively to be j(kappa_{n-1}). Define a measure mu_n on P(kappa_n) by letting mu(A) = 1 (for A subseteq P(kappa_n)) iff j"kappa_n \in j(A). If n >= m then there is a natural map of P(kappa_n) onto P(kappa_m) (X --> X \cap kappa_m). It is easy to check that this map projects mu_n onto mu_m. We say that the tower of measures <mu_i :i in omega> is complete if whenever <A_n : n in omega> is a sequence of sets such that mu_n(A_n) = 1 for all n in omega then there is a set X \subseteq lambda such that X \cap kappa_n is in A_n for all n in omega. Then the axiom Shipman mentions is equivalent to: There is a j:V_lambda --> V_lambda (non-trivial, elementary and with lambda of cof omega) such that the associated tower of measures is complete. (This is (ii) on Laver's list.) We can describe how the elementary embedding of the axiom arises from the tower thus. We have a direct limit: V --> Ult(V, mu_0) --> Ult(V, mu_1) ... The direct limit is well-founded iff the tower is complete. The obvious map of V into the direct limit is the elementary embedding of the axiom in question. --Bob Solovay	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9724382162094116, "perplexity": 4610.60912527181}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398462987.25/warc/CC-MAIN-20151124205422-00129-ip-10-71-132-137.ec2.internal.warc.gz"}
https://www.varsitytutors.com/algebra_ii-help/quadratic-inequalities	# Algebra II : Quadratic Inequalities ## Example Questions ### Example Question #352 : Quadratic Equations And Inequalities Solve the following quadratic inequality, and report your answer in interval form: Explanation: The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal. Negative x^2's are hard to work with, so we multiply through by -1. Now we can factor easily. By the zero product property, each of these factors will be equal to 0. Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality. Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as: ### Example Question #353 : Quadratic Equations And Inequalities Solve the following quadratic inequality: Explanation: First we want to rewrite the quadratic in standard form: Now we want to set it = 0 and factor and solve like normal. Using the zero product property, both factors produce a zero: So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality. Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints. Ergo, . ### Example Question #354 : Quadratic Equations And Inequalities What is the discriminant of the following quadratic equation: Explanation: The discriminant of a quadratic equation in form is equal to . The given equation is not in that form however, so we must first multiply it out to get it into that form. We therefore obtain: We therefore have , and . Our discriminant is therefore: The correct answer is therefore ### Example Question #355 : Quadratic Equations And Inequalities Solve the following quadratic inequality: and Explanation: 1. Rewrite the equation in standard form. 2. Set the equation equal to and solve by factoring. So, and are our zeroes. 3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing by plugging it into your original inequality.) Because the above statement is true, the solution is the interval between and . ### Example Question #356 : Quadratic Equations And Inequalities Solve this inequality. Explanation: Combine like terms first. Factor The zeroes are 3 and 8 so a number line can be divided into 3 sections. X<3 works, 3<x<8 does not work, and x>8 works ### Example Question #357 : Quadratic Equations And Inequalities Solve: Explanation: Start by setting the inequality to zero and by solving for . Now, plot these two points on to a number line. Notice that these two numbers effectively divide up the number line into three regions: , and Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true. For , let Since this is not less than , the solution to this inequality cannot lie in this region. For , let . Since this will make the inequality true, the solution can lie in this region. Finally, for , let Since this number is not less than zero, the solution cannot lie in this region. Thus, the solution to this inequality is ### Example Question #358 : Quadratic Equations And Inequalities Solve: The solution cannot be determined with the information given. Explanation: First, set the inequality to zero and solve for . Now, plot these two numbers on to a number line. Notice how these numbers divide the number line into three regions: Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true. For , let Since this is not less than zero, the solution to the inequality cannot be found in this region. For , let Since this is less than zero, the solution is found in this region. For , let Since this is not less than zero, the solution is not found in this region. Then, the solution for this inequality is ### Example Question #17 : Solving Quadratic Functions Solve: Explanation: Start by changing the less than sign to an equal sign and solve for . Now, plot these two numbers on a number line. Notice how the number line is divided into three regions: Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds. For , let Since this number is not less than zero, the solution cannot be found in this region. For , let Since this number is less than zero, the solution can be found in this region. For let . Since this number is not less than zero, the solution cannot be found in this region. Because the solution is only negative in the interval , that must be the solution. ### Example Question #359 : Quadratic Equations And Inequalities Solve: Explanation: First, set the inequality to zero and solve for . Now, plot these two numbers on to a number line. Notice how these numbers divide the number line into three regions: Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true. For , let Since this solution is greater than or equal to , the solution can be found in this region. For , let Since this is less than or equal to , the solution cannot be found in this region. For , let Since this is greater than or equal to , the solution can be found in this region. Because the solution can be found in every single region, the answer to this inequality is ### Example Question #18 : Solving Quadratic Functions Which value for would satisfy the inequality ? Not enough information to solve	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.849868655204773, "perplexity": 515.1200814524143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00462.warc.gz"}
https://www.physicsforums.com/threads/copenhagen-many-worlds-falsified.23049/	# Copenhagen & Many Worlds Falsified? 1. Apr 28, 2004 Staff Emeritus Professor John Cramer, author of the transactional interpretation of QM, recently gave a talk at Boskone, the famous science fiction con, that sugggests a new experiment may be about to falsify the predictions of the Copenhagen and Many Worlds interpretations(and also decoherence and Consistent Histories). Not surprisingly, since Cramer is giving the talk, the transactional interpretation is NOT falsified by the experiment. Here's the Power Point presentation from the talk 2. Apr 28, 2004 ### ZapperZ Staff Emeritus It would have been nice if he had included an exact reference to the Afshar experiment, especially since his whole argument rests solely on the validity of its result. It is also strange that he argued that his theory of TI of QM and the formalism of QM both are consistent with the Afshar experiment, but CI and MW aren't. I find it strange because both CI and MW are BASED on the formalism of QM, i.e. they are not separate from it. So how he came up with the conclusion that they offer a result that differ from the QM formalism itself is puzzling. This is because if that is true, then one can LOGICALLY falsify both interpretations without having to do any experiment. Zz. 3. Apr 28, 2004 Staff Emeritus Read the presentation again, note his distinction between an interpretation and a model, and also the different responses to Afshar's wire grid. It seems that in spite of the mantra about the equivalence of the interpretations there is indeed a difference, and that the difference is testable. At any rate, if this is not so, the negation needs to be supported. 4. Apr 28, 2004 ### DrChinese Apparently, Afshar gave a talk at Texas A&M yesterday on a forthcoming paper: Violation of Bohr's principle of complementarity in an optical "which-way" experiment. I could not locate much on him. Unless the HUP is shown to be violated, I am not sure how an experiment is going to detect differences between competing interpretations. Maybe the answer is in the powerpoint from Cramer. But I would object to drawing a distinction between a model and an interpretation. Either way, we are dealing with "scientific theory" regardless of what you call it. 5. Apr 28, 2004 ### slyboy I am not familiar with Ashfar's experiment, but there is certainly a difference between the predictions of Many-Worlds and the orthodox von-Neumann approach to quantum mechanics. I will not discuss Copenhagen here, because it probably has enough variants to evade any experimental test you might propose. In the orthodox approach, wavefunction collapse is postulated to happen when you make a measurement. Whether this collapse is a real physical process, or just an updating of the information represented by the wavefunction is deliberately not addressed. However, if we can agree that some particular physical process constitutes a measurement, then we should not be able to recover the original superposition after the measurement has taken place. In contrast, Many-Worlds postulates no-collapse, so we should be able to undo a measurement if we can control a sufficiently large number of degrees of freedom of the quantum system, apparatus, environment, etc. The difficulty of testing this is to actually agree on what constitutes a measurement. Different interpretations can evade the predictions of many proposed experimental tests can by defining the notion of measurement differently. I suspect that a similar issues might arise in the analysis of Ashfar's experiment, but I have to look at it in more detail before I make any further comment. Proponents of different interpretations have got very good at making it look like their point of view is the only reasonable one to take. Therefore, I would advise people to take Cramer's view with a pinch of salt until the implications of the experiment have been fully debated. It is worth noting that Copenhagen, Many-Worlds and the transactional interpretation are not the only options on the market and I don't think that the dividing line between theory and interpretation can be made as clear as he thinks it is. 6. Apr 28, 2004 ### ZapperZ Staff Emeritus Obviously, I'm not the only one who has never heard of this experiment till now. Considering that I troll most of the major physics journals regularly (daily?), I'm quite curious as to where this result was published. Considering the implication of the experiment, such measurement would qualify to be considered for publication in "The Big 3" of physics journals. If it wasn't, why? [.. and all the usual warning bells typically would go off here - can we say "cold fusion"?] If anyone can find the citation to this experiment, pass it along. Thanks! Zz. 7. Apr 29, 2004 ### slyboy It appears that this result has not been published yet - nor is it available as a preprint. I imagine that it is in the process of being written up. Ashfar has just been giving a few seminars about the result to selected audiences and this is apparently how Cramer knows about it. We will have to wait until the paper appears to make any sensible comments. I hope he posts it to arXiv/quant-ph soon. 8. May 1, 2004 ### RageSk8 I was just about to post a topic about this, but lucky for me I searched the forum first. Anyways.... I'll quote from my source: "It has been widely accepted that the rival interpretations of quantum mechanics, e.g., the Copenhagen Interpretation, the Many-Worlds Interpretation, and my father John Cramer's Transactional Interpretation, cannot be distinguished or falsified by experiment, because the experimental predictions come from the formalism that all such interpretations describe. However, the Afshar Experiment demonstrates in an interaction-free way that there is a loophole in this logic: if the interpretation is inconsistent with the formalism, then it can be falsified. In particular, the Afshar Experiment falsifies the Copenhagen Interpretation, which requires the absence of interference in a particle-type measurement. It also falsifies the Many-Worlds Interpretation which tells us to expect no interference between "worlds" that are physically distinguishable, e.g., that correspond to the photon's measured passage through one pinhole or the other." http://www.kathryncramer.com/wblog/archives/000530.html [Broken] Last edited by a moderator: May 1, 2017 9. May 2, 2004 ### ZapperZ Staff Emeritus The blog did not indicate two things: (1) that there's no references to the Afshar experiment and (2) that it is having problems in the refereeing stage. In fact, unconfirmed reports have indicated that the Afshar experiment report that was uploaded to the e-print archive was removed, something that is unheard of for arXiv. http://groups.yahoo.com/group/undernetphysics/message/907 Zz. Last edited by a moderator: May 1, 2017 10. Jun 11, 2004 ### Eye_in_the_Sky I see that this thread hasn't been moving for quite some time. Nor has it been read much, lately, either. So, I'll post what I have to say only in brief. This is how I see the situation: 1) The reported results of the Afshar experiment are just what the quantum formalism predicts. 2) The type of thinking that Cramer employs - in the name of "Copenhagen" - in order to arrive at a conclusion which contradicts the Afshar results is, in fact, a manner of thought which the Copenhagen Interpretation itself deems invalid. 3) This manner of thought results from taking a rather extreme point of view regarding a notion which must be invoked in any typical discussion of a Wheeler-Delayed-Choice scenario. This type of thinking is inconsistent with the quantum formalism. Last edited: Jun 13, 2004 11. Jun 13, 2004 ### Eye_in_the_Sky Afshar Setup The following summary of the Afshar experiment may prove helpful. -------------------------- Afshar experimental arrangement: A beam of light of fixed wavelength from a coherent source is incident upon the following sequential arrangement: - double-slit [1, 2] - wire gird [WG] - lens [L] - particle detectors [1' ,2'] See diagram. -------------------------- Some details: 1) The wire grid WG consists of wires placed at the positions of the (would-be) interference minima (for a screen placed at that distance from the double-slit). 2) The lens L is such that rays from slit 1 (2), in the absence of any obstacles, will be focused at detector 1' (2'). 3) With only one slit open, and WG in place, the detector experiences a 6% loss relative to the setup with no WG. WG is therefore said to have 6% opacity. -------------------------- Summary of reported results: a) No WG + both slits open --> no loss; b) WG + only one slit open --> 6% loss; c) WG + both slits open --> < 0.1% loss. -------------------------- Conclusion: Since the results of c) are very nearly the same as those of a) and, on the other hand, appreciably different from those of b), the situation in c) must be that of "wave-like" behavior (i.e. the quantum system interacts with both slits, and not just one slit). That is to say, two waves are propagating, one from each slit, and in the vicinity of WG these waves overlap and interfere to produce minima at the locations of the wires in WG. For this reason, WG is essentially transparent to the incident beam. #### Attached Files: • ###### Afshar_Setup.jpg File size: 16.3 KB Views: 153 Last edited: Aug 1, 2004 12. Jun 13, 2004 ### DrChinese Thanks for supplying this information. It doesn't seen so weird to me. There must be a lot of experimental setups which demonstrate the wave nature of light. For example, you can insert a polarizer lens between 2 other polarizer lenses, and at the proper angles get more light from 3 than from 2. This loosely compares to the Afshar setup, which also gets more light with 2 slits than with 1. Do you have any idea how Afshar concludes this separates one QM interpretation from another? I cannot picture how the formalisms would yield anything but identical predictions. 13. Jun 13, 2004 ### Eye_in_the_Sky I think it's fair to say that, in the literal sense, by definition, an "interpretation of a formalism" is something necessarily consistent with that formalism. Of course, in the process of formulating such an interpretation, one could err in some way, and consequently, end up with something which does, in fact, contradict the formalism. Regarding this, I would use the expression "the interpretation fails" in order to indicate that the said interpretation is inconsistent with the formalism. With the above in mind, what Afshar and Cramer have claimed[] regarding "Copenhagen" and "Many Worlds" amounts to the following: These interpretations fail on account of certain unwitting implications which contradict the quantum formalism. That is to say, Afshar and Cramer claim that the authors of these interpretations have unintentionally "stepped out of line" with the quantum formalism, making assertions which, when given the proper consideration, turn out to have implications which contradict the formalism itself.[] _____________________ [] This is only one of two parts of their claim. The other part is that these interpretations fail in relation to what the experimental facts happen to be (whence the significance of the Afshar results). [**] Does the content of this post serve to dispel the strangeness about which ZapperZ wrote (in the second post of this thread)? Last edited: Jun 16, 2004 14. Jun 14, 2004 ### Eye_in_the_Sky Something is bothering me in what I wrote above ... in things like: How can one calculate what the formalism predicts, unless, at least at some level and to some degree, one employs an interpretation to "guide" that calculation? ------------------------ I think it would be better to skip all of the generalizations and go right to the unambiguous particulars of what Cramer purports to be implied by "Copenhagen" and "Many Worlds". Doing that will certainly clear things up. 15. Jun 22, 2004 ### Eye_in_the_Sky Cramer, in the name of "Copenhagen" Let us now establish the particulars of Cramer's argument in relation to the Copenhagen Interpretation. -------------------------------------- Two Measurement Options: Take a look at the diagram. It shows a modification of the Afshar setup in which the wire grid WG has been removed. For this setup, consider the following two measurement options: (a) Before light can reach the lens, at a screen labeled sigma1, measure an interference pattern; (b) After light has passed through the lens, at a screen labeled sigma2, measure two distinct, separate images, situated at 1' and 2'. -------------------------------------- Cramer's Rendition: Here is what Cramer has to say, in the name of "Copenhagen", regarding these two options (see diagram): (C-a) Measure at sigma1, the interference pattern, giving the wavelength and the momentum of the photon; (C-b) Measure at sigma2, which slit the particle passed through, giving its position. -------------------------------------- Complementarity: By statements (C-a) and (C-b), Cramer purports that, according to the Copenhagen Interpretation, the following is true: P1: Options (a) and (b) are experiments which measure complementary attributes of the quantum system. -------------------------------------- Argument for (C-b): Cramer's statement (C-b) regarding option (b) is built upon the following type of argument: P2: A photon arriving at 1' (2') must have passed through slit 1 (2) and not 2 (1). Therefore, a measurement at sigma2 corresponds to a "determination" of which slit the photon went through. Since this type of argument serves as a basis for Cramer's statement (C-b) concerning (b), and (C-b) itself is said in the name of "Copenhagen", it must be that the precise sense in which an argument like P2 eventually leads to Cramer's (C-b) is seen by Cramer to be deemed valid by the Copenhagen Interpretation. -------------------------------------- Cramer's Conclusion: In a final step, the wire grid WG is put back into the arrangement of (b) to give, once again, the Afshar setup. With all of the above in mind, Cramer then concludes - in the name of "Copenhagen" - regarding the Afshar experiment (see diagram): The measurement-type forces particle-like behavior, so there should be no interference, and no minima ... This statement makes the intended meaning of Cramer's (C-b) absolutely clear. That intended meaning is equivalent to the following proposition: P3: In option (b), the measurement performed at sigma2 is physically equivalent to a nondemolition measurement which takes place at the site of the slits and determines which slit the photon passes through. To repeat, according to Cramer, P3 (like the propositions P1 and P2 above) is deemed correct by "Copenhagen". #### Attached Files: File size: 11.5 KB Views: 151 File size: 26.9 KB Views: 153 • ###### Afshar_Copen.jpg File size: 28.5 KB Views: 165 Last edited: Jul 30, 2004 16. Jun 22, 2004 ### Locrian So did this experiment ever get published? 17. Jul 26, 2004 ### Tony Wagstaff I can accept this, but how does Afshar's setup determine that photon's momentum when, excluding those points barred by the wires, it could have passed through the lens at any other point with varying probability? 18. Jul 26, 2004 ### FZ+ This has gotten into Newscientist as well. I can't claim to be an expert, but I am not convinced. Why must a photon arriving at 1' (2') have passed through slit 1 (2)? 19. Jul 26, 2004 ### Eye_in_the_Sky The Afshar setup doesn't determine the photon's momentum (... except for the magnitude, in that a lens with such-and-so index of refraction focuses the light as it does (... but this information, like that of the alleged "position measurement", is only retrodictive)). However, in order for Cramer to make his point, there is no need for a momentum determination. And what is Cramer's point? His point (erroneously made!) is: According to the Copenhagen Interpretation, proposition P3 is true. -------------------------------- The Copenhagenist is not convinced either. Here is his response: Proposition P2 is a "valid" statement in the following sense: Given that a photon has arrived at 1' (2'), then if a nondemolition measurement had been performed at the site of the slits, the photon would have been found to pass through slit 1 (2). However, in the absence of the actual performance of such a measurement, it is in general incorrect to say that, on account of our "knowing" what the result of such a measurement would have been, the quantum phenomena will be the same as those obtained when the measurement is in fact performed. That it is incorrect to say so follows from Bohr's "Principle of Nonseparability". In fact, with a bit of reflection on the two "measurement options", we see that they are not at all "complementary". They were: Option (a) is consistent with a "wave picture" but not a "particle picture", whereas, option (b), on the other hand, is consistent with both a "wave picture" and a "particle picture". The two options are therefore not complementary. All of this P2-like thinking is due to a "playing" of "Wheel-er's Game of Delayed Choice of Fortune". Last edited: Jul 27, 2004 20. Jul 27, 2004 ### Ontoplankton Is that really what the MWI tells us? I thought it wasn't. I think the claim that the MWI is inconsistent with the formalism of QM is rather strange, because essentially MWI is what happens when you take the formalism literally (it says there's a wave function, and nothing else). It looks to me like what has actually been falsified, if anything, is the standard strawman version of MWI that has world-splitting as a fundamental physical process. Am I correct? 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https://web2.0calc.com/questions/find-the-positive-difference-between-the-solutions	+0 # Find the positive difference between the solutions to the equation 6t^2 + 30 = 41t 0 160 1 Find the positive difference between the solutions to the equation 6t^2 + 30 = 41t Jul 20, 2021 #1 +35247 +1 re-write as 6t^2 - 41t + 30 = 0 Use Quadratic formula with a = 6 b=-41 c= 30 $$x = {-(-41) \pm \sqrt{(-41)^2-4(6)(30)} \over 2(6)}$$ to find x = 6 and x = 5/6 You can finish..... Jul 21, 2021	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9167992472648621, "perplexity": 4467.954200079799}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00070.warc.gz"}
https://stats.stackexchange.com/questions/437390/joint-models-with-time-to-progression	# Joint models with time to progression Consider a RCT (Randomized Controlled Trial) which aims at assessing the efficacy of a drug in patients suffering from a given cancer. In this trial, $$p$$ individuals are observed at several time points. At a certain point in time, some individuals are given a placebo and others are given the drug. Assume (out of simplicity) that each patient only has one tumor. The efficacy of the treatment (for a given individual $$i$$) can be assessed by measuring the tumor size $$M_i(t)$$. Longitudinal modeling can be used to gain insights on the way the tumor size evolves over time. The true time of disease progression, denoted by $$P_i$$ $$(1 \leq i \leq p$$), can be defined as the earliest time at which some criteria is met. In the following, we define progression time to be the time at which $$M_i(t) > c M_i(0)$$ (for some $$c > 1$$). This progression time is usually not observed as it may occur between two consecutive visits. We can define $$P_{i}^{\mathrm{obs}}$$ the earliest visit at which the criteria is met. In addition to this, this progression time can be right-censored as individuals may dropout before progression is actually observed. Let $$C_i$$ denote the censoring time. We actually observe $$T_i = \min(C_i, P_{i}^{\mathrm{obs}})$$ and the tumor sizes at each visit $$\mathbf{m}_i = (m_{i,j})_{1 \leq j \leq n_i}$$. In this situation, the time-to-event is censored but the observations are censored too. An individual who reaches "progression" is removed from the RCT. Joint models allow to combine survival analysis (here, the "time-to-event" is the time to progression) and longitudinal data analysis (modeling of tumor growth). As presented in this book by D. Rizopoulos), a joint model consists of a Cox proportional hazard model for the survival part and a Linear Mixed Effects (LME) model for the longitudinal part. Given that, for each individual, the observations are: $$(T_i, \delta_i, \mathbf{m}_i)$$ with $$\delta_i = \mathbb{1}_{P_{i}^{\mathrm{obs}} \leq C_i}$$, one can write the likelihood $$p(T_i, \delta_i, \mathbf{m}_i \mid \theta)$$ as: $$p(T_i, \delta_i, \mathbf{m}_i \mid \theta) = \int p(T_i, \delta_i, \mathbf{m}_i \mid \mathbf{b}_i, \theta) p(\mathbf{b}_i \mid \theta) \, d\mathbf{b}_i,$$ where $$\mathbf{b}_i$$ denote the random effects of the LME model and $$\theta$$ denote the model parameters. A key assumption is the following: conditionally on $$\mathbf{b}_i$$, the survival part and longitudinal part are independent. That is: $$p(T_i, \delta_i, \mathbf{m}_i \mid \mathbf{b}_i, \theta) = p(T_i, \delta_i \mid \mathbf{b}_i, \theta) p(\mathbf{m}_i \mid \mathbf{b}_i, \theta).$$ I am wondering whether this assumption (the conditional independence) actually holds in the situation I presented above. In many examples [used to illustrate joint models], there is no explicit relationship between the progression of the measurement and the time-to-event. For instance, there is no explicit relationship between time to death and the number of CD4 cells in patients suffering from AIDS. Still, in the present case, the relationship between the time-to-event (progression) and the longitudinal trajectory (tumor size) is explicit. How can I include this explicit relationship between the time-to-event and the longitudinal trajectory in a joint model? More specifically, should the survival part of the joint model depend on the threshold $$c$$? ## 2 Answers It seems to me that the only information is in the tumor size process $$M_i(t)$$, $$i = 1, \ldots, n$$. Either • you treat this as a longitudinal outcome evaluated at some follow-up times $$t_{ij}$$, $$j = 1, \ldots, n_i$$, and you fit an appropriate mixed effects model describing the average longitudinal evolutions; • or you are interested in the time until $$M_i(t) > c M_i(0)$$ that should be interval censored data, and you fit an appropriate survival model for it. It is not evident why you want to consider the same process twice in a joint model. At the point of model convergence, you will have (Pearson / Schoenfeld) residuals for the mixed and Cox models. Plot the residuals against each other. Inspect any possible trend. If the residuals show a trend, there are likely time-dependent effects of treatment on tumor size/status and a more sophisticated treatment should be handled such as time varying covariates. • I think I understand your question better. Basically, you are concerned because the time-to-PD is "explicitly" (completely) dependent on the tumor size, that simply adjusting for treatment assignment doesn't work, right? Are you using RECIST to measure PD? If so, then the only case that PD isn't a function of target or non-target lesion growth is the appearance of new lesions. Fit that as a separate event time. Does that answer your question? – AdamO Nov 22 '19 at 16:45 • Yes, time-to-PD depends on the tumor size. Indeed, "progression" is defined as the time at which the tumor becomes "too big". In a joint model, the survival part assumes that: $$h_i(t \mid M_i(t), w_i) = h_0(t) \exp\left( \gamma^{\top}w_i + \alpha m_i(t) \right),$$ and $$S_i(t) = \mathbb{P}\left( P_i^{\mathrm{obs}} > t \right) = \exp\left( \int_{0}^{+\infty} h_i(s) \, ds \right),$$ with $w_i$ some baseline covariates. I was under the impression that this could be modified to make the relationship between time-to-pd and tumor size explicit. – Pouteri Nov 22 '19 at 17:05 • @Pouteri could you clarify if disease progression is RECIST or not? – AdamO Nov 22 '19 at 17:08 • It could be another progression criteria than RECIST. We just assume that the progression time can be explicitly obtained from the time-varying covariates (RECIST is an example). – Pouteri Nov 22 '19 at 17:12 • @Pouteri basically, response assessment (change in tumor volume or area from baseline) is precisely what is used to determine whether PD occurred. So conditional on the longitudinal volume of a single-lesion tumor, there is 0 information added by the designation of "PD". – AdamO Nov 22 '19 at 17:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8134128451347351, "perplexity": 805.7637135179438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00007.warc.gz"}
http://www.mathworks.com/help/comm/ref/qpskdemodulatorbaseband.html?requestedDomain=www.mathworks.com&nocookie=true	# Documentation ### This is machine translation Translated by Mouse over text to see original. Click the button below to return to the English verison of the page. # QPSK Demodulator Baseband Demodulate QPSK-modulated data ## Library PM, in Digital Baseband sublibrary of Modulation ## Description The QPSK Demodulator Baseband block demodulates a signal that was modulated using the quaternary phase shift keying method. The input is a baseband representation of the modulated signal. The input must be a complex signal. This block accepts a scalar or column vector input signal. For information about the data types each block port supports, see Supported Data Types. ### Algorithm Hard-Decision QPSK Demodulator Signal Diagram for Trivial Phase Offset (odd multiple of ) Hard-Decision QPSK Demodulator Floating-Point Signal Diagram for Nontrivial Phase Offset Hard-Decision QPSK Demodulator Fixed-Point Signal Diagram for Nontrivial Phase Offset The exact LLR and approximate LLR cases (soft-decision) are described in Exact LLR Algorithm and Approximate LLR Algorithm in the Communications System Toolbox User's Guide. ## Dialog Box The phase of the zeroth point of the signal constellation. Constellation ordering Determines how the block maps each integer to a pair of output bits. Output type Determines whether the output consists of integers or bits. If the Output type parameter is set to `Integer` and Constellation ordering is set to `Binary`, then the block maps the point exp(jθ + jπm/2) to m, where θ is the Phase offset parameter and m is 0, 1, 2, or 3. The reference page for the QPSK Modulator Baseband block shows the signal constellations for the cases when Constellation ordering is set to either `Binary` or `Gray`. If the Output type is set to `Bit`, then the output contains pairs of binary values if Decision type is set to `Hard decision`. The most significant bit (i.e. the left-most bit in the vector), is the first bit the block outputs. If the Decision type is set to ```Log-likelihood ratio``` or `Approximate log-likelihood ratio`, then the output contains bitwise LLR or approximate LLR values, respectively. Decision type Specifies the use of hard decision, LLR, or approximate LLR during demodulation. This parameter appears when you select `Bit` from the Output type drop-down list. The output values for Log-likelihood ratio and Approximate log-likelihood ratio decision types are of the same data type as the input values. For integer output, the block always performs Hard decision demodulation. See Exact LLR Algorithm and Approximate LLR Algorithm in the Communications System Toolbox User's Guide for algorithm details. Noise variance source This field appears when ```Approximate log-likelihood ratio``` or `Log-likelihood ratio` is selected for Decision type. When set to `Dialog`, the noise variance can be specified in the Noise variance field. When set to `Port`, a port appears on the block through which the noise variance can be input. Noise variance This parameter appears when the Noise variance source is set to `Dialog` and specifies the noise variance in the input signal. This parameter is tunable in normal mode, Accelerator mode and Rapid Accelerator mode. If you use the Simulink® Coder™ rapid simulation (RSIM) target to build an RSIM executable, then you can tune the parameter without recompiling the model. This is useful for Monte Carlo simulations in which you run the simulation multiple times (perhaps on multiple computers) with different amounts of noise. The LLR algorithm involves computing exponentials of very large or very small numbers using finite precision arithmetic and would yield: • `Inf` to `-Inf` if Noise variance is very high • `NaN` if Noise variance and signal power are both very small In such cases, use approximate LLR, as its algorithm does not involve computing exponentials. Data Types Pane for Hard-Decision Output For bit outputs, when Decision type is set to `Hard decision`, the output data type can be set to `'Inherit via internal rule'`, ```'Smallest unsigned integer'```, `double`, `single`, `int8`, `uint8`, `int16`, `uint16`, `int32`, `uint32`, or `boolean`. For integer outputs, the output data type can be set to ```'Inherit via internal rule'```, `'Smallest unsigned integer'`, `double`, `single`, `int8`, `uint8`, `int16`, `uint16`, `int32`, or `uint32`. When this parameter is set to ```'Inherit via internal rule'``` (default setting), the block will inherit the output data type from the input port. The output data type will be the same as the input data type if the input is a floating-point type (`single` or `double`). If the input data type is fixed-point, the output data type will work as if this parameter is set to `'Smallest unsigned integer'`. When this parameter is set to `'Smallest unsigned integer'`, the output data type is selected based on the settings used in the Hardware Implementation pane of the Configuration Parameters dialog box of the model. If `ASIC/FPGA` is selected in the Hardware Implementation pane, and Output type is `Bit`, the output data type is the ideal minimum one-bit size, i.e., `ufix(1)`. For all other selections, it is an unsigned integer with the smallest available word length large enough to fit one bit, usually corresponding to the size of a char (e.g., `uint8`). If `ASIC/FPGA` is selected in the Hardware Implementation pane, and Output type is `Integer`, the output data type is the ideal minimum two-bit size, i.e., `ufix(2)`. For all other selections, it is an unsigned integer with the smallest available word length large enough to fit two bits, usually corresponding to the size of a char (e.g., `uint8`). Derotate factor This parameter only applies when the input is fixed-point and Phase offset is not an even multiple of $\frac{\pi }{4}$. You can select `Same word length as input` or ```Specify word length```, in which case you define the word length using an input field. Data Types Pane for Soft-Decision For bit outputs, when Decision type is set to `Log-likelihood ratio` or ```Approximate log-likelihood ratio```, the output data type is inherited from the input (e.g., if the input is of data type `double`, the output is also of data type `double`). ## Examples expand all Modulate and demodulate a noisy QPSK signal. Open the QPSK demodulation model. Run the simulation. The results are saved to the base workspace, where the variable `ErrorVec` is a 1-by-3 row vector. The BER is found in the first element. Display the error statistics. For the Eb/No provided, 4.3 dB, the resultant BER is approximately 0.01. Your results may vary slightly. ```ans = 0.0112 ``` Increase the Eb/No to 7 dB. Rerun the simulation, and observe that the BER has decreased. ```ans = 1.0000e-03 ``` ## Supported Data Types PortSupported Data Types Input • Double-precision floating point • Single-precision floating point • Signed fixed-point when: • Output type is `Integer` • Output type is `Bit` and Decision type is `Hard-decision` Var • Double-precision floating point • Single-precision floating point Output • Double-precision floating point • Single-precision floating point • Boolean when Output type is `Bit` and Decision type is `Hard-decision` • 8-, 16-, 32- bit signed integers • 8-, 16-, 32- bit unsigned integers • ufix(1) in ASIC/FPGA when Output type is `Bit` • ufix(2) in ASIC/FPGA when Output type is `Integer` ## HDL Code Generation This block supports HDL code generation using HDL Coder™. HDL Coder provides additional configuration options that affect HDL implementation and synthesized logic. For more information on implementations, properties, and restrictions for HDL code generation, see QPSK Demodulator Baseband in the HDL Coder documentation. ## Pair Block QPSK Modulator Baseband	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.855696976184845, "perplexity": 2444.329920572353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988722653.96/warc/CC-MAIN-20161020183842-00165-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.helpteaching.com/lessons/147/electromagnetic-spectrum	• ### Browse All Lessons ##### Assign Lesson Help Teaching subscribers can assign lessons to their students to review online! Tweet # Electromagnetic Spectrum Introduction: The electromagnetic spectrum represents the range of wavelengths or frequencies on which that electromagnetic radiation exists. Electromagnetic radiation on this spectrum is important to consider, especially with increasing concerns over Earth and changing climates. The use of fireworks, for example, destroys the ozone layer, leading to an increased amount of ultraviolet radiation entering and leading to increases in cases of skin cancer in humans. The electromagnetic spectrum can be arranged from lowest to highest frequency as follows: Radio waves, microwaves, infrared radiation, visible light (ROYGBIV), ultraviolet radiation, X-rays, and gamma rays. In this arrangement, as frequency increases going from radio waves to gamma rays, the wavelength will decrease and the energy of the electromagnetic radiation will increase. In general, electromagnetic radiation with higher frequencies and energies tend to be more dangerous. This is why destroying the ozone layer causes a major problem for humans, since ultraviolet radiation entering the Earth has high energy and high frequencies associated with it. It is also worth noting that the visible spectrum, in order from lowest to highest frequency, can be arranged as follows: red, orange, yellow, green, blue, indigo, and violet. Required Video Related Worksheets:	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9081031084060669, "perplexity": 715.2569854555111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00132.warc.gz"}
http://mathhelpforum.com/math-software/84263-matlab-probability.html	## [MATLAB] Probability Hello, I am currently really new with MATLAB and my lectures didn't really provide sufficient practice and exposure to how to function a MATLAB and I have an assignment due really soon so I hope I could get you guys to teach me how to deal with these questions. 1. The dice game craps is played as follows. The player throws two dice, and if the sum is seven or eleven, then he wins. If the sum is two, three of twelve, then he loses. If the sum is anything else, then he continues throwing until he either throws that number again (in which case he wins) or he throws a seven (in which case he loses). Estimate the probability that the player wins. You need to add commands to estimate the probability. The code provided is, clear all nreps=100000; Event_W = 0; for n=1:nreps a= %Simulate the throw of two dice using the function 'randunifd' if %Add code such that player obtain a 7 or 11 Event_W = Event_W +1; elseif %Add code such that player obtains a 2,3 or 12 else Throw = 1; while %Add condition such that we keep playing until player either wins or loses b= randunifd(1,6)+randunifd(1,6); if %Add condition such that player wins Event_W = Event_W+1; Throw=0; elseif %Add condition such that player loses Throw=0; end a=b; end end end RelFreq = Event_W/nreps I really have no idea how to do this at all. Your help would be greatly appreciated. 2. Consider the following experiment. We toss a fair coin 20 times. Let X be the length of the longest sequence of Heads. (i) Estimate the probability function p of X. That is for x = 1, ....,20, estimate $p(x) = P(X=x)$ (ii) Give an estimate of the mean value of X. You need to add commands where indicated in the MATLAB file. Here's the MATLAB file I got : clear all nreps=10000; p=zeros(1,20); %p is a vector of size 20 such that p(i)=P(X=i) for n=1:nreps n_heads = 0; %Length of current Head sequence max_heads= 0; %Length of maximum length sequence so far for %Add code to toss a coin 20 times if %Add condition to obtain a Head n_heads = n_heads +1; else n_heads = 0; end if %Add condition to see if current sequence is the longest max_heads = n_heads; end end %Add code to increment p(max_heads) by 1 end RelFreq = p/nreps %Add code to estimate the mean of X Thanks for your help! (:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9257701635360718, "perplexity": 1580.994329960237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257836399.81/warc/CC-MAIN-20160723071036-00161-ip-10-185-27-174.ec2.internal.warc.gz"}
https://brilliant.org/problems/extrinsic-semiconductor/	# Extrinsic Semiconductor The above are schematic diagrams of extrinsic semiconductors, into which doping agents of Boron(B) and phosphorus(P) have been introduced, respectively. Which of the following statements is correct? a) Silicon(Si) has $$4$$ valence electrons. b) Phosphorus(P) has $$3$$ valence electrons. c) The above left are N-type semiconductors and the above right are P-type semiconductors. d) Electrons are the majority charge carriers of the above right semiconductors, and holes are the majority charge carriers of the above left semiconductors. ×	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.929071307182312, "perplexity": 3489.1016569426683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549428325.70/warc/CC-MAIN-20170727162531-20170727182531-00645.warc.gz"}
http://math.stackexchange.com/questions/126250/difference-in-probability-distributions-from-two-different-kernels	# Difference in probability distributions from two different kernels I wonder if the probability kernels of Markov processes on the same state space are close enough, does it also hold for the probabilities of the event that depend only on first $n$ values of the process. More formally, let $(E,\mathscr E)$ be a measurable space and put $(E^n,\mathscr E^n)$, where $\mathscr E^n$ is the product $\sigma$-algebra. We say that $P$ is a stochastic kernel on $E$ if $$P:E\times\mathscr E\to [0,1]$$ is such that $P(x,\cdot)$ is a probability measure on $(E,\mathscr E)$ for all $x\in E$ and $x\mapsto P(\cdot,A)$ is a measurable function for all $A\in \mathscr E$. On the space $b\mathscr E$ of real-valued bounded measurable functions with a sup-norm $\\|f\\| = \sup\limits_{x\in E}\|f(x)\|$ we define the operator $$Pf(x) = \int\limits_E f(y)P(x,dy).$$ Its induced norm is given by $\\|P\\| = \sup\limits_{f\in b\mathscr E\setminus 0}\frac{\\|Pf\\|}{\\|f\\|}.$ Furthermore, for any stochastic kernel $P$ we can assign the family of probability measures $(\mathsf P_x)_{x\in E}$ on $(E^n,\mathscr E^n)$ which is defined uniquely by $$\mathsf P_x(A_0\times A_1\times \dots\times A_n) = 1_{A_0}(x)\int\limits_{A_n}\dots \int\limits_{A_1}P(x,dx_1)\dots P(x_{n-1},dx_n).$$ Let us consider another kernel $\tilde P$ which as well defines the operator on $b\mathscr E$ and the family of probability measures $\tilde{\mathsf P}_x$ on $(E^n,\mathscr E^n)$. I wonder what is the upper-bound on $$\sup\limits_{x\in E}\sup\limits_{F\in \mathscr E^n}\|\tilde{\mathsf P}_x(F) - \mathsf P_x(F)\|.$$ By induction it is easy to prove that $$\sup\limits_{x\in E}\|\tilde{\mathsf P}_x(A_0\times A_1\times \dots\times A_n)-\mathsf P_x(A_0\times A_1\times \dots\times A_n)\|\leq n\cdot\\|\tilde P - P\\|$$ but I am not sure if this result can be extended to any subset of $\mathscr E^n$. - The notation can be a bit cumbersome because of the nested integrals, but this solution relies only on very basic properties of integration and is direct (no induction). Consider the difference $a(x_0,F)=\mathsf P'_{x_0}(F)-\mathsf P_{x_0}(F)$. By uniqueness of measure it follows from the definition of $\mathsf P_{x_0}$ that: \begin{align} a(x_0,F) =&\int_E\dots\int_E 1_F(x_1,\dots x_n) P'(x_{n-1},dx_n)\dots P'(x_0,dx_1)\\ &- \int_E\dots\int_E 1_F(x_1,\dots x_n) P(x_{n-1},dx_n)\dots P(x_0,dx_1) \end{align} By introducing intermediate telescoping terms we can split this into a sum of $n$ terms $a(x_0,F)=\sum_{j=1}^n a_j(x_0,F)$ where \begin{align} a_j(x_0,F) =& \int_E\dots\int_E 1_F(x_1,\dots x_n) P'(x_{n-1},dx_n) \dots P'(x_{j-1},dx_j) P(x_{j-2},dx_{j-1})\dots P(x_0,dx_1)\\ &- \int_E\dots\int_E 1_F(x_1,\dots x_n) P'(x_{n-1},dx_n) \dots P'(x_j,dx_{j+1}) P(x_{j-1},dx_j)\dots P(x_0,dx_1) \end{align} The innermost part (consisting of $n-j$ nested integrals) is common to both terms, to make things more readable we factor it into $$g_j(x_1,\dots x_j) = \displaystyle\int_E\dots\int_E 1_F(x_1,\dots x_n) P'(x_{n-1},dx_n)\dots P'(x_j,dx_{j+1})$$ Then by linearity and $\|\int f\|\le\int \|f\|$: \begin{align} \|a_j(x_0,F)\|\le& \int_E\dots\int_E\left\|\int_E g_j(x_1,\dots x_j) \left(P'(x_{j-1},dx_j)-P(x_{j-1},dx_j)\right)\right\| P(x_{j-2},dx_{j-1})\dots\\ \le& \int_E\dots\int_E\\|P'-P\\| P(x_{j-2},dx_{j-1})\dots P(x_0,dx_1)\\ =& \\|P'-P\\|\\ \|a(x_0,F)\|\le& n\cdot\\|P'-P\\| \end{align} (the bound of the integral by $\\|P'-P\\|$ comes from considering the function $g_j(x_1,\dots x_{j-1},\cdot)$) - Thanks for your effort - but I tried to state a problem in a rigorous way and was willing to get the answer given in the same way. Conditioning on $X_{j-1} = x$, say, does not look rigorous. – Ilya Apr 19 '12 at 22:21 You're right, the notation was a bit sloppy, I updated the answer with a simpler solution that keeps your original notation. – Generic Human Apr 21 '12 at 12:07 +1 Now I understand your idea and I liked it, but it should be further formalized. Namely, the first formula does not seem to be obvious and requires proof as I made in step 2. of my answer. – Ilya Apr 22 '12 at 11:15 I'm not sure which formula you're referring to. The first line is a direct consequence of unicity: each term in the RHS clearly defines a measure on $\mathscr E^n$ and it matches the LHS for cartesian products (factor $1_F$ into the individual $1_{A_i}(x_i)$ which you can now move next to the corresponding integral sign), so the equality holds. In the second line, the fact that $a$ is the sum of the $a_j$ is due to the fact that all terms cancel out except the first and the last one: you don't need to look at the contents of the integrals. – Generic Human Apr 23 '12 at 0:14 I hope I have a solution for the problem, so I post it here. I'll be happy if you comment on the solution if it is correct or maybe provide more short and neat one. 1. First of all, I change a notation a bit and use $\mathsf P_x^n$ instead of $\mathsf P_x$ in OP to denote the probability measure on the space $(E^n,\mathcal E^n)$, just to mention the dependence on $n$ expolicitly. Then for all measurable rectangles $B = A_1\times A_2\times\dots\times A_n\in \mathcal E^{n-1}$ and the set $A_0\in \mathcal E$ it holds that $$\mathsf P_x^n(A_0\times B) = 1_{A_0}(x)\int\limits_{A_1}\dots \int\limits_{A_n}P(x_{n-1},dx_n)\dots P(x,dx_1) = 1_{A_0}(x)\int\limits_E \mathsf P_{y}^{n-1}(B)P(x,dy).$$ By the uniqueness of the probability measure $\mathsf P_x^n$ the same result holds for any $B\in \mathcal E^{n-1}$: $$\mathsf P_x^n(A_0\times B) = 1_{A_0}(x)\int\limits_E \mathsf P_{y}^{n-1}(B)P(x,dy). \tag{1}$$ 2. For any set $C\in \mathcal E^n = \mathcal E\times\mathcal E^{n-1}$ we can show that $$\mathsf P_x^n(C) = \int\limits_E \mathsf P_y^{n-1}(C_x)P(x,dy) \tag{2}$$ where $C_x = \{y\in E^{n-1}:(x,y)\in C\}\in\mathcal E^{n-1}$. To prove it we first verify $(2)$ for measurable rectangles $C = A\times B$ using $(1)$, hence $C_x = B$ if $x\in A$ and $\emptyset$ otherwise. By the advise of @tb this result further extends to all $C\in \mathcal E^n$ by $\pi$-$\lambda$ theorem. 3. The inequality $\left\|\tilde{\mathsf P}_x^n(C) - \mathsf P_x^n(C)\right\|\leq n\\|\tilde P-P\\|$ can be proved then by induction: it clearly holds for $n=1$ $$\left\|\tilde{\mathsf P}^1_x(C) - \mathsf P^1_x(C)\right\| = \left\|\tilde P(x,C_x) - P(x,C_x)\right\|\leq 1\cdot\\|\tilde P - P\\|.$$ If the same inequality holds for $n-1$, we have \begin{align} \left\|\tilde{\mathsf P}_x^n(C) - \mathsf P_x^n(C)\right\| &= \left\|\int\limits_E \tilde{\mathsf P}_y^{n-1}(C_x)\tilde P(x,dy)-\int\limits_E \mathsf P_y^{n-1}(C_x)P(x,dy)\right\| \\ &\leq \left\|\int\limits_E \left(\tilde{\mathsf P}_y^{n-1}(C_x) - \mathsf P_y^{n-1}(C_x)\right)\tilde P(x,dy)\right\|+\left\|\langle \tilde P(x,\cdot) - P(x,\cdot),\mathsf P_{(\cdot)}^{n-1}(C_x)\rangle\right\| \\ &\leq (n-1)\\|\tilde P - P\\|+\\|\tilde P - P\\| = n\\|\tilde P - P\\| \end{align} where $\langle m(\cdot),f(\cdot)\rangle = \int\limits_E f(y)\mu(dy)$ for all measurable bounded $f$ and all finite signed measures $\mu$. Since the RHS in the bound derived does not depend on $x,C$ it means that we proved desired bounds. - Great solution! – Generic Human Apr 21 '12 at 12:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9929066300392151, "perplexity": 278.9459334137524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042986444.39/warc/CC-MAIN-20150728002306-00048-ip-10-236-191-2.ec2.internal.warc.gz"}
https://dsp.stackexchange.com/questions/15190/how-to-create-a-convolution-matrix-with-a-variable-condition-number-cn/17595	How to Create a Convolution Matrix with a Variable Condition Number (CN) I want to know the performance of a deconvolution algorithm with different CN, so I'm convolving my signal with different convolution matrices(different CNs) and then applying the deconvolution algorithm, then the error between the original and reconstructed signals is measured. Is there a proper way to create the convolution matrix with variable CN. • Can you please review my answer? If it satisfies you, please mark it. If not, let me know what's missing and I will improve it. – Royi Oct 9 at 11:27 2 Answers Let's try to think of it intuitively. Given the LPF what would make the inverse "Hard"? The sections we won't be able to recover are where the LPF is 0, since there is no inverse for that we can multiply the result by. Real world LPF won't reach "Real" zero as usually. But what would make it hard to recover is where there are big ratios between the biggest and the smallest magnitude. As close and fast your LPF goes to zero (And has magnitudes which are big) the harder the recovery will be. Now just build analog LPF which this requirements at different scales, digitize it, create the Convolution Matrix and there you have it... A practical example is given in my answer to 1D Deconvolution with Gaussian Kernel (MATLAB). The condition number (CN) of a matrix (in $$L_2$$ norm) is the square root of of the ratio of the largest eigenvalue to the smallest eigenvalue. We see here that CN is related to the extreme (largest and smallest) eigenvalues. It looks like we can vary the CN by varying the eigenvalues. Convolution matrix is a Toeplitz matrix. There are some special Toeplitz matrices such as tri-diagonal and penta-diagonal matrices whose eigenvalues in terms of matrix entries are well-known [1-2]. In these case, by varying the matrix entries we can vary the eigenvalues and in turn the CN. [1] S. Noschese, L. Pasquini, and L. Reichel, "Tridiagonal Toeplitz matrices: properties and novel applications," Numer. Linear Algebra Appl., 20 (2013), pp. 302-326 [2] G. D. Smith, Numerical Solution of Partial Diﬀerential Equations, 2nd ed., Clarendon Press, Oxford, 1978.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9023100137710571, "perplexity": 809.7386043775905}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00167.warc.gz"}
http://mathhelpforum.com/pre-calculus/148275-ln-x-e-x-find-x.html	# Math Help - ln(x)=e^-x find x 1. ## ln(x)=e^-x find x ln(x)=e^(-x) find x... 2. Originally Posted by ice_syncer ln(x)=e^(-x) find x... Hi ice_syncer, you could use Newton's Method for finding roots. $f(x)=e^{-x}-lnx$ The solution to $lnx=e^{-x}$ is the root of $f(x)=0$ Take an initial shot at the solution as x=1.6 Then $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=1.6-\frac{e^{-1.6}-ln(1.6)}{-e^{-1.6}-\frac{1}{1.6}}=1.275767$ A few more iterations would close in on the root, approximately 1.31 Attached Thumbnails 3. Originally Posted by ice_syncer ln(x)=e^(-x) find x... Please post the whole question (in particular the background to it that probably includes something like "Find the approximate solution of ....") 4. how did you think of the initial shot ie x = 1.6 5. Originally Posted by ice_syncer how did you think of the initial shot ie x = 1.6 $e^{-x}$ is a decreasing function, while $lnx$ is increasing. $e^{-1}=0.368$ $ln1=0$ $e^{-2}=0.135$ $ln2=0.693$ Hence these cross between x=1 and x=2. You could choose any value of x in the vicinity as a starting point, say x=1.5 or so. Choosing x=1 or x=2 would also suffice. The attachment shows Newton's method homing in on the solution. Attached Thumbnails	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053463339805603, "perplexity": 2447.1394064400197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928923.21/warc/CC-MAIN-20150521113208-00214-ip-10-180-206-219.ec2.internal.warc.gz"}
http://mathhelpforum.com/pre-calculus/91959-small-limit-problem-print.html	# Small Limit Problem • June 5th 2009, 08:29 PM Kasper Small Limit Problem Hey, so I have this limit question, and I'm pretty sure that the limit does not exist, but I get my back up, because anyone could come to an answer like that from not knowing a way to simplify it. Can anyone see a way to simplify this and get a value? Thanks for any help! $lim_{x->1}\frac{x^3+x^2}{x-1}$ Also, sorry for the dumb looking "approaches" arrow in the limit, I couldn't find the code for a good arrow in the LaTeX tutorial. :( • June 5th 2009, 09:18 PM mr fantastic Quote: Originally Posted by Kasper Hey, so I have this limit question, and I'm pretty sure that the limit does not exist, but I get my back up, because anyone could come to an answer like that from not knowing a way to simplify it. Can anyone see a way to simplify this and get a value? Thanks for any help! $lim_{x->1}\frac{x^3+x^2}{x-1}$ Also, sorry for the dumb looking "approaches" arrow in the limit, I couldn't find the code for a good arrow in the LaTeX tutorial. :( $lim_{x \rightarrow 1^+} \frac{x^3+x^2}{x-1} = + \infty$ $lim_{x \rightarrow 1^-} \frac{x^3+x^2}{x-1} = - \infty$ Since the left hand and right hand limits are not equal, the limit does not exist. • June 5th 2009, 10:50 PM Kasper Ah right, I was looking at it all wrong. I keep trying to just simplify f(x) to find a way to sub in x and try to algebraically find an answer, I gotta start working on thinking of how the limit approaches rather than trying to find it when it's at the point in question. Thanks for the clarification!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.927401065826416, "perplexity": 429.31906324596986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783393463.1/warc/CC-MAIN-20160624154953-00104-ip-10-164-35-72.ec2.internal.warc.gz"}
https://zbmath.org/?q=an:0815.11053	× ## Galois groups with prescribed ramification.(English)Zbl 0815.11053 Childress, Nancy (ed.) et al., Arithmetic geometry. Conference on arithmetic geometry with an emphasis on Iwasawa theory, March 15-18, 1993, Arizona State Univ., Tempe, AZ, USA. Providence, RI: American Mathematical Society. Contemp. Math. 174, 35-60 (1994). Let $$K$$ be an algebraic number field or a function field in one variable $$x$$. The author considers normal extensions of $$K$$ with prescribed ramification. The background for this talk is the conjecture of Abhyankar, proved by the author, that in the case of an algebraically closed constant field of characteristic $$p > 0$$ and $$n$$ ramified places one can realize all finite groups $$G$$ as Galois groups over $$K$$ with the following property: Let $$p(G)$$ be the normal subgroup of $$G$$ generated by the $$p$$-Sylow groups of $$G$$. Then $$G/p (G)$$ has to have not more than $$2g + n - 1$$ generators, where $$g$$ denotes the genus of $$K$$ [Invent. Math. (to appear)]. In section 1 the author considers the case of a function field with algebraically closed or finite constant field. In section 2 he considers number fields mostly in connection with analogs of Abhyankar’s conjecture. There are many interesting examples. For the entire collection see [Zbl 0802.00017]. Reviewer: H.Koch (Berlin) ### MSC: 11R32 Galois theory 14H30 Coverings of curves, fundamental group 12F12 Inverse Galois theory 12F10 Separable extensions, Galois theory	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212955832481384, "perplexity": 239.51086817217762}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00519.warc.gz"}
https://www.techyv.com/questions/insert-picture-microsoft-word/	## Insert a picture into Microsoft word Asked By 0 points N/A Posted on - How to insert a picture into Microsoft word and I have .jpg files insert into my document. SHARE Answered By 0 points N/A #162506 ## Insert a picture into Microsoft word Hi 1. Start Microsoft Word, and then open the text that you want. 2. Get on to place the placing point at the place in the document where you want to put in the picture. On the Insert menu, point to Picture, and then click From File. 3. Look through to the folder that hold the picture that you desire, click the picture file, and then click Insert. 4. Click the inserted picture, and then resize the picture, if essential. Pull the rotation handle to rotate the picture, if required. 5. Apply the tools on the Picture toolbar to modify the quality of the picture. Thanks Answered By 0 points N/A #162505 ## Insert a picture into Microsoft word Hello, This is a really easy task and there are few methods available for do this. First method: 1. Open Microsoft windows. 2. Then click on the insert panel and click on the picture button. 3. Then browse your .jpg picture file from the open window and click on the insert button. 4. Picture will come to your word document page. Second method: 1. Open the Microsoft word and same time open your picture folder. 2. Restore down your picture folder and drag your .jpg picture into the Microsoft word page. Thank You, John Major. Answered By 0 points N/A #162507 ## Insert a picture into Microsoft word Dear, Inserting a picture is an easy task. First, you have to open the Microsoft Word. Then, on the menu bar at upper left corner, click the word Insert. After that, click the picture button and then find the ".jpg" file that you wanted to insert. Next, click insert. You will notice that the file you wanted is already in the document and that's it.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9278436303138733, "perplexity": 3149.532124065714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00359.warc.gz"}
https://www.reporterherald.com/2023/01/24/loveland-weather-for-tuesday-mostly-cloudy-and-32/	# Weather \| Loveland weather for Tuesday: mostly cloudy and 32 Tuesday is expected to be mostly cloudy, with a high near 32, according to the National Weather Service. Areas of freezing fog are possible before 9 a.m. The overnight low will be near 16, with a 50% chance of snow, mainly before 1 a.m. New snow accumulation of less than a half inch is possible. Wednesday is expected to be mostly cloudy, with a high near 32, a 30% chance of snow before 11 a.m. and wind gusts as high as 23 mph. New snow accumulation of less than a half inch is possible. The overnight low will be near 11 with wind gusts as high as 16 mph. Thursday is expected to be mostly sunny, with a high near 35 and wind gusts as high as 20 mph. The overnight low will be near 20. Friday is expected to be mostly cloudy, with a high near 37. The overnight low will be near 16, with a slight chance of snow after 11 p.m. Saturday is expected to be mostly cloudy, with a high near 28 and a chance of snow. The overnight low will be near 6, with a chance of snow. Sunday is expected to be cloudy, with a high near 17 and a slight chance of snow. The overnight low will be near zero with a chance of snow. ## National Weather Service See what the National Weather service is predicting here ## 24-Hour satellite Watch NOAA’s 24-hour satellite image here ## More in Weather Brought to you by Prairie Mountain Publishing	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8093233704566956, "perplexity": 2450.621385925032}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00704.warc.gz"}
https://www.groundai.com/project/training-of-photonic-neural-networks-through-in-situ-backpropagation/	Training of photonic neural networks through in situ backpropagation Training of photonic neural networks through in situ backpropagation Tyler W. Hughes, Momchil Minkov, Yu Shi, Shanhui Fan Ginzton Laboratory, Stanford University, Stanford, CA, 94305. July 25, 2019 Abstract Recently, integrated optics has gained interest as a hardware platform for implementing machine learning algorithms. Of particular interest are artificial neural networks, since matrix-vector multiplications, which are used heavily in artificial neural networks, can be done efficiently in photonic circuits. The training of an artificial neural network is a crucial step in its application. However, currently on the integrated photonics platform there is no efficient protocol for the training of these networks. In this work, we introduce a method that enables highly efficient, in situ training of a photonic neural network. We use adjoint variable methods to derive the photonic analogue of the backpropagation algorithm, which is the standard method for computing gradients of conventional neural networks. We further show how these gradients may be obtained exactly by performing intensity measurements within the device. As an application, we demonstrate the training of a numerically simulated photonic artificial neural network. Beyond the training of photonic machine learning implementations, our method may also be of broad interest to experimental sensitivity analysis of photonic systems and the optimization of reconfigurable optics platforms. I Introduction Artificial neural networks (ANNs), and machine learning in general, are becoming ubiquitous for an impressively large number of applications LeCun et al. (2015). This has brought ANNs into the focus of research in not only computer science, but also electrical engineering, with hardware specifically suited to perform neural network operations actively being developed. There are significant efforts in constructing artificial neural network architectures using various electronic solid-state platforms Merolla et al. (2014); Prezioso et al. (2015), but ever since the conception of ANNs, a hardware implementation using optical signals has also been considered Abu-Mostafa and Pslatis (1987); Jutamulia (1996). In this domain, some of the recent work has been devoted to photonic spike processing Rosenbluth et al. (2009); Tait et al. (2014) and photonic reservoir computing Brunner et al. (2013); Vandoorne et al. (2014a), as well as to devising universal, chip-integrated photonic platforms that can implement any arbitrary ANN Shainline et al. (2017); Shen et al. (2017). Photonic implementations benefit from the fact that, due to the non-interacting nature of photons, linear operations – like the repeated matrix multiplications found in every neural network algorithm – can be performed in parallel, and at a lower energy cost, when using light as opposed to electrons. A key requirement for the utility of any ANN platform is the ability to train the network using algorithms such as error backpropagation Rumelhart et al. (1986). Such training typically demands significant computational time and resources and it is generally desirable for error backpropagation to implemented on the same platform. This is indeed possible for the technologies of Refs. Merolla et al. (2014); Graves et al. (2016); Hermans et al. (2015) and has also been demonstrated e.g. in memristive devices Alibart et al. (2013); Prezioso et al. (2015). In optics, as early as three decades ago, an adaptive platform that could approximately implement the backpropagation algorithm experimentally was proposed Wagner and Psaltis (1987); Psaltis et al. (1988). However, this algorithm requires a number of complex optical operations that are difficult to implement, particularly in integrated optics platforms. Thus, the current implementation of a photonic neural network using integrated optics has been trained using a model of the system simulated on a regular computer Shen et al. (2017). This is inefficient for two reasons. First, this strategy depends entirely on the accuracy of the model representation of the physical system. Second, unless one is interested in deploying a large number of identical, fixed copies of the ANN, any advantage in speed or energy associated with using the photonic circuit is lost if the training must be done on a regular computer. Alternatively, training using a brute force, in situ computation of the gradient of the objective function has been proposed Shen et al. (2017). However, this strategy involves sequentially perturbing each individual parameter of the circuit, which is highly inefficient for large systems. In this work, we propose a procedure, which we label the time-reversal interference method (TRIM), to compute the gradient of the cost function of a photonic ANN by use of only in situ intensity measurements. Our procedure works by physically implementing the adjoint variable method (AVM), a technique that has typically been implemented computationally in the optimization and inverse design of photonic structures Georgieva et al. (2002); Veronis et al. (2004); Hughes et al. (2017). Furthermore, the method scales in constant time with respect to the number of parameters, which allows for backpropagation to be efficiently implemented in a hybrid opto-electronic network. Although we focus our discussion on a particular hardware implementation of a photonic ANN, our conclusions are derived starting from Maxwellâs equations, and may therefore be extended to other photonic platforms. The paper is organized as follows: In Section II, we introduce the working principles of a photonic ANN based on the hardware platform introduced in Ref. Shen et al. (2017). We also derive the mathematics of the forward and backward propagation steps and show that the gradient computation needed for training can be expressed as a modal overlap. Then, in Section III we discuss how the adjoint method may be used to describe the gradient of the ANN cost function in terms of physical parameters. In Section IV, we describe our procedure for determining this gradient information experimentally using in situ intensity measurements. We give a numerical validation of these findings in Section V and demonstrate our method by training a model of a photonic ANN in Section VI. We provide final comments and conclude in Section VII. Ii The Photonic Neural Network In this Section, we introduce the operation and gradient computation of a feed-forward photonic ANN. In its most general case, a feed-forward ANN maps an input vector to an output vector via an alternating sequence of linear operations and element-wise nonlinear functions of the vectors, also called ‘activations’. A cost function, , is defined over the outputs of the ANN and the matrix elements involved in the linear operations are tuned to minimize over a number of training examples via gradient-based optimization. The ‘backpropagation algorithm’ is typically used to compute these gradients analytically by sequentially utilizing the chain rule from the output layer backwards to the input layer. Here, we will outline these steps mathematically for a single training example, with the procedure diagrammed in Fig. 1a. We focus our discussion on the photonic hardware platform presented in Shen et al. (2017), which performs the linear operations using optical interference units (OIUs). The OIU is a mesh of controllable Mach-Zehnder interferometers (MZIs) integrated in a silicon photonic circuit. By tuning the phase shifters integrated in the MZIs, any unitary operation on the input can be implemented Reck et al. (1994); Clements et al. (2016), which finds applications both in classical and quantum photonics Carolan et al. (2015); Harris et al. (2017). In the photonic ANN implementation from Ref. Shen et al. (2017), an OIU is used for each linear matrix-vector multiplication, whereas the nonlinear activations are performed using an electronic circuit, which involves measuring the optical state before activation, performing the nonlinear activation function on an electronic circuit such as a digital computer, and preparing the resulting optical state to be injected to the next stage of the ANN. We first introduce the notation used to describe the OIU, which consists of a number, , of single-mode waveguide input ports coupled to the same number of single-mode output ports through a linear and lossless device. In principle, the device may also be extended to operate on a different number of inputs and outputs. We further assume directional propagation such that all power flows exclusively from the input ports to the output ports, which is a typical assumption for the devices of Refs. Miller (2013a); Shen et al. (2017); Harris et al. (2017); Carolan et al. (2015); Reck et al. (1994); Miller (2013b); Clements et al. (2016). In its most general form, the device implements the linear operation ^WXin=Zout, (1) where and are the modal amplitudes at the input and output ports, respectively, and , which we will refer to as the transfer matrix, is the off-diagonal block of the system’s full scattering matrix, (XoutZout)=(0^WT^W0)(XinZin). (2) Here, the diagonal blocks are zero because we assume forward-only propagation, while the off-diagonal blocks are the transpose of each other because we assume a reciprocal system. and correspond to the input and output modal amplitudes, respectively, if we were to run this device in reverse, i.e. sending a signal in from the output ports. Now we may use this notation to describe the forward and backward propagation steps in a photonic ANN. In the forward propagation step, we start with an initial input to the system, , and perform a linear operation on this input using an OIU represented by the matrix . This is followed by the application of a element-wise nonlinear activation, , on the outputs, giving the input to the next layer. This process repeats for the each layer until the output layer, . Written compactly, for Xl=fl(^WlXl−1)≡fl(Zl). (3) Finally, our cost function is an explicit function of the outputs from the last layer, . This process is shown in Fig. 1(a). To train the network, we must minimize this cost function with respect to the linear operators, , which may be adjusted by tuning the integrated phase shifters within the OIUs. While a number of recent papers have clarified how an individual OIU can be tuned by sequential, in situ methods to perform an arbitrary, pre-defined operation Miller (2013b, a, 2015); Annoni et al. (2017), these strategies do not straightforwardly apply to the training of ANNs, where nonlinearities and several layers of computation are present. In particular, the training of ANN requires gradient information which is not provided directly in the methods of Ref. Miller (2013b, a, 2015); Annoni et al. (2017). In Ref. Shen et al. (2017), the training of the ANN was done ex situ on a computer model of the system, which was used to find the optimal weight matrices for a given cost function. Then, the final weights were recreated in the physical device, using an idealized model that relates the matrix elements to the phase shifters. Ref. Shen et al. (2017) also discusses a possible in situ method for computing the gradient of the ANN cost function through a serial perturbation of every individual phase shifter (‘brute force’ gradient computation). However, this gradient computation has an unavoidable linear scaling with the number of parameters of the system. The training method that we propose here operates without resorting to an external model of the system, while allowing for the tuning of each parameter to be done in parallel, therefore scaling significantly better with respect to the number of parameters when compared to the brute force gradient computation. To introduce our training method we first use the backpropagation algorithm to derive an expression for the gradient of the cost function with respect to the permittivities of the phase shifters in the OIUs. In the following, we denote as the permittivity of a single, arbitrarily chosen phase shifter in layer , as the same derivation holds for each of the phase shifters present in that layer. Note that has an explicit dependence on , but all field components in the subsequent layers also depend implicitly on . As a demonstration, we take a mean squared cost function L (4) where is a complex-valued target vector corresponding to the desired output of our system given input . Starting from the last layer in the circuit, the derivative of the cost function with respect to the permittivity of one of the phase shifters in the last layer is given by dLdϵL =R{(XL−T)†dXLdϵL} (5) (6) ≡R{δTLd^WLdϵLXL−1}, (7) where is element-wise vector multiplication, defined such that, for vectors and , the -th element of the vector is given by . gives the real part, is the derivative of the th layer activation function with respect to its (complex) argument. We define the vector in terms of the error vector . For any layer , we may use the chain rule to perform a recursive calculation of the gradients Γl =^WTl+1δl+1 (8) δl =Γl⊙fl′(Zl) (9) dLdϵl =R{δTld^WldϵlXl−1}. (10) Figure 1(b) diagrams this process, which computes the vectors sequentially from the output layer to the input layer. A treatment for non-holomorphic activations is derived Appendix A. We note that the computation of requires performing the operation , which corresponds physically to sending into the output end of the OIU in layer . In this way, our procedure ‘backpropagates’ the vectors and physically through the entire circuit. In the previous Section, we showed that the crucial step in training the ANN is computing gradient terms of the form , which contain derivatives with respect to the permittivity of the phase shifters in the OIUs. In this Section, we show how this gradient may be expressed as the solution to an electromagnetic adjoint problem. The OIU used to implement the matrix , relating the complex mode amplitudes of input and output ports, can be described using first-principles electrodynamics. This will allow us to compute its gradient with respect to each , as these are the physically adjustable parameters in the system. Assuming a source at frequency , at steady state Maxwell’s equations take the form (11) which can be written more succinctly as ^A(ϵr)e=b. (12) Here, describes the spatial distribution of the relative permittivity (), is the free-space wavenumber, is the electric field distribution, is the electric current density, and due to Lorentz reciprocity. Eq. (12) is the starting point of the finite-difference frequency-domain (FDFD) simulation technique Shin and Fan (2012), where it is discretized on a spatial grid, and the electric field is solved given a particular permittivity distribution, , and source, . To relate this formulation to the transfer matrix , we now define source terms , , that correspond to a source placed in one of the input or output ports. Here we assume a total of input and output waveguides. The spatial distribution of the source term, , matches the mode of the -th single-mode waveguide. Thus, the electric field amplitude in port is given by , and we may establish a relationship between and , as Xin,i=b Tie (13) for over the input port indices, where is the -th component of . Or more compactly, Xin≡^Pine, (14) Similarly, we can define Zout,i=b Ti+Ne (15) for over the output port indices, or, Zout≡^Poute, (16) and, with this notation, Eq. (1) becomes ^W^Pine=^Poute (17) We now use the above definitions to evaluate the cost function gradient in Eq. (10). In particular, with Eqs. (10) and (17), we arrive at (18) Here is the modal source profile that creates the input field amplitudes at the input ports. The key insight of the adjoint variable method is that we may interpret this expression as an operation involving the field solutions of two electromagnetic simulations, which we refer to as the ‘original’ (og) and the ‘adjoint’ (aj) ^Aeog =bx,l−1 (19) ^Aeaj =^PToutδ, (20) where we have made use of the symmetric property of . Eq. (18) can now be expressed in a compact form as If we assume that this phase shifter spans a set of points, in our system, then, from Eq. (11), we obtain where is the Kronecker delta. Inserting this into Eq. (21), we thus find that the gradient is given by the overlap of the two fields over the phase-shifter positions dLdϵl=k20R⎧⎨⎩∑r∈rϕeaj(r)e% og(r)⎫⎬⎭. (23) This result now allows for the computation in parallel of the gradient of the loss function with respect to all phase shifters in the system, given knowledge of the original and adjoint fields. We now introduce our time-reversal interference method (TRIM) for computing the gradient from the previous section through in situ intensity measurements. This represents the most significant result of this paper. Specifically, we wish to generate an intensity pattern with the form , matching that of Eq. (23). We note that interfering and directly in the system results in the intensity pattern: I=\|eog\|2+\|eaj\|2+2R{eogeaj}, (24) the last term of which matches Eq. (23). Thus, the gradient can be computed purely through intensity measurements if the field can be generated in the OIU. The adjoint field for our problem, , as defined in Eq. (20), is sourced by , meaning that it physically corresponds to a mode sent into the system from the output ports. As complex conjugation in the frequency domain corresponds to time-reversal of the fields, we expect to be sent in from the input ports. Formally, to generate , we wish to find a set of input source amplitudes, , such that the output port source amplitudes, , are equal to the complex conjugate of the adjoint amplitudes, or . Using the unitarity property of transfer matrix for a lossless system, along with the fact that for output modes, the input mode amplitudes for the time-reversed adjoint can be computed as X∗TR=^WTlδl. (25) As discussed earlier, is the transfer matrix from output ports to input ports. Thus, we can experimentally determine by sending into the device output ports, measuring the output at the input ports, and taking the complex conjugate of the result. We now summarize the procedure for experimentally measuring the gradient of an OIU layer in the ANN with respect to the permittivities of this layer’s integrated phase shifters: 1. Send in the original field amplitudes and measure and store the intensities at each phase shifter. 2. Send into the output ports and measure and store the intensities at each phase shifter. 3. Compute the time-reversed adjoint input field amplitudes as in Eq. (25). 4. Interfere the original and the time-reversed adjoint fields in the device, measuring again the resulting intensities at each phase shifter. 5. Subtract the constant intensity terms from steps 1 and 2 and multiply by to recover the gradient as in Eq. (23). This procedure is also illustrated in Fig. 2. We numerically demonstrate this procedure in Fig. 3 with a series of FDFD simulations of an OIU implementing a unitary matrix Reck et al. (1994). These simulations are intended to represent the gradient computation corresponding to one OIU in a single layer, , of a neural network with input and delta vector . In these simulations, we use absorbing boundary conditions on the outer edges of the system to eliminate back-reflections. The relative permittivity distribution is shown in Fig. 3(a) with the positions of the variable phase shifters in blue. For demonstration, we simulate a specific case where , with unit amplitude in the bottom port and we choose . In Fig. 3(b), we display the real part of , corresponding to the original, forward field. The real part of the adjoint field, , corresponding to the cost function is shown in Fig. 3(c). In Fig. 3(d) we show the real part of the time-reversed copy of as computed by the method described in the previous section, in which is sent in through the input ports. There is excellent agreement, up to a constant, between the complex conjugate of the field pattern of (c) and the field pattern of (d), as expected. In Fig. 3(e), we display the gradient of the objective function with respect to the permittivity of each point of space in the system, as computed with the adjoint method, described in Eq. (23). In Fig. 3(f), we show the same gradient information, but instead computed with the method described in the previous section. Namely, we interfere the field pattern from panel (b) with the field pattern from panel (d), subtract constant intensity terms, and multiply by the appropriate constants. Again, (b) and (d) agree with good precision. We note that in a realistic system, the gradient must be constant for any stretch of waveguide between waveguide couplers because the interfering fields are at the same frequency and are traveling in the same direction. Thus, there should be no distance dependence in the corresponding intensity distribution. This is largely observed in our simulation, although small fluctuations are visible because of the proximity of the waveguides and the sharp bends, which were needed to make the structure compact enough for simulation within a reasonable time. In practice, the importance of this constant intensity is that it can be detected after each phase shifter, instead of inside of it. Finally, we note that this numerically generated system experiences a total power loss of 41% due to scattering caused by very sharp bends and stair-casing of the structure in the simulation. We also observe approximately 5-10% mode-dependent loss, as determined by measuring the difference in total transmitted power corresponding to injection at different input ports. Minimal amounts of reflection are also visible in the field plots. Nevertheless, TRIM still reconstructs the adjoint sensitivity with very good fidelity. Vi Example of ANN training Finally, we use the techniques from the previous Sections to numerically demonstrate the training of a photonic ANN to implement a logical XOR gate, defined by the following input to target () pairs [0 0]T→0, [0 1]T→1, [1 0]T→1, [1 1]T→0. (26) This problem was chosen as demonstration of learning a nonlinear mapping from input to output Vandoorne et al. (2014b) and is simple enough to be solved with a small network with only four training examples. As diagrammed in Fig. 6a, we choose a network architecture consisting of two unitary OIUs. On the forward propagation step, the binary representation of the inputs, , is sent into the first two input elements of the ANN and a constant value of is sent into the third input element, which serves to introduce artificial bias terms into the network. These inputs are sent through a unitary OIU and then the element-wise activation is applied. The output of this step is sent to another OIU and sent through another activation of the same form. Finally, the first output element is taken to be our prediction, , ignoring the last two output elements. Our network is repeatedly trained on the four training examples defined in Eq. (26) and using the mean-squared cost function presented in Eq. (4). For this demonstration, we utilized a matrix model of the system, as described in Reck et al. (1994); Clements et al. (2016), with mathematical details described in Appendix B. This model allows us to compute an output of the system given an input mode and the settings of each phase shifter. Although this is not a first-principle electromagnetic simulation of the system, it provides information about the complex fields at specific reference points within the circuit, which enables us to implement training using the backpropagation method as described in Section II, combined with the adjoint gradient calculation from Section III. Using these methods, at each iteration of training we compute the gradient of our cost function with respect to the phases of each of the integrated phase shifters, and sum them over the four training examples. Then, we perform a simple steepest-descent update to the phase shifters, in accordance with the gradient information. This is consistent with the standard training protocol for an ANN implemented on a conventional computer. Our network successfully learned the XOR gate in around 400 iterations. The results of the training are shown in Fig. 6b-d. We note that this is meant to serve as a simple demonstration of using the in-situ backpropagation technique for computing the gradients needed to train photonic ANNs. However, our method may equally be performed on more complicated tasks, which we show in the Appendix C. Vii Discussion and Conclusion Here, we justify some of the assumptions made in this work. Our strategy for training a photonic ANN relies on the ability to create arbitrary complex inputs. We note that a device for accomplishing this has been proposed and discussed in Miller (2017). Our recovery technique further requires an integrated intensity detection scheme to occur in parallel and with virtually no loss. This may be implemented by integrated, transparent photo-detectors, which have already been demonstrated in similar systems Annoni et al. (2017). Furthermore, as discussed, this measurement may occur in the waveguide regions directly after the phase shifters, which eliminates the need for phase shifter and photodetector components at the same location. Finally, in our procedure for experimentally measuring the gradient information, we suggested running isolated forward and adjoint steps, storing the intensities at each phase shifter for each step, and then subtracting this information from the final interference intensity. Alternatively, one may bypass the need to store these constant intensities by introducing a low-frequency modulation on top of one of the two interfering fields in Fig. 2(c), such that the product term of Eq. (24) can be directly measured from the low-frequency signal. A similar technique was used in Annoni et al. (2017). We now discuss some of the limitations of our method. In the derivation, we had assumed the operator to be unitary, which corresponds to a lossless OIU. In fact, we note that our procedure is exact in the limit of a lossless, feed-forward, and reciprocal system. However, with the addition of any amount of uniform loss, is still unitary up to a constant, and our procedure may still be performed with the added step of scaling the measured gradients depending on this loss (see a related discussion in Ref. Miller (2017)). Uniform loss conditions are satisfied in the OIUs experimentally demonstrated in Refs. Shen et al. (2017); Miller (2013b). Mode-dependent loss, such as asymmetry in the MZI mesh layout or fabrication errors, should be avoided as its presence limits the ability to accurately reconstruct the time-reversed adjoint field. Nevertheless, our simulation in Fig. 3 indicates that an accurate gradient can be obtained even in the presence of significant mode-dependent loss. In the experimental structures of Refs. Shen et al. (2017); Miller (2013b), the mode-dependent loss is made much lower due to the choice of the MZI mesh. Thus we expect our protocol to work in practical systems. Our method, in principle, computes gradients in parallel and scales in constant time. In practice, to get this scaling would require careful design of the circuits controlling the OIUs. Conveniently, since our method does not directly assume any specific model for the linear operations, it may gracefully handle imperfections in the OIUs, such as deviations from perfect 50-50 splits in the MZIs. Lastly, while we chose to make an explicit distinction between the input ports and the output ports, i.e. we assume no backscattering in the system, this requirement is not strictly necessary. Our formalism can be extended to the full scattering matrix. However, this would require special treatment for subtracting the backscattering. The problem of overfitting is one that must be addressed by ‘regularization’ in any practical realization of a neural network. Photonic ANNs of this class provide a convenient approach to regularization based on ‘dropout’ Srivastava et al. (2014). In the dropout procedure, certain nodes are probabilistically and temporarily ‘deletedâ from the network during train time, which has the effect of forcing the network to find alternative paths to solve the problem at hand. This has a strong regularization effect and has become popular in conventional ANNs. Dropout may be implemented simply in the photonic ANN by ‘shutting offâ channels in the activation functions during training. Specifically, at each time step and for each layer and element , one may set with some fixed probability. In conclusion, we have demonstrated a method for performing backpropagation in an ANN based on a photonic circuit. This method works by physically propagating the adjoint field and interfering its time-reversed copy with the original field. The gradient information can then be directly measured out as an in-situ intensity measurement. While we chose to demonstrate this procedure in the context of ANNs, it is broadly applicable to any reconfigurable photonic system. One could imagine this setup being used to tune phased arrays Sun et al. (2013), optical delivery systems for dielectric laser accelerators Hughes et al. (2018), or other systems that rely on large meshes of integrated optical phase shifters. Furthermore, it may be applied to sensitivity analysis of photonic devices, enabling spatial sensitivity information to be measured as an intensity in the device. Our work should enhance the appeal of photonic circuits in deep learning applications, allowing for training to happen directly inside the device in an efficient and scalable manner. Furthermore, this method is broadly applicable to integrated and adaptive optical systems, enabling the possibility for automatic self-configuration and optimization without resorting to brute force gradient computation or model-based methods, which often do not perfectly represent the physical system. Funding Information Gordon and Betty Moore Foundation (GBMF4744); Schweizerischer Nationalfonds zur FÃ¶rderung der Wissenschaftlichen Forschung (P300P2_177721); Air Force Office of Scientific Research (FA9550-17-1-0002). References • LeCun et al. (2015) Y. LeCun, Y. Bengio, and G. Hinton, Nature 521, 436 (2015). • Merolla et al. (2014) P. A. Merolla, J. V. Arthur, R. Alvarez-Icaza, A. S. Cassidy, J. Sawada, F. Akopyan, B. L. Jackson, N. Imam, C. Guo, Y. Nakamura, B. Brezzo, I. Vo, S. K. Esser, R. 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Fan, Journal of Computational Physics 231, 3406 (2012). • Vandoorne et al. (2014b) K. Vandoorne, P. Mechet, T. Van Vaerenbergh, M. Fiers, G. Morthier, D. Verstraeten, B. Schrauwen, J. Dambre, and P. Bienstman, Nature Communications 5, 3541 (2014b). • Miller (2017) D. A. B. Miller, Opt. Express 25, 29233 (2017). • Srivastava et al. (2014) N. Srivastava, G. Hinton, A. Krizhevsky, I. Sutskever, and R. Salakhutdinov, Journal of Machine Learning Research 15, 1929 (2014). • Sun et al. (2013) J. Sun, E. Timurdogan, A. Yaacobi, E. S. Hosseini, and M. R. Watts, Nature 493, 195 (2013). • Hughes et al. (2018) T. W. Hughes, S. Tan, Z. Zhao, N. V. Sapra, Y. J. Lee, K. J. Leedle, H. Deng, Y. Miao, D. S. Black, M. Qi, O. Solgaard, J. S. Harris, J. Vuckovic, R. L. Byer, and S. Fan, Physical Review Applied (Accepted 2018). Appendix A Non-holomorphic Backpropagation In the previous derivation, we have assumed that the functions are holomorphic. For each element of input , labeled , this means that the derivative of with respect to its complex argument is well defined, or the derivative dfldz=limΔz→0fl(z+Δz)−fl(z−Δz)2Δz (27) does not depend on the direction that approaches in the complex plane. Here we show how to extend the backpropagation derivation to non-holomorphic activation functions. We first examine the starting point of the backpropagation algorithm, considering the change in the mean-squared loss function with respect to the permittivity of a phase shifter in the last layer OIU, as written in Eq. (7) of the main text as dLdϵL=R{ΓTLdXLdϵL} (28) Where we had defined the error vector for simplicity and is the output of the final layer. To evaluate this expression for non-holomorphic activation functions, we split and its argument into their real and imaginary parts fL(Z)=u(α,β)+iv(α,β), (29) where is the imaginary unit and and are the real and imaginary parts of , respectively. We now wish to evaluate , which gives the following via the chain rule dfdϵ=dudα⊙dαdϵ+dudβ⊙dβdϵ+idvdα⊙dαdϵ+idvdβ⊙dβdϵ, (30) where we have dropped the layer index for simplicity. Here, terms of the form correspond to element-wise differentiation of the vector with respect to the vector . For example, the -th element of the vector is given by . Now, inserting into Eq. (28), we have dLdϵL=R{ ΓTL⊙(dudα+idvdα)TdαdϵL (31) + ΓTL⊙(dudβ+idvdβ)TdβdϵL}. (32) We now define real and imaginary parts of as and , respectively. Inserting the definitions of and in terms of and and doing some algebra, we recover dLdϵL=R{ (ΓR⊙dudα)Td^WLdϵLXL−1 (33) − (ΓI⊙dvdα)Td^WLdϵLXL−1 (34) −i (ΓR⊙dudβ)Td^WLdϵLXL−1 (35) +i (ΓI⊙dvdβ)Td^WLdϵLXL−1}. (36) Finally, the expression simplifies to dLdϵL=R{[ ΓR⊙(dudα−idudβ) (37) + ΓI⊙(−dvdα+idvdβ)]Td^WLdϵLXL−1}. (38) As a check, if we insert the conditions for to be holomorphic, namely dudα=dvdβ, and dudβ=−dvdα, (39) Eq. (36) simplifies to dLdϵL =R{[ΓR⊙(dudα+idvdα)+ (40) ΓI⊙(−dvdα+idudα)]Td^WLdϵLXL−1} (41) =R{[ΓL⊙(dudα+idvdα)]Td^WLdϵLXL−1} (42) =R{[ΓL⊙fl′(ZL)]Td^WLdϵLXL−1} (43) =R{δTLd^WLdϵLXL−1} (44) as before. This derivation may be similarly extended to any layer in the network. For holomorphic activation functions, whereas we originally defined the vectors as δl=Γl⊙fl′(Zl), (45) for non-holomorphic activation functions, the respective definition is δl=ΓR⊙(dudα−idudβ)+ΓI⊙(−dvdα+idvdβ), (46) where and are the respective real and imaginary parts of , and are the real and imaginary parts of , and and are the real and imaginary parts of , respectively. We can write this more simply as δl=R{Γl⊙dfdα}−iR{Γl⊙dfdβ}. (47) In polar coordinates where and , this equation becomes δl=exp(−iϕ)(R{Γl⊙dfdr}−iR{Γl⊙1rdfdϕ}) (48) where all operations are element-wise. Appendix B Photonic neural network simulation In Sections 4 and 5 of the main text, we have shown, starting from Maxwell’s equations, how the gradient information defined for an arbitrary problem can be obtained through electric field intensity measurements. However, since the full electromagnetic problem is too large to solve repeatedly, for the purposes of demonstration of a functioning neural network, in Section 6 we use the analytic, matrix representation of a mesh of MZIs as described in Ref. Clements et al. (2016). Namely, for an even , the matrix of the OIU is parametrized as the product of unitary matrices: ^W=^RN^RN−1…^R2^R1^D, (49) where each implements a number of two-by-two unitary operations corresponding to a given MZI, and is a diagonal matrix corresponding to an arbitrary phase delay added to each port. This is schematically illustrated in Fig. 5(a) for . For the ANN training, we need to compute terms of the form dLdϕ=R{YTd^WdϕX}, (50) for an arbitrary phase and vectors and defined following the steps in the main text. Because of the feed-forward nature of the OIU-s, the matrix can also be split as ^W=^W2^Fϕ^W1, (51) where is a diagonal matrix which applies a phase shift in port (the other elements are independent of ), while and are the parts that precede and follow the phase shifter, respectively (Fig. 5(b)). Thus, Eq. (50) becomes R{YTd^WdϕX} =R{YT^W2d^Fϕdϕ^W1X} (52) =−I{(^WT2Y)ieiϕ(^W1X)i}, where is the -th element of the vector , and denotes the imaginary part. This result can be written more intuitively in a notation similar to the main text. Namely, if is the field amplitude generated by input from the right, measured right after the phase shifter corresponding to , while is the field amplitude generated by input from the right, measured at the same point, then By recording the amplitudes in all ports during the forward and the backward propagation, we can thus compute in parallel the gradient with respect to every phase shifter. Notice that, within this computational model, we do not need to go through the full procedure outlined in Section 4 of the main text. However, this procedure is crucial for the in situ measurement of the gradients, and works even in cases which cannot be correctly captured by the simplified matrix model used here. Appendix C Training demonstration In the main text we show how the in-situ backpropagation method may be used to train a simple XOR network. Here we demonstrate training on a more complex problem. Specifically, we generate a set of one thousand training examples represented by input and target pairs. Here, where and are the independent inputs, which we constrain to be real for simplicity, and represents a mode added to the third port to make the norm of the same for each training example. In this case, we choose . Each training example has a corresponding label, which is encoded in the desired output, , as and for and respectively. For a given and , we define and as the magnitude and phase of the vector in the 2D-plane, respectively. To generate the corresponding class label, we first generate a uniform random variable between 0 and 1, labeled , and then set if exp(−(r−r0−Δsin(2ϕ))22σ2)+0.1 U>0.5. (54) Otherwise, we set . For the demonstration, , , and . The underlying distribution thus resembles an oblong ring centered around , with added noise. As diagrammed in Fig. 6(a), we use a network architecture consisting of six layers of unitary OIUs, with an element-wise activation after each unitary transformation except for the last in the series, which has an activation of . After the final activation, we apply an additional ‘softmax’ activation, which gives a normalized probability distribution corresponding to the predicted class of . Specifically, these are given by , where is the first/second element of the output vector of the last activation (the other two elements are ignored). The ANN prediction for the input is set as the larger one of these two outputs, while the total cost function is defined in the cross-entropy form L=1MM∑m=1L(m)=1MM∑m=1−log(s(zm,t)), (55) where is the cost function of the -th example, the summation is over all training examples, and is the output from the target port, , as defined by the target output of the -th example. We randomly split our generated examples into a training set containing 75% of the originally generated training examples, while the remaining 25% are used as a test set to evaluate the performance of our network on unseen examples. As in the XOR demonstration, we utilized our matrix model of the system described in Section B. As in the main text, at each iteration of training we compute the gradient of the cost function with respect to the phases of each of the integrated phase shifters, and sum this over each of the training examples. For the backpropagation through the activation functions, since and are non-holomorphic, we use eq. 48 from Section A, to obtain δL =2Z∗L⊙R{ΓL} (56) δl =exp(−iϕl)⊙R{Γl}, (57) where is a vector containing the phases of and is given by the derivative of the cross-entropy loss function for a single training example ΓL=∂L(m)∂zm,i=s(zm,i)−δi,t, (58) where is the Kronecker delta. With this, we can now compute the gradient of the loss function of eq. 55 with respect to all trainable parameters, and perform a parallel, steepest-descent update to the phase shifters, in accordance with the gradient information. Our network successfully learned the this task in around 4000 iterations. The results of the training are shown in Fig. 6(b). We achieved a training and test accuracy of 91% on both the training and test sets, indicating that the network was not overfitting to the dataset. This can also be confirmed visually from Fig. 6(c). The lack of perfect predictions is likely due to the inclusion of noise. You are adding the first comment! 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The feedback must be of minimum 40 characters and the title a minimum of 5 characters	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8646154999732971, "perplexity": 579.4831237402052}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00434.warc.gz"}
http://math.stackexchange.com/questions/266816/why-open-and-closed-boxes-are-measurable	# Why open and closed boxes are measurable Let $B := \prod_{i=1}^n (a_i,b_i) := \{(x_1,\cdots,x_n) \in \mathbb R^n \mid \forall i: x_i \in (a_i,b_i) \}$. Can someone help me to show that $B$ is a measurable set, i.e. that if $A \subseteq \mathbb R^n$ then $m^(A) \geq m^(A \cap B) + m^(A \cap B^c)$ where $$m^(E) = \inf \left \{ \sum_{j=0}^\infty vol(B_j) : E \subseteq \bigcup_{i=0}^\infty B_i \text{ where } (B_i)_{i=0}^\infty \text{ at most countable } \right \}$$ and the $B_i$ have to be open boxes. Further is $vol(B) := \prod_{i=1}^n (b_i-a_i)$ for each open box $B$. - First see if you can do the case $n=1$. – GEdgar Dec 29 '12 at 1:50 Yes, I can :D How can I proceed after that ? Can I show that the Cartesian product is measurable ? I.e. $B = (a_1,b_1) \times \cdots \times (a_n,b_n)$ where each $(a_i,b_i)$ is measurable. – Epsilon Dec 29 '12 at 2:12 It would be nice to show that $(a,\infty)^n$ is measurable for all $a \in \mathbb R$.Then an open box $B$ is an intersection of finitly many boxes of the form $(a,\infty)^n$ but I already have proven that finite intersection preserves measurability. – Epsilon Dec 29 '12 at 2:40 It would be nice to show that $(a,\infty)^n$ is measurable for all $a \in \mathbb R$ and $(-\infty,a)^n$, too .Then an open box $B$ is an intersection of finitly many boxes of the form $(a,\infty)^n$ or $(-\infty,a)^n$ but I already have proven that finite intersection preserves measurability. (Edit of the above comment) – Epsilon Dec 29 '12 at 2:46 For the case $n=1$ – leo Dec 29 '12 at 17:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9601149559020996, "perplexity": 173.3446903528993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049277475.33/warc/CC-MAIN-20160524002117-00163-ip-10-185-217-139.ec2.internal.warc.gz"}
https://www.cheenta.com/test-of-mathematics-solution-subjective-65-minimum-value-of-quadratic/	How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances? # Test of Mathematics Solution Subjective 65 - Minimum Value of Quadratic This is a Test of Mathematics Solution Subjective 65 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance. ## Problem Show that for all real x, the expression ${ax^2}$ + bx + C ( where a, b, c are real constants with a > 0), has the minimum value ${\frac{4ac - b^2}{4a}}$ . Also find the value of x for which this minimum value is attained. ## solution: f (x) ${ax^2}$ + bx + c Now minimum derivative = 0 & 2nd order derivative > 0. ${\frac{df(x)}{dx}}$ = 2ax + b Or ${\frac{d^2f(x)}{dx^2}}$ =2a Now 2a> so 2nd order derivative > 0 so ${\frac{d^2f(x)}{dx^2}}$ = 2. So minimum occurs when ${\frac{df(x)}{d(x)}}$ = 0 or 2ax + b = 0 or 2ax = -b or x = ${\frac{-b}{2a}}$ (ans) At x = ${\frac{-b}{2a}}$ ${ax^2}$ + bx + c = ${a\times {\frac{b^4}{4a^2}}}$ + ${b\times {\frac{-b}{2a}}}$ + c = ${\frac{4ac-b^2}{4a}}$ (proved)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8071150183677673, "perplexity": 1689.1306779434444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00288.warc.gz"}
http://lilypond.org/doc/v2.19/Documentation/learning/absolute-note-names.en.html	### 2.4.3 Absolute note names So far we have used `\relative` to define pitches. This is usually the fastest way to enter most music. Without `\relative`, pitches are interpreted in absolute mode. In this mode, LilyPond treats all pitches as absolute values. A `c'` will always mean middle C, a `b` will always mean the note one step below middle C, and a `g,` will always mean the note on the bottom staff of the bass clef. ```{ \clef "bass" c'4 b g, g, \| g,4 f, f c' \| } ``` Writing a melody in the treble clef involves a lot of quote `'` marks. Consider this fragment from Mozart: ```{ \key a \major \time 6/8 cis''8. d''16 cis''8 e''4 e''8 \| b'8. cis''16 b'8 d''4 d''8 \| } ``` Common octave marks can be indicated just once, using the command `\fixed` followed by a reference pitch: ```\fixed c'' { \key a \major \time 6/8 cis8. d16 cis8 e4 e8 \| b,8. cis16 b,8 d4 d8 \| } ``` With `\relative`, the previous example needs no octave marks because this melody moves in steps no larger than three staff positions: ```\relative { \key a \major \time 6/8 cis''8. d16 cis8 e4 e8 \| b8. cis16 b8 d4 d8 \| } ``` If you make a mistake with an octave mark (`'` or `,`) while working in `\relative` mode, it is very obvious – many notes will be in the wrong octave. When working in absolute mode, a single mistake will not be as visible, and will not be as easy to find. However, absolute mode is useful for music which has large intervals, and is extremely useful for computer-generated LilyPond files. When cutting and pasting melody fragments, absolute mode preserves the original octave. Sometimes music is arranged in more complex ways. If you are using `\relative` inside of `\relative`, the outer and inner relative sections are independent: ```\relative { c'4 \relative { f'' g } c } ``` To use absolute mode inside of `\relative`, put the absolute music inside `\fixed c { … }` and the absolute pitches will not affect the octaves of the relative music: ```\relative { c'4 \fixed c { f'' g'' } c \| c4 \fixed c'' { f g } c } ``` Other languages: català, česky, deutsch, español, français, magyar, italiano, 日本語, nederlands.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9148290753364563, "perplexity": 2866.0807394278377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257826773.17/warc/CC-MAIN-20160723071026-00158-ip-10-185-27-174.ec2.internal.warc.gz"}
https://cerncourier.com/a/atlas-hunts-for-new-physics-with-dibosons/	# ATLAS hunts for new physics with dibosons 22 September 2017 Beyond the Standard Model of particle physics (SM), crucial open questions remain such as the nature of dark matter, the overabundance of matter compared to antimatter in the universe, and also the mass scale of the scalar sector (what makes the Higgs boson so light?). Theorists have extended the SM with new symmetries or forces that address these questions, and many such extensions predict new resonances that can decay into a pair of bosons (diboson), for example: VV, Vh, Vγ and γγ, where V stands for a weak boson (W and Z), h for the Higgs boson, and γ is a photon. The ATLAS collaboration has a broad search programme for diboson resonances, and the most recent results using 36 fb–1 of proton–proton collision data at the LHC taken at a centre-of-mass energy of 13 TeV in 2015 and 2016 have now been released. Six different final states characterised by different boson decay modes were considered in searches for a VV resonance: 4, ℓℓνν, ℓℓqq, νqq, ννqq and qqqq, where , ν and q stand for charged leptons (electrons and muons), neutrinos and quarks, respectively. For the Vh resonance search, the dominant Higgs boson decay into a pair of b-quarks (branching fraction of 58%) was exploited together with four different V decays leading to ℓℓbb, νbb, ννbb and qqbb final states. A Zγ resonance was sought in final states with two leptons and a photon. A new resonance would appear as an excess (bump) over the smoothly distributed SM background in the invariant mass distribution reconstructed from the final-state particles. The left figure shows the observed WZ mass distribution in the qqqq channel together with simulations of some example signals. An important key to probe very high-mass signals is to identify high-momentum hadronically decaying V and h bosons. ATLAS developed a new technique to reconstruct the invariant mass of such bosons combining information from the calorimeters and the central tracking detectors. The resulting improved mass resolution for reconstructed V and h bosons increased the sensitivity to very heavy signals. No evidence for a new resonance was observed in these searches, allowing ATLAS to set stringent exclusion limits. For example, a graviton signal predicted in a model with extra spatial dimensions was excluded up to masses of 4 TeV, while heavy weak-boson-like resonances (as predicted in composite Higgs boson models) decaying to WZ bosons are excluded for masses up to 3.3 TeV. Heavier Higgs partners can be excluded up to masses of about 350 GeV, assuming specific model parameters.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9793265461921692, "perplexity": 2521.5777476783637}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00694.warc.gz"}
https://www.physicsforums.com/threads/stuck-on-the-following-integral.129708/	# Stuck on the following integral 1. Aug 25, 2006 ### suspenc3 Hi, I am kinda stuck on the following integral: $$\int\sqrt{\frac{1+x}{1-x}}dx$$ any hints? 2. Aug 25, 2006 ### TD If you let $$y^2 = \frac{{1 + x}}{{1 - x}} \Leftrightarrow x = \frac{{y^2 - 1}}{{y^2 + 1}}$$ The integral will become fraction of rationals, losing the square root. 3. Aug 25, 2006 ### suspenc3 so are you saying to substitute that for x? 4. Aug 25, 2006 ### suspenc3 or by making this substitution, the square root will be taken away 5. Aug 25, 2006 ### neutrino Or you could do the trig substitution $$x = \sin\theta$$ 6. Aug 25, 2006 ### TD Yes, use that substitution to lose the square root. I already solved for x as well, which allows you to easily find dx in terms of dy by differentiating both sides.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9738191366195679, "perplexity": 3449.437090252776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00495-ip-10-171-10-108.ec2.internal.warc.gz"}
https://www.hartleygroup.org/publication/ferroelectric-liquid-crystals-induced-by-atropisomeric-biphenyl-dopants-the-effect-of-chiral-perturbations-on-achiral-dopants/	# Ferroelectric liquid crystals induced by atropisomeric biphenyl dopants: the effect of chiral perturbations on achiral dopants C. Scott Hartley and Robert P. Lemieux* Liq. Cryst. 2004, 31, 1101–1108 https://doi.org/10.1080/02678290410001715999 ## Abstract The addition of the achiral biphenyl dopant 2,2′,6,6′-tetramethyl-4,4′-bis(4-n-nonyloxybenzoyloxy)biphenyl (3) or its dithionoester or dithioester analogue (45) to a 4 mol % mixture of the atropisomeric biphenyl dopant (R)-2,2′,6,6′-tetramethyl-3,3′-dinitro-4,4′-bis(4-n-nonyloxybenzoyloxy)biphenyl, (R)-1, in the phenylpyrimidine SmC host PhP1 produces a significant amplification of the spontaneous polarization induced by (R)-1. This amplification may be due to a chiral perturbation by (R)-1 which causes a shift in the equilibrium between enantiomeric conformations of the achiral dopant. The degree of polarization amplification afforded by the achiral dopant, as expressed by the polarization amplification factor PAF, varies with the nature of the linking group. This may be ascribed to different rotational distributions of the core transverse dipole moments relative to the polar axis of the SmC* phase and/or to differences in lateral bulk of the polar linking groups. The latter may affect the degree of chiral molecular recognition achieved by 35 in the binding site of the SmC* phase.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8046143651008606, "perplexity": 4346.871221658779}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141171126.6/warc/CC-MAIN-20201124053841-20201124083841-00693.warc.gz"}
http://hellenicaworld.com/Science/Mathematics/en/MeyerhoffManifold.html	### - Art Gallery - In hyperbolic geometry, the Meyerhoff manifold is the arithmetic hyperbolic 3-manifold obtained by ( 5 , 1 ) {\displaystyle (5,1)} {\displaystyle (5,1)} surgery on the figure-8 knot complement. It was introduced by Robert Meyerhoff (1987) as a possible candidate for the hyperbolic 3-manifold of smallest volume, but the Weeks manifold turned out to have slightly smaller volume. It has the second smallest volume $${\displaystyle V_{m}=12\cdot (283)^{3/2}\zeta _{k}(2)(2\pi )^{-6}=0.981368\dots }$$ of orientable arithmetic hyperbolic 3-manifolds, where $$\zeta _{k}$$ is the zeta function of the quartic field of discriminant $${\displaystyle -283}$$. Alternatively, $${\displaystyle V_{m}=\Im ({\rm {{Li}_{2}(\theta )+\ln \|\theta \|\ln(1-\theta ))=0.981368\dots }}} where \( {\displaystyle {\rm {{Li}_{n}}}}$$ is the polylogarithm and \|x\| is the absolute value of the complex root$$\theta$$ (with positive imaginary part) of the quartic $${\displaystyle \theta ^{4}+\theta -1=0}$$. Ted Chinburg (1987) showed that this manifold is arithmetic. Gieseking manifold Weeks manifold References Chinburg, Ted (1987), "A small arithmetic hyperbolic three-manifold", Proceedings of the American Mathematical Society, 100 (1): 140–144, doi:10.2307/2046135, ISSN 0002-9939, JSTOR 2046135, MR 0883417 Chinburg, Ted; Friedman, Eduardo; Jones, Kerry N.; Reid, Alan W. (2001), "The arithmetic hyperbolic 3-manifold of smallest volume", Annali della Scuola Normale Superiore di Pisa. Classe di Scienze. Serie IV, 30 (1): 1–40, ISSN 0391-173X, MR 1882023 Meyerhoff, Robert (1987), "A lower bound for the volume of hyperbolic 3-manifolds", Canadian Journal of Mathematics, 39 (5): 1038–1056, doi:10.4153/CJM-1987-053-6, ISSN 0008-414X, MR 0918586 Mathematics Encyclopedia World Index	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9691330790519714, "perplexity": 2191.2307850439784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00284.warc.gz"}
https://mathhothouse.me/category/cnennai-math-institute-entrance-exam/	## Category Archives: Cnennai Math Institute Entrance Exam ### The animals went in which way The animals may have gone into Noah’s Ark two by two, but in which order did they go in? Given the following sentence (yes, sentence! — I make no apologies for the punctuation), what was the order in which the animals entered the Ark? The monkeys went in before the sheep, swans, chickens, peacocks, geese, penguins and spiders, but went in after the horses, badgers, squirrels and tigers, the latter of which went in before the horses, the penguins, the rabbits, the pigs, the donkeys, the snakes and the mice, but the mice went before the leopards, the leopards before the squirrels, the squirrels before the chickens, the chickens before the penguins, spiders, sheep, geese and the peacocks, the peacocks before the geese and the penguins, the penguins before the spiders and after the geese and the horses, the horses before the donkeys, the chickens and the leopards, the leopards after the foxes and the ducks, the ducks before the goats, swans, doves, foxes and badgers before the chickens, horses, squirrels and swans and after the lions, tigers, foxes, squirrels and ducks, the ducks after the lions, elephants, rabbits and otters, the otters before the elephants, tigers, chickens and beavers, the beavers after the elephants, the elephants before the lions, the lions before the tigers, the sheep before the peacocks, the swans before the chickens, the pigs before the snakes, the snakes before the foxes, the pigs after the rabbits, goats, tigers and doves, the doves before the chickens, horses, goats, donkeys and snakes, the snakes after the goats, and the donkeys before the mice and the squirrels. 🙂 🙂 🙂 Nalin Pithwa. ### Announcement: Scholarships for RMO Training Mathematics Hothouse. ### Can anyone have fun with infinite series? Below is list of finitely many puzzles on infinite series to keep you a bit busy !! 🙂 Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series. Puzzle 1: A grandmother’s vrat (fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her vrat? Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that $p+2$ is also a prime number. Then, the first number theorist would pipette out $\frac{1}{p}$ litres of chemical and the second $\frac{1}{(p+2)}$ litres. How many times do they have to play this game to empty the flask completely? Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table? Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations. Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using $t_{i}$ for $i^{th}$ tap and $d_{j}$ for $j^{th}$ drain, can be written as follows: $\{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}$. When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, $n_{A}<1$. Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: $\{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}$. At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, $0.5n_{A}$. How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences? I hope u do have fun!! -Nalin Pithwa. ### Logicalympics — 100 meters!!! Just as you go to the gym daily and increase your physical stamina, so also, you should go to the “mental gym” of solving hard math or logical puzzles daily to increase your mental stamina. You should start with a laser-like focus (or, concentrate like Shiva’s third eye, as is famous in Hindu mythology/scriptures!!) for 15-30 min daily and sustain that pace for a month at least. Give yourself a chance. Start with the following: The logicalympics take place every year in a very quiet setting so that the competitors can concentrate on their events — not so much the events themselves, but the results. At the logicalympics every event ends in a tie so that no one goes home disappointed 🙂 There were five entries in the room, so they held five races in order that each competitor could win, and so that each competitor could also take his/her turn in 2nd, 3rd, 4th, and 5th place. The final results showed that each competitor had duly taken taken their turn in finishing in each of the five positions. Given the following information, what were the results of each of the five races? The five competitors were A, B, C, D and E. C didn’t win the fourth race. In the first race A finished before C who in turn finished after B. A finished in a better position in the fourth race than in the second race. E didn’t win the second race. E finished two places behind C in the first race. D lost the fourth race. A finished ahead of B in the fourth race, but B finished before A and C in the third race. A had already finished before C in the second race who in turn finished after B again. B was not first in the first race and D was not last. D finished in a better position in the second race than in the first race and finished before B. A wasn’t second in the second race and also finished before B. So, is your brain racing now to finish this puzzle? Cheers, Nalin Pithwa. PS: Many of the puzzles on my blog(s) are from famous literature/books/sources, but I would not like to reveal them as I feel that students gain the most when they really try these questions on their own rather than quickly give up and ask for help or look up solutions. Students have finally to stand on their own feet! (I do not claim creating these questions or puzzles; I am only a math tutor and sometimes, a tutor on the web.) I feel that even a “wrong” attempt is a “partial” attempt; if u can see where your own reasoning has failed, that is also partial success! ### Pick’s theorem to pick your brains!! Pick’s theorem: Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as $(B/2)+I-1$ square units, where B is number of lattice points on the boundary, I is number of lattice points in the interior of the polygon. Prove this theorem! Do you like this challenge? Nalin Pithwa. ### Limits that arise frequently We continue our presentation of basic stuff from Calculus and Analytic Geometry, G B Thomas and Finney, Ninth Edition. My express purpose in presenting these few proofs is to emphasize that Calculus, is not just a recipe of calculation techniques. Or, even, a bit further, math is not just about calculation. I have a feeling that such thinking nurtured/developed at a young age, (while preparing for IITJEE Math, for example) makes one razor sharp. We verify a few famous limits. Formula 1: If $\|x\|<1$, $\lim_{n \rightarrow \infty}x^{n}=0$ We need to show that to each $\in >0$ there corresponds an integer N so large that $\|x^{n}\|<\in$ for all n greater than N. Since $\in^{1/n}\rightarrow 1$, while $\|x\|<1$. there exists an integer N for which $\in^{1/n}>\|x\|$. In other words, $\|x^{N}\|=\|x\|^{N}<\in$. Call this (I). This is the integer we seek because, if $\|x\|<1$, then $\|x^{n}\|<\|x^{N}\|$ for all $n>N$. Call this (II). Combining I and II produces $\|x^{n}\|<\in$ for all $n>N$, concluding the proof. Formula II: For any number x, $\lim_{n \rightarrow \infty}(1+\frac{x}{n})^{n}=e^{x}$. Let $a_{n}=(1+\frac{x}{n})^{n}$. Then, $\ln {a_{n}}=\ln{(1+\frac{x}{n})^{n}}=n\ln{(1+\frac{x}{n})}\rightarrow x$, as we can see by the following application of l’Hopital’s rule, in which we differentiate with respect to n: $\lim_{n \rightarrow \infty}n\ln{(1+\frac{x}{n})}=\lim_{n \rightarrow \infty}\frac{\ln{(1+x/n)}}{1/n}$, which in turn equals $\lim_{n \rightarrow \infty}\frac{(\frac{1}{1+x/n}).(-\frac{x}{n^{2}})}{-1/n^{2}}=\lim_{n \rightarrow \infty}\frac{x}{1+x/n}=x$. Now, let us apply the following theorem with $f(x)=e^{x}$ to the above: (a theorem for calculating limits of sequences) the continuous function theorem for sequences: Let $a_{n}$ be a sequence of real numbers. If $\{a_{n}\}$ be a sequence of real numbers. If $a_{n} \rightarrow L$ and if f is a function that is continu0us at L and defined at all $a_{n}$, then $f(a_{n}) \rightarrow f(L)$. So, in this particular proof, we get the following: $(1+\frac{x}{n})^{n}=a_{n}=e^{\ln{a_{n}}}\rightarrow e^{x}$. Formula 3: For any number x, $\lim_{n \rightarrow \infty}\frac{x^{n}}{n!}=0$ Since $-\frac{\|x\|^{n}}{n!} \leq \frac{x^{n}}{n!} \leq \frac{\|x\|^{n}}{n!}$, all we need to show is that $\frac{\|x\|^{n}}{n!} \rightarrow 0$. We can then apply the Sandwich Theorem for Sequences (Let $\{a_{n}\}$, $\{b_{n}\}$ and $\{c_{n}\}$ be sequences of real numbers. if $a_{n}\leq b_{n}\leq c_{n}$ holds for all n beyond some index N, and if $\lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}c_{n}=L$,, then $\lim_{n\rightarrow \infty}b_{n}=L$ also) to conclude that $\frac{x^{n}}{n!} \rightarrow 0$. The first step in showing that $\|x\|^{n}/n! \rightarrow 0$ is to choose an integer $M>\|x\|$, so that $(\|x\|/M)<1$. Now, let us the rule (formula 1, mentioned above), so we conclude that:$(\|x\|/M)^{n}\rightarrow 0$. We then restrict our attention to values of $n>M$. For these values of n, we can write: $\frac{\|x\|^{n}}{n!}=\frac{\|x\|^{n}}{1.2 \ldots M.(M+1)(M+2)\ldots n}$, where there are $(n-M)$ factors in the expression $(M+1)(M+2)\ldots n$, and the RHS in the above expression is $\leq \frac{\|x\|^{n}}{M!M^{n-M}}=\frac{\|x\|^{n}M^{M}}{M!M^{n}}=\frac{M^{M}}{M!}(\frac{\|x\|}{M})^{n}$. Thus, $0\leq \frac{\|x\|^{n}}{n!}\leq \frac{M^{M}}{M!}(\frac{\|x\|}{M})^{n}$. Now, the constant $\frac{M^{M}}{M!}$ does not change as n increases. Thus, the Sandwich theorem tells us that $\frac{\|x\|^{n}}{n!} \rightarrow 0$ because $(\frac{\|x\|}{M})^{n}\rightarrow 0$. That’s all, folks !! Aufwiedersehen, Nalin Pithwa. ### Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule Reference: Thomas, Finney, 9th edition, Calculus and Analytic Geometry. Continuing our previous discussion of “theoretical” calculus or “rigorous” calculus, I am reproducing below the proof of the finite limit case of the stronger form of l’Hopital’s Rule : L’Hopital’s Rule (Stronger Form): Suppose that $f(x_{0})=g(x_{0})=0$ and that the functions f and g are both differentiable on an open interval $(a,b)$ that contains the point $x_{0}$. Suppose also that $g^{'} \neq 0$ at every point in $(a,b)$ except possibly at $x_{0}$. Then, $\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac{f^{x}}{g^{x}}$ ….call this equation I, provided the limit on the right exists. The proof of the stronger form of l’Hopital’s Rule is based on Cauchy’s Mean Value Theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s Rule. Cauchy’s Mean Value Theorem: Suppose that the functions f and g are continuous on $[a,b]$ and differentiable throughout $(a,b)$ and suppose also that $g^{'} \neq 0$ throughout $(a,b)$. Then there exists a number c in $(a,b)$ at which $\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$…call this II. The ordinary Mean Value Theorem is the case where $g(x)=x$. Proof of Cauchy’s Mean Value Theorem: We apply the Mean Value Theorem twice. First we use it to show that $g(a) \neq g(b)$. For if $g(b)$ did equal to $g(a)$, then the Mean Value Theorem would give: $g^{'}(c)=\frac{g(b)-g(a)}{b-a}=0$ for some c between a and b. This cannot happen because $g^{'}(x) \neq 0$ in $(a,b)$. We next apply the Mean Value Theorem to the function: $F(x) = f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]$. This function is continuous and differentiable where f and g are, and $F(b) = F(a)=0$. Therefore, there is a number c between a and b for which $F^{'}(c)=0$. In terms of f and g, this says: $F^{'}(c) = f^{'}(c)-\frac{f(b)-f(a)}{g(b)-g(a)}[g^{'}(c)]=0$, or $\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$, which is II above. QED. Proof of the Stronger Form of l’Hopital’s Rule: We first prove I for the case $x \rightarrow x_{o}^{+}$. The method needs no change to apply to $x \rightarrow x_{0}^{-}$, and the combination of those two cases establishes the result. Suppose that x lies to the right of $x_{o}$. Then, $g^{'}(x) \neq 0$ and we can apply the Cauchy’s Mean Value Theorem to the closed interval from $x_{0}$ to x. This produces a number c between $x_{0}$ and x such that $\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}$. But, $f(x_{0})=g(x_{0})=0$ so that $\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)}{g(x)}$. As x approaches $x_{0}$, c approaches $x_{0}$ because it lies between x and $x_{0}$. Therefore, $\lim_{x \rightarrow x_{0}^{+}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(c)}{g^{'}(c)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(x)}{g^{'}(x)}$. This establishes l’Hopital’s Rule for the case where x approaches $x_{0}$ from above. The case where x approaches $x_{0}$ from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval $[x,x_{0}]$, where $x< x_{0}$QED. ### The Sandwich Theorem or Squeeze Play Theorem It helps to think about the core concepts of Calculus from a young age, if you want to develop your expertise or talents further in math, pure or applied, engineering or mathematical sciences. At a tangible level, it helps you attack more or many questions of the IIT JEE Advanced Mathematics. Let us see if you like the following proof, or can absorb/digest it: Reference: Calculus and Analytic Geometry by Thomas and Finney, 9th edition. The Sandwich Theorem: Suppose that $g(x) \leq f(x) \leq h(x)$ for all x in some open interval containing c, except possibly at $x=c$ itself. Suppose also that $\lim_{x \rightarrow c}g(x)= \lim_{x \rightarrow c}h(x)=L$. Then, $\lim_{x \rightarrow c}f(x)=c$. Proof for Right Hand Limits: Suppose $\lim_{x \rightarrow c^{+}}g(x)=\lim_{x \rightarrow c^{+}}h(x)=L$. Then, for any $\in >0$, there exists a $\delta >0$ such that for all x, the inequality $c implies $L-\in and $L-\in ….call this (I) These inequalities combine with the inequality $g(x) \leq f(x) \leq h(x)$ to give $L-\in $L-\in $-\in ….call this (II) Therefore, for all x, the inequality $c implies $\|f(x)-L\|<\in$. …call this (III) Proof for LeftHand Limits: Suppose $\lim_{x \rightarrow c^{-}} g(x)=\lim_{x \rightarrow c^{-}}=L$. Then, for $\in >0$ there exists a $\delta >0$ such that for all x, the inequality $c-\delta implies $L-\in and $L-\in …call this (IV). We conclude as before that for all x, $c-\delta implies $\|f(x)-L\|<\in$. Proof for Two sided Limits: If $\lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x)=L$, then $g(x)$ and $h(x)$ both approach L as $x \rightarrow c^{+}$ and as $x \rightarrow c^{-}$ so $\lim_{x \rightarrow c^{+}}f(x)=L$ and $\lim_{x \rightarrow c^{-}}f(x)=L$. Hence, $\lim_{x \rightarrow c}f(x)=L$. QED. Let me know your feedback on such stuff, Nalin Pithwa ### Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem Lagrange’s Mean Value Theorem: If a function $f(x)$ is continuous on the interval $[a,b]$ and differentiable at all interior points of the interval, there will be, within $[a,b]$, at least one point c, $a, such that $f(b)-f(a)=f^{'}(c)(b-a)$. Cauchy’s Generalized Mean Value Theorem: If $f(x)$ and $phi(x)$ are two functions continuous on an interval $[a,b]$ and differentiable within it, and $phi(x)$ does not vanish anywhere inside the interval, there will be, in $[a,b]$, a point $x=c$, $a, such that $\frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}$. Some questions based on the above: Problem 1: Form Lagrange’s formula for the function $y=\sin(x)$ on the interval $[x_{1},x_{2}]$. Problem 2: Verify the truth of Lagrange’s formula for the function $y=2x-x^{2}$ on the interval $[0,1]$. Problem 3: Applying Lagrange’s theorem, prove the inequalities: (i) $e^{x} \geq 1+x$ (ii) $\ln (1+x) , for $x>0$. (iii) $b^{n}-a^{n} for $b>a$. (iv) $\arctan(x) . Problem 4: Write the Cauchy formula for the functions $f(x)=x^{2}$, $phi(x)=x^{3}$ on the interval $[1,2]$ and find c. More churnings with calculus later! Nalin Pithwa. ### Could a one-sided limit not exist ? 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http://mathoverflow.net/questions/73878/differential-equation-with-some-constraints	# Differential equation with some constraints I posted this to stackexchange, and after some hours got a comment that was so pessimistic about finding some neat orderly solution, that I'm posting it here too. (In case anyone cares, this is related to this question, which I posted here earlier.) I'd like $\alpha,\beta,\gamma$ as functions of $t$, satisfying the following conditions: \begin{align} \alpha+\beta+\gamma & = 0 \\ \sin^2\alpha + \sin^2\beta + \sin^2\gamma & = c^2 \\ \left\| \frac{d}{dt}(\sin\alpha,\sin\beta,\sin\gamma)\right\| & = 1 \end{align} I'm thinking of $c^2$ as small. At the very least that means $<2$, and intuitively it means $\ll 2$. Some geometry shows that there is a qualitative change in the nature of the solutions when $c^2$ goes from $<2$ to $>2$. Later edit: The question above asks for a parametrization by arc-length. Here's an ugly parametrization by something quite remote from arc length: $$\beta = \frac{\arccos\left(\frac{1 + \sin^2\alpha - c^2}{\cos\alpha}\right)-\alpha}{2}$$ And then $\gamma = \pi - \alpha - \beta$. In order to get the whole curve, you'd need a multiple-valued arccosine and then you'd pick the right value for the particular point on the curve. One thing that fails to be obvious to me just from the way the function above is written, giving $\beta$ as a function of $\alpha$, is that that function is its own inverse. So here's a less demanding question that the one above: Is there some nice pleasant way of parametrizing the curve that, if not by arc-length, at least treats $\alpha$, $\beta$, and $\gamma$ equally, so that it's perfectly self-evident from the way it's written that the whole expression is symmetric in $\alpha,\beta,\gamma$? - BTW, the first two constraints imply that $0 \le c^2 \le 9/4$. – Michael Hardy Aug 28 '11 at 19:29 A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2.$$ More precisely, it's a quarter of $S_1$, because $S_1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S_1$ with the sphere $\Sigma_c: \\|\vec v\\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve. The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2.$$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds to $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circle, then at the end speed up the resulting arc-length parametrizations by a factor $1/c$ to undo the scaling. - The hour is late and I will look at this tomorrow. But even before digesting everything above, I've up-voted it because the identity following "$S_1$" looks just like what I wrote in another stackexchange posting (except for a factor of 2.....): math.stackexchange.com/questions/59508/trigonometric-identity And that one arose from a product of three sines, whereas this one arose from a sum of products of two sines, and there ought to be certain connections. So are you suggesting that there is a nice neat solution after all? Or only that there is one in a limiting case? – Michael Hardy Aug 28 '11 at 3:35 The arc-length parametrization is the inverse function of an arc-length integral, which involves a square root and can only rarely have an elementary formula (already for an ellipse we famously get elliptic integral). I see that Robert Bryant already worked out what happens here, and verified that there's almost never an elementary formula; so it won't get any more nice or neat than inverting the integral of the square root of a rational function. Did you have a reason to expect or hope for a particularly nice form here? – Noam D. Elkies Aug 28 '11 at 17:18 @Noam: Well, now I've started looking at this answer while awake. First I wondered how you got $S_1$ from the first constraint. Then I saw how it can be done, if not how you did it, which I suspect is different. To be continued....... – Michael Hardy Aug 29 '11 at 17:55 Building on Noam's suggestion, you could try using the inherent symmetry of the problem: Set $\sigma_1 = x^2 + y^2 + z^2$, $\sigma_2 = x^2y^2+y^2z^2+z^2x^2$, and $\sigma_3 = x^2y^2z^2$. Then your conditions become $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$. Meanwhile, you have $$dt^2 = dx^2 + dy^2 + dz^2 = \frac{d(x^2)^2}{4x^2}+\frac{d(y^2)^2}{4y^2}+\frac{d(z^2)^2}{4z^2},$$ and this latter expression, being symmetric in $x^2,y^2,z^2$, can be expressed as a differential expression in $\sigma_1, \sigma_2, \sigma_3$. I won't write out the details, but a short computation (using Maple) shows that, taking advantage of the relations $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$, this leads to the relation $$dt^2 = \frac{\bigl(c^6(c^2{-}2)-4(36{-}52c^2{+}21c^4{-}2c^6)\sigma_3+16(c^2{-}1){\sigma_3}^2\bigr)}{16\sigma_3\bigl(c^6(2{-}c^2)-4(27{-}18c^2{+}2c^4)\sigma_3-16{\sigma_3}^2\bigr)}\bigl(d\sigma_3\bigr)^2.$$ Now, for example, you can see why $c^2=2$ is special. The integral that gives $t$ will simplify dramatically in this case; in fact, it becomes an elementary integral. For general values of $c$, though, this is a hyperelliptic integral, and you won't find any simple relation between $t$ and $\sigma_3$, so, a fortiori, none between $t$ and $x$, $y$, and $z$. There are various special values of $c$ for which the roots and poles of the rational expression cancel, such as $c=0$, $c = \pm\sqrt{2}$, and $c = \pm\frac32$, and, for these, you'd expect the integral to simplify considerably, but, otherwise, you don't expect any nice relation. Added remark: By the way, you can get from this more directly to the relation between $t$ and $x$, $y$, and $z$, since, for example, letting $u$ represent any one of $x^2$, $y^2$, or $z^2$, one has the relation $u^3 - c^2 u^2 + (\sigma_3 + \tfrac14 c^4)u - \sigma_3 = 0$, which can clearly be solved for $\sigma_3$ as a rational function of $u$. Substituting this into the above relation gives a differential equation directly relating $t$ and, say, $u = x^2$. It's not a particularly nice relation, though. Ultimately, this gives a relation of the form $x = F(t,c)$ where $F$ is some function periodic of period $3\tau(c)$ in the first variable for some $\tau(c)>0$. Then one finds that $y = F(t + \tau(c),c)$ and $z = F(t-\tau(c),c)$. This is, of course, a very symmetric expression, though it's not explicit. - That $c^2 = 0$ and $c^2 = (3/2)^2$ are the two opposite extreme values of the sum of squares of the three sines follows from the first constraint. That $c^2 = 2$ gives an exceptionally well-behaved situation is actually seen just be thinking about secondary-school-level trigonometry. It's striking how a little trigonometry problem leads straight into moderately exotic (by comparison to this sort of trigonometry) functions, but I suppose the same can be said of other things that we've all seen. I'm going to print out the two answers posted so far and think about them before saying much more. – Michael Hardy Aug 28 '11 at 19:38 OK, the "less demanding" question does seem more tractable; a few possible answers follow, though none is clearly the most "nice pleasant way of parametrizing" your curve. One direction leads to the trigonometric solution of a cubic equation with all roots real; the other leads to an elliptic curve with 6-torsion, and even to an extremal elliptic K3 surface! Which if any of these is best for you is a matter of taste and of what you're trying to do with these curves. Let $(X,Y,Z) = (\sin^2 \alpha, \phantom.\sin^2 \beta, \phantom.\sin^2 \gamma)$. Then $(X,Y,Z)$ are coordinates of an algebraic curve $$E_c : X+Y+Z = c^2, \phantom{=} X^2+Y^2+Z^2 - 2(YZ+ZX+XY) + 4XYZ = 0.$$ So far we've preserved the $S_3$ symmetry, and can recover the original variables via $\alpha = \arcsin X^{1/2} = \frac12 \arccos(1-2X)$ and likewise for $\beta,\gamma$. But this begs the question of what $E_c$ looks like, and leaves us with multivalued arcsines or arccosines. The latter problem seems inherent in another symmetry of the equation: we can translate $\alpha,\beta,\gamma$ by $a\pi,b\pi,c\pi$ for any integers $a,b,c$ with $a+b+c=0$. But we can try to do more with $E_c$. One direction is to express everything in terms of elementary symmetric functions $\sigma_1,\sigma_2,\sigma_3$ of $X,Y,Z$, as R.Bryant did: the first equation says $\sigma_1=c^2$, and the second says $\sigma_1^2 = 4 (\sigma_3 - \sigma_2)$; so $(\sigma_1,\sigma_2,\sigma_3)$ are parametrized in terms of $\sigma_3$, and then $X,Y,Z$ are the three roots of $$0 = u^3 - \sigma_1 u^2 + \sigma_2 u - \sigma_3 = u^3 - c^2 u^2 + (\sigma_3 + \tfrac14 c^4)u - \sigma_3.$$ This is still manifestly symmetric but rather implicit. We we can solve the cubic; since it has three real roots the solution will involve trisecting some auxiliary angle $\theta$, itself given as the arccosine of some explicit but complicated algebraic function of $c$ and $\sigma_3$. The roots will then be given in terms of $c$, $\sigma_3$, and the cosines of $\theta/3$, $(\theta+2\pi)/3$, and $(\theta+4\pi)/3$, and the action of $S_3$ will correspond to replacing $\theta$ by the equivalent $\pm (\theta+2\pi n)$ for some $n \bmod 3$. This will be far from nice and pleasant (compare with the formulas for constructing a regular 13-gon using an angle trisector, as in p.192 of Gleason's Monthly article), but it will have the advantage of leaving the symmetry close to the surface. Another direction is to consider $E_c$ on its own terms. It is an elliptic curve, so rational functions on it like $x,y,z$ can be parametrized by elliptic functions like $\wp$ and $\wp'$. Moreover $E_c$ inherits the $S_3$ action so the resulting formulas must retain this symmetry; and the periodicity of $\wp,\wp'$ may even cancel out the ambiguity in the arcsine or arccosine. That's great if you love elliptic curves, not so great if you regard $\wp$ as yet another obscure transcendental function... At least these elliptic curves are rather nice: the cyclic permutations of $X,Y,Z$ are translations by 3-torsion points of $E_c$, and there's also a 2-torsion point because switching two of the variables, say $Y \leftrightarrow Z$, has a rational fixed point where the third variable vanishes (this corresponds to taking $\alpha = 0$ and $\beta+\gamma=0$ in the original equation). So $E_c$ actually has 6-torsion. If I did this right, an equivalent equation for $E_c$ is $$y^2 = x^3 + ((c^2-3) x - (c^2-2)^2)^2,$$ which exhibits the 3-torsion points where $x=0$, and has 2-torsion at $(x,y) = -((c^2-2)^2,0)$. As it happens $E_c$ is not far from the universal elliptic curve with a 6-torsion point, which is given by $y^2 = x^3 + ((h-3)x - (h-2)^2)^2$. What's more, our substitution $h=c^2$ produces an elliptic K3 surface whose fiber $E_c$ becomes singular at the familiar points $c = 0, \phantom.\pm \sqrt2, \phantom.\pm \frac32$, and also $c=\infty$ — and the multiplicities at $c=0,\phantom.\pm\sqrt2,\phantom.\infty$ are large enough that this elliptic K3 surface is "extremal" (finite Mordell-Weil group, maximal Picard number)! Such surfaces have attracted considerable attention over the years, starting with the Miranda-Persson list of semistable extremal surfaces (Math. Z. 201 (1989), 339–361), which includes ours with multiplicity vector $[1,1,4,6,6,6]$. This makes your family of curves very nice in that context, even if it doesn't do much to answer your motivating question... - I tried to include this link to the Miranda-Persson paper, which however does not seem to work in the above answer: springerlink.com/content/u30268141h636x04/fulltext.pdf – Noam D. Elkies Aug 29 '11 at 5:48 Very nice, Noam! I especially like the relation with the K3 surface. I knew that the curve on $x^2$, $y^2$, $z^2$ was an elliptic curve with some symmetries, but I didn't figure out the properties of the ($8$-fold) branched cover that represents the original $xyz$-curve or the branched cover of that on which $dt$ is actually a meromorphic differential. – Robert Bryant Aug 29 '11 at 11:54 Thanks, Robert! Yes, the original curve, to say nothing of the $dt$ double cover, looks much more complicated, but M.Hardy seems willing to extract square roots of a function of one variable for free in his setting $-$ in any case he'll have to take some inverse trig function to recover his original variables $\alpha,\beta,\gamma$. – Noam D. Elkies Aug 29 '11 at 12:57 It's more a case of my not having decided what I should be willing to do........ – Michael Hardy Aug 29 '11 at 18:07 .....but I am willing to use $\wp$ and things like Jacobi's elliptic functions. If we momentarily adopt Euler's willingness to say that if $\alpha$ is an infinitely small positive number then $\sin\alpha=\alpha$, then when $c$ is an infinitely small positive number, the curve can be parametrized using the ordinary sine and cosine functions. So if $c$ is merely very small, then one should get periodic functions that are approximately those. – Michael Hardy Aug 29 '11 at 18:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.941217303276062, "perplexity": 237.44909751509596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164026971/warc/CC-MAIN-20131204133346-00036-ip-10-33-133-15.ec2.internal.warc.gz"}
http://mathhelpforum.com/algebra/227821-nth-roots-unit-proofs-question.html	# Math Help - nth roots of unit proofs question 1. ## nth roots of unit proofs question For 4i) i get to $\theta = ± k \frac{2 \pi}{n}$ k is an integer $k = ±1 {z}_{1} = {e}^{i2\pi)} , {z}_{2} = {e}^{-i2\pi}$ I just proved it like that, not sure if its right ii) for even n ${(-z)}^{n} = {z}^{n} = 1$ but for odd n ${(-z)}^{n} = -{z}^{n} = -1$ for the sum, i got zero for the product: for even n i got $(-1)\frac{n}{2} . (1)\frac{n}{2} = -\frac{{n}^{2}}{4}$ since you have (-1) half the time and 1 half the time for odd n $(1)\frac{n}{2} . (-1)(\frac{n}{2} +1) = -(\frac{{n}^{2}}{4} + \frac{n}{2})$ since you have one more -1 than 1 2. ## Re: nth roots of unit proofs question Originally Posted by Applestrudle Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation. $e^z=\exp(z)=\|z\|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$ This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$ For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$ Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$. Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$. For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$ Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$. 3. ## Re: nth roots of unit proofs question Originally Posted by Plato Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation. $e^z=\exp(z)=\|z\|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$ This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$ For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$ Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$. Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$. For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$ Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$. for the sum z1 +z2 + z3+ z4+ .... i did the sum from k =1 to k=n of ${e}^{\frac{i2\pi k}{n}}$ and i got $\frac{1-{e}^{\frac{2 \pi k}{n}}}{1 - {e}^{\frac{i2 \pi}{n}}}$ i used geometric series x + x^2 and x^3 + x^4 .... x^n = (1-x^n)/(1-x) for the product z1.z2.z3.z4.z5 ... I used the geometric series in the exponential since ${z}_{k} = {e}^{\frac{2 \pi k}{n}}$ and i got the sum of the product as equal to ${e}^{i2\pi \frac{1-n}{n}}$ for the nth roots of i I got for the product ${w}_{k} = {e}^{i(\frac{pi}{2n}+\frac{2\pik}{n})}$ so the product is e^(ipi/2n + sum of 2pik/n) for k from 1 to n for the sum, using the geometric series, I got (2pi/n)(-n), giving -2pi ? then the final answer for the product is ${e}^{i(\frac{\pi}{2n} - 2\pi)}$ 4. ## Re: nth roots of unit proofs question Originally Posted by Plato Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation. $e^z=\exp(z)=\|z\|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$ This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$ For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$ Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$. Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$. For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$ Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$. for iv) i got the product P equals P = w1.w2.w3.w4.w5.w6....wn (the 1,2,3...n are subs) since -w* is also a solution, you can say wk = -wk* so they repeat themselves (is this logic correct?) and P = (w1)^2 (w2)^2 (w3)^2....(w n/2)^2 $P = {e}^{i( \pi\(frac{n}{2}) + 4\pi 8\pi +12\pi ....}$ in the exponent there is the sum from k =1 to k = (n/2) of 4k.pi but i don't know how to simplify the sum, do i just leave it as a sum, is this even correct?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9953600168228149, "perplexity": 1029.1141194311479}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500824391.7/warc/CC-MAIN-20140820021344-00376-ip-10-180-136-8.ec2.internal.warc.gz"}
http://aas.org/archives/BAAS/v30n3/dps98/356.htm	Session 29. Comets I Contributed Oral Parallel Session, Wednesday, October 14, 1998, 2:00-3:20pm, Madison Ballroom C ## [29.08] First Maps of Comet Hale-Bopp at 60 and 175\mu m S.B. Peschke (MPI f\"ur Kernphysik), M. Stickel (MPI f\"ur Astronomie), I. Heinrichsen (IPAC), H. B\"ohnhardt (ESO), C. M. Lisse (University of Maryland), E. Gr\"un (MPI f\"ur Kernphysik), D. J. Osip (MIT) First maps of a comet at 60 and 175\mu m were obtained using ISOPHOT, the photometer of the Infrared Space Observatory(ISO). The observations were carried out on December 30, 1997 , mapping an area of 9'\times9' centered on comet Hale-Bopp at both filters. Each measurement consisted of 3 individual submaps offset by a third of a pixel in both directions to increase the final resolution of the maps. The final maps were composed of the submaps with the use of a drizzle algorithm. Within the same orbit 3-175\mu m filter photometry on comet Hale-Bopp was performed as well as multi-aperture photometry near the peak wavelength of thermal emission. The same photometric sequence was repeated as 'shadow observation' at the same position as that tracked in the initial sequence for precise background subtraction. Quasi-simultaneous observations in the near-IR were obtained with the 3.6m at La Silla/Chile. >From the 60 and 175\mu m, radial intensity profiles have been derived which are compared to the ones obtained from the near-IR data and to the results of multi-aperture photometry. Since dust grains have the highest thermal emitting efficiency closest to their own size, the emission in the maps observed with the two filters are dominated by the the thermal emission of different size grains. From the comparison of the different wavelength maps, indications on perferred concentration of different grain sizes can be derived. Grain size distribution modeling has been carried out for the spectral energy distribution derived with multi-filter photometry to get an indication of the coma composition with will in turn be used as input for dynamical modeling. First results will be presented. [Previous] \| [Session 29] \| [Next]	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9513021111488342, "perplexity": 4651.291933201574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510261771.50/warc/CC-MAIN-20140728011741-00381-ip-10-146-231-18.ec2.internal.warc.gz"}
https://www.nag.com/numeric/nl/nagdoc_26.2/nagdoc_fl26.2/html/f01/f01fdf.html	# NAG Library Routine Document ## 1Purpose f01fdf computes the matrix exponential, ${e}^{A}$, of a complex Hermitian $n$ by $n$ matrix $A$. ## 2Specification Fortran Interface Subroutine f01fdf ( uplo, n, a, lda, Integer, Intent (In) :: n, lda Integer, Intent (Inout) :: ifail Complex (Kind=nag_wp), Intent (Inout) :: a(lda,) Character (1), Intent (In) :: uplo #include <nagmk26.h> void f01fdf_ (const char uplo, const Integer n, Complex a[], const Integer lda, Integer ifail, const Charlen length_uplo) ## 3Description ${e}^{A}$ is computed using a spectral factorization of $A$ $A = Q D QH ,$ where $D$ is the diagonal matrix whose diagonal elements, ${d}_{i}$, are the eigenvalues of $A$, and $Q$ is a unitary matrix whose columns are the eigenvectors of $A$. ${e}^{A}$ is then given by $eA = Q eD QH ,$ where ${e}^{D}$ is the diagonal matrix whose $i$th diagonal element is ${e}^{{d}_{i}}$. See for example Section 4.5 of Higham (2008). ## 4References Higham N J (2005) The scaling and squaring method for the matrix exponential revisited SIAM J. Matrix Anal. Appl. 26(4) 1179–1193 Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA Moler C B and Van Loan C F (2003) Nineteen dubious ways to compute the exponential of a matrix, twenty-five years later SIAM Rev. 45 3–49 ## 5Arguments 1: $\mathbf{uplo}$ – Character(1)Input On entry: if ${\mathbf{uplo}}=\text{'U'}$, the upper triangle of the matrix $A$ is stored. If ${\mathbf{uplo}}=\text{'L'}$, the lower triangle of the matrix $A$ is stored. Constraint: ${\mathbf{uplo}}=\text{'U'}$ or $\text{'L'}$. 2: $\mathbf{n}$ – IntegerInput On entry: $n$, the order of the matrix $A$. Constraint: ${\mathbf{n}}\ge 0$. 3: $\mathbf{a}\left({\mathbf{lda}},\right)$ – Complex (Kind=nag_wp) arrayInput/Output Note: the second dimension of the array a must be at least ${\mathbf{n}}$. On entry: the $n$ by $n$ Hermitian matrix $A$. • If ${\mathbf{uplo}}=\text{'U'}$, the upper triangular part of $A$ must be stored and the elements of the array below the diagonal are not referenced. • If ${\mathbf{uplo}}=\text{'L'}$, the lower triangular part of $A$ must be stored and the elements of the array above the diagonal are not referenced. On exit: if ${\mathbf{ifail}}={\mathbf{0}}$, the upper or lower triangular part of the $n$ by $n$ matrix exponential, ${e}^{A}$. 4: $\mathbf{lda}$ – IntegerInput On entry: the first dimension of the array a as declared in the (sub)program from which f01fdf is called. Constraint: ${\mathbf{lda}}\ge {\mathbf{n}}$. 5: $\mathbf{ifail}$ – IntegerInput/Output On entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details. For environments where it might be inappropriate to halt program execution when an error is detected, the value is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value is used it is essential to test the value of ifail on exit. On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6). ## 6Error Indicators and Warnings If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf). Errors or warnings detected by the routine: ${\mathbf{ifail}}>0$ The computation of the spectral factorization failed to converge. If ${\mathbf{ifail}}=i$, the algorithm to compute the spectral factorization failed to converge; $i$ off-diagonal elements of an intermediate tridiagonal form did not converge to zero (see f08fnf (zheev)). ${\mathbf{ifail}}=-1$ On entry, ${\mathbf{uplo}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{uplo}}=\text{'L'}$ or $\text{'U'}$. ${\mathbf{ifail}}=-2$ On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{n}}\ge 0$. ${\mathbf{ifail}}=-3$ ${\mathbf{ifail}}=-4$ On entry, ${\mathbf{lda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{lda}}\ge {\mathbf{n}}$. ${\mathbf{ifail}}=-99$ See Section 3.9 in How to Use the NAG Library and its Documentation for further information. ${\mathbf{ifail}}=-399$ Your licence key may have expired or may not have been installed correctly. See Section 3.8 in How to Use the NAG Library and its Documentation for further information. ${\mathbf{ifail}}=-999$ Dynamic memory allocation failed. See Section 3.7 in How to Use the NAG Library and its Documentation for further information. ## 7Accuracy For an Hermitian matrix $A$, the matrix ${e}^{A}$, has the relative condition number $κA = A2 ,$ which is the minimal possible for the matrix exponential and so the computed matrix exponential is guaranteed to be close to the exact matrix. See Section 10.2 of Higham (2008) for details and further discussion. ## 8Parallelism and Performance f01fdf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library. f01fdf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information. Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information. The integer allocatable memory required is n, the real allocatable memory required is n and the complex allocatable memory required is approximately $\left({\mathbf{n}}+\mathit{nb}+1\right)×{\mathbf{n}}$, where nb is the block size required by f08fnf (zheev). The cost of the algorithm is $O\left({n}^{3}\right)$. As well as the excellent book cited above, the classic reference for the computation of the matrix exponential is Moler and Van Loan (2003). ## 10Example This example finds the matrix exponential of the Hermitian matrix $A = 1 2+2i 3+2i 4+3i 2-2i 1 2+2i 3+2i 3-2i 2-2i 1 2+2i 4-3i 3-2i 2-2i 1 .$ ### 10.1Program Text Program Text (f01fdfe.f90) ### 10.2Program Data Program Data (f01fdfe.d) ### 10.3Program Results Program Results (f01fdfe.r)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 74, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9816107749938965, "perplexity": 2037.1069422139478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00574.warc.gz"}
https://www.physicsforums.com/threads/momentum-exchange-of-virtual-pions.728808/	# Momentum exchange of virtual pions 1. ### gildomar 75 I know that the strong force is viewed as the exchange of virtual pions between two nucleons, with the mass and range of them confirmed by the energy-time uncertainty principle. But if the momentum of the pion is transferred from one nucleon to the other in the interaction, wouldn't that give an equivalent repulsive force between them instead of an attractive one? 2,375 3. ### gildomar 75 Thanks; guess I didn't look far enough back in the past topics. 4. ### K^2 2,470 Virtual pion exchange is just one of the contributions. It's the dominant effect, but there are other things going on. Nucleons can exchange gluons directly, as well as exchange other mesons. Pions happen to be the lightest of mesons and not restricted by confinement, so they end up being better mediators for nuclear forces, but not the only ones there. 5. ### ChrisVer 2,375 gluons don't exist in the nuclei level due to confinement, so that's why in fact (effectively) you get the puons. The main quarks you can make your "effective particle" consist of, are up and down, because I think (from the deep inelastic scattering on protons) we already know that strange is not favorable at all...it's almost not existing in the sea particles... 6. ### RGevo 90 How can you draw any strong interaction without involving gluons? 7. ### Bill_K 4,157 From www.phy.ohiou.edu/~elster/lectures/fewblect_2.pdf‎: 8. ### K^2 2,470 That's precisely what I said. Pions dominate interaction because they are not confined. Other processes are non-dominant either due to confinement or higher masses of mediator particles. Where's the problem? 9. ### Bill_K 4,157 Please don't take offense when someone agrees with you.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9019763469696045, "perplexity": 1412.4497421506644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737833893.68/warc/CC-MAIN-20151001221713-00090-ip-10-137-6-227.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/96944/are-all-subrings-of-the-rationals-euclidean-domains	Are all subrings of the rationals Euclidean domains? This is a purely recreational question -- I came up with it when setting an undergraduate example sheet. Let's go with Wikipedia's definition of a Euclidean domain. So an ID $R$ is a Euclidean domain (ED) if there's some $\phi:R\backslash\{0\}\to\mathbf{Z}_{\geq0}$ or possibly $\mathbf{Z}_{>0}$ (I never know what $\mathbf{N}$ means, and the Wikipedia page (at the time of writing) uses $\mathbf{N}$ as the target of $\phi$, but in this case it doesn't matter, because I can just add one to $\phi$ if necessary) such that the usual axioms hold. Now onto subrings of the rationals. The subrings of the rationals turn out to be in bijection with the subsets of the prime numbers. If $X$ is a set of primes, then define $\mathbf{Z}_X$ to be the rationals $a/b$ with $b$ only divisible by primes in $X$. Different sets $X$ give different subrings, and all subrings are of this form. This needs a little proof, but a little thought, or a little googling, leads you there. If $X$ is empty, then $\mathbf{Z}_X=\mathbf{Z}$, which is an ED: the usual $\phi$ taken is $\phi(x)=\|x\|$. If $X$ is all the primes then $\mathbf{Z}_X=\mathbf{Q}$ and this is an ED too (at least according to Wikipedia -- I think some sources demand that an ED is not a field, but let's not go there); we can just let $\phi$ be constant. If $X$ is all but one prime, say $p$, then $\mathbf{Z}_X$ is the localisation of $\mathbf{Z}$ at $(p)$, and $\phi$ can be taken to be the $p$-adic valuation (if we're allowing $\phi$ to take the value zero, which we may as well). Note however that this is a rather different "style" of $\phi$ to the case $X$ empty: this $\phi$ is "non-archimedean" in origin, whereas in the case of $X$ empty we used an "archimedean" $\phi$. This sort of trick generalises to the case where $X$ is all but a finite set of primes -- see the "Dedekind domain with only finitely many non-zero primes" example on the Wikipedia page. Of course the question is: if $X$ is now an arbitrary set of primes, is $\mathbf{Z}_X$ an ED? - What happens when $X$ is finite? In particular, when $X=\{2\}$ and $X=\{2,3\}$. – lhf Jan 6 '12 at 14:53 I think that in general one strategy for constructing $\phi$, for any ID, is this: you let $A_0$ be zero, you let $A_1$ be the units, and for $n\geq2$ you let $A_n$ be the elements $r$ of $R$ not in any earlier $A_i$ but such that the map $\cup_{j<i}A_j\to R/(r)$ is surjective. The idea is that $\phi(r)=i$ for $r\in A_i$ and you hope that the union of the $A_i$ is $R$: you've then proved $R$ is an ED, and I think that what I sketch here is basically an iff. – Kevin Buzzard Jan 6 '12 at 15:24 So if $X=\{2\}$ then $A_2$ contains (amongst other things) the primes such that 2 is a primitive root, and then $A_3$ contains (amongst other things) the primes such that either 2 or one of these primes in $A_2$ is a primitive root and etc etc. And then you just have to hope that you get everything shrug. – Kevin Buzzard Jan 6 '12 at 15:24 Given Alex's (very nice, +1) answer below, I wonder if you can follow-up to ask for a more "coherent" system of $\phi_X$'s such that $\phi_X$ agrees with the $p$-adic valuation when $X$ is the complement of a prime, and/or that $\phi_X$ behaves nicely with respect to shrinking/enlarging $X$. – Cam McLeman Jan 6 '12 at 17:11 Yes. Let $\phi(a/b) = \|a\|$ where $a/b$ is written in lowest terms. To see that this is a Euclidean function, let $a/b,c/d\in \mathbb{Z}_X$ be nonzero and in lowest terms and write $$\frac{a}{b}=\frac{nd}{b}\cdot \frac{c}{d}+\frac{s}{t}$$ which means that $\phi(s/t)=\phi((a-nc)/b)\leq \|a-nc\|$ which for a suitable value of $n$ is less than $\phi(a/b) = \|a\|$. Very nice. Note that you implicitly use the fact that if $a/b$ is in the subring then so is $m/b$ for any integer $m$. I wonder if you've proved that any subring of the field of fractions of an ED is an ED?? – Kevin Buzzard Jan 6 '12 at 16:07 PS I now see that my mistake was to be too hung up on the $p$-adic side of the story. – Kevin Buzzard Jan 6 '12 at 16:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9774937629699707, "perplexity": 135.7418628275303}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500800767.23/warc/CC-MAIN-20140820021320-00066-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.amdainternational.com/3vv8wv/thermal-conductivity-dimension-c26795	. is the mean free path, which measures the average distance a molecule travels between collisions. The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. 3 The presence of the reciprocal lattice wave vector implies a net phonon backscattering and a resistance to phonon and thermal transport resulting finite λL,[45] as it means that momentum is not conserved. Ω A The thermal conductivity of a given material often depends on the temperature and even the direction of heat transfer. → → Thermal Conductivity - k - is used in the Fourier's equation. k / [31][32] For rigid elastic spheres, {\displaystyle {\rm {W/K}}} v {\displaystyle {\vec {q}}} Dimension of L.A-1 is equal to that of L-1 actually. ω the Green-Kubo relations, are difficult to apply in practice, typically consisting of averages over multiparticle correlation functions. T Temperature dependence of the mean free path has an exponential form T {\displaystyle f} W ) / 1 So the only temperature-dependent quantity is the heat capacity c, which, in this case, is proportional to T. So. Elementary calculations then lead to the expression, where P The thermal conductivity of a material depends on its temperature, density and moisture content. Λ {\displaystyle \beta } is independent of with k0 a constant. For monatomic gases, such as the noble gases, the agreement with experiment is fairly good. still holds. [29] Since Thermal conductivity is a measure of a substance’s ability to transfer heat through a material by conduction. The exact mechanisms of thermal conduction are poorly understood in liquids: there is no molecular picture which is both simple and accurate. ∝ If A is constant as well the expression can be integrated with the result, where TH and TL are the temperatures at the hot end and the cold end respectively, and L is the length of the bar. Therefore, these phonons have to possess energy of {\displaystyle P\propto {e}^{-E/kT}} [46][failed verification] This was done by assuming that the relaxation time τ decreases with increasing number of atoms in the unit cell and then scaling the parameters of the expression for thermal conductivity in high temperatures accordingly.[45]. Some possible ways to realize these interfaces are nanocomposites and embedded nanoparticles/structures. {\displaystyle {\rm {K/W}}} [50], In an isotropic medium, the thermal conductivity is the parameter k in the Fourier expression for the heat flux. 1 m ; Input the cross-sectional area (m 2)Add your materials thickness (m)Enter the hot side temperature (°C)Enter the cold side temperature (°C) {\displaystyle \Omega (T)} Power is the rate of heat flow, (i.e.) k T Thermal conductivity of the stainless steel is 16.26 W/m-K . G Thermal conductivity has dimensions of $\mathrm{Power / (length * temperature)}$. q {\displaystyle \mu } {\displaystyle \sim k\Theta /2} In alloys the density of the impurities is very high, so l and, consequently k, are small. e [48] Therefore, specific thermal conductivity is calculated as: Δ . the system approaches a vacuum, and thermal conduction ceases entirely. , which is a significant fraction of Debye energy that is needed to generate new phonons. / {\displaystyle \lambda } is the volume of a mole of liquid, and This transport mechanism is theorized to be limited by the elastic scattering of acoustic phonons at lattice defects. This equation is a result of combining the four previous equations with each other and knowing that Θ For this reason a vacuum is an effective insulator. However, thermal conductivity, which is its reciprocal, is frequently given in specific units of ∇ The use of one million computational cells made it possible to establish a numerical error of less than 0.1%. {\displaystyle \lambda } This is achieved by introducing interface scattering mechanism, which requires structures whose characteristic length is longer than that of impurity atom. Short wavelength phonons are strongly scattered by impurity atoms if an alloyed phase is present, but mid and long wavelength phonons are less affected. Insulation Material Thermal Conductivity Chart . {\displaystyle 0} In this post we will work on the derivation of thermal conductivity formula first, then we will find the dimension of thermal conductivity as well. Another approach is to use analytic models or molecular dynamics or Monte Carlo based methods to describe thermal conductivity in solids. , not deviating by more than + , we get the equation which converts from specific thermal conductivity to absolute thermal conductivity: Again, since thermal conductivity and resistivity are reciprocals of each other, it follows that the equation to convert specific thermal conductivity to absolute thermal resistance is: The thermal conductivity of T-Global L37-3F thermal conductive pad is given as 1.4 W/(mK). To incorporate more complex interparticle interactions, a systematic approach is necessary. 2 At higher temperatures (10 K < T < Θ), the conservation of energy c Thermal conductivity is defined as the transportation of energy due to the random movement of molecules across the temperature gradient. Absolute thermal conductivity, in contrast, is a component property used to compare the heat-transfer ability of different components (i.e., an extensive property). Thermal Conductivity. {\displaystyle \Lambda =v\tau } ⋅ f 1 is small compared with macroscopic (system) dimensions. . eval(ez_write_tag([[250,250],'physicsteacher_in-large-mobile-banner-2','ezslot_3',154,'0','0']));Putting the dimension of Work in equation 2, Dimension of Thermal Conductivity (k) = (ML2)(T-3) L-1 θ-1 = M1 L1 T -3 θ -1 ______ (4), In the next part of this tutorial, let’s find out the values of k for a few selected materials. The entirety of this section assumes the mean free path k = [Q L] / [A (T1-T2) t ] …………………… (1). R-values per inch given in SI and Imperial units (Typical values are approximations, based on the average of available results. , Thermal conductivity is a material property. C or This is particularly useful, for example, when calculating the maximum power a component can dissipate as heat, as demonstrated in the example calculation here. It is convenient to introduce the thermal-conductivity integral, If the temperature difference is small, k can be taken as constant. 3 or {\displaystyle \mathbf {q} _{1}=\mathbf {q} _{2}+\mathbf {q} _{3}+\mathbf {G} } It is a measure of a substances ability to transfer heat through a material by conduction. 0 He is an avid Blogger who writes a couple of blogs of different niches. Mean free path is one factor that determines the temperature dependence for λL, as stated in the following equation, where Λ is the mean free path for phonon and W−1). Power. {\displaystyle \left\langle v_{x}^{2}\right\rangle ={\frac {1}{3}}v^{2}} for a variety of interparticle force laws. T energy flow in a given time. ∂ {\displaystyle \lambda _{A}} What is the Law of Conservation of Energy and how to derive its equation? When steady state conditions are assumed the total time derivate of phonon number is zero, because the temperature is constant in time and therefore the phonon number stays also constant. Alternately, the approximate expression Larger grain dimensions will reduce or eliminate the effect of edge states on the thermal conductivity of the two-dimensional carbon-based material, since the direction of heat flow is perpendicular to the irregularly shaped edges of the monolayer graphene ribbon, as defined previously. W and very close to In that case, Conversion from specific to absolute units, and vice versa. 1 Additional optical modes could also be caused by the presence of internal structure (i.e., charge or mass) at a lattice point; it is implied that the group velocity of these modes is low and therefore their contribution to the lattice thermal conductivity λL ( {\displaystyle {\frac {P}{\Delta T}}} In physics, thermal conductivity is the ability of a material to conduct heat. / T c μ P is the speed of sound in the liquid. {\displaystyle \lambda _{A}} Ranges are marked with "–". [34], For gases whose molecules are not spherically symmetric, the expression So we can write the expression in this way, In the next part of this tutorial, let’s find out th, Thermal Conductivity Derivation\| Dimension of thermal conductivity, Thermal conductivity definition, formula, and…, Zeroth Law of Thermodynamics and thermal equilibrium, Poisson's ratio, Strain energy & Thermal Stress -…, What is the Law Of Conservation Of Momentum? {\displaystyle k} It is denoted by k. The inverse of thermal conductivity is thermal resistivity. A ) Ultimately, as the density goes to v Plug this into your thermal conductivity equation. T k Thermal conductivity is the amount of heat that is lost over time. From these ideas, it can be concluded that increasing crystal complexity, which is described by a complexity factor CF (defined as the number of atoms/primitive unit cell), decreases λL. These processes include the scattering of phonons by crystal defects, or the scattering from the surface of the crystal in case of high quality single crystal. {\displaystyle k_{\text{B}}} v Now we will derive the Thermal Conductivity expression. Put your thermometer in an unobtrusive area of your sample. is a numerical constant of order Describing anharmonic effects is complicated because an exact treatment as in the harmonic case is not possible, and phonons are no longer exact eigensolutions to the equations of motion. T These findings not only expand the basic understanding of thermal transport in complex oxides, but also provide a path to dynamically control the thermal conductivity. {\displaystyle k=f\mu c_{v}} T Use a thermometer to measure the amount of heat passing through the sample from the warm side to the cool side to get your thermal conductivity constant. where ω At higher temperatures the mean free path is limited by the phonons, so the thermal conductivity tends to decrease with temperature. Existing models suffer from the lack of experimental data for the thermal properties of the polymer resist ﬁlms. (here is independent of W λ Any expressions for thermal conductivity which are exact and general, e.g. 1 {\displaystyle c_{v}} Depending on the molecular substructure of ammonium cations and owing to the weaker interactions in the layered structures, the thermal conductivities of our two-dimensional hybrid perovskites range from 0.10 to 0.19 W m –1 K –1, which is drastically lower than that of their three-dimensional counterparts. λ The thermal conductivity of steel is about 1700 times higher than that of mineral wool, which may cause problems with the numer- ical accuracy. by making the following approximation T c = absolute thermal conductivity (W/K, or W/°C). 2 This has been confirmed by the experiments of Chang and Jones on commercial glasses and glass ceramics, where the mean free paths were found to be limited by "internal boundary scattering" to length scales of 10−2 cm to 10−3 cm.[37][38]. Dimension of L.A-1 is equal to that of L-1 actually. / − x < Under these assumptions, an elementary calculation yields for the thermal conductivity. {\displaystyle \mu } electronvolt – what is electronvolt(eV) and how is eV related to Joule? / ⟩ {\displaystyle b=2} ℏ ) A value of 200,000 is predicted for 99.999% 12C at 80 K, assuming an otherwise pure crystal.[26]. {\displaystyle \Delta T} How does the heat transfer conduction calculator works? An explicit treatment of this effect is difficult in the Chapman-Enskog approach. More complex interaction laws introduce a weak temperature dependence. / ⟨ Thermal conductivities of PW, CW, TDCW measured at axial direction are all higher than that at radial direction and the thermal conductivity of TDCW is 0.669 Wm −1 K −1 at 50 °C, which is 114% higher than thermal conductivity of pure TD at 50 °C. 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Sensitivity are required for the thermal bridge of a rod with unknown....	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9384014010429382, "perplexity": 1394.2664774364714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00160.warc.gz"}
http://www.ck12.org/statistics/Applications-of-Normal-Distributions/lecture/Normal-Distribution%3A-Finding-the-Mean-and-Standard-Deviation-Part-2/r1/	<meta http-equiv="refresh" content="1; url=/nojavascript/"> # Applications of Normal Distributions ## Using computational skill and technology to sketch and shade appropriate area under the normal curve 0% Progress Practice Applications of Normal Distributions Progress 0% Normal Distribution: Finding the Mean and Standard Deviation (Part 2) Explains the mean and standard deviation in terms of a normal distribution.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8901944160461426, "perplexity": 2441.5599486282795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736675795.7/warc/CC-MAIN-20151001215755-00144-ip-10-137-6-227.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/1351020/expected-value-when-die-is-rolled-n-times/1351043	# Expected value when die is rolled $N$ times Suppose we have a die with $K$ faces with numbers from 1 to $K$ written on it, and integers $L$ and $F$ ($0 < L \leq K$). We roll it $N$ times. Let $a_i$ be the number of times (out of the $N$ rolls) that a face with number $i$ written on it came up as the top face of the die. I need to find the expectation of the value $a_1^F \times a_2^F \times \cdots a_L^F$ For example, let $N=2, K=6, L=2$ and $F=1$ Then, we roll the $6$-face die $2$ times, and we are interested in the value $a_1 \times a_2$. The only two possible scenarios when this value is not zero are $(1, 2)$ and $(2, 1)$. Both of them have $a_1 \times a_2 = 1$ and happen with probability $1 / 36$ each. So $P / Q = (1 + 1) / 36 = 1 / 18$ Let $A_i$ denote the number of times number $i$ appears (each number is equally likely to appear) and $\mathcal{A}$ be the set of all possible combinations of $a\equiv(a_1,\dots,a_K)$ s.t. $\sum_{k=1}^Ka_k=N$ and each $a_k\ge 0$. Then for $a\in \mathcal{A}$ $$P\{A_1=a_1,\dots,A_K=a_K\}=\frac{N!}{\prod_{k=1}^Ka_k!}K^{-\sum_{k=1}^Na_k}=\frac{K^{-N}N!}{\prod_{k=1}^Ka_k!}$$ and $$\mathbb{E}\left[\prod_{k=1}^LA_k^F\right]=K^{-N}N!\times\sum_{\mathcal{A}}\frac{\prod_{k=1}^La_k^F}{\prod_{k=1}^Ka_k!}$$ Edit: The above formula can be simplified. Assume that $F=1$, $N\ge L$, and the relevant probabilities are $(p_1,\dots,p_K)$. Then $$\prod_{k=1}^Lp_k \times\frac{\partial^L}{\partial p_1\cdots \partial p_L} \left(\prod_{k=1}^Lp_k^{a_k} \right)=\prod_{k=1}^L a_k p_k^{a_k}$$ Since $$(p_1+\cdots+p_K)^N=\sum_{\mathcal{A}}\binom{N}{a_1,\dots,a_K}\prod_{k=1}^Kp^{a_k}$$ differentiating the LHS and noticing that $\sum_{k=1}^Kp_k=1$ yields $$\prod_{k=1}^Lp_k \times\frac{\partial^L}{\partial p_1\cdots \partial p_L}(p_1+\cdots+p_K)^N=\prod_{k=1}^Lp_k\times \prod_{n=0}^{L-1}(N-n)$$ Consequently, since $p_k=K^{-1}$, $k=1,\dots,K$, $$\mathbb{E}\left[\prod_{k=1}^LA_k\right]=K^{-L}\prod_{n=0}^{L-1}(N-n)$$ For $N<L$ this expectation is $0$ because $a_k=0$ for some $k=1,\dots,L$. • Can you provide some example to show your approach – mat7 Jul 6 '15 at 7:30 • You can apply your own example... – d.k.o. Jul 6 '15 at 7:37 • Can't we make use of the fact that faces are numbered from 1 to K. So we can make use of it to find combinations that sum up to N – mat7 Jul 6 '15 at 7:38 • Its pretty simple example to show this formula correctness. So perhaps it will be great if you quote some good one – mat7 Jul 6 '15 at 7:38 • It doesn't matter how faces are labeled. Still, it's easy to write a simple program (in your favourite language) to calculate the expectation... – d.k.o. Jul 6 '15 at 7:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8668333888053894, "perplexity": 208.4456475902287}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257920.67/warc/CC-MAIN-20190525064654-20190525090654-00516.warc.gz"}
https://openreview.net/forum?id=jNB6vfl_680	## Global Magnitude Pruning With Minimum Threshold Is All We Need Published: 28 Jan 2022, 22:06, Last Modified: 13 Feb 2023, 23:23ICLR 2022 SubmittedReaders: Everyone Keywords: Pruning, Model Compression, One-shot, Global Magnitude Pruning Abstract: Neural network pruning remains a very important yet challenging problem to solve. Many pruning solutions have been proposed over the years with high degrees of algorithmic complexity. In this work, we shed light on a very simple pruning technique that achieves state-of-the-art (SOTA) performance. We showcase that magnitude based pruning, specifically, global magnitude pruning (GP) is sufficient to achieve SOTA performance on a range of neural network architectures. In certain architectures, the last few layers of a network may get over-pruned. For these cases, we introduce a straightforward method to mitigate this. We preserve a certain fixed number of weights in each layer of the network to ensure no layer is over-pruned. We call this the Minimum Threshold (MT). We find that GP combined with MT when needed, achieves SOTA performance on all datasets and architectures tested including ResNet-50 and MobileNet-V1 on ImageNet. Code available on github. One-sentence Summary: Global magnitude pruning along with minimum threshold is a very simple pruning technique and at the same time sufficient to obtain SOTA pruning performance. 19 Replies	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8109785318374634, "perplexity": 1793.7431035249522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00573.warc.gz"}
https://brilliant.org/problems/constructing-a-tetrahedron/	# Constructing a tetrahedron For this problem, we will define a "unit polyhedron" to be a polyhedron of edge length one. The picture to the right shows how a larger tetrahedron can be built from four unit tetrahedra and one unit octahedron. Amanda wishes to construct a much larger tetrahedron with many more unit tetrahedra and octahedra. In the end, she builds one using exactly 364 unit octahedra and some unit tetrahedra. How many unit tetrahedra did she use? Assumption: The final construction is one solid tetrahedron, with nothing extra sticking out, and nothing missing (no holes). No unit octahedra or tetrahedra are cut in any way. Image credit: http://www.matematicasvisuales.com/ ×	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8635006546974182, "perplexity": 2772.8436356694533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588341.58/warc/CC-MAIN-20211028131628-20211028161628-00600.warc.gz"}
https://artofproblemsolving.com/wiki/index.php?title=Sieve_of_Sundaram&oldid=118479	# Sieve of Sundaram ## Description The Sieve of Sundaram is a sieve for a range of odd prime numbers.The sieve starts out listing the natural numbers to like the following example () ### Step 2 Then it finds all numbers of form ( fancy way of saying start at and increase by , for each not already eliminated ) and eliminates them (empty cells): ### Step 3 Double each number and add one: ## Basis The basis of the algorithm is that for (without loss of generality, as the form we double and add 1 to is made up of commutative operations); with both factors nontrivial. ## Optimization The equivalent in Step 2 above, of the Sieve of Eratosthenes starting the at the next prime squared, is starting at for next index ## Generalization for prime finding Looking close, if you know modular arithmetic, you'll note the remainder 1 forms the modular multiplicative group modulo 2. This generalizes to the modular multiplicative group mod any natural number( with loss of factors in the prime list). This generalization either needs copies of the same numbers ( one for each class), or (triangle numbers) colors . The class of remainder 1, is the only class where variable must stay nonzero to introduce a nontrivial factor by default ( negative representatives increase variable value required by 1). Modulo 30 is sketched out below (8 classes): $\begin{array}{ccccccccc} c\slash d&1&7&11&13&17&19&23&29\\ 1&a+b&7a+b&11a+b&13a+b&17a+b&19a+b&23a+b&29a+b\\ 7&a+7b&7(a+b)+1&11a+7b+2&13a+7b+3&17a+7b+3&19a+7%b+4&23a+7b+5&29a+7b+6\\ 11&a+11b&7a+11b+2&11(a+b)+4&13a+11b+4&17a+11b+6&19a+11b+6&23a+11b+8&29a+11b+10\\ 13&a+13b&7a+13b+3&11a+13b+4&13(a+b)+5&17a+13b+7&19a+13b+8&23a+13b+9&29a+13b+12\\ 17&a+17b&7a+17b+3&11a+17b+6&13a+17b+7&17(a+b)+9&19a+17b+10&23a+17b+13&29a+17b+16\\ 19&92&93&94&95&96&97&98\\ 23&107&108&109&110&111&112&113&114\\ 29&122&123&124&125&126&127&128\\ \end{array}$ (Error compiling LaTeX. ! Extra alignment tab has been changed to \cr.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8271015286445618, "perplexity": 1521.8532809279598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00629.warc.gz"}
http://arxiv-export-lb.library.cornell.edu/abs/2208.06924v1	physics.hist-ph (what is this?) # Title: On the Map-Territory Fallacy Fallacy Abstract: This paper presents a meta-theory of the usage of the free energy principle (FEP) and examines its scope in the modelling of physical systems. We consider the so-called map-territory fallacy' and the fallacious reification of model properties. By showing that the FEP is a consistent, physics-inspired theory of inferences of inferences, we disprove the assertion that the map-territory fallacy contradicts the principled usage of the FEP. As such, we argue that deploying the map-territory fallacy to criticise the use of the FEP and Bayesian mechanics itself constitutes a fallacy: what we call the {\it map-territory fallacy fallacy}. In so doing, we emphasise a few key points: the uniqueness of the FEP as a model of particles or agents that model their environments; the restoration of convention to the FEP via its relation to the principle of constrained maximum entropy; the Jaynes optimality' of the FEP under this relation; and finally, the way that this meta-theoretical approach to the FEP clarifies its utility and scope as a formal modelling tool. Taken together, these features make the FEP, uniquely, {\it the} ideal model of generic systems in statistical physics. Comments: 23 pages Subjects: History and Philosophy of Physics (physics.hist-ph); Statistical Mechanics (cond-mat.stat-mech) Cite as: arXiv:2208.06924 [physics.hist-ph] (or arXiv:2208.06924v1 [physics.hist-ph] for this version)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8906964063644409, "perplexity": 2767.6654286285197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00285.warc.gz"}
https://darrenjw.wordpress.com/tag/independent/	## One-way ANOVA with fixed and random effects from a Bayesian perspective This blog post is derived from a computer practical session that I ran as part of my new course on Statistics for Big Data, previously discussed. This course covered a lot of material very quickly. In particular, I deferred introducing notions of hierarchical modelling until the Bayesian part of the course, where I feel it is more natural and powerful. However, some of the terminology associated with hierarchical statistical modelling probably seems a bit mysterious to those without a strong background in classical statistical modelling, and so this practical session was intended to clear up some potential confusion. I will analyse a simple one-way Analysis of Variance (ANOVA) model from a Bayesian perspective, making sure to highlight the difference between fixed and random effects in a Bayesian context where everything is random, as well as emphasising the associated identifiability issues. R code is used to illustrate the ideas. ### Example scenario We will consider the body mass index (BMI) of new male undergraduate students at a selection of UK Universities. Let us suppose that our data consist of measurements of (log) BMI for a random sample of 1,000 males at each of 8 Universities. We are interested to know if there are any differences between the Universities. Again, we want to model the process as we would simulate it, so thinking about how we would simulate such data is instructive. We start by assuming that the log BMI is a normal random quantity, and that the variance is common across the Universities in question (this is quite a big assumption, and it is easy to relax). We assume that the mean of this normal distribution is University-specific, but that we do not have strong prior opinions regarding the way in which the Universities differ. That said, we expect that the Universities would not be very different from one another. ### Simulating data A simple simulation of the data with some plausible parameters can be carried out as follows. set.seed(1) Z=matrix(rnorm(10008,3.1,0.1),nrow=8) RE=rnorm(8,0,0.01) X=t(Z+RE) colnames(X)=paste("Uni",1:8,sep="") Data=stack(data.frame(X)) boxplot(exp(values)~ind,data=Data,notch=TRUE) Make sure that you understand exactly what this code is doing before proceeding. The boxplot showing the simulated data is given below. ### Frequentist analysis We will start with a frequentist analysis of the data. The model we would like to fit is $y_{ij} = \mu + \theta_i + \varepsilon_{ij}$ where i is an indicator for the University and j for the individual within a particular University. The “effect”, $\theta_i$ represents how the ith University differs from the overall mean. We know that this model is not actually identifiable when the model parameters are all treated as “fixed effects”, but R will handle this for us. > mod=lm(values~ind,data=Data) > summary(mod) Call: lm(formula = values ~ ind, data = Data) Residuals: Min 1Q Median 3Q Max -0.36846 -0.06778 -0.00069 0.06910 0.38219 Coefficients: Estimate Std. Error t value Pr(>\|t\|) (Intercept) 3.101068 0.003223 962.244 < 2e-16 indUni2 -0.006516 0.004558 -1.430 0.152826 indUni3 -0.017168 0.004558 -3.767 0.000166 * indUni4 0.017916 0.004558 3.931 8.53e-05 * indUni5 -0.022838 0.004558 -5.011 5.53e-07 * indUni6 -0.001651 0.004558 -0.362 0.717143 indUni7 0.007935 0.004558 1.741 0.081707 . indUni8 0.003373 0.004558 0.740 0.459300 --- Signif. codes: 0 ‘*’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1019 on 7992 degrees of freedom Multiple R-squared: 0.01439, Adjusted R-squared: 0.01353 F-statistic: 16.67 on 7 and 7992 DF, p-value: < 2.2e-16 We see that R has handled the identifiability problem using “treatment contrasts”, dropping the fixed effect for the first university, so that the intercept actually represents the mean value for the first University, and the effects for the other Univeristies represent the differences from the first University. If we would prefer to impose a sum constraint, then we can switch to sum contrasts with options(contrasts=rep("contr.sum",2)) and then re-fit the model. > mods=lm(values~ind,data=Data) > summary(mods) Call: lm(formula = values ~ ind, data = Data) Residuals: Min 1Q Median 3Q Max -0.36846 -0.06778 -0.00069 0.06910 0.38219 Coefficients: Estimate Std. Error t value Pr(>\|t\|) (Intercept) 3.0986991 0.0011394 2719.558 < 2e-16 ind1 0.0023687 0.0030146 0.786 0.432048 ind2 -0.0041477 0.0030146 -1.376 0.168905 ind3 -0.0147997 0.0030146 -4.909 9.32e-07 * ind4 0.0202851 0.0030146 6.729 1.83e-11 * ind5 -0.0204693 0.0030146 -6.790 1.20e-11 * ind6 0.0007175 0.0030146 0.238 0.811889 ind7 0.0103039 0.0030146 3.418 0.000634 * --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1019 on 7992 degrees of freedom Multiple R-squared: 0.01439, Adjusted R-squared: 0.01353 F-statistic: 16.67 on 7 and 7992 DF, p-value: < 2.2e-16 This has 7 degrees of freedom for the effects, as before, but ensures that the 8 effects sum to precisely zero. This is arguably more interpretable in this case. ### Bayesian analysis We will now analyse the simulated data from a Bayesian perspective, using JAGS. #### Fixed effects All parameters in Bayesian models are uncertain, and therefore random, so there is much confusion regarding the difference between “fixed” and “random” effects in a Bayesian context. For “fixed” effects, our prior captures the idea that we sample the effects independently from a “fixed” (typically vague) prior distribution. We could simply code this up and fit it in JAGS as follows. require(rjags) n=dim(X)[1] p=dim(X)[2] data=list(X=X,n=n,p=p) init=list(mu=2,tau=1) modelstring=" model { for (j in 1:p) { theta[j]~dnorm(0,0.0001) for (i in 1:n) { X[i,j]~dnorm(mu+theta[j],tau) } } mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) autocorr.plot(output) pairs(as.matrix(output)) crosscorr.plot(output) On running the code we can clearly see that this naive approach leads to high posterior correlation between the mean and the effects, due to the fundamental lack of identifiability of the model. This also leads to MCMC mixing problems, but it is important to understand that this computational issue is conceptually entirely separate from the fundamental statisticial identifiability issue. Even if we could avoid MCMC entirely, the identifiability issue would remain. A quick fix for the identifiability issue is to use “treatment contrasts”, just as for the frequentist model. We can implement that as follows. data=list(X=X,n=n,p=p) init=list(mu=2,tau=1) modelstring=" model { for (j in 1:p) { for (i in 1:n) { X[i,j]~dnorm(mu+theta[j],tau) } } theta[1]<-0 for (j in 2:p) { theta[j]~dnorm(0,0.0001) } mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) autocorr.plot(output) pairs(as.matrix(output)) crosscorr.plot(output) Running this we see that the model now works perfectly well, mixes nicely, and gives sensible inferences for the treatment effects. Another source of confusion for models of this type is data formating and indexing in JAGS models. For our balanced data there was not problem passing in data to JAGS as a matrix and specifying the model using nested loops. However, for unbalanced designs this is not necessarily so convenient, and so then it can be helpful to specify the model based on two-column data, as we would use for fitting using lm(). This is illustrated with the following model specification, which is exactly equivalent to the previous model, and should give identical (up to Monte Carlo error) results. N=n*p data=list(y=Data$values,g=Data$ind,N=N,p=p) init=list(mu=2,tau=1) modelstring=" model { for (i in 1:N) { y[i]~dnorm(mu+theta[g[i]],tau) } theta[1]<-0 for (j in 2:p) { theta[j]~dnorm(0,0.0001) } mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) As suggested above, this indexing scheme is much more convenient for unbalanced data, and hence widely used. However, since our data is balanced here, we will revert to the matrix approach for the remainder of the post. One final thing to consider before moving on to random effects is the sum-contrast model. We can implement this in various ways, but I’ve tried to encode it for maximum clarity below, imposing the sum-to-zero constraint via the final effect. data=list(X=X,n=n,p=p) init=list(mu=2,tau=1) modelstring=" model { for (j in 1:p) { for (i in 1:n) { X[i,j]~dnorm(mu+theta[j],tau) } } for (j in 1:(p-1)) { theta[j]~dnorm(0,0.0001) } theta[p] <- -sum(theta[1:(p-1)]) mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) Again, this works perfectly well and gives similar results to the frequentist analysis. #### Random effects The key difference between fixed and random effects in a Bayesian framework is that random effects are not independent, being drawn from a distribution with parameters which are not fixed. Essentially, there is another level of hierarchy involved in the specification of the random effects. This is best illustrated by example. A random effects model for this problem is given below. data=list(X=X,n=n,p=p) init=list(mu=2,tau=1) modelstring=" model { for (j in 1:p) { theta[j]~dnorm(0,taut) for (i in 1:n) { X[i,j]~dnorm(mu+theta[j],tau) } } mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) taut~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","taut","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) The only difference between this and our first naive attempt at a Bayesian fixed effects model is that we have put a gamma prior on the precision of the effect. Note that this model now runs and fits perfectly well, with reasonable mixing, and gives sensible parameter inferences. Although the effects here are not constrained to sum-to-zero, like in the case of sum contrasts for a fixed effects model, the prior encourages shrinkage towards zero, and so the random effect distribution can be thought of as a kind of soft version of a hard sum-to-zero constraint. From a predictive perspective, this model is much more powerful. In particular, using a random effects model, we can make strong predictions for unobserved groups (eg. a ninth University), with sensible prediction intervals based on our inferred understanding of how similar different universities are. Using a fixed effects model this isn’t really possible. Even for a Bayesian version of a fixed effects model using proper (but vague) priors, prediction intervals for unobserved groups are not really sensible. Since we have used simulated data here, we can compare the estimated random effects with the true effects generated during the simulation. > apply(as.matrix(output),2,mean) mu tau taut theta[1] theta[2] 3.098813e+00 9.627110e+01 7.015976e+03 2.086581e-03 -3.935511e-03 theta[3] theta[4] theta[5] theta[6] theta[7] -1.389099e-02 1.881528e-02 -1.921854e-02 5.640306e-04 9.529532e-03 theta[8] 5.227518e-03 > RE [1] 0.002637034 -0.008294518 -0.014616348 0.016839902 -0.015443243 [6] -0.001908871 0.010162117 0.005471262 We see that the Bayesian random effects model has done an excellent job of estimation. If we wished, we could relax the assumption of common variance across the groups by making tau a vector indexed by j, though there is not much point in persuing this here, since we know that the groups do all have the same variance. #### Strong subjective priors The above is the usual story regarding fixed and random effects in Bayesian inference. I hope this is reasonably clear, so really I should quit while I’m ahead… However, the issues are really a bit more subtle than I’ve suggested. The inferred precision of the random effects was around 7,000, so now lets re-run the original, naive, “fixed effects” model with a strong subjective Bayesian prior on the distribution of the effects. data=list(X=X,n=n,p=p) init=list(mu=2,tau=1) modelstring=" model { for (j in 1:p) { theta[j]~dnorm(0,7000) for (i in 1:n) { X[i,j]~dnorm(mu+theta[j],tau) } } mu~dnorm(0,0.0001) tau~dgamma(1,0.0001) } " model=jags.model(textConnection(modelstring),data=data,inits=init) update(model,n.iter=1000) output=coda.samples(model=model,variable.names=c("mu","tau","theta"),n.iter=100000,thin=10) print(summary(output)) plot(output) This model also runs perfectly well and gives sensible inferences, despite the fact that the effects are iid from a fixed distribution and there is no hard constraint on the effects. Similarly, we can make sensible predictions, together with appropriate prediction intervals, for an unobserved group. So it isn’t so much the fact that the effects are coupled via an extra level of hierarchy that makes things work. It’s really the fact that the effects are sensibly distributed and not just sampled directly from a vague prior. So for “real” subjective Bayesians the line between fixed and random effects is actually very blurred indeed… ## Introduction to the particle Gibbs sampler ### Introduction Particle MCMC (the use of approximate SMC proposals within exact MCMC algorithms) is arguably one of the most important developments in computational Bayesian inference of the 21st Century. The key concepts underlying these methods are described in a famously impenetrable “read paper” by Andrieu et al (2010). Probably the most generally useful method outlined in that paper is the particle marginal Metropolis-Hastings (PMMH) algorithm that I have described previously – that post is required preparatory reading for this one. In this post I want to discuss some of the other topics covered in the pMCMC paper, leading up to a description of the particle Gibbs sampler. The basic particle Gibbs algorithm is arguably less powerful than PMMH for a few reasons, some of which I will elaborate on. But there is still a lot of active research concerning particle Gibbs-type algorithms, which are attempting to address some of the deficiencies of the basic approach. Clearly, in order to understand and appreciate the recent developments it is first necessary to understand the basic principles, and so that is what I will concentrate on here. I’ll then finish with some pointers to more recent work in this area. ### PIMH I will adopt the same approach and notation as for my post on the PMMH algorithm, using a simple bootstrap particle filter for a state space model as the SMC proposal. It is simplest to understand particle Gibbs first in the context of known static parameters, and so it is helpful to first reconsider the special case of the PMMH algorithm where there are no unknown parameters and only the state path, $x$ of the process is being updated. That is, we target $p(x\|y)$ (for known, fixed, $\theta$) rather than $p(\theta,x\|y)$. This special case is known as the particle independent Metropolis-Hastings (PIMH) sampler. Here we envisage proposing a new path $x_{0:T}^\star$ using a bootstrap filter, and then accepting the proposal with probability $\min\{1,A\}$, where $A$ is the Metropolis-Hastings ratio $\displaystyle A = \frac{\hat{p}(y_{1:T})^\star}{\hat{p}(y_{1:T})},$ where $\hat{p}(y_{1:T})^\star$ is the bootstrap filter’s estimate of marginal likelihood for the new path, and $\hat{p}(y_{1:T})$ is the estimate associated with the current path. Again using notation from the previous post it is clear that this ratio targets a distribution on the joint space of all simulated random variables proportional to $\displaystyle \hat{p}(y_{1:T})\tilde{q}(\mathbf{x}_0,\ldots,\mathbf{x}_T,\mathbf{a}_0,\ldots,\mathbf{a}_{T-1})$ and that in this case the marginal distribution of the accepted path is exactly $p(x_{0:T}\|y_{1:T})$. Again, be sure to see the previous post for the explanation. ### Conditional SMC update So far we have just recapped the previous post in the case of known parameters, but it gives us insight in how to proceed. A general issue with Metropolis independence samplers in high dimensions is that they often exhibit “sticky” behaviour, whereby an unusually “good” accepted path is hard to displace. This motivates consideration of a block-Gibbs-style algorithm where updates are used that are always accepted. It is clear that simply running a bootstrap filter will target the particle filter distribution $\tilde{q}(\mathbf{x}_0,\ldots,\mathbf{x}_T,\mathbf{a}_0,\ldots,\mathbf{a}_{T-1})$ and so the marginal distribution of the accepted path will be the approximate $\hat{p}(x_{0:T}\|y_{1:T})$ rather than the exact conditional distribution $p(x_{0:T}\|y_{1:T})$. However, we know from consideration of the PIMH algorithm that what we really want to do is target the slightly modified distribution proportional to $\displaystyle \hat{p}(y_{1:T})\tilde{q}(\mathbf{x}_0,\ldots,\mathbf{x}_T,\mathbf{a}_0,\ldots,\mathbf{a}_{T-1})$, as this will lead to accepted paths with the exact marginal distribution. For the PIMH this modification is achieved using a Metropolis-Hastings correction, but we now try to avoid this by instead conditioning on the previously accepted path. For this target the accepted paths have exactly the required marginal distribution, so we now write the target as the product of the marginal for the current path times a conditional for all of the remaining variables. $\displaystyle \frac{p(x_{0:T}^k\|y_{1:T})}{M^T} \times \frac{M^T}{p(x_{0:T}^k\|y_{1:T})} \hat{p}(y_{1:T})\tilde{q}(\mathbf{x}_0,\ldots,\mathbf{x}_T,\mathbf{a}_0,\ldots,\mathbf{a}_{T-1})$ where in addition to the correct marginal for $x$ we assume iid uniform ancestor indices. The important thing to note here is that the conditional distribution of the remaining variables simplifies to $\displaystyle \frac{\tilde{q}(\mathbf{x}_0,\ldots,\mathbf{x}_T,\mathbf{a}_0,\ldots,\mathbf{a}_{T-1})} {\displaystyle p(x_0^{b_0^k})\left[\prod_{t=0}^{T-1} \pi_t^{b_t^k}p\left(x_{t+1}^{b_{t+1}^k}\|x_t^{b_t^k}\right)\right]}$. The terms in the denominator are precisely the terms in the numerator corresponding to the current path, and hence “cancel out” the current path terms in the numerator. It is therefore clear that we can sample directly from this conditional distribution by running a bootstrap particle filter that includes the current path and which leaves the current path fixed. This is the conditional SMC (CSMC) update, which here is just a conditional bootstrap particle filter update. It is clear from the form of the conditional density how this filter must be constructed, but for completeness it is described below. The bootstrap filter is run conditional on one trajectory. This is usually the trajectory sampled at the last run of the particle filter. The idea is that you do not sample new state or ancestor values for that one trajectory. Note that this guarantees that the conditioned on trajectory survives the filter right through to the final sweep of the filter at which point a new trajectory is picked from the current selection of $M$ paths, of which the conditioned-on trajectory is one. Let $x_{1:T} = (x_1^{b_1},x_2^{b_2},\ldots,x_T^{b_T})$ be the path that is to be conditioned on, with ancestral lineage $b_{1:T}$. Then, for $k\not= b_1$, sample $x_0^k \sim p(x_0)$ and set $\pi_0^k=1/M$. Now suppose that at time $t$ we have a weighted sample from $p(x_t\|y_{1:t})$. First resample by sampling $a_t^k\sim \mathcal{F}(a_t^k\|\boldsymbol{\pi}_t),\ \forall k\not= b_t$. Next sample $x_{t+1}^k\sim p(x_{t+1}^k\|x_t^{a_t^k}),\ \forall k\not=b_t$. Then for all $k$ set $w_{t+1}^k=p(y_{t+1}\|x_{t+1}^k)$ and normalise with $\pi_{t+1}^k=w_{t+1}^k/\sum_{i=1}^M w_{t+1}^i$. Propagate this weighted set of particles to the next time point. At time $T$ select a single trajectory by sampling $k'\sim \mathcal{F}(k'\|\boldsymbol{\pi}_T)$. This defines a block Gibbs sampler which updates $2(M-1)T+1$ of the $2MT+1$ random variables in the augmented state space at each iteration. Since the block of variables to be updated is random, this defines an ergodic sampler for $M\geq2$ particles, and we have explained why the marginal distribution of the selected trajectory is the exact conditional distribution. Before going on to consider the introduction of unknown parameters, it is worth considering the limitations of this method. One of the main motivations for considering a Gibbs-style update was concern about the “stickiness” of a Metropolis independence sampler. However, it is clear that conditional SMC updates also have the potential to stick. For a large number of time points, particle filter genealogies coalesce, or degenerate, to a single path. Since here we are conditioning on the current path, if there is coalescence, it is guaranteed to be to the previous path. So although the conditional SMC updates are always accepted, it is likely that much of the new path will be identical to the previous path, which is just another kind of “sticking” of the sampler. This problem with conditional SMC and particle Gibbs more generally is well recognised, and quite a bit of recent research activity in this area is directed at alleviating this sticking problem. The most obvious strategy to use is “backward sampling” (Godsill et al, 2004), which has been used in this context by Lindsten and Schon (2012), Whiteley et al (2010), and Chopin and Singh (2013), among others. Another related idea is “ancestor sampling” (Lindsten et al, 2014), which can be done in a single forward pass. Both of these techniques work well, but both rely on the tractability of the transition kernel of the state space model, which can be problematic in certain applications. ### Particle Gibbs sampling As we are working in the context of Gibbs-style updates, the introduction of static parameters, $\theta$, into the problem is relatively straightforward. It turns out to be correct to do the obvious thing, which is to alternate between sampling $\theta$ given $y$ and the currently sampled path, $x$, and sampling a new path using a conditional SMC update, conditional on the previous path in addition to $\theta$ and $y$. Although this is the obvious thing to do, understanding exactly why it works is a little delicate, due to the augmented state space and conditional SMC update. However, it is reasonably clear that this strategy defines a “collapsed Gibbs sampler” (Lui, 1994), and so actually everything is fine. This particular collapsed Gibbs sampler is relatively easy to understand as a marginal sampler which integrates out the augmented variables, but then nevertheless samples the augmented variables at each iteration conditional on everything else. Note that the Gibbs update of $\theta$ may be problematic in the context of a state space model with intractable transition kernel. In a subsequent post I’ll show how to code up the particle Gibbs and other pMCMC algorithms in a reasonably efficient way.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 46, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8768958449363708, "perplexity": 629.7348132894087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737152.0/warc/CC-MAIN-20200807025719-20200807055719-00336.warc.gz"}
https://www.caltech.edu/campus-life-events/calendar/math-graduate-student-seminar-19	Thursday, February 14, 2019 12:00pm to 1:00pm Linde Hall 255 Small Gaps Between Primes Alex Perozim De Faveri, Department of Mathematics, Caltech, One of the most famous open problems in number theory is the twin prime conjecture, which says that there are infinitely many prime numbers at distance two. I will introduce some of the tools used to deal with this problem, such as the Selberg sieve and the Bombieri-Vinogradov theorem, and outline the new ideas that led to the breakthroughs of Zhang and Maynard in 2013, who independently proved that the gap between primes is bounded infinitely often. If time permits, I will touch on the known obstructions towards twin primes, such as the parity problem in sieve theory.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9078130125999451, "perplexity": 353.4337342434424}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00187.warc.gz"}
https://blog.evanchen.cc/2015/09/05/some-notes-on-valuations/	# Some Notes on Valuations There are some notes on valuations from the first lecture of Math 223a at Harvard. ## 1. Valuations Let ${k}$ be a field. Definition 1 A valuation $\displaystyle \left\lvert - \right\rvert : k \rightarrow \mathbb R_{\ge 0}$ is a function obeying the axioms • ${\left\lvert \alpha \right\rvert = 0 \iff \alpha = 0}$. • ${\left\lvert \alpha\beta \right\rvert = \left\lvert \alpha \right\rvert \left\lvert \beta \right\rvert}$. • Most importantly: there should exist a real constant ${C}$, such that ${\left\lvert 1+\alpha \right\rvert < C}$ whenever ${\left\lvert \alpha \right\rvert \le 1}$. The third property is the interesting one. Note in particular it can be rewritten as ${\left\lvert a+b \right\rvert < C\max\{ \left\lvert a \right\rvert, \left\lvert b \right\rvert \}}$. Note that we can recover ${\left\lvert 1 \right\rvert = \left\lvert 1 \right\rvert \left\lvert 1 \right\rvert \implies \left\lvert 1 \right\rvert = 1}$ immediately. Example 2 (Examples of Valuations) If ${k = \mathbb Q}$, we can take the standard absolute value. (Take ${C=2}$.) Similarly, the usual ${p}$-adic evaluation, ${\nu_p}$, which sends ${p^a t}$ to ${p^{-a}}$. Here ${C = 1}$ is a valid constant. These are the two examples one should always keep in mind: with number fields, all valuations look like one of these too. In fact, over ${\mathbb Q}$ it turns out that every valuation “is” one of these two valuations (for a suitable definition of equality). To make this precise: Definition 3 We say ${\left\lvert - \right\rvert_1 \sim \left\lvert - \right\rvert_2}$ (i.e. two valuations on a field ${k}$ are equivalent) if there exists a constant ${k > 0}$ so that ${\left\lvert \alpha \right\rvert_1 = \left\lvert \alpha \right\rvert_2^k}$ for every ${\alpha \in k}$. In particular, for any valuation we can force ${C = 2}$ to hold by taking an equivalent valuation to a sufficient power. In that case, we obtain the following: Lemma 4 In a valuation with ${C = 2}$, the triangle inequality holds. Proof: First, observe that we can get $\displaystyle \left\lvert \alpha + \beta \right\rvert \le 2 \max \left\{ \left\lvert \alpha \right\rvert, \left\lvert \beta \right\rvert \right\}.$ Applying this inductively, we obtain $\displaystyle \left\lvert \sum_{i=1}^{2^r} a_i \right\rvert \le 2^r \max_i \left\lvert a_i \right\rvert$ $\displaystyle \sum_{i=1}^{n} a_i \le 2n\max_i \left\lvert a_i \right\rvert.$ From this, one can obtain $\displaystyle \left\lvert \alpha+\beta \right\rvert^n \le \left\lvert \sum_{j=0}^n \binom nj \alpha^j \beta^{n-j} \right\rvert \le 2(n+1) \sum_{j=0}^n \left\lvert \binom nj \right\rvert \left\lvert \alpha \right\rvert^j \left\lvert \beta \right\rvert^{n-j} \le 4(n+1)\left( \left\lvert \alpha \right\rvert+\left\lvert \beta \right\rvert \right)^n.$ Letting ${n \rightarrow \infty}$ completes the proof. $\Box$ Next, we prove that Lemma 5 If ${\omega^n=1}$ for some ${n}$, then${\left\lvert \omega \right\rvert = 1}$. In particular, on any finite field the only valuation is the trivial one which sends ${0}$ to ${0}$ and all elements to ${1}$. Proof: Immediate, since ${\left\lvert \omega \right\rvert^n = 1}$. $\Box$ ## 2. Topological field induced by valuations Let ${k}$ be a field. Given a valuation on it, we can define a basis of open sets $\displaystyle \left\{ \alpha \mid \left\lvert \alpha - a \right\rvert < d \right\}$ across all ${a \in K}$, ${d \in \mathbb R_{> 0}}$. One can check that the same valuation gives rise to the same topological spaces, so it is fine to assume ${C = 2}$ as discussed earlier; thus, in fact we can make ${k}$ into a metric space, with the valuation as the metric. In what follows, we’ll always assume our valuation satisfies the triangle inequality. Then: Lemma 6 Let ${k}$ be a field with a valuation. Viewing ${k}$ as a metric space, it is in fact a topological field, meaning addition and multiplication are continuous. Proof: Trivial; let’s just check that multiplication is continuous. Observe that \displaystyle \begin{aligned} \left\lvert (a+\varepsilon_1)(b+\varepsilon_2) - ab \right\rvert & \le \left\lvert \varepsilon_1\varepsilon_2 \right\rvert + \left\lvert a\varepsilon_2 \right\rvert + \left\lvert b\varepsilon_1 \right\rvert \\ &\rightarrow 0. \end{aligned} $\Box$ Now, earlier we saw that two valuations which are equivalent induce the same topology. We now prove the following converse: Proposition 7 If two valuations ${\left\lvert - \right\rvert_1}$ and ${\left\lvert - \right\rvert_2}$ give the same topology, then they are in fact equivalent. Proof: Again, we may safely assume that both satisfy the triangle inequality. Next, observe that ${\left\lvert a \right\rvert < 1 \iff a^n \rightarrow 0}$ (according to the metric) and by taking reciprocals, ${\left\lvert a \right\rvert > 1 \iff a^{-n} \rightarrow 0}$. Thus, given any ${\beta}$, ${\gamma}$ and integers ${m}$, ${n}$ we derive that $\displaystyle \left\lvert \beta^n\gamma^m \right\rvert_1 < 1 \iff \left\lvert \beta^n\gamma^m \right\rvert < 1$ with similar statements holding with “${<}$” replaced by “${=}$”, “${>}$”. Taking logs, we derive that $\displaystyle n \log\left\lvert \beta \right\rvert_1 + m \log \left\lvert \gamma \right\rvert_1 < 0 \iff n \log\left\lvert \beta \right\rvert_2 + m \log \left\lvert \gamma \right\rvert_1 < 0$ and the analogous statements for “${=}$”, “${>}$”. Now just choose an appropriate sequence of ${m}$, ${n}$ and we can deduce that $\displaystyle \frac{\log \left\lvert \beta_1 \right\rvert}{\log \left\lvert \beta_2 \right\rvert} = \frac{\log \left\lvert \gamma_1 \right\rvert}{\log \left\lvert \gamma_2 \right\rvert}$ so it equals a fixed constant ${c}$ as desired. $\Box$ ## 3. Discrete Valuations Definition 8 We say a valuation ${\left\lvert - \right\rvert}$ is discrete if its image around ${1}$ is discrete, meaning that if ${\left\lvert a \right\rvert \in [1-\delta,1+\delta] \implies \left\lvert a \right\rvert = 1}$ for some real ${\delta}$. This is equivalent to requiring that ${\{\log\left\lvert a \right\rvert\}}$ is a discrete subgroup of the real numbers. Thus, the real valuation (absolute value) isn’t discrete, while the ${p}$-adic one is. ## 4. Non-Archimedian Valuations Most importantly: Definition 9 A valuation ${\left\lvert - \right\rvert}$ is non-Archimedian if we can take ${C = 1}$ in our requirement that ${\left\lvert a \right\rvert \le 1 \implies \left\lvert 1+a \right\rvert \le C}$. Otherwise we say the valuation is Archimedian. Thus the real valuation is Archimedian while the ${p}$-adic valuation is non-Archimedian. Lemma 10 Given a non-Archimedian valuation ${\left\lvert - \right\rvert}$, we have ${\left\lvert b \right\rvert < \left\lvert a \right\rvert \implies \left\lvert a+b \right\rvert = \left\lvert a \right\rvert}$. Proof: We have that $\displaystyle \left\lvert a \right\rvert = \left\lvert (a+b)-b \right\rvert \le \max\left\{ \left\lvert a+b \right\rvert, \left\lvert b \right\rvert \right\}.$ On the other hand, ${\left\lvert a+b \right\rvert \le \max \{ \left\lvert a \right\rvert, \left\lvert b \right\rvert\}}$. $\Box$ Given a field ${k}$ and a non-Archimedian valuation on it, we can now consider the set $\displaystyle \mathcal O = \left\{ a \in k \mid \left\lvert a \right\rvert \le 1 \right\}$ and by the previous lemma, this turns out to be a ring. (This is the point we use the fact that the valuation is non-Archimedian; without that ${\mathcal O}$ need not be closed under addition). Next, we define $\displaystyle \mathcal P = \left\{ a \in k \mid \left\lvert a \right\rvert < 1 \right\} \subset \mathcal O$ which is an ideal. In fact it is maximal, because ${\mathcal O/\mathcal P}$ is the set of units in ${\mathcal O}$, and is thus necessarily a field. Lemma 11 Two valuations are equal if they give the same ring ${\mathcal O}$ (as sets, not just up to isomorphism). Proof: If the valuations are equivalent it’s trivial. For the interesting converse direction (they have the same ring), the datum of the ring ${\mathcal O}$ lets us detect whether ${\left\lvert a \right\rvert < \left\lvert b \right\rvert}$ by simply checking whether ${\left\lvert ab^{-1} \right\rvert < 1}$. Hence same topology, hence same valuation. $\Box$ We will really only work with valuations which are obviously discrete. On the other hand, to detect non-Archimedian valuations, we have Lemma 12 ${\left\lvert - \right\rvert}$ is Archimedian if ${\left\lvert n \right\rvert \le 1}$ for every ${n = 1 + \dots + 1 \in k}$. Proof: Clearly Archimedian ${\implies}$ ${\left\lvert n \right\rvert \le 1}$. The converse direction is more interesting; the proof is similar to the analytic trick we used earlier. Given ${\left\lvert a \right\rvert \le 1}$, we wish to prove ${\left\lvert 1+a \right\rvert \le 1}$. To do this, first assume the triangle inequality as usual, then $\displaystyle \left\lvert 1+a \right\rvert^n < \sum_j \left\lvert \binom nj \right\rvert\left\lvert a \right\rvert^j \le \sum_{j=0}^n \left\lvert a \right\rvert^j \le \sum_{j=0}^n 1 = n+1.$ Finally, let ${n \rightarrow \infty}$ again. $\Box$ In particular, any field of finite characteristic in fact has ${\left\lvert n \right\rvert = 1}$ and thus all valuations are non-Archimedian. ## 5. Completions We say that a field ${k}$ is complete with respect to a valuation ${\left\lvert - \right\rvert}$ if it is complete in the topological sense. Theorem 13 Every field ${k}$ is with a valuation ${\left\lvert - \right\rvert}$ can be embedded into a complete field ${\overline{k}}$ in a way which respects the valuation. For example, the completion of ${\mathbb Q}$ with the Euclidean valuation is ${\mathbb R}$. Proof: Define ${\overline{k}}$ to be the topological completion of ${k}$; then extend by continuity; $\Box$ Given ${k}$ and its completion ${\overline{k}}$ we use the same notation for the valuations of both. Proposition 14 A valuation ${\left\lvert - \right\rvert}$ on ${\overline{k}}$ is non-Archimedian if and only if the valuation is non-Archimedian on ${k}$. Proof: We saw non-Archimedian ${\iff}$ ${\left\lvert n \right\rvert \le 1}$ for every ${n = 1 + \dots + 1}$. $\Box$ Proposition 15 Assume ${\left\lvert - \right\rvert}$ is non-Archimedian on ${k}$ and hence ${\overline{k}}$. Then the set of values achieved by ${\left\lvert - \right\rvert}$ coincides for ${k}$ and ${\overline{k}}$, i.e. ${\{ \left\lvert k \right\rvert \} = \{ \left\lvert \overline{k} \right\rvert \}}$. Not true for Archimedian valuations; consider ${\left\lvert \sqrt2 \right\rvert = \sqrt2 \notin \mathbb Q}$. Proof: Assume ${0 \neq b \in \overline{k}}$; then there is an ${a \in k}$ such that ${\left\lvert b-a \right\rvert < \left\lvert b \right\rvert}$ since ${k}$ is dense in ${\overline{k}}$. Then, ${\left\lvert b \right\rvert \le \max \{ \left\lvert b-a \right\rvert, \left\lvert a \right\rvert \}}$ which implies ${\left\lvert b \right\rvert = \left\lvert a \right\rvert}$. $\Box$ ## 6. Weak Approximation Theorem Proposition 16 (Weak Approximation Theorem) Let ${\left\lvert-\right\rvert_i}$ be distinct nontrivial valuations of ${k}$ for ${i=1,\dots,n}$. Let ${k_i}$ denote the completion of ${k}$ with respect to ${\left\lvert-\right\rvert_i}$. Then the image $\displaystyle k \hookrightarrow \prod_{i=1}^n k_i$ is dense. This means that distinct valuations are as different as possible; for example, if ${\left\lvert-\right\rvert _1 = \left\lvert-\right\rvert _2}$ then we might get, say, a diagonal in ${\mathbb R \times \mathbb R}$ which is as far from dense as one can imagine. Another way to think of this is that this is an analogue of the Chinese Remainder Theorem. Proof: We claim it suffices to exhibit ${\theta_i \in k}$ such that $\displaystyle \left\lvert \theta_i \right\rvert_j \begin{cases} > 1 & i = j \\ < 1 & \text{otherwise}. \end{cases}$ Then $\displaystyle \frac{\theta_i^r}{1+\theta_i^r} \rightarrow \begin{cases} 1 & \text{ in } \left\lvert-\right\rvert_i \\ 0 & \text{ otherwise}. \end{cases}$ Hence for any point ${(a_1, \dots, a_n)}$ we can take the image of ${\sum \frac{\theta_i^r}{1+\theta_i^r} a_i \in k}$. So it would follow that the image is dense. Now, to construct the ${\theta_i}$ we proceed inductively. We first prove the result for ${n=2}$. Since the topologies are different, we exhibit ${\alpha}$, ${\beta}$ such that ${\left\lvert \alpha_1 \right\rvert < \left\lvert \alpha_2 \right\rvert}$ and ${\left\lvert \beta_1 \right\rvert > \left\lvert \beta_2 \right\rvert}$, and pick ${\theta=\alpha\beta^{-1}}$. Now assume ${n \ge 3}$; it suffices to construct ${\theta_1}$. By induction, there is a ${\gamma}$ such that $\displaystyle \left\lvert \gamma \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \gamma \right\rvert_i < 1 \text{ for } i = 2, \dots, n-1.$ Also, there is a ${\psi}$ such that $\displaystyle \left\lvert \delta \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \delta \right\rvert_n < 1.$ Now we can pick $\displaystyle \theta_1 = \begin{cases} \gamma & \left\lvert \gamma \right\rvert_n < 1 \\ \phi^r\gamma & \left\lvert \gamma \right\rvert_n = 1 \\ \frac{\gamma^r}{1+\gamma^r} & \left\lvert \gamma \right\rvert_n > 1 \\ \end{cases}$ for sufficiently large ${r}$. $\Box$ ## 2 thoughts on “Some Notes on Valuations” 1. I could not refrain from commenting. Exceptionally well written! Like	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 170, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870467185974121, "perplexity": 291.77987811992557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00092.warc.gz"}
https://donnate.github.io/publication/2018-01-22-distance	# Tracking network distances: an overview Published in Annals of Applied Statistics 12.2 (2018): 971-1012, 2018 Recommended citation: Donnat, Claire and Holmes, Susan (2018). "Tracking network distances: an overview." Annals of Applied Statistics 12.2 (2018): 971-1012. Graphs have emerged as one of the most powerful frameworks for encapsulating information about evolving interactions or similarities between a set of agents: in such studies, the data typically consist of a set of graphs tracking the state of a system at different times. A critical step in the data analysis process thus lies in the selection of an appropriate distance between networks: how can we devise a metric that is bost robust to small perturbations of the graph structure and sensitive to the properties that make two graphs similar? In this review, we thus propose to provide an overview of some of the existing distances and to introduce a few alternative ones. In particular, we will try to provide ground and principles for choosing an appropriate distance over another, and highlight these properties on both a real-life microbiome application as well as synthetic examples. Finally, we extend our study to the analysis spatial dynamics, and show the performance of our method on a recipe network.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8396750688552856, "perplexity": 661.025907703875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655895944.36/warc/CC-MAIN-20200707204918-20200707234918-00278.warc.gz"}
https://math.stackexchange.com/questions/742686/taylor-series-maclaurin-series	# Taylor Series, Maclaurin series From Rogawski ET 2e section 10.7, exercise 4. Find the Maclaurin series for $f(x) = \dfrac{x^2}{1-8x^8}$. $$\dfrac{x^2}{1-8x^8} = \sum_{n=0}^{\infty} [\textrm{_______________}]$$ Hi! I am working on some online Calc2 homework problem and I am not quite sure how to go about solving this Taylor series. I know I should substitute $8x^8$ for $x$ in the Maclaurin series for $1 \over (1-x)$, but the $x^2$ in the numerator of the problem is throwing me off. If someone could help me find the Maclaurin series and on what interval the expansion is valid, I would greatly appreciate it! Notice $\frac{1}{1-x} = \sum x^n$ . Hence $$\frac{ x^2}{1 - 8x^8} = x^2 \sum (8x^8)^n = \sum 8^n x^{8n + 2}$$ and this is valid for $\| 8x^8 \| < 1$ • Thank you so much, I have more more question related to your answer…how would I write \|x\|<(1/8)^(1/8) in interval notation? – user124539 Apr 6 '14 at 21:59 You can just write $$\frac{x^2}{1-8x^8}=x^2\left(1+8x^8+(8x^8)^2+\cdots\right)=x^2+8x^{10}+64x^{18}+\cdots$$ The interval of convergence will be for all $x$ such that $$\|8x^8\|<1$$ which is equivalent to $$\|x\|<\left(\frac{1}{8}\right)^{{1\over 8}}$$ • Thank you so much, I have more more question related to your answer…how would I write \|x\|<(1/8)^(1/8) in interval notation? – user124539 Apr 6 '14 at 22:00 • It would be $$\left(-\left(\frac{1}{8}\right)^{{1\over 8}},\left(\frac{1}{8}\right)^{{1\over 8}}\right)$$ – user138335 Apr 6 '14 at 22:03 What you have got is almost the right answer. $$\frac{x^2}{1-8x^8} = x^2 \sum_{n=0}^{\infty}(8x^8)^n = \sum_{n=0}^{\infty}8^nx^{8n+2}$$, which is valid if $\|8x^8\|<1$,hence, $\|x\|<\frac{1}{8^{1/8}}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9462524652481079, "perplexity": 288.3774158233372}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00449.warc.gz"}
http://math.stackexchange.com/questions/172669/for-what-value-of-h-the-set-is-linearly-dependent	# For what value of h the set is linearly dependent? For what value of $h$ set $(\vec v_1 \ \vec v_2 \ \vec v_3)$ is linearly dependent? $$\vec v_1=\left[ \begin{array}{c} 1 \\ -3 \\ 2 \end{array} \right];\ \vec v_2=\left[ \begin{array}{c} -3 \\ 9 \\ -6 \end{array} \right] ;\ \vec v_3=\left[ \begin{array}{c} 5 \\ -7 \\ h \end{array} \right]$$ Attempt: After row reducing the augmented matrix of $A\vec x=\vec 0$ where $A=(\vec v_1 \ \vec v_2 \ \vec v_3)$: $$\begin{bmatrix} 1 & -3 & 5 & 0 \\ -3 & 9 & -7 & 0 \\ 2 & -6 & h & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 5 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & h-10 & 0 \end{bmatrix}$$ I am not sure whether the set is linearly dependent when $h=10$ or for any $h$. Help please. - The set is always linearly dependent since $v_2 = -3v_1.$ – user2468 Jul 19 '12 at 2:27 @J.D. so it is enough for the set of three vectors to have two vectors that are collinear to be a linearly dependent set, right? – Koba Jul 19 '12 at 3:39 Indeed. A quick geometric reminder for yourself: the basis in $\Bbb{R}^3.$ If you pick two vectors collinear in the direction of the $x$-axis & a vector in the $z$ direction, would you be able to describe every vector in $\Bbb{R}^3$? Of course not. – user2468 Jul 19 '12 at 3:45 That reduced matrix shows you that the set of vectors is linearly dependent for every value of $h$. If $h\ne 10$, the system has no solution, and if $h=10$, it has infinitely many, so there is no value of $h$ that gives it exactly one solution. Indeed, you can see this directly from the vectors themselves: $v_2=-3v_1$. I think you may have confused the data, @Brian: if $\,h=10\,$ then the third row becomes all zero and, thus, the original set of three vectors is linearly dependent, as asked. True, if $\,h\neq 10\,$ then the homogeneous system is inconsistent, but we don't really care about that as nothing was asked about solutions of linear systems, homogeneous or non-homog. – DonAntonio Jul 19 '12 at 2:17 In fact, I think the OP confused himself by writing down an augmented matrix as if he wanted to solve some linear system, whereas a $\,3\times 3\,$ matrix with the vectors' components is enough to find out whether they're l.i. or not. And then yes, as you wrote: for any value of $\,h\,$ the three vectors are l.d. This is also easy to check calculating the easy determinant of that square matrix, which is zero no matter what $\,h\,$ is. – DonAntonio Jul 19 '12 at 2:21 @DonAntonio: I suspect that you’re right about the confusion, but you missed the point of my answer. If the three vectors were linearly independent for some $h$, then for that $h$ the homogeneous system would have only the trivial solution. But there is no value of $h$ for which this is the case, so for every $h$ the vectors must be linearly dependent. – Brian M. Scott Jul 19 '12 at 2:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9284145832061768, "perplexity": 151.08234224200757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931008720.43/warc/CC-MAIN-20141125155648-00060-ip-10-235-23-156.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/57067/hausdorff-dimension-of-non-recurrent-walks/57074	# Hausdorff dimension of non-recurrent walks Preface: I am fairly new to the concept of Hausdorff dimension, so I don't know how interesting a question this is. Identify walks on $\mathbb{Z}$ with infinite binary sequences (say $0$ means moving left, $1$ means moving right). It is then well-known that, in the Cantor space $2^\mathbb{N}$ under the Lebesgue measure, the set $A$ of non-recurrent walks -- i.e. those sequences $x$ for which $\frac{\sum_{k=1}^n x(k)}{n} = \frac{1}{2}$ for only finitely many $n$ -- is null. I am curious as to the Hausdorff dimension of this set, but I do not see how to figure this. Thus my question: What is the Hausdorff dimension of the set $A$ of non-recurrent walks? Perhaps this too is already well-known, but if so I could not locate the result. I hope this question is not trivial, though if it is at least I will have learned that. I look forward to any replies; this seems like a marvelous site! - First of all the dimension of the space. You didn't give the metric you want to use, so I'll use my favourite one: two points are at distance $2^{-n}$ if they first disagree in the $n$th symbol. The 1-Hausdorff measure agrees with coin-flipping measure so that the full space has Hausdorff dimension 1. Now for the non-recurrent subset. It's also of Hausdorff dimension 1. Probably the simplest way to see this is to use some technology: consider the measure $\mu_\alpha$ that is coin-flipping with weights $\alpha$ and $1-\alpha$. Clearly for $\alpha\ne1/2$ this measure is supported on the non-recurrent set. But the Hausdorff dimension of the measure is $(-\alpha\log\alpha-(1-\alpha)\log(1-\alpha))/\log 2$. This is a lower bound for the Hausdorff dimension of the non-recurrent set. But as $\alpha\to1/2$, this lower bound converges to 1. In case you have an aversion to dynamical systems there is a way to do this with a single (non-invariant) measure and get the result without taking any limits. Let $n_1 < n_2 < \ldots$ be a sequence of density 0 such that the number of terms up to $N$ ($T(N)$ say) grows faster than $\sqrt{3N\log\log N}$ (e.g. $(n_i)$ is the sequence $\lfloor i^2/\log i\rfloor$). Now build the measure that is fair coin-tossing at each $n$ except at the $n_i$ when you always put a 1. Since the fair part of the process puts you in the range $\pm\sqrt{(2+\epsilon)N\log\log N}$ for all sufficiently large $N$ (by the Law of the Iterated Logarithm), the unfair part of the process guarantees that you only return to 0 finitely many times. Hence this measure is supported on the non-recurrent set. This measure has Hausdorff dimension 1: the measure of a $2^{-N}$ neighbourhood of a typical point is $2^{-N+T(N)}$. The Hausdorff dimension is the limit of $\log(2^{-N+T(N)})/\log(2^{-N})$. The fact that $T(N)=o(N)$ guarantees that the Hausdorff dimension is 1. - Yes, you were right about the metric I intended. Thank you for the nice succinct solution(s). – Michel Mar 2 '11 at 12:39 Anthony's answer settles the matter, but I'll say a few words about relevant terminology and references that are too long to fit in the comment box. (And go a bit beyond what you actually asked, but may give things some context.) This is essentially a question in multifractal analysis. Given an asymptotic property such as "the asymptotic frequency of ones is bounded away from 1/2", one can study the set of points with this property in different ways. If you study the measure of this set, you're doing ergodic theory; if you study the dimension of this set, you're doing multifractal analysis. In your setting, a natural thing to do is to fix $\alpha \in [0,1]$ and to consider the set $K_\alpha = \{ x \mid \frac 1n \sum_{k=1}^n x(k) \to \alpha \}$. (Each set $K_\alpha$ is contained in your non-recurrent set.) Then $2^{\mathbb{N}} = (\bigcup_{\alpha\in [0,1]} K_\alpha) \cup \hat K$, where $\hat K$ is the set of points $x$ for which $\lim \frac 1n \sum_{k=1}^n x(k)$ does not exist. This is an example of a multifractal decomposition. One can show that the measures $\mu_\alpha$ in Anthony's answer have the property that $\mu_\alpha(K_\alpha)=1$ and $$\dim_H K_\alpha = \dim_H \mu_\alpha = \frac{-\alpha\log \alpha - (1-\alpha)\log (1-\alpha)}{\log 2}.$$ Thus the function $\alpha \mapsto K_\alpha$ is an analytic and concave function of $\alpha$; this is an example of a multifractal spectrum. There are lots of these, associated to various asymptotic quantities, and you can find a lot of subtle behaviour. (For example, one can ask how big the set $\hat K$ is, and it turns out that even though it is null for every shift-invariant measure on $2^\mathbb{N}$, it still has full Hausdorff dimension.) For more on this, you can see the book "Dimension Theory in Dynamical Systems", by Yakov Pesin -- the first few chapters are very abstract and difficult to follow if you're not already pretty familiar with some of the basic ideas in dimension theory, but you can also skip straight to the chapters on multifractal analysis and get an idea of what's going on. There's also a survey paper by Barreira, Pesin, and Schmeling from 1997 or thereabouts that is a good introduction. - Thanks for taking the time to add this; I find it helpful. And the references are much appreciated. – Michel Mar 2 '11 at 12:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9696692228317261, "perplexity": 125.32267074743272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098468.93/warc/CC-MAIN-20150627031818-00214-ip-10-179-60-89.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/388147/induction-of-emf-by-motion-of-a-conductor-motional-emf	# Induction of emf by motion of a conductor (Motional EMF) Does emf get induced for any conductor moving perpendicularly to a magnetic field? In the textbook explanation, they gave an example of a fixed metal frame over which a metal rod can roll. The field goes into the area of the frame and the rod moves horizontally over the frame, such that its motion is perpendicular to the field. In this case, the area bounded by the frame and the rod keeps changing, hence the flux changes, and an emf is induced. But consider a case where there is no frame, and the rod is just moving perpendicular to a magnetic field with constant velocity. Is emf still induced across the ends of the rod? Also, is the shape of the conductor of any significance? For example, replace the straight rod with a semicircular one in the above question. • The answer is yes to an emf being induced as you do not need a complete conducting circuit to induced the emf. Feb 23, 2018 at 9:28 • @Farcher But isn't emf induced only when there is a change in flux? If the object is moving inside a uniform magnetic field, then the number of field lines passing through the conductor is always the same until the instant it just leaves the field. Why then should an emf be induced at all, regardless of whether it is a complete conducting circuit or not? Feb 23, 2018 at 11:20 • Your question is probably a duplicate. If you put in motional emf into this site's search engine you will find a number of answers relating to induced emf and motional emf. Here is one to read physics.stackexchange.com/q/239741 Feb 23, 2018 at 12:13 Induced emf is not defined w.r.t change in flux. It is defined as, $$\xi=\int \vec{f}_{mag}.\vec{dl}$$ where where $f_{mag}$ is the total force per unit charge which drives the current around the circuit. This can be due to the battery or can be due to a non electrostatic electric field or a magnetic field. The source cannot be an electrostatic field since the line integral of an electrostatic field over the entire circuit is $0$ since it is a conservative field. Now if there is a single rod moving in a region with perpendicular magnetic field, there will be seperation of charges due to the Lorentz force, $\vec{f}_{mag}=\vec{v} \times\vec{B}$. Thus an emf will be induced according to the previous formula. Since the circuit is not complete, there won't be any current flow. For your second question, the answer is a clear no. Induced emf only depends on the line integral. It can be calculated for any kind of shape you can come up with. If you still have doubts regarding induced emf, you can refer to the book by David J. Griffiths. • I'd add that for motional emf, the seat of the emf is the magnetic Lorentz force, as explained in the above answer. $\mathscr E=-\frac{d\Phi}{dt}$ is just a convenient way of calculating the motional emf. Its advantage is that it automatically sums the emfs induced in different parts of the circuit that may be moving. Jun 28, 2019 at 16:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8747087717056274, "perplexity": 190.97552275948655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335303.67/warc/CC-MAIN-20220929003121-20220929033121-00740.warc.gz"}
https://chem.libretexts.org/Courses/Eastern_Wyoming_College/EWC%3A_Introductory_Chemistry_(Budhi)/13%3A_Solutions/13.07%3A_Solution_Dilution	# 13.7: Solution Dilution Skills to Develop • Explain how concentrations can be changed in the lab • Understand how stock solutions are used in the laboratory We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms - dilute and concentrated. • dilute solution is one in which there is a relatively small amount of solute dissolved in the solution. • concentrated solution contains a relatively large amount of solute. These two terms do not provide any quantitative information (actual numbers) - but they are often useful in comparing solutions in a more general sense. These terms also do not tell us whether or not the solution is saturated or unsaturated, or whether the solution is "strong" or "weak". These last two terms will have special meanings when we discuss acids and bases, so be careful not to confuse these terms. ### Stock Solutions It is often necessary to have a solution whose concentration is very precisely known. Solutions containing a precise mass of solute in a precise volume of solution are called stock (or standard) solutions. To prepare a standard solution a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL are are carefully calibrated to a single volume. On the narrow stem is a calibration mark. The precise mass of solute is dissolved in a bit of the solvent and this is added to the flask. Then enough solvent is added to the flask until the level reaches the calibration mark. Often it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution as described in the previous section. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration. ### Dilutions of Stock (or Standard) Solutions Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next we willl dilute this solution - we do that by adding more water, not more salt: $$\rightarrow$$ Before Dilution After Dilution The molarity of solution 1 is $M_1 = \dfrac{\text{moles}_1}{\text{liter}_1}$ and the molarity of solution 2 is $M_2 = \dfrac{\text{moles}_2}{\text{liter}_2}$ rearrange the equations to find moles: $\text{moles}_1 = M_1 \text{liter}_1$ and $\text{moles}_2 = M_2 \text{liter}_2$ What stayed the same and what changed between the two solutions? By adding more water, we changed the volume of the solution. Doing so also changed it's concentration. However, the number of moles of solute did not change. So, $moles_1 = moles_2$ Therefore $\boxed{M_1V_1= M_2V_2 } \label{diluteEq}$ where • $$M_1$$ and $$M_2$$ are the concentrations of the original and diluted solutions and • $$V_1$$ and $$V_2$$ are the volumes of the two solutions Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are easy to do. Suppose that you have $$100. \: \text{mL}$$ of a $$2.0 \: \text{M}$$ solution of $$\ce{HCl}$$. You dilute the solution by adding enough water to make the solution volume $$500. \: \text{mL}$$. The new molarity can easily be calculated by using the above equation and solving for $$M_2$$. $M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}$ The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves deciding how much of a highly concentrated solution is required to make a desired quantity of solution of lesser concentration. The highly concentrated solution is typically referred to as the stock solution. Example $$\PageIndex{1}$$: Diluting NITRIC ACID Nitric acid $$\left( \ce{HNO_3} \right)$$ is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is $$16 \: \text{M}$$. How much of the stock solution of nitric acid needs to be used to make $$8.00 \: \text{L}$$ of a $$0.50 \: \text{M}$$ solution? SOLUTION Steps for Problem Solving Identify the "given"information and what the problem is asking you to "find." Given: M1, Stock $$\ce{HNO_3} = 16 \: \text{M}$$ $$V_2 = 8.00 \: \text{L}$$ $$M_2 = 0.50 \: \text{M}$$ Find: Volume stock $$\ce{HNO_3} \left( V_1 \right) = ? \: \text{L}$$ List other known quantities none Plan the problem First, rearrange the equation algebraically to solve for $$V_1$$. $V_1 = \frac{M_2 \times V_2}{M_1}$ Calculate and cancel units Now substitute the known quantities into the equation and solve. $V_1 = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}$ Think about your result. $$250 \: \text{mL}$$ of the stock $$\ce{HNO_3}$$ needs to be diluted with water to a final volume of $$8.00 \: \text{L}$$. The dilution is by a factor of 32 to go from $$16 \: \text{M}$$ to $$0.5 \: \text{M}$$. Exercise $$\PageIndex{1}$$ A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8149338960647583, "perplexity": 745.1595362403518}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00226.warc.gz"}
https://www.physicsforums.com/threads/number-theory-find-two-smallest-integers-with-same-remainders.613244/	# Homework Help: Number theory find two smallest integers with same remainders 1. Jun 11, 2012 ### Wildcat 1. The problem statement, all variables and given/known data Find the two smallest positive integers(different) having the remainders 2,3, and 2 when divided by 3,5, and 7 respectively. 2. Relevant equations 3. The attempt at a solution I got 23 and 128 as my answer. I tried using number theory where I started with 7x +2 as the number, then if divided by 5 the remainder would be 3 so subtract 3 from 7x+2 to get 7x-1 when I use this method and stop there by having x=0,1,2,3,... 3 works 7(3)-1 is divisible by 5 so put 3 in the original 7(3) +2 =23 Then I just used a trial and error method to find 128. Is my answer correct?? I know my procedure is sketchy because I have never been exposed to number theory. 2. Jun 11, 2012 ### Fightfish Your answers are correct. There are of course, more formal methods of solving it. In number theory, we usually use the method of taking modulos. Let me illustrate this for the question below: From the remainders, we have: a == 2 (mod 3) - (1) a == 3 (mod 5) - (2) a == 2 (mod 7) - (3) From (3), the numbers must have the form a = 7k+2, where k is any positive integer. Using (1): 7k + 2 == 2 (mod 3) This implies that 7k == 0 (mod 3), quite a useful result! Thus k = 3n, where n is any positive integer, and so our numbers a = 21n + 2. Using (2): 21n + 2 == 3 (mod 5) This implies that 21 n == 1 (mod 5). Since 21 == 1 (mod 5), n == 1 (mod 5) as well for the equation to hold. Thus the numbers a that satisfy the conditions are of the form 21n + 2, n = 1,6,11,16,21... The first two numbers are thus 21(1) + 2 = 23 and 21(6) + 2 = 128 3. Jun 18, 2012 ### Wildcat Thanks!! I need to take a class on number theory!	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.896325409412384, "perplexity": 575.2112646520516}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00384.warc.gz"}
http://mathhelpforum.com/discrete-math/189538-statistical-recurrence-relation.html	## statistical recurrence relation I've a simple recurrence relation of functions: $a(x)_{n+1} = a(x)_{n} f(y_{n})$ where the distribution of y is known, and given by, say, p(y). the function f is also given. i'm looking for a (statistical) solution for the relation	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9829529523849487, "perplexity": 1898.0661179911172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988650.6/warc/CC-MAIN-20150728002308-00255-ip-10-236-191-2.ec2.internal.warc.gz"}
https://www.arxiv-vanity.com/papers/1703.01159/	# Manipulation of entanglement sudden death in an all-optical setup Ashutosh Singh Light and Matter Physics Group, Raman Research Institute, Sadashivanagar, Bangalore 560080, India. Siva Pradyumna A.R.P. Rau Department of Physics and Astronomy, Louisiana State University - Baton Rouge, LA 70803, USA. Urbasi Sinha ###### Abstract The unavoidable and irreversible interaction between an entangled quantum system and its environment causes decoherence of the individual qubits as well as degradation of the entanglement between them. Entanglement sudden death (ESD) is the phenomenon wherein disentanglement happens in finite time even when individual qubits decohere only asymptotically in time due to noise. Prolonging the entanglement is essential for the practical realization of entanglement-based quantum information and computation protocols. For this purpose, the local NOT operation in the computational basis on one or both qubits has been proposed. Here, we formulate an all-optical experimental set-up involving such NOT operations that can hasten, delay, or completely avert ESD, all depending on when it is applied during the process of decoherence. Analytical expressions for these are derived in terms of parameters of the initial state’s density matrix, whether for pure or mixed entangled states. After a discussion of the schematics of the experiment, the problem is theoretically analyzed, and simulation results of such manipulations of ESD are presented. ## 1 Introduction Quantum entanglement [1,2] is a non-classical correlation shared among quantum systems which could be non-local [3,4] in some cases. It is a fundamental trait of quantum mechanics. Like classical correlations, entanglement also decays with time in the presence of noise in the ambient environment. The decay of entanglement depends on the initial state and the type and amount of noise (Amplitude damping, Phase damping, etc.) acting on the system [5-7]. The entangled states: and (maximally entangled “Bell states“ for ) being the simplest and most useful entangled states in quantum information processing receive special attention. The maximally entangled states and undergo asymptotic decay of entanglement in the presence of an amplitude damping channel (ADC). The non maximally entangled states always undergo asymptotic decay of entanglement, whereas undergo asymptotic decay for and a finite time end called entanglement sudden death (ESD) for in the presence of ADC. On the other hand, a pure phase damping channel (PDC) causes entanglement to always decay asymptotically. Two different initial states (, where ), which share the same amount of initial entanglement (measured through Negativity) being affected by the same type of noise may follow very different trajectories of entanglement decay. In the presence of multiple stochastic noises, although the decoherence of individual qubits follows the additive law of relaxation rates, the decay of entanglement, does not. In fact, entanglement may not decay asymptotically at all, and disentanglement can happen in finite time (ESD). ESD has been experimentally demonstrated in atomic [8] and photonic systems [9,10]. Since entanglement is a resource in quantum information processing [11-13], manipulation that prolongs entanglement will help realize protocols that would otherwise suffer due to short entanglement times. Also, entanglement purification [14] and distillation [15] schemes could possibly recover the initial correlation from the ensemble of noise-degraded correlation so long as the system has not completely disentangled. Therefore, the delay or avoidance of ESD is important. Several proposals exist to suppress the decoherence; for example, decoherence-free subspaces [16-19], quantum error correction [20,21], dynamical decoupling [22-24], quantum Zeno effect [25-27], quantum measurement reversal [28-33], and delayed-choice decoherence suppression [34]. Protecting entanglement using weak measurement and quantum measurement reversal [32,33], and delayed choice decoherence suppression [34] have been experimentally demonstrated. Both of these schemes, however, have the limitation that the success probability of decoherence suppression decreases as the strength of the weak interaction increases. The practical question we want to address here is; whether, given a two-qubit entangled state in the presence of amplitude damping channel which causes disentanglement in finite time, can we alter the time of disentanglement by a suitable operation during the process of decoherence? A theoretical proposal exists in the literature for such manipulation of ESD [35] through a local unitary operation (NOT operation in computational basis: ) performed on the individual qubits which swaps their population of ground and excited states. Depending on the time of application of this NOT operation, it can avoid, delay, or hasten the ESD. Based on this proposal [35], we have extended the experimental set up [9] for ESD and propose here an all-optical experimental set up for manipulating ESD involving the NOT operation on one or both the qubits of a bipartite entangled state in a photonic system. The system consists of polarization-entangled photons () produced in the sandwich configuration Type-I SPDC (spontaneous parametric down conversion)[38]. These photons are sent to two displaced identical Sagnac interferometers, where ADC is simulated using rotating HWPs (half wave plates) placed in the path of incoming photons (See Fig.2). The HWP selectively causes a polarized photon to “decay“ to ( and serve as ground and excited states of the system, the two states of a qubit)[9]. The NOT operation is implemented by a HWP with fast axis at relative to , placed right after the ADC. This HWP is followed by PBS (polarizing beam splitter) to segregate the and polarizations and, with subsequent ADC after the NOT operation implemented by a set of secondary HWPs acting on only. Such a set of secondary HWPs simulating the ADC (or evolution of qubits in noisy environment) is essential to our study as the ADC (for example, spontaneous emission in case of a two-level atomic system) continues to act even after the NOT operation is applied [35] and these secondary HWPs simulate it in our proposed experiment. The orientation of the HWP () plays the role of time () with () analogous to () for a two-level atomic system decaying to the ground state due to spontaneous emission. We use Negativity as a measure of entanglement. It is defined as the sum of absolute values of negative eigen values of the partially transposed density matrix [36,37]. We find that our simulation results for the manipulation of ESD involving NOT operations on one or both the qubits of a polarization entangled photonic system in presence of ADC are completely consistent with the theoretical predictions of the reference [35] which has analyzed an atomic system. The merit of our scheme is that it can delay or avoid ESD (provided the NOT operation is performed sufficiently early) unlike previous experiments [32-34] where success probabilities scaled with the strength of the weak interaction. Since the photonic system is time independent and noise is simulated using HWPs, it gives experimentalists complete freedom to study and manipulate the disentanglement dynamics in a controlled manner. In this, our photonic system through a controllable HWP offers an advantage over others such as atomic states where the decay occurs through noise sources lying outside experimental control. The NOT operation that we apply through a HWP is the analogue of flipping spin in a nuclear magnetic system, achieved through what is referred to as a pulse. The paper is organized as follows: In section (2), we discuss the all-optical implementation of the proposed ESD experiment and analyze it theoretically using the Kraus operator formalism. In section (3) and (4), we discuss and theoretically analyze the proposed ESD-manipulation experiment involving NOT operation on both or on only one of the qubits, respectively. In section (5), we give analytical expressions for probabilities and also for ESD and its manipulation curves in terms of the parameters of the initial state (density matrix). The first of these is the setting (“time“) for ESD, the next setting for the NOT marking the border between hastening and delay; that is, if the NOT is applied after (and of course before ), it actually hastens, ESD happening before , whereas application before delays ESD to stretch past to larger but still finite value less than one. The third, , marks the border between delaying or completely avoiding ESD. Applying the NOT after delays to a larger value whereas applying before avoids ESD altogether. In section (6), we summarize the results of manipulation of ESD for different pure and mixed initial entangled states, giving numerical values of , and . Section (7) concludes with pros and cons of the proposed scheme for the manipulation of ESD and the future scope of this work. ## 2 Proposed experimental set up for ESD and its analysis The proposed experimental setup for ESD is shown in figure (1), which is a generalization of the scheme used in reference [9]. The type-1 polarization entangled photons () can be prepared by standard methods [38]; the amplitudes and and relative phase are controlled by the HWP and QWP (quarter wave plate). These entangled photons are sent to two displaced Sagnac interferometers with HWPs simulating the ADC, where decoherence takes place, and finally these photons are sent for tomographic reconstruction of the quantum state [39]. The and polarizations of the photon serve as the ground and excited states of the analogous atomic system, while output spatial modes of the reservoir serve as the ground and excited states of the reservoir. Asymmetry between degenerate polarization states of a photon ( ) is introduced by the HWP rotation such that it selectively causes an incident polarization to “decay“ to , while leaving polarization intact. Thus the and polarization states are analogous to the ground and excited states, respectively, of a two-level atomic system. The ADC is implemented using two HWPs: oriented at respectively, such that incident polarization amplitude “decays“ to . For different fixed orientations of , evolution in ADC is completed by rotating . The PBS is used to segregate the and polarization amplitudes such that the HWP is applied only to polarization for it to serve as excited state of the system and leaving polarization (ground state of the system) undisturbed. The single qubit Kraus operators for ADC are given by, M1=(100√1−p) , M2=(0√p00), (1) where for ADC mimicked by HWP in a photonic system [9]. These operators satisfy the completeness condition, M†1M1+M†2M2=I, (2) where is the identity matrix. The Kraus operators for the two qubits are obtained by taking appropriate tensor products of single qubit Kraus operators as follows, Mij=Mi⊗Mj ; i,j=1,2. (3) Label another set of Kraus operators by , with variable replaced by () to distinguish it from the former, with the form of Kraus operators remaining similar to that in (1). Such a splitting into two angles or two values of probability will prove convenient for later applications in the section of manipulation using optical elements in between. Let the initial state of the system be \|ψ⟩=\|α\|\|HH⟩+\|β\|exp(ιδ)\|VV⟩, (4) with a corresponding density matrix given by, ρ(0,0)=⎛⎜ ⎜ ⎜⎝u00v00000000v∗00x⎞⎟ ⎟ ⎟⎠, (5) where and . In general, if , this represents a pure entangled state, otherwise a mixed entangled state. A more general mixed state with non-zero entries in the other two diagonal positions is considered in appendix [B]. The initial state of the system (5) in the presence of ADC (due to at ) evolves as follows, ρ(1)(p,0)=∑i,jMij ρ(0,0) M†ij ; i,j=1,2. (6) Apply the Kraus operators to complete the evolution in the presence of ADC (due to at ) as follows, ρ(1)(p,p′)=∑i,jM′ij ρ(p,0) M′†ij ; i,j=1,2 , (7) =⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ρ(1)11(p,p′)00ρ(1)14(p,p′)0ρ(1)22(p,p′)0000ρ(1)33(p,p′)0ρ(1)41(p,p′)00ρ(1)44(p,p′)⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠, where, ρ(1)11(p,p′) =u+p2x+p′2(1−p)2x+2p′(1−p)px, (8) ρ(1)22(p,p′) =(1−p′)p′(1−p)2x+(1−p′)(1−p)px, ρ(1)33(p,p′) =(1−p′)p′(1−p)2x+(1−p′)(1−p)px, ρ(1)44(p,p′) =(1−p′)2(1−p)2x, ρ(1)14(p,p′) =(1−p′)(1−p)v, ρ(1)41(p,p′) =(1−p′)(1−p)v∗. ## 3 Proposed experimental set up for manipulation of ESD using the NOT operation on both qubits of a bipartite entangled state The proposed experimental set up for manipulation of ESD based on the local NOT operation performed on both the qubits of a bipartite entangled state (5) is shown in figure (2). The HWP acts as ADC for incident polarized photon and then NOT operation is performed by at , which swaps the and amplitudes, which are then segregated by PBS . The ADC is continued by synchronous rotation of oriented at , which causes the swapped amplitude to “decay“ to . The photons from the output spatial modes of the interferometer are sent for tomographic reconstruction of the quantum state [39]. The initial state of the system (5) in the presence of ADC (due to at ) evolves as follows, ρ(2)(p,0)=∑i,jMij ρ(0,0) M†ij ; i,j=1,2. (9) Apply the NOT operation on both the qubits at as follows, ρ(2)(pn,0)=(^σx⊗^σx)ρ(2)(p,0) (^σx⊗^σx)†, (10) where is the Pauli matrix. This amounts to switching the elements and and and and interchanging (complex conjugation) the off-diagonal elements. Apply the Kraus operators to complete the evolution of the system in the presence of ADC (due to and at ) after the NOT operation as follows, ρ(2)(pn,p′)=∑i,jM′ij ρ(2)(pn,0) M′†ij ; i,j=1,2, (11) with entries now in the form of (7) given by, ρ(2)11(pn,p′) =(1−pn)2x+2p′(1−pn)pnx+p′2(u+p2nx), (12) ρ(2)22(pn,p′) =(1−p′)(1−pn)pnx+(1−p′)p′(u+p2nx), ρ(2)33(pn,p′) =(1−p′)(1−pn)pnx+(1−p′)p′(u+p2nx), ρ(2)44(pn,p′) =(1−p′)2(u+p2nx), ρ(2)14(pn,p′) =(1−p′)(1−pn)v∗, ρ(2)41(pn,p′) =(1−p′)(1−pn)v. ## 4 Effect of the NOT operation applied on only one of the qubits The proposed experimental set up for studying the effect of a NOT operation applied on only one of the qubits of a bipartite entangled state in the presence of ADC on the dynamics of entanglement is to retain one half, say the lower, as in figure (1) and have only the upper half as in figure (2), the optical elements and occurring only in the upper arm. The initial state of the system (5) in the presence of ADC (due to at ) evolves as follows, ρ(3)(p,0)=∑i,jMij ρ(0,0) M†ij ; i,j=1,2. (13) Apply the NOT operation on only one of the qubits by at , let us say first qubit, at as follows, ρ(3)(pn,0)=(^σx⊗^I) ρ(3)(p,0) (^σx⊗^I)†. (14) Apply next the Kraus operators to complete the evolution of the system in the presence of ADC (due to at ) after the NOT operation to give, ρ(3)(pn,p′)=∑i,jM′ij ρ3(pn,0) M′†ij ; i,j=1,2, (15) with entries now in the form of (7) given by, ρ(3)11(pn,p′) =p′(1−pn)2x+(1−pn)pnx+p′2(1−pn)pnx+p′(u+p2nx), (16) ρ(3)22(pn,p′) =(1−p′)(1−pn)2x+(1−p′)p′(1−pn)pnx, ρ(3)33(pn,p′) =(1−p′)p′(1−pn)pnx+(1−p′)(u+p2nx), ρ(3)44(pn,p′) =(1−p′)2(1−pn)pnx, ρ(3)23(pn,p′) =(1−p′)(1−pn)v∗, ρ(3)32(pn,p′) =(1−p′)(1−pn)v. ## 5 Some analytical expressions Let the two polarization entangled qubits constitute the system, as given by eqn (5), and the action of the rotating HWPs simulate the ADC. This causes a polarized photon to probabilistically “decay“ to with probability (), where () is the angle between the fast axis of the HWP and . The ADC probability at which ESD happens, depends on the initial state parameters of the entangled system and the ADC setting of first HWP . The criterion for ESD as indicated by a switch in sign of the eigenvalues of the partial transpose of Eq(7) is given by . For the initial state (5), the condition for ESD is obtained by computing the Negativity of the state (7) and equating it to zero. The condition for ESD is given by, p′0=\|v\|−xpx(1−p). (17) Let us denote the effective end of entanglement due to combined evolution through two HWPs by . The involves a multiplication of survival probabilities to give, 1−pend=(1−p)(1−p′0) with pend=\|v\|/x. (18) depending only on the initial state parameters in (5). For the manipulation of ESD using NOT operation on both the qubits, this operation switches and , and , and interchanges the off-diagonal elements and in Eq. (7). With subsequent evolution, the criterion for when ESD now happens, can be used to determine the value of that marks the boundary between hastening or not relative to , and similarly the value that is the boundary between delaying past or averting ESD completely. We get, pA=1−2u2(1−u) , pB=\|v\|−u1+\|v\|−u. (19) For the manipulation of ESD using NOT operation on only one of the qubits, this operation now switches and , and , and moves into the position. Following subsequent evolution, the ESD criterion through the partial transpose matrix now becomes . We now get, pA=\|v\|u+2\|v\| , pB=\|v\|2\|v\|2−u+1. (20) These simple expressions defining the time for ESD, and the times for NOT operation that define the delay/hasten and avert/delay boundaries may also be given for a more general mixed state density matrix with also non-zero entries in the two other diagonal position in (5) and are recorded in Appendix [B]. The Appendix [C] records similar expressions for a density matrix with non-zero values in the other off-diagonal position as in [35]. The NOT operation applied on both the qubits at of a bipartite entangled state leads to the end of entanglement given by, (21) When NOT operation is applied on only one of the qubits at , the end of entanglement is given by, pend=[4x(pn−1)pn+4(pn−1)2\|v\|2+1]1/2+2xpn−12xpn. (22) ## 6 Results and Discussion As an example, we choose the initial state with and and report the results for ESD, and ESD-manipulation using the NOT operation on one or both the qubits of the bipartite entangled state. For ESD using two HWPs, the disentanglement happens for at and for any other combination of and , follows the non-linear eqn (17) in , and the effective end due to two HWPs is given by non-linear eqn (18) with . The plot of Negativity N vs. probability of decay of qubits () for the state (7) is shown in figure (3). The plot of Purity (defined as ) vs. probability of decay of qubits () for ESD is shown in figure (4). The two qubit entangled state (5), initially in a pure state, gets mixed at intermediate stages of amplitude damping and finally becomes pure again when both the qubits have decohered down to the ground state () at or . However, at ESD for , it ends as a mixed disentangled state. For the manipulation of ESD using NOT operation on both the qubits: we get , and . The corresponding plot of Negativity vs. ADC probability () for the state (11) is shown in figure (5). For the manipulation of ESD using NOT operation on only one of the qubits: we get . The corresponding plot of Negativity vs. ADC probability () for the state (15) is shown in figure (6). The plot of vs. for eqn (21) and (22) such that the NOT operations applied on both (only one of) the qubits at leads to disentanglement at is shown by dashed (solid) blue curve in figure (7). In the avoidance range , the vs. curves are cut off at to signify the asymptotic decay with probabilities remaining in the physical domain. For comparison, we have also included the results of ESD; eqn (17) and a rendering of (18), for every , giving the value of , the compounding of them giving the flat line at as shown by dotted red curve and dot-dashed red line in figure (7). The role of NOT operation on manipulation of ESD is evident as for (i) , we get avoidance of ESD with in this range for NOT operation on only one or both the qubits (ii) ( ), we get delay of ESD as the dashed (solid) blue curve lies above 0.5 but less than 1, and (iii) (), we get hastening of ESD as the dashed (solid) blue curve dips below 0.5. The discussion so far, and figures , pertain to the choice , and result in for NOT applied to both whereas when applied to just one qubit. This is an example when and both lie in the physically relevant interval (). All three phenomena, of hastening (), delaying (), and averting () ESD then occur. The appearance of the various manipulation regimes (hastening, delay, and avoidance of ESD) critically depends on the choice of the parameters of the initial state (density matrix) of the system as expressed in eqn (17-20). Consider a general initial state (5), with , which captures pure as well as mixed entangled states. The condition for the existence of hastening regime is for manipulation of ESD using single or double NOT operation. The condition for existence of avoidance regime is () for manipulation of ESD using single (double) NOT operation. For the pure entangled state (4), the condition for a physically relevant is . Thus, pure entangled states (4) with give rise to hastening, delay as well as avoidance of ESD, whereas states with give rise to delay and avoidance of ESD only. For all values of initial parameters, the analytical expressions in (17-20) provide for ESD, and and . When these lie within the domain , all three regimes are realized. Otherwise, one may have only two or one of the three regimes of avoidance, delay or hastening of ESD. More general expressions for a wider class of density matrices than (5) are given in Appendix [B,C]. Consider another example of pure entangled state of the form (5) with and . For this state, , () and does not exist in the physical domain for single (double) NOT operation. Therefore, we get only delay and avoidance of ESD. The corresponding plot of vs. is shown by solid (dashed) blue curve in figure(8). Next, consider an example of mixed entangled state of the form (5) with and . For this state, , ( does not exist), and does not exist in the physical domain for single (double) NOT operation. Therefore, NOT operation applied on only one (both) of the qubits delays as well avoids (only delays) the ESD. The corresponding plot of vs. is shown by solid (dashed) blue curve in figure (9). ## 7 Conclusions and future work We have proposed an all-optical experimental setup for the demonstration of hastening, delay, and avoidance of ESD in the presence of ADC in a photonic system. The simulation results of the manipulation of ESD considering a photonic system, when NOT operations are applied on one or both the qubits, are completely consistent with the theoretical predictions of reference [35] for the two-level atomic system where spontaneous emission is the ADC. We give analytical expressions for which depend on the parameters of the density matrix of the system for both the forms considered here in (5) and that in [35]. Our proposal also has an advantage over decoherence suppression using weak measurement and quantum measurement reversal, and delayed choice decoherence suppression. There, as the strength of weak interaction increases, the success probability of decoherence suppression decreases. In our scheme, however, we can manipulate the ESD, in principle, with unit success probability as long as we perform the NOT operation at the appropriate wave plate angle which is analogous to time in the atomic system. Delay and avoidance of ESD, in particular, will find application in the practical realization of quantum information and computation protocols which might otherwise suffer a short lifetime of entanglement. Also, it will have implications towards such control over other physical systems. The advantage of the manipulation of ESD in a photonic system is that one has complete control over the damping parameters, unlike in most atomic systems. An experimental realization of our proposal will be important for practical noise engineering in quantum information processing, and is under way. Further work in the future could study the dynamics of entanglement in the presence of the generalized ADC [40-43] and the squeezed generalized ADC [44] and the possible schemes for manipulation of entanglement sudden death in the presence of such damping channels. ## 8 Acknowledgments US and AS acknowledge Prof. Ujjwal Sen for his comments on the manuscript. 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A 77, 012318 (2008). ## Appendix A An alternative approach to analyze the ESD and manipulation experiments We provide here an alternative and intuitive approach to analyze the ESD and its manipulation set up by tagging the photon polarization states with the spatial modes of the interferometer upon the action of each of optical component encountered in the photon’s path. The evolution of system plus reservoir is represented by a unitary operator . The degrees of freedom of the reservoir can be traced out from to get the Kraus operators which govern the evolution of the system by itself. Consider the experimental set up for ESD as shown in fig(1). An incident polarized photon is transmitted though the PBS and traverses the interferometer in a counter-clockwise direction, returns to and is transmitted into spatial mode of the reservoir. The corresponding quantum map is given by, USR\|H⟩S\|a⟩R→\|H⟩S\|a⟩R. (23) An incident polarized photon is reflected by PBS and traverses the interferometer in a clockwise direction. The action of HWPs and and PBS and are represented by the quantum map, USR\|V⟩S\|a⟩R H1@θ−−−→P2√1−p\|V⟩S\|a′⟩R+√p\|H⟩S\|b⟩R, (24) H2@θ′−−−−−−−→in \|V⟩ arm√1−p[√1−p′ \|V⟩S\|a′⟩R +√p′ \|H⟩S\|b′⟩R]+√p\|H⟩S\|b⟩R], where Consider the experimental setup for ESD manipulation using double NOT operation as shown in figure (2). An incident polarized photon is transmitted through PBS and traverses the interferometer in a counter-clockwise direction where the NOT operation is applied by HWP and ADC afterwards is simulated by . The corresponding quantum map is given by, USR\|H⟩S\|a⟩R H5@45−−−−→\|V⟩S\|b⟩R , (25) H6@θ′−−−→√1−p′\|V⟩S\|b⟩R+√p′\|H⟩S\|a⟩R . An incident polarized photon is reflected by PBS and traverses the interferometer in a clockwise direction where ADC is introduced by HWP followed by NOT operation by HWP and then ADC is continued by HWP . The corresponding quantum map is given by, USR\|V⟩S\|a⟩R H1@θ−−−→√1−p\|V⟩S\|a⟩R+√p\|H⟩S\|b⟩R , (26) P2−−−−→H5@45√1−p\|H⟩S\|b⟩R+√p\|V⟩S\|a′⟩R , H2@θ′−−−−−−−→in \|V⟩ arm√1−p\|H⟩S\|b⟩R +√p[√1−p′\|V⟩S\|a′⟩R+√p′\|H⟩S\|b′⟩R] . ## Appendix B Analytical expressions for X-state with non-zero entries in the other diagonal terms As discussed in Section 5, we again use the negativity criterion for the partial transposed density matrix to determine the occurrence of ESD. By following, as in [35], the evolution of the parameters in Eq.(2), and double NOT switching and , and , and swapping the off-diagonal terms and , whereas a single NOT at one end switches and , and , and moves and inward along the anti-diagonal, we again obtain analytical expressions for the various (or equivalently, and ) of interest. Consider the initial mixed entangled state with the density matrix given in a form more general than (5): ρ2=⎛⎜ ⎜ ⎜⎝a00z0b0000c0z∗00d⎞⎟ ⎟ ⎟⎠, (27) where . As per the convention for ground and excited states in the reference [35], we choose as the population of both qubits being in excited states and being the population of both qubits in the ground states unlike the (reversed) convention in the main body of this paper. In the presence of ADC, the condition for ESD is given by, p0=−b−c+[(b−c)2+4\|z\|2]1/22a. (28) For manipulation of ESD using NOT operation on both the qubits, the condition for hastening ESD is given by, pA=a−d1+a−d. (29) The condition for avoidance of ESD is given by, pB=1−2a+b+c−[(b−c)2+4\|z\|2)]1/22[(a+b)(a+c)−\|z\|2]. (30) For manipulation of ESD using NOT operation on only one of the qubits, the condition for hastening ESD is given by, pA=1−(c+a)[(c+a)(1−p0)−1](a+b){(a+b)−[(b−c)2+4\|z\|2]1/2}−a. (31) and the condition for avoidance of ESD is given by, pB=\|z\|2−c\|z\|2+a. (32) ## Appendix C Analytical expressions for the X-state in Reference [35] Consider the initial mixed entangled state with the form of density matrix as in reference [35], ρ1=⎛⎜ ⎜ ⎜⎝a0000bz00z∗c0000d⎞⎟ ⎟ ⎟⎠. (33) For simplicity, the 1/3 factor in [35] has been absorbed into the density matrix elements. In the presence of ADC, the condition for ESD is given by, p0=−b−c+[(b+c+2a)2−4(a−\|z\|2)]1/22a, (34) where , is the spontaneous decay rate of the two level atomic qubit as introduced in [35]. For manipulation of ESD using NOT operation on both the qubits, the condition for hastening ESD is given by, pA=a−d1+a−d. (35) The condition for avoidance of ESD is given by, pB=2(a−\|z\|2)−(2a+b+c)+[(b+c+2a)2−4(a−\|z\|2)]1/22(a−\|z\|2). (36) For manipulation of ESD using NOT operation on only one of the qubits, the condition for hastening ESD is given by, pA=(c+a)[2a(1−p0)−(c+a)(1−p0)+c+d]−a(c+a)[2a(1−p0)−(b+a)]−a. (37) The condition for avoidance of ESD is given by, pB=1−a+c(a+b)(a+c)+\|z\|2. (38) For hastening, delay and avoidance to exist in a physical region, the corresponding parameters must satisfy the condition . As an example, the choice gives , , and an unphysical negative value of . This means that neither hastening nor averting ESD is possible, only delaying it by applying NOT between and .	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8814756870269775, "perplexity": 1887.7924512363054}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00393.warc.gz"}
https://testbook.com/learn/maths-derivatives-of-logarithmic-functions/	# Derivatives of Logarithmic Functions with Formulas, Proof and Solved Examples 0 Save We are going to learn the key concepts of derivatives of logarithmic functions with definitions, important formulas with proof, properties and graphical representation of derivatives. We have also added a few solved examples for the derivatives of logarithmic functions that candidates will find beneficial in their exam preparation. ## What are Derivatives? Derivatives of a function is a concepts in mathematics of real variables that measure the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). They are a part of differential calculus. There are various methods of log differentiation. Derivative of a function of a real variable measures the sensitivity to change the function value (output value) with respect to a change in its argument (input value). Derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. You should also learn about the application of derivatives after differentiating log function. Derivative of a function is represented by $${dy\over{dx}}$$ or f′(x), represents the limit of the secant’s slope as h approaches zero. Example of derivative: The derivative of a displacement function is velocity. ## What are Derivatives of Logarithmic Functions? Derivatives of Logarithmic functions of the variable with respect to itself are equal to its reciprocal. Derivatives of Logarithmic Functions are a series of formulae that can be used to differentiate logarithmic functions quickly. $${d\over{dx}}{logx}={1\over{x}}$$ Derivatives of logarithmic functions are used to find solutions to differential equations. ## Derivatives of Logarithmic Functions Formula The Derivatives of Logarithmic Functions Formula by using the normal method is as follows: If x > 0 and y= ln⁡x, then $${dy\over{dx}} = {1\over{x}}$$ If x≠0 and y=ln\|x\|, then $${dy\over{dx}} = {1\over{x}}$$ If the natural log is not just x but instead is g(x), a differentiable function. Now, using the chain rule, we get a more general derivative: for all values of x for which g(x) > 0, the derivative of f(x) = ln(g(x)) is given by $$f'(x) = {1\over{g(x)}}g’(x)$$ ### Derivatives of Logarithmic Functions using Graphical Representation To learn the derivatives of logarithmic functions using graphical representation, let’s look at a graph of the log function with base e: f(x) = log_e(x). The tangent at x = 2 is included on the graph. The slope of the tangent of y = ln x at $$\displaystyle{x}={2}$$ is $$\displaystyle\frac{1}{{2}}.$$ We can observe this from the graph, by looking at the ratio rise/run. If y = ln x, x 1 2 3 4 5 The slope of the graph 1 1/2 1/3 1/4 1/5 1/x 1 1/2 1/3 1/4 1/5 Learn about Derivative of Cos3x and Differentiation and Integration ## Derivatives of Logarithmic Functions Proof According to the definition of the derivatives of logarithmic functions proof by first principle of derivative, the logarithmic differentiation of f(x)=x with respect to x can be written in limited operation form as follows: $$\begin{matrix} f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}} f(x)=logx\\ f(x+h)=log(x+h)\\ f(x+h)–f(x)=log(x+h)-logx=log({x+h\over{x}})=log(1+{h\over{x}})\\ {f(x+h)–f(x)\over{h}}={log(1+{h\over{x}})\over{h}}\\ \text{Let’s put (h\x) as (1\n)}\\ {f(x+h)–f(x)\over{h}}=({1\over{h}}){log(1+{1\over{n}})}=({n\over{x}}){log(1+{1\over{n}})}=({1\over{x}})({log(1+{1\over{n}})})^n\\ \text{As h tends to zero (1\n) tends to}\infty\\ \lim _{h{\rightarrow}\infty}{f(x+h)–f(x)\over{h}}=\lim _{h{\rightarrow}\infty}{(1\over{x}})({log(1+{1\over{n}})})^n={(1\over{x}})\lim _{h{\rightarrow}\infty}({log(1+{1\over{n}})})^n\\ \text{The value of the limit is equal to}\\ \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n} = e \approx 2.718281828459 \ldots\\ \left( {{{\log }_a}x} \right)^\prime = \frac{1}{x}{\log _a}e\\ {\log _a}e = \frac{{\ln e}}{{\ln a}} = \frac{1}{{\ln a}}\\ f’(x)={d\over{dx}}{logx}={1\over{x}} \end{matrix}$$ Learn about First Principles of Derivatives and Derivative of Root x ## Solved Examples of Derivatives of Logarithmic Functions Example: Find the derivative of y = ln 2x We use the log law: log ab = log a + log b We can write our question as: y = ln 2x = ln 2 + ln x Now, the derivative of a constant is 0, so $${d\over{dx}} ln 2 = 0$$ So we are left with (from our formula above) $$y’ = {d\over{dx}}lnx = {1\over{x}}$$ Example: Find the derivative of $$y = lnx^2$$ We use the log law: $$log a^n = n log a$$ So we can write the question as $$y = ln x^2 = 2 ln x$$ The derivative will be simply 2 times the derivative of ln x. $$y’ = 2\times{d\over{dx}}lnx = {2\over{x}}$$ Example: Find the derivative of $$f(x) = ln (x^3+3x−4)$$ $$f(x) = ln (x^3+3x−4)$$ We use the formula $$f'(x) = {1\over{g(x)}}g’(x)$$ where g(x) is the composite function of x. Therefore we get, $$f'(x) = {1\over{(x^3+3x−4)}}[3x^2 + 3]$$ Example: Find the derivative of $$f(x) = ln({x^2sinx\over{2x+1}})$$ $$f(x)=ln({x^2sinx\over{2x+1}}) = 2lnx + ln(sinx) − ln(2x+1)$$ We use the formula $$f'(x) = {1\over{g(x)}}g’(x)$$ where g(x) is the composite function of x. Therefore we get, $$\begin{matrix} f'(x) = {2\over{x}} + {1\over{sinx}}.cosx – {1\over{2x+1}}.2\\ f'(x) = {2\over{x}} + cotx – {2\over{2x+1}} \end{matrix}$$ Example: Differentiate: $$f(x)=ln(3x+2)^5$$ $$f(x)=ln(3x+2)^5$$ We use the log law: $$log a^n = n log a$$ f(x) = 5ln(3x+2) We use the formula $$f'(x) = {1\over{g(x)}}g’(x)$$ where g(x) is the composite function of x. Therefore we get, $$\begin{matrix} f'(x) = 5\times {1\over{(3x + 2}} \times3\\ f'(x) = {15\over{(3x + 2}} \end{matrix}$$ $$f(x) = {3^x\over{3^x + 2}}$$ $$\begin{matrix} f(x) = {3^x\over{3^x + 2}}\\ f’(x) = {3^x ln 3(3^x + 2) – 3^x ln 3(3^x)\over{(3^x + 2)^2}}\\ f’(x) = {2⋅3^xln3\over{(3^x + 2)^2}} \end{matrix}$$ Hope this article on the Derivatives of Logarithmic Functions was informative. Get some practice of the same on our free Testbook App. Download Now! If you are checking Derivatives of Logarithmic Functions article, also check the related maths articles: Logarithmic Functions Derivative of log x Applications of Derivatives Derivative Rules Logarithmic Function MCQs Derivatives of Algebraic Functions ## Derivatives of Logarithmic Functions FAQs Q.1 What are Derivatives? Ans.1 The derivative of a function of real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The derivative of a function, represented by $${dy\over{dx}}$$ or f′(x), represents the limit of the secant’s slope as h approaches zero. Example: The derivative of a displacement function is velocity. Q.2 What is the derivative of a logarithmic function? Ans.2 The derivative of a logarithmic function of the variable with respect to itself is equal to its reciprocal. $${d\over{dx}}{logx}={1\over{x}}$$ Q.3 What is derivative of log 2x? Ans.3 f(x) = log 2x We use the log law: log ab = log a + log b We can write our question as: y = log 2x = log 2 + ln x Now, the derivative of a constant is 0, so {d\over{dx}} log 2 = 0 So we are left with (from our formula above) y’ = {d\over{dx}}logx = {1\over{x}} Q.4 What is the derivative of exponential functions? Ans.4 The derivative of an exponential term, which contains a variable as a base and a constant as power, is called the constant power derivative rule. Suppose a and x represent a constant and a variable respectively then the exponential function is written as a^x in mathematics. The derivative of a raised to the power of x with respect to x is written in the following form in calculus. $${d\over{dx}}a^x=a^x.\ln{a}$$ Q.5 What is the derivative of f(x) = loga? Ans.5 a is constant. Derivative of any function of constant is 0. Therefore, f’(x) = o. Q.5 What is the derivative of log a? Ans.5 The derivative of loga(x): y=loga(x)x=ay1=ddx(ay)1=ayln(a)dydxdydx=1ayln(a)dydx=1xln(a). Q.6 What is the differentiation of log 2? Ans.6 The derivative of log 2(x) log 2 ( x ) with respect to x is 1xln(2) 1 x ln (2).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9464746713638306, "perplexity": 430.22502133748213}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00121.warc.gz"}
http://math.stackexchange.com/users/33789/probabilitnator?tab=summary	# Probabilitnator less info reputation 7 bio website location age member for 1 year, 10 months seen Mar 28 at 14:06 profile views 15 # 4 Questions 2 A certain family of continuous functions on $[0,1]^2$ the closure of which linear span is $\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ 2 Approximation of bounded measurable functions with continuous functions 1 Applicability of Itô's Lemma for $g\in \mathcal{C}^2((0,1)^2)\cap \mathcal{C}_0([0,1]^2)$ 0 The $\alpha$-Potential-Operator (Definition and resolvent Equation) # 38 Reputation +10 A certain family of continuous functions on $[0,1]^2$ the closure of which linear span is $\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ +16 Approximation of bounded measurable functions with continuous functions +5 Applicability of Itô's Lemma for $g\in \mathcal{C}^2((0,1)^2)\cap \mathcal{C}_0([0,1]^2)$ This user has not answered any questions # 10 Tags 0 probability × 2 0 special-functions 0 stochastic-processes × 2 0 stochastic-analysis 0 analysis × 2 0 potential-theory 0 measure-theory 0 operator-theory 0 functions 0 stochastic-integrals # 7 Accounts Quantitative Finance 1,294 rep 13 MathOverflow 179 rep 9 Stack Overflow 118 rep 8 Area 51 101 rep 1 Mathematics 38 rep 7	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8211734294891357, "perplexity": 2804.3622521854454}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00011-ip-10-147-4-33.ec2.internal.warc.gz"}
https://puzzling.stackexchange.com/questions/85679/how-do-you-make-prime-computers	# How do you make Prime “COMPUTERS”? $$Given$$: $$COMPUTERS$$ is the smallest Pan Digital containing all the digits 1 to 9 occurring only once. $$COMPUTERSV$$ is a Prime only when one of the correct digit ($$V$$)is added at the end. Also, $$COMPUTERSE$$ = $$CE$$ * $$EOTOCTPC$$ What is the digit $$V$$ that has to be added to make $$COMPUTERS$$ A Prime? • Enough info given... – Uvc Jun 29 '19 at 10:39 $$1$$ $$COMPUTERS=123456789$$, $$V\ne\{2,4,6,8\}$$ (even), $$V\ne\{3,6,9\}$$ (multiple of $$3$$), $$V\ne\{5\}$$ (multiple of $$5$$). So $$V=1$$ or $$V=7$$. But $$E=7$$ and this is not prime, as given later. Therefore $$V=1$$. And indeed $$1234567891$$ is prime.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.910769522190094, "perplexity": 480.89126690514837}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00244.warc.gz"}
http://www.cs.nyu.edu/pipermail/fom/2013-August/017551.html	# [FOM] First Order Logic MartDowd at aol.com MartDowd at aol.com Sat Aug 31 14:12:25 EDT 2013 The second-order axiom of induction is $\forall S ( 0\in S \wedge \forall n ( n\in S\Rightarrow n+1\in S ) \Rightarrow \forall n ( n\in S ))$ It is provable in ZFC that $N$ (the integers with 0,1,+,x) is the only structure satisfying Peano's axioms with the second order induction axiom. However, the statements in the language of number theory which can be proved in ZFC to hold in this structure are recursively enumerable. The true statements are not. Any attempt to remedy the situation by means of providing axioms for second order validity cannot succeed. - Martin Dowd In a message dated 8/30/2013 2:29:24 P.M. Pacific Daylight Time, hewitt at concurrency.biz writes: I am having trouble understanding why the proponents of first-order logic think that second-order systems are unusable. [Dedekind 1888] and [Peano 1889] thought they had achieved success because they had presented axioms for natural numbers and real numbers such that models of these axioms are unique up to isomorphism with a unique isomorphism -------------- next part -------------- An HTML attachment was scrubbed... URL: </pipermail/fom/attachments/20130831/168abbac/attachment.html>	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8954846858978271, "perplexity": 3700.8232159404283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999669442/warc/CC-MAIN-20140305060749-00069-ip-10-183-142-35.ec2.internal.warc.gz"}
http://link.springer.com/chapter/10.1007%2F978-1-4757-1915-4_18	1986, pp 348-370 Measurements of Total Carbon Dioxide and Alkalinity in the North Atlantic Ocean in 1981 * Final gross prices may vary according to local VAT. Abstract The ocean uptake of fossil fuel CO, has long been recognized as the principal modulator of the rising atmospheric CO, level. If we are to observe and understand this effect, then an essential step is the accurate measurement of the CO, properties of the ocean. Historically, this has been quite difficult to achieve. Although measurements of some kind date back to the late 19th century, complete, documented, and verifiable measurements are scarce indeed. This chapter describes and documents the series of total CO, and alkalinity measurements of seawater made on the North Atlantic Ocean during the Transient Tracers in the Ocean (TTO) expedition in 1981, and presents briefly the signals these data reveal.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8537163138389587, "perplexity": 1504.5726423734302}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657133485.50/warc/CC-MAIN-20140914011213-00011-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"}
https://brilliant.org/discussions/thread/problem-in-rating-system/	× # Problem in rating system Initially, this week, in Number theory and Algebra, I had a nice rating almost equal to level 5. I had solved all questions, except 1 question viewed but unattempted in both subjects. Since Tuesday, these 2 questions were undone. And the problem is, my rating in both the subjects is decreasing intensively, that is, almost 200 points till now! I don't even know why has it happened. If it is supposed to be the same way, then Brilliant staff could tell me why. But if this some sort of a glitch, staff must fix it, as it may further reduce and demote me! I have already lost extreme number of rating. Had those 200 rating not vanished, I would have been on level 5! I hope this will surely be explained to me by the Birlliant staff, waiting for a soon reply, especially before next week, so that the problem doesn't get delayed to one more week! Note by Akshat Jain 3 years, 2 months ago Sort by: Hi Akshat, It looks like your rating has been dropping because you have been viewing lots of problems outside your level. When you view these problems it affects your rating exactly like . We will fix it to better indicate this by displaying ratings on problems that aren't in your problem set. You can recover your rating, and improve it, by solving the problems you viewed. Staff · 3 years, 2 months ago Sir, till date i m solving only problems of my level. How can i get acess to see and solve problems of higher level... · 3 years, 2 months ago What! :O I have viewed many many problems outside my level, because only 4 problems in each subject every week is not enough food for my mind! Though I have solved most of them, the ones unsolved have got me. I personally feel this should be changed. I mean, rating should not decrease on viewing a problem outside the problem set, though it should be reduced on giving the wrong answer. I would really like to suggest the staff to make this change! · 3 years, 2 months ago I didn't know about this at all. So just like every week, I have been solving problems outside my level and didn't understand why my rating was going down. Just like you, my rating has gone down by 400-500. But now that I know, I hope it won't go down again. · 3 years, 2 months ago	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8203195333480835, "perplexity": 1156.686035433998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00141-ip-10-171-10-70.ec2.internal.warc.gz"}
http://tex.stackexchange.com/questions/130641/how-to-put-the-qed-symbol-of-a-proof-at-the-right-place-inside-align	# How to put the QED symbol of a proof at the right place inside align? When a proof ends with a formula in equation or equation* environment, putting \qedhere after the equation would cause the QED symbol to appear in the right place. I.e. at the end of the line in which the equation appears. For example: $$x = y+z \qedhere$$ However, I cannot get the same result when the formula is inside an align/align* environment using the same method: \begin{align} x &= a+b \\ &= y+z \qedhere \end{align} The square appears just after (y+z) and is not shifted to the right boundary of the page. Here is a MWE: \documentclass[a4paper]{article} \usepackage{hyperref} \usepackage{amsthm} \usepackage[cmex10]{amsmath} \interdisplaylinepenalty=2500 \usepackage{amssymb} \title{Example to Show QED is Misplaced} \author{} \begin{document} \begin{proof} This proof is typeset correctly: \begin{equation} x = y + z \qedhere \end{equation} \end{proof} \begin{proof} But this one not! \begin{align} x & = u + v \\ & = y + z \qedhere \end{align} \end{proof} \end{document} - Why don't you use the proof environment of the acm packages? – Willem Van Onsem Aug 29 '13 at 9:50 ntheorem might also be worth a look – moewe Aug 29 '13 at 9:54 It appears at the right margin in my experiment (but it has the side effect of removing the equation number). Can you show a minimal working example (MWE)? – egreg Aug 29 '13 at 9:58 I added a MWE to the end of my question. I am sorry, I don't know how can I upload the MWE file directly. – Mani Bastani Parizi Aug 29 '13 at 11:30 @ManiBastaniParizi The cmex10 option to amsmath may be needed only with very old TeX installations; don't use it unless you get a Math formula deleted error. In this case, first try updating your TeX distribution. – egreg Aug 29 '13 at 16:32 ## 2 Answers Change the order, this works just fine, amsthm after amsmath, otherwise it might be a bit hard for it to hook into align* \documentclass[a4paper]{article} \usepackage{amsmath} \usepackage{amsthm} \interdisplaylinepenalty=2500 \usepackage{amssymb} \usepackage{hyperref} \title{Example to Show QED is Misplaced} \author{} \begin{document} \begin{proof} This proof is typeset correctly: \begin{equation} x = y + z \qedhere \end{equation} \end{proof} \begin{proof} But this one not! \begin{align} x & = u + v \\ & = y + z \qedhere \end{align} \end{proof} \end{document} - Thank you very much :-)! – Mani Bastani Parizi Aug 29 '13 at 12:10 with the package order amsthm before amsmath, a warning is issued: Package amsthm Warning: The \qedhere command may not work correctly here on input line .... it's also documented in the first section of amsthdoc that "amsthm must be loaded after amsmath, not before." – barbara beeton Aug 29 '13 at 14:30 Another caveat, be sure to have all the & in the last line. If the preceding lines contained for example there & don't forget to use all of them on the last line too. – Yrogirg Oct 20 '15 at 1:58 Section 5 of the amsthm package documentation contains the following. When used with the amsmath package, version 2 or later, \qedhere will position the QED symbol flush right; with earlier versions, the symbol will be spaced a quad away from the end of the text or display. If \qedhere produces an error message in an equation, try using \mbox{\qedhere} instead. However, when I tried this with your example I got a QED symbol one quad away from the end of the display, despite the fact that my distribution contains amsmath version 2.13. However, using \tag*{\qedhere} instead solved the problem. - @daleif's solution is preferable, though! – Ian Thompson Aug 29 '13 at 11:55 see the second paragraph of section "1 introduction" of amsthdoc. (i had to check, because if it didn't already say that the loading order is important, i would have to add it to the list of bugs. glad i don't have to.) – barbara beeton Aug 29 '13 at 14:32 @barbarabeeton --- indeed, I overlooked the issue of loading order. My hack first, consult manual later approach wasn't the best here! – Ian Thompson Aug 29 '13 at 15:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9615054130554199, "perplexity": 1827.0168753386295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464053379198.78/warc/CC-MAIN-20160524012939-00057-ip-10-185-217-139.ec2.internal.warc.gz"}
http://mathhelpforum.com/geometry/210916-vertex-triangle-equilateral-print.html	the vertex of the triangle to be equilateral • January 7th 2013, 06:41 AM rcs the vertex of the triangle to be equilateral A triangle has two vertices at ( -a, 0 ) and (a , 0 ). What must be the third vertex of the triangle to be equilateral? Thanks • January 7th 2013, 06:54 AM Plato Re: the vertex of the triangle to be equilateral Quote: Originally Posted by rcs A triangle has two vertices at ( -a, 0 ) and (a , 0 ). What must be the third vertex of the triangle to be equilateral? The third vertex will be $(0,b)$ where the distance $\mathcal{D}[(0.b);(a,0)]=2\|a\|$ • January 7th 2013, 07:07 AM skeeter 1 Attachment(s) Re: the vertex of the triangle to be equilateral Quote: Originally Posted by rcs A triangle has two vertices at ( -a, 0 ) and (a , 0 ). What must be the third vertex of the triangle to be equilateral? ... note that half an equilateral triangle is a 30-60-90 special right triangle.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8122163414955139, "perplexity": 696.8814794118399}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00172-ip-10-164-35-72.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/108121/computing-the-general-form-of-the-coefficients-of-a-power-series	# Computing the general form of the coefficients of a power series This question is related to at least two previous questions: Finding the power series of a rational function and Computing the nth coefficient of the power series representing a given rational function However, mine goes in a slightly different (and perhaps more general) direction. I want to obtain the general form of the coefficients of a power series representing some rational function. As it has been observed before, this can be done mechanically, and the Mathematica function SeriesCoefficient does the magic. E.g. I ask SeriesCoefficient[-2 ((1 + x)^2) (1 + x + x^2)/((1 - x)^4 (x^2 - 1)), {x, 0, n}] and I get the answer $(1+n)^2(4+2n+n^2)/2$. Now, does anybody know how Mathematica does it? Ultimately, what I want to know is whether you can trust that answer blindly, i.e. whether it is not necessary to PROVE that result in a paper, say. Thank you very much in advance. - "what I want to know is whether you can trust that answer blindly" - never trust the output of any computer program blindly. Any complex piece of software can and will have bugs. – Ｊ. Ｍ. Feb 11 '12 at 13:36 You can blindly distrust that answer. All the coeffieciets of you series must be even (because of the factor $-2$ and the constant term $-1$ of the denominator) but your answer does not have this property (it gives $7$ for $n=1$ for instance). Also note you can simplify your fraction by a factor $x+1$. – Marc van Leeuwen Feb 11 '12 at 14:34 @Marc: Yes, OP copied the output of Mathematica wrong, as the actual result returned has the factor $(1+n)^2$ as opposed to $1+n^2$... – Ｊ. Ｍ. Feb 11 '12 at 14:38 @J.M. You're right, thanks. I will fix the error. – Manolito Pérez Feb 11 '12 at 14:41 In general (and quite simply in your case) you can break your rational function into partial fractions, then get the series for the partial fractions and add them. I am no where near awake enough to do this by hand right now. I have no idea if this is what Mathematica does. – deinst Feb 11 '12 at 14:59 If I have done this correctly, $$\frac{-2 ((1 + x)^2) (1 + x + x^2)}{((1 - x)^4 (x^2 - 1))}=-\frac{2}{(x-1)^2}-\frac{10}{(x-1)^3}-\frac{18}{(x-1)^4}-\frac{12}{(x-1)^5}$$ From this and the fact that the power series of $\frac{1}{(x-1)^n}$ is $\sum_i\binom{n-1+i}{n-1}x^i$ you can get the result. I am not going to slog through the algebra, Mathematica is better at that sort of thing. - OK, thanks. I assume you can do that for any arbitrary rational function, right? – Manolito Pérez Feb 11 '12 at 15:32 Yes. It gets fiddly, but still straightforward when you need to deal with complex roots, the kind of thing best left to Mathematica (or something similar) – deinst Feb 11 '12 at 16:00 Then that means that in principle I could accept Mathematica's answer without further proof, right? – Manolito Pérez Feb 11 '12 at 16:59 @ManolitoPérez Probably. I'd say that it is more likely that Mathematica will be correct than anything you do by hand. – deinst Feb 11 '12 at 17:09 Whether you can get an explicit expression at all for the coefficients at all depends on your capability of factoring the denominator. For instance, for the Fibonacci numbers, their explicit (Binet's) formula involves the roots to the reversed polynomial $x^2-x-1$ of the denominator $1-x-x^2$ of the rational expression $\frac x{1-x-x^2}$ giving their generating function (the reversal induces the involution $\alpha\mapsto\alpha^{-1}$ on the roots, so the reversed denominator can be factored if and only if the denominator can). Here and in your example the denominator is easily factored, but there is no method that will factor arbitrary polynomials. You may try what SeriesCoefficient gives you for rational functions with denominators that are known not to be solvable by radicals (try $1-x^4+x^5$), or for those that are but whose roots are given by horrendous expressions (try a quartic polynomial with a not too special form). So while you can probably trust a computer algebra system when it does give a nice expression (and then it's not hard either to check the correctness by hand), you should not trust it to in general give you a nice expression at all. - Thanks, I will try that. – Manolito Pérez Feb 12 '12 at 11:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8398056030273438, "perplexity": 457.0775949780392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510276584.58/warc/CC-MAIN-20140728011756-00299-ip-10-146-231-18.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/178474/taylor-polynomial-of-fx-1-1-cos-x	Taylor polynomial of $f(x) = 1/(1+\cos x)$ I'm trying to solve a problem from a previous exam. Unfortunately there is no solution for this problem. So, the problem is: Calculate the Taylor polynommial (degree $4$) in $x_0 = 0$ of the function: $$f(x) = \frac{1}{1+\cos(x)}$$ What I tried so far: • calculate all $4$ derivatives • $1+\cos(x) = 2\cos^2(\frac{x}{2})$ and work with this formula • $\int\frac{1}{1+\cos(x)}dx = \tan(\frac{x}{2})$ and then use the Taylor series of $\tan(\frac{x}{2})$ • $\frac{1}{1 + \cos(x)} = \frac{1}{1 + \left(1 + \frac{x^2}{2!} + \cdots\right)}$ What do you think, is there a good way to calculate the Taylor polynomial of this function or is there just the hard way (derivatves)? - Let's do this in a couple of different ways. We want to expand $\dfrac{1}{1 + \cos x}$. Method 1 If it's not obvious what the best method is, we might just choose one and go. A fourth degree Taylor Expansion isn't so bad - it's only a few derivatives. So you might just go, suffer through a few derivatives, and save yourself the trouble of deciding the best way. Alternately, if you happen to know the series for $\tan x$, then that's a great way to proceed (referring to your idea of using the series expansion for $\tan (x/2)$ Method 2 If we are certain it has a Taylor expansion, and we are comfortable then we know it will look like $a_0 + a_1x + a_2x^2/2 + \ldots$ We know that $\cos x = 1 - x^2/2 + x^4/4! + \ldots$, so that $\dfrac{1}{1 + \cos x} = \dfrac{1}{2 - x^2/2 + x^4/4! + \dots}$ So we consider $\dfrac{1}{2 - x^2/2 + x^4/4! + \dots} = a_0 + a_1x + a_2x^2/2 + \ldots$, or equivalently $$(a_0 + a_1x + a_2x^2/2 + \ldots)(2 - x^2/2 + x^4/4! + \dots) = 1$$ By equating the coefficients of powers of $x$ on the left and on the right (which are all $0$ except for $x^0$, which has coefficient $1$), we get that $2a_0 = 1$, $a_1 = 0$, $a_0(-x^2/2) + (a_2x^2/2)(2) = 0$, etc. This isn't too bad, and is just a set of linear equations. - wow, +1! thanks for this solution! – lee.O Aug 3 '12 at 16:58 this way of finding a taylorpolynom is really good to know, thanks a lot! – lee.O Aug 3 '12 at 17:18 I'm glad you like it - – mixedmath Aug 3 '12 at 17:23 There are some things that can make it easier, since you are expanding around $x_0 = 0$. Let $f(x) = \frac{1}{1+\cos x} = \left(1+\cos x\right)^{-1}$. We can compute $df/dx$ as $\frac{df}{dx} = -\left(1+\cos x\right)^{-2}\frac{d (1+\cos x)}{dx} = -\left(1+\cos x\right)^{-2}\sin x$. Now, we have a product rule situation going to higher derivatives. This makes our life a little easier, since $\sin x_0 = \sin 0 = 0$. In other words, we can simply "not care" about higher-order derivatives that have a $\sin$ term in them. So, in short, the best way to compute the Taylor expansion quickly for a few terms on an exam, in my opinion, is to compute the derivatives, and note that since we need to compute $\frac{d^nf}{dx^n}\mid_{x = 0}$, to just skip the rigor of computing in detail any term that gets multiplied by $\sin x$. - In fact, since the function to be expanded is an even function, one only needs to worry about computing the coefficients of the even-order terms... – Ｊ. Ｍ. Aug 3 '12 at 17:41 @J.M. Indeed. I'm hoping the OP recognizes the pattern -- it certainly helps in doing such computations quickly within a time limit. – Arkamis Aug 3 '12 at 17:53 Thanks for your answer, I had to try it with your approach :). But @J.M. , what do you mean by "only needs to worry about the even-order-terms"? – lee.O Aug 3 '12 at 20:01 @lee.O He means that "odd" derivatives will always wash out. Consider $$f'(x) = -\left(1+\cos x\right)^{-2}\sin x,$$ but $$f''(x) = 2\left(1+\cos x\right)^{-3}\sin x - \left(1+\cos x\right)^{-2}\cos x.$$ Compute these for $x=0$, and the $\sin x$ term makes $f'(0) = 0$ in the first derivative. For the second derivative, however, the $\sin x$ term only wipes out part of the answer. – Arkamis Aug 3 '12 at 20:59 now i got it, thanks for your explanation! :) – lee.O Aug 3 '12 at 21:23 $$\cos x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}=1+\cos x= 2-\frac{x^2}{2}+\frac{x^4}{24}-...\Longrightarrow$$ $$\frac{1}{1+\cos x}=\frac{1}{2-\frac{x^2}{2}+\frac{x^4}{24}-...}=\frac{1}{2}\frac{1}{\left[1-\left(\frac{x}{2}\right)^2\right]\left(1-\frac{x^2}{24}+...\right)}=$$ $$=\frac{1}{2}\left(1+\frac{x^2}{4}+\frac{x^4}{8}+...\right)$$by taking the development of $$\frac{1}{1-x^2}$$ - When I start with: $\frac{1}{2}\dfrac{1}{\left[ 1 - (x/2)^2\right](1-x^2/24 + \dots)} = \frac{1}{2}(1 + x^2/4 + x^4/8 + \dots) \cdot \dfrac{1}{(1 - x^2/24 + \dots)}$ That is - what happened to the $(1 - x^2/24 + \dots)$ of the denominator? – mixedmath Aug 3 '12 at 16:56 There's still need to develop it, of course. I'd say that in a similar was as was done above, and multiply: above, it is "swallowed" in the ...part within the parentheses. It all depends, of course, what polynomial approximation is wanted. – DonAntonio Aug 3 '12 at 22:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9033166170120239, "perplexity": 409.91656240310886}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663060.18/warc/CC-MAIN-20140930004103-00345-ip-10-234-18-248.ec2.internal.warc.gz"}
https://www.groundai.com/project/reconstruction-of-the-core-convex-topology-and-its-applications-in-vector-optimization-and-convex-analysis/	Reconstruction of the core convex topology and its applications in vector optimization and convex analysis Reconstruction of the core convex topology and its applications in vector optimization and convex analysis Wayne State University, Detroit, MI 48202; Majid Soleimani-damaneh University of Tehran, Tehran, Iran; [email protected] April 2017 Abstract In this paper, the core convex topology on a real vector space , which is constructed just by operators, is investigated. This topology, denoted by , is the strongest topology which makes into a locally convex space. It is shown that some algebraic notions existing in the literature come from this topology. In fact, it is proved that algebraic interior and vectorial closure notions, considered in the literature as replacements of topological interior and topological closure, respectively, in vector spaces not necessarily equipped with a topology, are actually nothing else than the interior and closure with the respect to the core convex topology. We reconstruct the core convex topology using an appropriate topological basis which enables us to characterize its open sets. Furthermore, it is proved that is not metrizable when X is infinite-dimensional, and also it enjoys the Hine-Borel property. Using these properties, -compact sets are characterized and a characterization of finite-dimensionality is provided. Finally, it is shown that the properties of the core convex topology lead to directly extending various important results in convex analysis and vector optimization from topological vector spaces to real vector spaces. Keywords: Core convex topology, Functional Analysis, Vector optimization, Convex Analysis. 1 Introduction Convex Analysis and Vector Optimization under real vector spaces, without any topology, have been studied by various scholars in recent years [1, 2, 3, 8, 16, 18, 19, 20, 29, 30]. Studying these problems opens new connections between Optimization, Functional Analysis, and Convex analysis. Since (relative) interior and closure notions play important roles in many convex analysis and optimization problems [4, 18, 21], due to the absence of topology, we have to use some algebraic concepts. To this end, the concepts of algebraic (relative) interior and vectorial closure have been investigated in the literature, and many results have been provided invoking these algebraic concepts; see e.g. [1, 2, 3, 11, 16, 18, 20, 23, 24, 29, 30] and the references therein. The main aim of this paper is to unifying vector optimization in real vector spaces with vector optimization in topological vector spaces. In this paper, core convex topology (see [10, 19]) on an arbitrary real vector space, , is dealt with. Core convex topology, denoted by , is the strongest topology which makes a real vector space into a locally convex space (see [10, 19]). The topological dual of under coincides with its algebraic dual [10, 19]. It is quite well known that when a locally convex space is given by a family of seminorms, the locally convex topology is deduced in a standard way and vice versa. In this paper, is reconstructed by a topological basis. It is known that algebraic (relative) interior of a convex set is a topological notion which can be derived from core convex topology [10, 19]. We provide a formula for -interior of an arbitrary (nonconvex) set with respect to the algebraic interior of its convex components. Furthermore, we show that vectorial closure is also a topological notion coming from core convex topology (under mild assumptions). According to these facts, various important results, in convex analysis and vector optimization can be extended easily from topological vector spaces (TVSs) to real vector spaces. Some such results are addressed in this paper. After providing some basic results about open sets in , it is proved that, is not metrizable under topology if it is infinite-dimensional. Also, it is shown that enjoys the Hine-Borel property. A characterization of open sets in terms of there convex components is given. Moreover, -convergence as well as -compactness are characterized. The rest of the paper unfolds as follows. Section 2 contains some preliminaries and Section 3 is devoted to the core convex topology. Section 4 concludes the paper by addressing some results existing in vector optimization and convex analysis literature which can be extended from TVSs to real vector spaces, utilizing the results given in the present paper. 2 Preliminaries Throughout this paper, is a real vector space, is a subset of , and is a nontrivial nonempty ordering convex cone. is called pointed if . , , and denote the cone generated by , the convex hull of , and the affine hull of , respectively. For two sets and a vector , we use the following notations: A±B:={a±b: a∈A, b∈B},¯a±A:={¯a±a: a∈A},A∖B:={a∈A: a∉B}. is the set of all subsets of and for , ∪Γ:={x∈X: ∃A∈Γ; x∈A} The algebraic interior of , denoted by , and the relative algebraic interior of , denoted by , are defined as follows [16, 19]: cor(A):={x∈A: ∀x′∈X,∃λ′>0; ∀λ∈[0,λ′], x+λx′∈A},icr(A):={x∈A: ∀x′∈L(A),∃λ′>0; ∀λ∈[0,λ′], x+λx′∈A}, where is the linear hull of . When we say that is solid; and we say that is relatively solid if . The set is called algebraic open if . The set of all elements of which do not belong to and is called the algebraic boundary of . The set is called algebraically bounded, if for every and every there is a such that x+ty∉A ∀t∈[λ,∞). If is convex, then there is a simple characterization of as follows: if and only if for each there exists such that for all Lemma 2.1. Let be a vector basis for , and be nonempty and convex. if and only if for each there exists scalar such that Proof. Assume that for each there exists scalar such that Let . There exist a finite set and positive scalars such that d=∑j∈Jλjej−∑j∈Jμjej, a±δjej∈A, ∀j∈J. Let and . Considering , we have a+2mδλjej∈[a,a+δjej], a−2mδμjej∈[a−μjej,a], ∀j∈J, where stands for the line segment joining ,. Since is convex, a+2mδλjej∈A, a−2mδμjej∈A, ∀j∈J, and then, due to the convexity of again, a2+∑j∈Jδλjej∈A2, a2−∑j∈Jδμjej∈A2. This implies a+∑j∈Jδλjej−∑j∈Jδμjej∈A, which means Furthermore, the convexity of guarantees that for all Thus The converse is obvious. ∎ Some basic properties of the algebraic interior are summarized in the following lemmas. The proof of these lemmas can be found in the literature; see e.g. [1, 2, 10, 16, 19]. Lemma 2.2. Let be a nonempty set in real vector space . Then the following propositions hold true: 1. If is convex, then , 2. , 3. for each 4. for each 5. If , then is absorbing (i.e. ). Lemma 2.3. Let be a convex cone. Then the following propositions hold true: i. If , then is a convex cone, ii. , iii. If are convex and relatively solid, then iv. If is a convex (concave) function, then is continuous. Although the (relative) algebraic interior is usually defined in vector spaces without topology, in some cases it might be useful under TVSs too. It is because the algebraic (relative) interior can be nonempty while (relative) interior is empty. The algebraic (relative) interior preserves most of the properties of (relative) interior. Let be a real topological vector space (TVS) with topology . We denote this space by The interior of with respect to topology is denoted by . A vector is called a relative interior point of if there exists some open set such that The set of relative interior points of is denoted by Lemma 2.4. Let be a real topological vector space (TVS) and . Then . If furthermore is convex and , then . The algebraic dual of is denoted by , and exhibits the duality pairing, i.e., for and we have . The nonnegative dual and the positive dual of are, respectively, defined by K+:={l∈X′: ⟨l,a⟩≥0, ∀a∈K},K+s:={l∈X′: ⟨l,a⟩>0, ∀a∈K∖{0}}. If is a convex cone with nonempty algebraic interior, then The vectorial closure of , which is considered instead of closure in the absence of topology, is defined by [1] vcl(A):={b∈X:∃x∈X;∀λ′>0,∃λ∈[0,λ′];b+λx∈A}. is called vectorially closed if . 3 Main results This section is devoted to constructing core convex topology via a topological basis. Formerly, the core convex topology was constructed via a family of separating semi-norms on ; see [19]. In this section, we are going to construct core convex topology directly by characterizing its open sets. The first step in constructing a topology is defining its basis. The following definition and two next lemmas concern this matter. Definition 3.1. [22] Let be a subset of , where stands for the power set of . Then, is called a topological basis on if and moreover, the intersection of each two members of can be represented as union of some members of . The following lemma shows how a topology is constructed from a topological basis. Lemma 3.1. If is a topological basis on , then the collection of all possible unions of members of is a topology on . Lemma 3.2 provides the basis of the topology which we are looking for. The proof of this lemma is clear according to Lemma 2.2. Lemma 3.2. The collection B:={A⊆X : cor(A)=A, conv(A)=A} is a topological basis on . Now, we denote the topology generated by B:={A⊆X : cor(A)=A, conv(A)=A} by ; more precisely τc:={∪Γ∈P(X) : Γ⊆B}. The following theorem shows that is the strongest topology which makes into a locally convex TVS. This theorem has been proved in [19] using a family of semi-norms defined on . Here, we provide a different proof. Theorem 3.1. i. is a locally convex TVS; ii. is the strongest topology which makes into a locally convex space. Proof. By Lemmas 3.1 and 3.2, is a topology on . Proof of part i: To prove this part, we should show that is a Hausdorff space, and two operators addition and scalar multiplication are -continuous. Continuity of addition: Let and let be a -open set containing . We should find two -open sets and containing and , respectively, such that . Since is a basis for , there exists such that x+y∈A⊆V. Defining Vx:=12(A−x−y)+x \textmdand Vy:=12(A−x−y)+y, by Lemma 2.2, we conclude that and Convexity of , implies that and are the desired -open sets, and hence the addition operator is -continuous. Continuity of scalar multiplication: Let , , and be a -open set containing . without lose of generality, assume that . We must show that there exist and a -open set containing such that (α−ε,α+ε)Vx⊆V. Since , by considering in the definition of algebraic interior, there exists such that αx+λx∈V, λ∈(−δ,δ). Define U:=(V−αx)∩−(V−αx). We get , , and by Lemma 2.2, . Furthermore, is balanced (i.e. for each ), because is convex and . Now, we claim that (α−δ2,α+δ2)(12∣α∣+δU+x)⊆V. (1) To prove (1), let with . Therefore ∣α+t2∣α∣+δ∣=12∣α+t∣α∣+δ2∣≤12∣α∣+∣t∣∣α∣+δ2≤12 Thus, α+t2∣α∣+δU⊆12U. Hence, (α+t)(12∣α∣+δU+x)=α+t2∣α∣+δU+αx+tx⊆12U+αx+tx ⊆12(V−αx)+αx+tx=12V+12(αx+2tx)⊆12V+12V=V. This proves (1). Setting and proves the continuity of the scalar multiplication operator. Now, we show that is a Hausdorff space. To this end, suppose . Consider such that , and set and . It is not difficult to see that and while . This implies that is a Hausdorff space. ii. Let be an arbitrary topology on which makes into a locally convex space. Let be a locally convex basis of topology For each we have (by Lemma 2.4) and hence . Thus we have , which leads to and completes the proof.∎ The interior of with respect to topology is denoted by The following theorem shows that the algebraic interior (i.e. ) for convex sets is a topological interior coming from . Theorem 3.2. [19, Proposition 6.3.1] Let be a convex set. Then intc(A)=cor(A). Proof. Since is a TVS, ; see Lemma 2.4. Since is convex, is also convex, and furthermore (by Lemma 2.2). Hence, . Therefore, , because in the biggest subset of belonging to . Thus , and the proof is completed. ∎ Notice that the convexity assumption in Theorem 3.2 is essential; see Example 3.1. The proof of the following result is similar to that of Theorem 3.2. Theorem 3.3. If is a convex subset of , then , where denotes the relative interior of with respect to the topology . It is seen that the convexity assumption plays a vital role in Theorems 3.2 and 3.3. In the following two results, we are going to characterize the -interior of an arbitrary (nonconvex) nonempty set with respect to the of its convex components. Since and is a basis for the set could be written as union of some subsets of which are algebraic open. Lemma 3.3. Let be a nonempty subset of real vector space Then could be uniquely decomposed to the maximal convex subsets of , i.e. where are non-identical maximal convex subsets (not necessary disjoint) of (Here, sets are called convex components of ). Proof. For every , let be the set of all maximal convex subsets of containing (the nonemptiness of such is derived from Zorn lemma). Set the index set (this type of defining enables us to avoid repetition). For every define Hence, where are non-identical maximal convex subsets of To prove the uniqueness of s, suppose such that are non-identical maximal convex subsets of Let and ; then , and hence there exists such that This means , and the proof is completed.∎ Theorem 3.4. Let be a nonempty subset of real vector space . Then intc(A)=⋃i∈Icor(Ai), where are convex components of Proof. Let There exists such that Set Clearly and hence . Furthermore it is easy to verify that each chain (totaly ordered subset) in has an upper bound within . Therefore, using Zorn lemma, has a maximal element. Let be maximal element of . Obviously, is a convex component of . It leads to the existence of an such that . Therefore, To prove the other side, suppose for some Since is convex, by Theorem 3.2, Example 3.1. Consider A:={(x,y)∈R2:y≥x2}∪{(x,y)∈R2: y≤−x2}∪{(x,y)∈R2: y=0} as a subset of It can be seen that while However, where are convex components of as follows A1={(x,y)∈R2:y≥x2}, A2={(x,y)∈R2: y≤−x2}, A3={(x,y)∈R2: y=0}, A4={(x,y)∈R2: x=0}. The following theorem shows that the topological dual of is the algebraic dual of . In the proof of this theorem, we use the topology which induces on . This topology, denoted by , is as follows: τ0={∪Γ∈P(X) : Γ⊆Ψ}, where Ψ={A⊆X : A=f−11(I1)∩f−12(I2)∩...∩f−1n(In)\textmdforsomen∈N,\textmdsome \textmdopenintervalsI1,I2,...,In⊆R\textmdandsomef1,f2,...,fn∈X′}. [10] Proof. Let denote the topology which induces on . By [26, Theorem 3.10], makes into a locally convex TVS and . By Theorem 3.1, . Hence .∎ The following result provides a characterization of finite-dimensional spaces utilizing and the topology induced by on . Theorem 3.6. is finite-dimensional if and only if Proof. Assume that is finite-dimensional. Since there is only one topology on which makes this space a TVS, we have To prove the converse, by indirect proof assume that . Let be an ordered basis of ; and denote the vector of coordinates of with respect to the basis . It is easy to show that, for each , where l1(I)={{ti}i∈I⊆R : ∥{ti}∥1=∑i∈I∣ti∣<∞}. Define by Since is a norm on , thus is also a norm on We denote this norm by . Since is the strongest locally convex topology on , we have , where stands for the unit ball with On the other hand, every -open set containing origin, contains an infinite-dimensional subspace of (see [26]). Hence, This implies , and the proof is completed. ∎ Theorem 3.7 demonstrates that the vectorial closure () for relatively solid convex sets, in vector space , is a topological closure coming from . The closure of with respect to is denoted by Theorem 3.7. Let be a convex and relatively solid set. Then Proof. Since is a TVS, it is easy to verify that To prove the converse, let and . Without loss of generality, we assume , and then we have and , where stands for the algebraic interior of with respect to the subspace . We claim that To show this, assume that ; then there exists a functional such that and for each . Therefore the set U:={z∈X: f(z)>12} is a open neighborhood of with which contradicts Now, we restrict our attention to subspace . By Theorem 3.2, , and thus, by [18, Lemma 1.32], , which means . Therefore , and the proof is completed. ∎ Corollary 3.1. Let be a nonempty subset of with finite many convex components. Then clc(A)=n⋃i=1vcl(Ai) where are convex components of . Proof. According to Theorem 3.7, we have clc(A)=n⋃i=1clc(Ai)=n⋃i=1vcl(Ai). The equality may not be true in general; even if each convex component, , is relatively solid. For example, consider as the set of rational numbers in . The convex components of are singleton, which are relatively solid. Therefore , while ⋃q∈Qvcl({q})=⋃q∈Q{q}=Q≠R=clc(Q). A TVS is called metrizable if there exists a metric such that -open sets in coincide with -open sets in . Now, we are going to show that is not metrizable when is infinite-dimensional. Lemma 3.4 helps us to prove it. This lemma shows that every algebraic basis of the real vector space is far from the origin with respect to topology. Lemma 3.4. Let be an algebraic basis of . Then Proof. Define A:={∑i∈Fλixi−∑i∈Fμixi:∑i∈F(λi+μi)=1, λi,μi>0; ∀i∈F, \textmd and F\textmdisa\textmdfinitesubsetofI}. We claim that the following propositions hold true; a. , b. , c. . To prove (a), assume that and Then there exist positive scalars and finite sets such that; x=∑i∈Fλixi−∑i∈Fμixi, y=∑i∈S¯¯¯λixi−∑i∈S¯¯¯μixi, and . Hence, tx+(1−t)y=∑i∈F∪Sˆλixi−∑i∈F∪Sˆμixi, where ˆλi=⎧⎪ ⎪⎨⎪ ⎪⎩tλi+(1−t)¯¯¯λi i∈F∩Stλi i∈F∖S(1−t)¯¯¯λi i∈S∖F ˆμi=⎧⎪⎨⎪⎩tμi+(1−t)¯¯¯μi i∈F∩Stμi i∈F∖S(1−t)¯¯¯μi i∈S∖F Furthermore, ∑i∈F∪Sˆλi+ˆμi=∑i∈F∩S[t(λi+μi)+(1−t)(¯¯¯λi+¯¯¯μi)]+∑i∈F∖St(λi+μi)+∑i∈S∖F(1−t)(¯¯¯λi+¯¯¯μi)=∑i∈Ft(λi+μi)+∑i∈S(1−t)(¯¯¯λi+¯¯¯μi)=t+(1−t)=1. Therefore , and hence is a convex set. To prove (b), first notice that for each ; then due to the convexity of , from (a) we get , because of Lemma 2.1. We prove assertion (c) by indirect proof. If , then xj=∑i∈Fλixi−∑i∈Fμixi for some , and some finite set . Also, and values are positive and . Since is a linear independent set, we have , and also . Hence which is a contradiction. This completes the proof of assertion (c). Now, we prove the lemma invoking . Since is convex, by theorem 3.2 and claim (b), is a open neighborhood of . On the other hand, (c) implies Thus, Theorem 3.8. If is infinite-dimensional, then is not metrizable. Proof. Suppose that is an infinite-dimensional metrizable TVS with metric Then for every choose such that and is linear independent. This process generates a linear independent sequence such that, (i.e. is convergent to zero). Furthermore, this sequence can be extended to a basis of , say . This makes a contradiction, according to Lemma 3.4, because In every topological vector space, compact sets are closed and bounded, while the converse is not necessarily true. A topological vector space in which every closed and bounded set is compact, is called a Hiene-Borel space. Although it is almost rare that an infinite-dimensional locally convex space be a Hine-Borel space (for example, normed spaces), the following result proves that is a Hine-Borel space. Theorem 3.9. enjoys the Hine-Borel property. Moreover, every compact set in lies in some finite-dimensional subspace of Proof. Let be a -closed and -bounded subset of . First we claim that the linear space , i.e. the smallest linear subspace containing , is finite-dimensional. To see this, by indirect proof, assume that is an infinite-dimensional subspace of ; then there exists a linear independent sequence in . Thus, the sequence is a linear independent set in and also is -bounded. Hence, . Furthermore, has a basis , containing This makes a contradiction, according to Lemma 3.4, because Hence, is a closed and bounded set contained in a finite-dimensional space . Therefore, by traditional Hine-Borel theorem in finite-dimensional spaces, is a compact set in , and hence is a -compact set in One of the important methods to realize the behavior of a topology defined on a nonempty set is to perceive (characterize) the convergent sequences. Assume that a sequence is -convergent to some (i.e, ). Thus, is a -compact set. Therefore, by Theorem 3.9, lies in a finite-dimensional subspace of . This shows that, convergent sequences of are exactly norm-convergent sequences contained in finite-dimensional subspaces of . Hence, every sequence with infinite-dimensional could not be convergent. So, the convergence in is almost strict; however, this is not surprising because strongest-topology () contains more number of open sets than any other topology which makes into a locally convex space. 4 Applications In this section, we address some important results in convex analysis and optimization under topological vector spaces which can be directly extended to real vector spaces, utilizing the main results presented in the previous section. Some of these results can be found in the literature with different complicated proofs. Subsection 4.1 is devoted to some Hahn-Banach type separation theorems. 4.1 Separation Theorem 4.1. Assume that are two disjoint convex subsets of , and Then there exist some and some scalar such that f(a)≤α≤f(b), ∀a∈A,b∈B. (2) Furthermore, f(a)<α, ∀a∈cor(A). (3) Proof. Since is a TVS and , by a standard separation theorem on topological vector spaces (see [26]), there exists some and some scalar satisfying (2) and (3). This completes the proof according to Theorem 3.5.∎ Theorem 4.2. Two disjoint convex sets	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9889529943466187, "perplexity": 384.8593262552897}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00467.warc.gz"}
https://eccc.weizmann.ac.il/keyword/17923/	Under the auspices of the Computational Complexity Foundation (CCF) REPORTS > KEYWORD > UNCERTAINTY PRINCIPLE: Reports tagged with Uncertainty principle: TR12-031 \| 4th April 2012 Tom Gur, Omer Tamuz #### Testing Booleanity and the Uncertainty Principle Revisions: 1 Let $f:\{-1,1\}^n \to \mathbb{R}$ be a real function on the hypercube, given by its discrete Fourier expansion, or, equivalently, represented as a multilinear polynomial. We say that it is Boolean if its image is in $\{-1,1\}$. We show that every function on the hypercube with a ... more >>> TR18-016 \| 25th January 2018 Naomi Kirshner, Alex Samorodnitsky #### On $\ell_4$ : $\ell_2$ ratio of functions with restricted Fourier support Revisions: 1 Given a subset $A\subseteq \{0,1\}^n$, let $\mu(A)$ be the maximal ratio between $\ell_4$ and $\ell_2$ norms of a function whose Fourier support is a subset of $A$. We make some simple observations about the connections between $\mu(A)$ and the additive properties of $A$ on one hand, and between $\mu(A)$ and ... more >>> ISSN 1433-8092 \| Imprint	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9302639961242676, "perplexity": 1022.7526358996577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655899931.31/warc/CC-MAIN-20200709100539-20200709130539-00484.warc.gz"}
http://math.stackexchange.com/questions/52481/how-do-we-prove-that-a-sphere-maximizes-the-volume-enclosed-among-all-simple-clo	How do we prove that a sphere maximizes the volume enclosed among all simple closed surfaces of given surface area? How do we prove that among all closed surfaces with a given surface area, the sphere is the one that encloses the largest volume, and not do it by cases? so far I've tried is that I know the formula for the surface of the sphere and volume of sphere - You certainly cannot prove this result simply by examining a bunch of closed surfaces and comparing volume enclosed. There are infinitely many different closed surfaces, so you cannot test them all. The problem is not entirely trivial (compare with the 2-dimensional case, with closed simple curves, area, and length). – Arturo Magidin Jul 19 '11 at 20:02 The is a really good elementary discussion of this problem at cut-the-knot.org/do_you_know/isoperimetric.shtml – John M Jul 20 '11 at 19:03 Perhaps use the calculus of variations and the Euler-Lagrange formula. – asmeurer Nov 14 '12 at 6:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9269478917121887, "perplexity": 370.9013345523405}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438043062723.96/warc/CC-MAIN-20150728002422-00094-ip-10-236-191-2.ec2.internal.warc.gz"}
http://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/StronglyConnectedBlocks	LinearAlgebra - Maple Programming Help Home : Support : Online Help : Mathematics : Linear Algebra : LinearAlgebra Package : Solvers : LinearAlgebra/StronglyConnectedBlocks LinearAlgebra StronglyConnectedBlocks compute the strongly connected blocks of a square Matrix Calling Sequence StronglyConnectedBlocks(M, opts) Parameters M - n x n square matrix with some sparsity opts - options controlling the output Description • The StronglyConnectedBlocks(M) function computes and returns a list containing the nonzero strongly connected blocks contained in the input Matrix M, i.e. [M_1, ..., M_r]. The strongly connected blocks are square sub-matrices of the input Matrix after a sequence of row and column exchanges have been performed to minimize the size of the blocks. Note that these sub-matrices do not contain all the entries present in the input Matrix, but rather only those needed to compute the determinant, characteristic polynomial, or eigenvalues of the Matrix. In addition, any zero blocks are not output. • In order for StronglyConnectedBlocks to provide a benefit, the input Matrix should have some sparsity. If the input Matrix is fully dense, the command will output a list containing only the input Matrix (without sparsity no break-down into blocks is possible). • For an input n x n Matrix M, if we let m = sum( Row/ColumnDimension(M_i), i=1..r ) for the output Matrix list [M_1, ..., M_r], then m <= n, where the inequality is needed for any zero blocks contained in the matrix. • The output blocks satisfy the following: Determinant(M) is zero if m < n and product(Determinant(M_i), i=1..r) otherwise. CharacteristicPolynomial(M,x) = x^(n-m) * product( CharacteristicPolynomial(M_i,x), i=1..r) • The option returnsingular is true by default, but when set to false, the block matrices will not be formed when the input Matrix is singular (i.e. has zero blocks). This is useful for efficient computation of a determinant (as a zero block means the determinant is zero, so there is no sense in forming the block matrices). • The option outputform is matrixlist by default, which returns a list of the strongly connected blocks in Matrix form as described above. If outputform is set to rowlist instead, it provides a list of the rows (columns) of the Matrix for each block. Examples > $A≔\mathrm{Matrix}\left(\left[\left[a,b,c,d,e\right],\left[f,g,h,i,j\right],\left[0,0,0,0,k\right],\left[0,0,0,v,w\right],\left[0,0,0,x,y\right]\right]\right):$ > $\mathrm{LinearAlgebra}:-\mathrm{StronglyConnectedBlocks}\left(A\right)$ $\left[\left[\begin{array}{cc}{y}& {x}\\ {w}& {v}\end{array}\right]{,}\left[\begin{array}{cc}{a}& {b}\\ {f}& {g}\end{array}\right]\right]$ (1) > $\mathrm{LinearAlgebra}:-\mathrm{StronglyConnectedBlocks}\left(A,\mathrm{returnsingular}=\mathrm{false}\right)$ ${0}$ (2) > $\mathrm{LinearAlgebra}:-\mathrm{StronglyConnectedBlocks}\left(A,\mathrm{outputform}=\mathrm{rowlist}\right)$ $\left[\left[{5}{,}{4}\right]{,}\left[{3}\right]{,}\left[{1}{,}{2}\right]\right]$ (3)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9238011240959167, "perplexity": 1114.2343934716987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257836399.81/warc/CC-MAIN-20160723071036-00322-ip-10-185-27-174.ec2.internal.warc.gz"}
https://dsp.stackexchange.com/questions/64342/the-complexity-of-such-function-run-in-matlab	# The complexity of such function run in Matlab The below function is representing an algorithm, so how can I get its complexity? I don't mean the time of running by using the tic .. toc, I mean how many operation (Additions and multiplications) are performed in this loop. for times=1:m; for col=1:N; product(col)=abs(T(:,col)'r_n); end [val,pos]=max(product); Aug_t=[Aug_t,T(:,pos)]; T(:,pos)=zeros(M,1); aug_y=(Aug_t'Aug_t)^(-1)Aug_t'Yy; r_n=Yy-Aug_taug_y; pos_array(times)=pos; end Size of parameters, m = 256 , N = 256, T= [256,256] and M = [256,1] You should know what each operator (i.e., ) and each called function (i.e. product) does. Then add up those operations. For instance, I'm pretty sure that in Matlab, the way you're building up Aug_t means that Aug_t' * Aug_t generates a vector dot-product, so for each element in Aug_t there's a multiply-accumulate.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8592966794967651, "perplexity": 1825.5759002975522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00149.warc.gz"}
https://www.physicsforums.com/threads/bose-equilibrium-distribution-and-atomic-units.855304/	# Bose Equilibrium Distribution and Atomic Units Tags: 1. Feb 2, 2016 ### Raptor112 1. The problem statement, all variables and given/known data For my project I need to compute the average the number of photons given by the expression: $\bar{n}= \frac{e^{-\bar{h}\omega/\kappa T}}{1-e^{-\bar{h} \omega / \kappa T}}$ where $\kappa$ is the Boltzmann constant and $\omega$ is the oscillator frequency. For the Hamiltonian in my project simulation, $\bar{h} =1$ so how would $\bar{n}$ be expressed? 2. Relevant equations Is it as simple as $\bar{h} =1$ in the expression of $\bar{n}$ so: $\bar{n}= \frac{e^{-\omega/\kappa T}}{1-e^{\omega / \kappa T}}$ but then doesn't the argument of the exponential has dimensions, as opposed to being dimensionless which is what it's supposed to be? 2. Feb 2, 2016 ### Staff: Mentor You have to express all quantities in atomic units. For instance, ω will be in units of the inverse of the atomic unit of time. There is no atomic unit of temperature, so T will still be in kelvin, but you have to calculate the correct value for the Boltzmann constant. Last edited: Feb 2, 2016 3. Feb 2, 2016 ### Raptor112 According to wikipedia it's just one by definition: https://en.wikipedia.org/wiki/Boltzmann_constant 4. Feb 2, 2016 ### Staff: Mentor Draft saved Draft deleted Similar Discussions: Bose Equilibrium Distribution and Atomic Units	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9173685312271118, "perplexity": 932.8675592481362}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948599156.77/warc/CC-MAIN-20171217230057-20171218012057-00631.warc.gz"}
https://www.physicsforums.com/threads/can-we-measure-time-and-accelaration-at-the-same-time.185383/	# Can we measure time and accelaration at the same time? 1. Sep 18, 2007 ### goksen 1. The problem statement, all variables and given/known data can we measure time and accelaration at the same time? what exactly acceleration operator is? 2. Relevant equations heisenberg's uncertainity principle 3. The attempt at a solution I guess acc. operator is dp/dt i.e. $$\frac{\partial^2 }{\partial t \partial x}$$(ignoring constants) and $$\delta a \delta t$$ is <p> which say we may not measure 2. Sep 18, 2007 ### dextercioby To the first question one cannot answer in the context of quantum mechanics. The accelerator operator is $$\hat{a}(t)=\frac{1}{m}\frac{d\hat{p}(t)}{dt}$$ in the Heisenberg picture.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9874891042709351, "perplexity": 3308.0877902378024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720153.61/warc/CC-MAIN-20161020183840-00270-ip-10-171-6-4.ec2.internal.warc.gz"}