How to add product details to Drupal Commerce order confirmation email

Question: Default email that is sent when user completes the checkout process contains order number and other details. I want to include additional product details and product type in this email. How can I do that?
Answer: you can do that by using php code inside text message body.
$wrapper = entity_metadata_wrapper('commerce_order', $commerce_order);
    $line_items =  $wrapper->commerce_line_items->value();
    foreach($line_items as $line_item){
        $product = commerce_product_load($line_item->commerce_product['und'][0]['product_id']);
            echo "Product Title :".$product->title;
            echo "Product Title :".$product->type;
     }

Answer: You want the Commerce Message module and it's dependencies. It includes customer emails with a full order details, including customer profiles and all the products purchased.
Comment: Hi joshmiller, I have installed the Commerce Message module, but I cannot find any evidence that it "includes customer emails with a full order details, including customer profiles and all the products purchased". Can you please advise on where and how I can set that up? Seems like Drupal Commerce would include a standard "hey, here's what you just ordered" email like every single other ecommerce website has. I am shocked that it doesn't.
Comment: Drupal 7 Commerce 1 doesn't include order details in their notification email. If you use Commerce Message, you first have to review rules / maybe enable rules that will trigger the email during the event of your choice. Additionally, you can edit the emails by editing the message templates in the UI by going to the Message's listing page.


coherent sheaves on affine formal schemes

Question: Let $\hat{X} = \text{Spf} \hat{A}$ be obtained as the formal completion of an affine scheme $X = \text{Spec} A$ where $A$ is an adic noetherian ring. Given a coherent sheaf $\mathfrak{F}$ on $\hat{X}$, is it always possible to find a coherent sheaf $\mathcal{F}$ on $X$ such that $\hat{\mathcal{F}} = \mathfrak{F}$?
Comment: Let me just make the orthogonal comment that this is true if $X$ is projective, by formal GAGA. Algebraicity in the affine case is pretty rare as you can see from ts3uji's example below.
Answer: According to EGA I (Springer edition), Theoreme (10.10.2), there is an equivalence of categories between the category of finitely generated $A$-modules and the category of coherent $O_{{\rm Spf}(A)}$-modules. In particular, the answer to your question is Yes.
Comment: But does this answer the original question? You show there is a f.g. $\hat A$-module, but is it induced by a f.g. $A$-module.
Comment: I think this is a problem of definitions. Usually an adic ring $A$ is already defined to be complete and thus $A = \hat A$. Then coherent modules over $O_{{\rm Spec}(A)}$ and $O_{{\rm Spf}(A)}$ are defined in a different way but EGAI tells you that there is not really a difference. If you define it to be only a topological ring endowed with some $I$-adic topology for some ideal $I$ - and thus $A \ne \hat A$ in general, then the answer is No as you explained.
Comment: @Wedhorn: I completely agree that there is a potential confusion. My interpretation was based on the fact that the OP uses two different letters $A$ and $\hat A$ for the rings.
Answer: Take $A=k[x,y]$, $\hat A=k[[x,y]]$, and suppose coherent sheaf corresponds to the $\hat A$-module $N=\hat A/(f)$, where $f\in k[[x,y]]$ is not "algebraic": say, $f=y-exp(x)$, assuming $k$ has characteristic zero. It is clear that $N$ does not come from completion of any f.generated $A$-module $M$. (Proof: Suppose $N=\hat M$. Since $N$ has no $A$-torsion, any torsion of
$M$ maps to zero, so we can replace $M$ with $M/(torsion)$ and assume $M$ is torsion-free. Then $M$ embeds in a locally free module (its second dual), but that would make $N$ embed in the free $\hat A$-module, which is false.)


Why is the Cryogenics Lab in New New York?

Question: Fry was frozen in Old New York's Cryogenics Lab and went 1000 years into the future. When he woke he was in New New York. Old New York is under New New York but the Cryogenics lab stays in the same spot?
Answer: Consider that the purpose of the Cryogenics Lab is that people be thawed when the right time has arrived.  It does not seem unreasonable that people would continue to want to be frozen through that millennium, nor that some would be thawed in less than one thousand years' time.  It is therefore quite possible that, as New New York was built over Old New York, the lab relocated to remain on the surface, where its customers were.  This does rely on the equipment being movable, but the worst case I can think of is that one would need to hook up a generator (which a cryogenics lab should have anyway in case of grid-power outages), then move the entire building, with the equipment inside.  Considering how much people are willing to pay for cryogenic freezing (it's an awful lot!), that shouldn't be too problematic, and would probably be good business sense.
Comment: In the first episode, you can see the city outside the Cryogenics Lab while Fry is frozen. Old New York is destroyed and New New York is built in the same spot. Nothing ever happened to the Lab.
Comment: Keep in mind the window view we get of Old New York being destroyed while Fry is frozen, does not show everything that happened in the 1000 years, and should perhaps be thought of as a discrete montage and not a continuous film. This opens up the possibility that in between one of the montage segments the cryo lab was moved to the new construction above ground while the rest of Old New York stayed below. Here's a question though: did the aliens destroy old or new New York?
Comment: @Cover202: In that case, a rather larger question is raised, namely that of how Old New York can be below New New York, when it was destroyed.  Chalk it up to a continuity error.  My point, though, was that there were reasonable explanations for the lab being in New New York, based on their desire for continued profit.
Answer: The building looked tall... it's possible that the ground of New New York was just at a different floor level than that of Old New York.  Maybe, for example, they changed the base of the building to be the 10th floor, and then sealed off the lower 9 that were now below ground level (and part of Old New York).


Is it possible to 'give prosperity'?

Question: Here's the sentence:

Who knows how the prosperity you give becomes the prosperity you get

Is it possible to 'give prosperity'?
Comment: Taking after *'The God of the heavens -- He doth **give prosperity** to us, ...* (Nehemiah 2:20 ?)
Answer: Prosperity is a noun; it is the state of being prosperous.
The synonyms are also nouns: success, profitability, affluence, wealth, opulence, luxury, fortune, ease, plenty, comfort, security, well-being
If one can give ease, fortune, wealth, security or comfort, one should also give prosperity.
In the sense of giving wealth, I have heard it more often used as a verb: he has plans to prosper you... (e.g. Jeremiah 29:11)
Answer: In giving money, he indirectly gave prosperity to his beneficiaries.'
I agree it is seldom expressed in that way. It sounds a bit awkward, but the meaning is clear and unambiguous.
'The discovery of oil has given prosperity to Ambrosia'. I realise it is more usual to say 'has brought prosperity', but I see no fundamental reason why one couldn't use 'give'.
Comment: The sinister down-voter casts his dastardly shadow yet again!
Comment: Seconded! Always great to have no feedback whatsoever on a downvote, isn't it! (NOT!)
Answer: I find the idea of giving someone 'a state' to be challenging in the case of prosperity to the point of informal, if not completely incorrect.
Many words which indicate a state or condition in someone else seem to be things that you cause in someone else, through them, as it were, not necessarily things you give to them. There are exceptions of course.
Consider a few example states other than prosperity:

nervousness
happiness
unconsciousness

"I give you nervousness."

Sounds wrong, while "I cause your nervousness" works.

"I give you happiness."

Sounds better, perhaps because it's more readily conceptualised independently of those experiencing the state, and hence seems less dependent on the subject, even though here it is simply the state of the receiver, and dependent on them.

"I give you unconsciousness."

Sounds wrong again, but arguably isn't completely wrong. "I caused your unconsciousness" we want to say (and past tense because... well, you can't chat with someone who's unconscious).
So prosperity, being a state that one has, I think "feels" more correct in the form of 'bringing about' or 'causing'.

the prosperity you give

somehow sounds more awkward than "the prosperity you share" or even "the prosperity that you bring". But if you ask me, the whole sentence sounds awkward and archaic. Although the latter, however, may be the intention.

Is it possible to 'give prosperity'?


Is it possible for a substance to absorb visible light but still appear white?

Question: This is a UV-vis spectrum for zinc oxide
Comment: No substance is perfectly transparent or reflective.
Comment: Most white solids seem to do that too
Answer: Your eyes are not sensitive to UV light so any absorption there will be invisible to you. The visible spectrum you give has fairly uniform absorbance across the visible. The whiteness you see is primarily due to scattering of white room lights or sunlight off the powdered zinc oxide. Light is scattered when there is a change in refractive index between two mediums, in this case air and zinc oxide. The effect is enhanced when the crystals are of a very irregular shape and so have many different scattering surfaces. Sugar and salt show the same effect when powdered  even though large crystals of each of them can be made that are clear and colourless. As the oxide has a more or less constant absorption no wavelength is very much more absorbed than any other, or alternatively, all wavelengths are equally scattered.
If you can find a solvent with similar refractive index to the oxide (and in which it does not dissolve) then the crystals will become far more transparent and effectively disappear. It is easier see this effect if you wet a piece of frosted glass with water as it will then become far more transparent. Alcohol or benzene would be better as they have refractive index closer to that of many types of glass. Zinc oxide has a value of 2 higher than any common solvent, such as diiodomethane approx 1.7.
Answer: Your impression of color is a result of physical (absorption spectrum of the substance, power spectral density of the illuminating light), biological (cone responses) and psychological (surround, adaptation, fatigue...) factors. From a purely physical point of view the posted spectra are not of a white substance (which would absorb light equally at all visible wavelengths) but of a bluish one as it absorbs less at the shorter visible wavelengths than at the longer ones. And yet you see it as white. This is because your brain averages everything in the scene and sets its white point based on that average much as the automatic white point setting in a video camera does. Thus a piece of white paper looks white when viewed under tungsten light (Coordinated Color Temperature 2400 K) as it does in shadow on a sunny day (CCT 8000 K).
There are three color receptors in the eye and their responses are approximately bell shaped curves located in the short, medium and long wavelength parts of the visible spectrum (400 - 700 nm). If those three are approximately equally stimulated by an object (or, more precisely, if an object stimulates them approximately in the same proportion as the scene) that object will appear white. Note that the spectrum of a white object does not at all need to be uniform or even approximately uniform in order to produce an impression of white. When we look at a TV, for example, what we see as a white screen is actually stimulating our eyes with three rather narrow band clumps of light energy corresponding to the primaries of the screen. It can all seem rather complex. If absorption of visible light occurs in a part of the spectrum (near 500 nm) where all three color matching functions are small and cross over one another one would not notice much change in the color of the material as each of the tristimulus values would change little. But if the absorption were at, respectively,  425 nm (peak of the short wavelength cmf) or 600 nm (peak of the long wavelength cmf) the, respectively, short or long tristimulus would change appreciably and the effect be noticeable.
The key word in your question is "appear" and the answer to it is "yes". We can go straight from your posted spectra to color coordinates in any of several systems and I'm confident that by playing around with illuminants and choosing absorbtion peaks properly we could find spectra distorted relative to yours that were imperceptibly close to the same color as the one yielded by the original spectra. These are called 'metamers'. Metamerism (i.e. producing something of one color that looks like a desired color even though the spectra are very different) is obviously the basis for color TV, color printing etc. In fact we could photograph your zinc oxide and look at it on a TV screen. It would still look white though the light coming from the screen would be missing several chunks of wavelengths.


What was the charge that was given by the Jewish court to Jesus?

Question: I need this for a bible trivia! So I can beat the game and it's rlly important to me, so please tell me.
Comment: Welcome to Christianity.SE! I hope you'll spend some time browsing the questions and answers here. Before asking further questions, I would suggest taking the [Site Tour](http://christianity.stackexchange.com/tour), and reading the article in the help section on [How do I ask a good question?](http://christianity.stackexchange.com/help/how-to-ask)
Comment: The problem with this question is its trivial nature. Read some sources, maybe *the* source, and you will surely find your answer. That said, it's not actually off-topic, but it does deserve its downvotes.
Comment: What have you done to look up the answer? Have you checked on Google? Wikipedia? In the Bible? If there's no apparent research effort, people will vote this down.
Comment: Jesus openly admitted to breaking the Sabbath and dietary laws, he took a more liberal view of the laws during a time it was taboo. But ultimately the Romans kill anyone who challenges their authority. Him claiming divinity posed a big threat to them since it could be the makings of a rebellion. The Romans make examples of people like this, hence the cross.
Answer: The events surrounding Jesus' death are rife with trickery and very clearly outside of normal legal proceedings for the time. The short answer is that the Jews believed Jesus committed blasphemy, but they had it out for him long before He supposedly did blaspheme.
Before Jesus was captured and executed, the High Priest and his assembly plotted to kill Jesus, seemingly because they did not like the kinds of things he said, but not because he actually broke any law:

Then the chief priests, the scribes, and the elders of the people assembled at the palace of the high priest, who was called Caiaphas, and plotted to take Jesus by trickery and kill Him.
Matthew 26:3-4 (NKJV)

You can see that they wanted him dead even if they had to use "trickery". They did arrest him, but seemingly without cause in the Garden of Gethsemane. He was then taken to the Sanhedrin where he was questioned. This was the trickery. Many false witnesses came out and accused him of many things, but Jesus remained silent. It was not until Jesus spoke that the High Priest declared what Jesus said as blasphemy then demanded that no further evidence was needed to warrant Jesus' execution.

And the high priest answered and said to Him, "I put You under oath by the living God: Tell us if You are the Christ, the Son of God!"
Jesus said to him, "It is as you said. Nevertheless, I say to you, hereafter you will see the Son of Man sitting at the right hand of the Power, and coming on the clouds of heaven."
Then the high priest tore his clothes, saying, "He has spoken blasphemy! What further need do we have of witnesses? Look, now you have heard His blasphemy! What do you think?"
They answered and said, "He is deserving of death."
Matthew 26:63-66 (NKJV)

They charged Jesus with blasphemy, a very serious crime to the Jews, but meaningless to the Romans, who had the sole authority over penal actions. Nevertheless, the angry mob handed Jesus over to Pontius Pilate, the governor of the area. Pilate questioned Jesus himself to ascertain the truth and what should be done. Pilate eventually determined that Jesus was blameless according to Roman law, so he attempted to have him released. There was apparently a custom where Pilate would release a single prisoner during the Passover feast based on the wishes of the multitude. Pilate presented them with two options.

And at that time they had a notorious prisoner called Barabbas. Therefore, when they had gathered together, Pilate said to them, "Whom do you want me to release to you? Barabbas, or Jesus who is called Christ?" For he knew that they had handed Him over because of envy ...
The governor answered and said to them, "Which of the two do you want me to release to you?"
They said, "Barabbas!"
Pilate said to them, "What then shall I do with Jesus who is called Christ?"
They all said to him, "Let Him be crucified!"
Then the governor said, "Why, what evil has He done?"
But they cried out all the more, saying, "Let Him be crucified!"
Matthew 27:16-18, 21-23 (NKJV)

If Pilate kept records, he probably just picked any number of the false accusations against Jesus, but in the minds of the Jewish mob, Jesus was a blasphemer and deserved death.
The answer to your trivia question is actually complicated and highly contextual. I would bet that the answer according to the game is "blasphemy".


Why is Power = Voltage $\times$ Current?

Question: So P = I*V because V is the amount of energy per coulomb and I is the amount of electrons going past a cross sectional area in one second.
So if we do the math, since V = J/C and I = C/sec, if we multiply those values we get J/sec which is the amount of energy turned to other energy from electric potential energy and the equation seemingly works out because the Coulombs disappear from equation and I can choose not to overthink about it . (But i want to really understand the whole thing)
But the thing confusing part is that it doesn't make sense intuitively because no matter how hard I try to come up with something meaningful,just multiplying the rate of flow which is the amount of coulombs passing through a cross sectional area by the potential difference seems very difficult to connect. (because for example lets say it takes an electron to move through a potential difference of 1V in 2 seconds, and if the current is 1A, the electron behind it will enter the resistor(light bulb) when it is halfway through the resistor)It only makes sense if let's say, 3 coulomb of charge moves through a potential difference of 1V in 3 seconds, which means it converted 3 joule of electrical energy in 3 seconds, therefore it has a power of 1J/sec = 1 watt. Does charge instantly get converted to other energy and the 2 points that has a potential difference is just a cross section?
Comment: "...because V is the amount of energy per coulomb". The potential difference, $V$, between two points is the work required per unit charge (work per coulomb of charge) to move the charge between the two points. Then the rate of work (power) is the voltage times the amount of coulombs per second (current).
Comment: As indicated [here](https://physics.stackexchange.com/help/how-to-ask), we encourage users to search and research before asking questions, and this feels like the kind of query that can be found [elsewhere](https://en.wikipedia.org/wiki/Electric_power).
Comment: @secavara I actually did a lot of research and was reading that wikipedia page exactly hours ago but if you actually see the contents of the page it doesn't really answer my question.
Comment: I isn't the amount of electrons going past a point but the amount if charge. Clearly since the basic unit of charge is that on the electron there is a relationship between amount of charge and number of electrons.
Answer:
multiplying the rate of flow which is the amount of coulombs passing through a cross sectional area by the potential difference seems very difficult to connect

Sorry, but there isn't much more to it than that.
It might help to consider that if we're calculating power from current and potential difference we're typically not just talking about a current through a cross-sectional area, but along a path with non-zero length. So there is a separation between the two points (or equipotential regions) that charge is moving between, and we assume that all the current that leaves the first point reaches the second point, so that we can talk about the current from one place to the other, rather than just across a surface.
But with that said, the potential difference tells you the energy transferred by each element of charge, so if you know the rate of charge being transferred you know the rate of energy being transferred. And "power" is the name we use for a rate of energy being transferred.
You describe it perfectly, but then say you don't understand it, so I think you really are just over-thinking it.
Comment: Yes but i thought like, potential difference is something constant(for example some has 12V, some has 1.5V, and if the current is going through a light bulb,  the amount of work done per coulomb(in this case Electrical PE to light and thermal energy), it must move some distance since deltaPE= qEd.If 1coulomb of charge goes through a 1V of potential difference(which is some distance) it dissipated 1 joule of energy, but what if the charge takes more than 1 second to go that distance, that must mean another 1 coulomb of charge will start to do work before the first wave of electrons do work?
Comment: @ikigai20, because that's the definition of "1 ampere". If it was more or less than 1 C/s, then the current would be more or less than 1 A.
Comment: Also we know that charge isn't stored in an (ideal) resistor. But charge is a conserved quantity. So the exact same current that goes in one terminal has to come out the other terminal. This is a form of Kirchhoff's current law (or, KCL is an application of the principle of conservation of charge).
Comment: If the 1 coulomb takes 2 seconds to go through the same potential difference then the power is just one half the previous value. This also means that the current is just one half the pevious value. And no matter how long it takes for a specific pack of charge to go through the 1 volt, other packs will enter and exit during the same time. At any given time the amount of charge moving between two points in a DC circuit is the same.
Comment: @ikigai20, Typically the charge flows continuously through the device, l like water through a hose. If I set the flow rate of my garden hose to deliver 1 liter per minute of water, it doesn't pop out 1 liter of water, then wait a minute with nothing coming out before another liter appears, it just flows in a slow trickle continuously, and if I consider any 1 cm segment of the hose, there will be a small amount of water in it at any time I choose to look (if I were to use a higher flow rate, like 1 liter per second, I'd find the hose entirely full of water at any moment I look).
Comment: @ThePhoton, oh that makes sense but the thing that is baffling me is that why we choose the entire voltage drop.(for example if 1A of current goes through 1 volt of difference, it has 1 watt power) 1 ampere of current could have gone through 0.5 volt in one second and still has electrical potential energy.
Comment: @ThePhoton It only makes sense to me if the resistor is a 2 dimensional cross sectional area that instantly turns electrical energy to other energy by just detecting the amount of charge that goes through it but in reality, i think resistor is something like a transistor that has some distance. I can understand that more current is proportionale to more energy per time but it makes little sense when I think about it in terms of electrons constantly flowing through a wire.
Comment: Everyone here is saying 1A going through 1V is 1J/sec but how do you know that 1A of current is going to get in and out of a resistor in exactly a second?
Answer: Consider a charge $q$ moving with speed $v$.  In $t$ seconds it will move a distance $d=vt$, so the current (charge per second passing by that cross section area) is $I =  q/t = qv/d$.
Electric field is the gradient of the potential, so the electric field that exists within that distance $d$ is $E=V/d$, so $V=Ed$  (imagine a short distance or a constant field so no derivative needed for this hand-waving argument).
Now multiply: $$VI = (Ed)(\frac{qv}{d}) = qEv = Fv$$  Force times velocity:  power.
Hope that helps!
Comment: Thanks for the answer but i do have one question. The V that you got from E=V/d must be some arbitrary number depending on the distance but in circuit problems, the voltage drop is already set before(for example 1A current going through 1V so the power must be 1W) but how do we know if that 1A of current goes through the distance of 1V in exactly 1 seconds?
Comment: Do you mean *1 C of charge* rather than *1 A of current*?
Comment: oh my bad i meant to say 1C of charge
Comment: Slightly abusing notation we know that the definition of power is $P = E/t$ or in the case of an electrostatic problem $P=qV/t$.  Turn that around $t=qV/P$.   You told me that $q=1$C, $V=1$V, and $P=1$W.    So we get $t=1$s.
Answer: Force on a charge $\lambda dl$
$f dl = \lambda \vec{E} dl$
Amount of work done by $\vec{E}$ moving a charge $\lambda dl$ with a velocity $\vec{v}$ in time dt
$dw = \lambda \vec{E}  \cdot \vec{v} dl  dt$
Rate at which this happens (power)
$dp = \lambda \vec{E}  \cdot \vec{v} dl$
Substitute in: $\vec{I} = \lambda \vec{v}$
$dp  =  \vec{E} \cdot \vec{I}  dl$
Integrate to find the total amount of work an electric field $\vec{E}$ does on a distribution
$ P = \int \vec{E} \cdot \vec{I}  dl$
Negative charges move opposite the field, so  $\vec{I}$ and dl are in the opppsite direction so can be swapped *** read bottom ***
$P= -\int  I \vec{E} \cdot \vec{dl}  $
When current is constant, it is independant of the integral and can be factored out.
$P= -I  \int   \vec{E} \cdot \vec{dl}$
Sub in $v = -\int \vec{E} \cdot \vec{dl}$
$$P=IV$$
This  makes sense , as a potential difference between A and B being
Postive , would mean positive  work is done on negative charge.
*** the reason I have said that I and dl are in opposite directions is a technical detail of my proof, in the beginning, I assumed that  dl = v dt, meaning I'm finding the work done  given the electrons move in the direction I'm evaluating my potential ( from negative to positive), which creates  a current opposing my dl vector***
Answer: To get a basic intuition for electrical equations I tend to think of a hydrological analogue. When water is moving through waterways, the difference in elevation corresponds to a potential difference, and the rate of water moving though a canal corresponds to the current. Water likes to move from high places to low places, like positive charges like to move from high electric potential to low potential; the fact that in most situations it is the negative charges (electrons) that actually move to produce current is an unfortunate detail that can be ignored (for electrical equations there is no distinction between current produced by positive charges moving one way or negative charges moving in the opposite direction, so no intuition is lost by thinking of positive charges moving).
The equation $P=V*I$ for the power dissipated in a resistor is then analogous to the hydrostatic power dissipated by a current of water flowing between points of different elevation (and which could be converted to other forms of energy as in the turbines of a hydroelectric plant). This power is the product of (gravitational potential change due to) the difference in elevation times the rate of flow (in mass of water displaced per unit of time). Each unit of water dissipates energy equal to the product of its mass and the change of its gravitational potential (the latter being height change times gravitational acceleration, but you can think of this as the "tension" between the reservoirs between which the water is flowing).
Different units of water dissipate energy independently and the total is formed by summing over all units of water displaced. It should not be intuitively difficult to see that the power dissipated is proportional both to the difference in elevation between source and sink (producing pressure for the water to flow), and to the rate of flow that actually takes place.
Answer: Along the lines of the answer by @garyp ,
it might be good to note that $VI=Fv$ in that answer
hints of a "mechanical analogy".
Thinking in terms of [generalized-]work (as mechanics or thermodynamics),
$d{\cal W}={\cal F}\ d{\cal x} = \ (\mbox{Electric Potential}\ V)(\mbox{displaced charge}\ dq)=V\ dq $,
the electric potential can be considered a generalized-force as a "work done per unit displaced-charge", thinking of charge as a configuration variable. (It may help to think of "charging up a capacitor" as displacing charge from one plate onto the other plate.)
Then $P=\frac{d{\cal W}}{dt}$.
In a different context, for some steady-state situation,
the associated instantaneous power (assuming that $V$ is constant in time)
is then $P=\frac{d{\cal W}}{dt}= V\frac{dq}{dt}=VI $ (analogous to $P=F\frac{dx}{dt}=Fv$).
Answer: Other answers address why power is given by the product of voltage and current, on grounds of physics. Regardless of the physical meaning we associate with this product, the sum of voltage-current products of all branches (or components) of a network (or circuit) is guaranteed to be zero by Tellegen's theorem:
$$\sum_k V_k I_k = 0. $$
where the sum is over all branches. In circuits where the voltage-current product corresponds to power, this is a statement of conservation of power (i.e. power supplied is equal to power consumed), which is reassuring. In other words, based on network theory alone, the voltage-current product is a suitable candidate to represent power because it is conserved.
Tellegen's theorem is actually much more general than this. In fact, the components of the circuit need not correspond to physical devices at all: the only conditions are that the branch voltages obey Kirchhoff's voltage law and the branch currents obey Kirchhoff's current law. Under these very general conditions, the total "power" will still be zero.
Answer: "Power" as a concept, on its own, doesn't really make a lot of sense unless you think about what is doing the work. That's because that's what it is: it is the rate of doing work or more generally, utilizing energy.
In this case, what is happening is the electric field is doing work on the electric charges, and thus expiring energy in the process. That field, in turn, is being powered by whatever power source is used in the circuit, which is pumping energy into it to fuel its doing the work.
When a charge interacts with that field, it feels a force: $F = qE$. When that charge is pushed by such force over a distance $d$, work $W = Fd$ is done. Depending on the nature of the intervening medium, that push may happen over a long or a short time and as a result, the rate ($\frac{W}{t}$ or, if variable, $\frac{dW}{dt}$) at which the electric force field does work on the charge will be different.
Answer: The force over the charge of a small volume in an electric field is: $\delta F = \delta q E = \rho A\delta x E$, where $\rho$ is the local density of movable charges and $A$ the wire cross section. If there is no magnetic induction effects acting, the electric field can be expressed as the derivative of a scalar potential $V$:
$$E = \frac{dV}{dx}\implies \delta F = \rho A\delta x \frac{dV}{dx}$$ when $\delta x \to 0$: $$dF = \rho A dV$$ Multiplying by velocity of the charge, the left side is the power by definition:
$$dP = \rho Av dV = IdV$$
Integrating over a length of wire :
$$P = VI$$
Answer: Simplify it. Power is energy in application of delivery to a new or transitioning it's system. The qualities of a engine, motor or conductor dictate how much energy shall be delivered as power in comparison of movement/resistance in a given time frame of that energy's conveyer. Laws of thermodynamics and thermal properties of what's carrying the energy. Waste heat is produced thru out all duration of  transit
Answer: Let's take water as an analogy. V is the height of fall of the water and I is the volume that falls on the mill wheel.
In electricity, V is the potential difference between the source and the sink and I is the number of electrons that are moved through an electric drive or another sink.


Health effects of DDT contamination of coffee

Question: Are coffee beans often contaminated with significant amount of DDT? And are those levels known to have any negative health effects?
Comment: I've tried to clarify your question, if I accidentally changed the meaning you can edit it again and correct it.
Comment: What's the claim here? Have you read / heard that DDT is a problem for coffee-drinkers?
Comment: Any Polish speakers with access to PubMed? http://www.ncbi.nlm.nih.gov/pubmed/12235669
Answer: Depending on where the coffee is grown, it is quite possible it could be contaminated with DDT. In a study by the Smithsonian (R. Rice, J. Ward, 1996), they examined the amounts of DDT and DDD (DDT's toxic metabolite) present in green coffee beans. The limit of detection was 0.001 ppm, and they reported amounts ranging from 0.001 to 0.019 ppm.
There is no 'allowed use' for DDT in the US/Canada as it is banned there (and in 33 other countries) and DDT is only used for malaria control as per the Stockholm Convention. It is a persistent organic pollutant (POP), and is known to both bioaccumulate and biomagnify.
DDT has been linked to poor semen quality in men, and a 5-fold increase in risk of breast cancer in women born after 1931, who were exposed to the highest amounts of DDT before its use was halted in 1972. Prenatal exposure has been linked to a decrease in cognitive skills in preschoolers when the cord serum contained DDT in excess of 2 ppb (ng/ml). Researchers are also studying the possibility of a link between pesticides and autism
Given the half-life of DDT when mixed into the soil (between 5-8 years, taking between 25-40 years for a 90% reduction in DDT levels), it is possible to still find residue on produce today.  The International Agency for Research on Cancer (IARC)'s report on DDT can be found here. They've classified it as a 2B - Possibly carcinogenic to humans. They also list the maximum allowed DDT residue limits for a number of countries.
Now - what constitutes a 'significant' amount?  It depends on the context.
The drinking water guidelines (according to the IARC report, above) list the WHO limits for DDT of 1 ug/L (1 ppb), and the US EPA allows 2.85 ug/L (2.85 ppb). The report doesn't list coffee limits for the US, but other beans for other countries are listed between 0.05 - 7 mg/kg (1 mg/kg = 1 ppm in soil).
For drinking water standards, the amount of DDT on coffee beans would be significant. For foodstuff standards, it is not considered significant.
As far as negative health effects - there is no definitive cancer link for human exposure to DDT (plenty of animal studies, however).  Toxnet lists a number of studies delineating the body burden of DDT - all of which are much greater in concentration than what is found in those coffee beans.
For healthy adults, the risk posed by those concentrations would currently be considered small, as exposure to DDT is not solely restricted to coffee beans.
There's still so much we don't know about the long term effects of DDT, to be honest - but then that's the case with many chemicals we use today.
Comment: What happens to DDT in the beans when they're roasted? Many chemicals will break down under high temperatures (whether roasting is high enough, I dunno). While a study on green coffee beans shows that DDT can be found in them, the real study should have been done on beans of various roastiness.
Comment: Honestly, there is a dearth of data regarding the contamination levels of coffee beans (which is, sadly quite common with regards to many chemical compounds - we simply don't have the data on them).  It's a moot point for DDT though really, since the amount present isn't significantly toxic to humans.  If I recall correctly, the amount is ~100mg/kg for oral exposure in humans.  I'll have to get out my Tox Bible and check though.
Comment: I should also add there was a study done at the U of Hawai'i regarding beans from East Timor (I think), however I was never able to retrieve the pdf file to see the results, and no abstract was listed.


Bush transplants bought too early

Question:
I bought some of these starters from a big box store and it says plant them when chance of frost is gone.  Since I'm probably still 1 month out how should I best keep them?  Should I water them in the plastic bags they are in or leave them dry?  Inside or out in cold shed?
Answer: I think the problem is that the instructions were written by a "writer", not a gardener. Plant them as soon as practical, all plants shown grow where temperatures are below freezing. I am sure I have planted blueberries and raspberries and countless trees, fruit and other wise, before the last frost. I can't remember when I planted grapes and strawberries. The greatest danger is direct sun on the bag when plants are still in them. I always try to plant woody plants when dormant which normally means freezing is still possible. Keep them in a cool place until you are ready to plant; that will be better than in the warm store.
Comment: You're probably correct about the instructions. I'd be willing to bet that the supplier has a single set of planting instructions that it uses for all plants, regardless of cultivar: "Plant after danger of frost at the same level as in the pot. Water well" etc.
Comment: Judging from the picture, the plants are still dormant. The biggest danger is that you keep them somewhere "warm" (i.e. above about 5C or 40F) and they start to grow *before* you have planted them. With no access to water, they will then be badly damaged, or die, very quickly (i.e. in hours, not days.) You can't plant anything if the *soil* is still frozen, because you can't firm frozen soil around the newly planted roots, but air temperatures below zero don't matter.

Bush transplants bought too early


Two Electric Ranges with the Same Issues

Question: I live in an apartment building that is 100 years old.  My electric range burners don't all work the same way and they have hot spots on the elements. I only use one element as the others can't be trusted. Even on the element I use, food has to be watched carefully as it will burn in one spot even on a medium setting. The oven is unpredictable even after calibrating the temperature as best I can.  I have been trying to figure this out.  The landlord replaced the very old electric range with another used range.  The problems are are the same on both ranges. The outlet for the range is hanging out from the wall.  I have no access to the circuit breakers or power box.  I tried to explain to the landlord that it is very unlikely that 2 ranges would have the same problem unless it is related to the power source.  Any electricians out there who knows the answer to this?  Thanks!
Comment: Sounds like the elements need to be replaced with new ones.  Most just pull out and new ones pushed in.  The hanging outlet is not code and needs to be fixed.
Comment: This isn't enough information for even a guess. Please add: the country you're located in, the name and model of the two ranges, and pictures of both the outlet where the range(s) connect, as well as the range's plug itself. It'd also help to know if the ranges in other apartments operate correctly on all elements.
Comment: Not the power source, just old, broken ranges.
Comment: Picture of the range would be nice.  Does it turn red unevenly?  What do the elements do that can't be trusted?  Might be the cookware?
Comment: What shape/size are the hotspots on the elements, and where are they located WRT the center of the element?  (You may want to rent an IR camera from the local tool-rental shop for this, even, although you can do a rough measurement with a pyrometer and some time/patience)
Comment: Sounds like you live in a rent regulated apartment.
Comment: The ranges were swapped out right in front of me.  There were no part exchanges.  The hot spot on the one element that I use is in the area where the source power comes in and the hot spot on the old range was in the same place.  I have experience with manufacturing electronics, but not with household electric.  I was thinking that there is a phase missing? ... something that would make both ranges fail in the same places with the same issues.  Such as; the same element problem and the fact that both ovens took a long time to pre-heat and the temperature varies in the same way...
Answer: Electric coil elements with "hot spots" are failing and should be replaced with new ones, not a different range with old ones (possibly the same old ones, if they swapped parts from one to the other, as they are pretty interchangeable and your landlord reads as "cheap".) When they fail all the way, the arc can blow a hole in the bottom of a pot...just not a fun thing to let happen, particularly for a part that's typically $10-$12 or so.
An "unpredictable" oven is vague as heck, but typically you need an oven thermometer to hit temperatures reliably on any oven you don't already know the mapping of "dial to reality" for, as they are rarely accurate. I've had an oven element blow once, which was quite the show, since it arced wonderfully without exceeding the amperage of the range breaker, so it just kept arcing. I was happy that I was in the the kitchen and heard, then saw, it so I could shut it down. It was not subtle.
The range outlet hanging from the wall (or the power supply in general) would not affect either of these, but does reinforce the "cheap or outright slumlord" impression you're giving of your landlord. Often the only real solution to that is finding a new place to rent, though sometimes you can get a tenants rights organization to help you in attempting to fight with them, but it's often a losing (or pyrrhic) battle.
Comment: Some jurisdictions have laws which require landlords who provide cooking facilities (here, stoves) to provide _working_ appliances. Such laws sometimes give tenants considerable leverage to force compliance.
Comment: Not only that, but coil "burners" are *consumable* parts, very cheap and easily replaced... and it's the position of some landlords that these are like light bulbs and tenants should be responsible for changing them as they fail.


Olden version of "psychopath"

Question: Apparently, the term "psychopath" was coined in 1888, and at that point, it might not have even been used by the laypeople. So, I*m wondering about a word used for people that display psychopathic characteristics that was used in the 1850s. In the book im writing, the main character is from Birmingham, UK, so the word will have to have been common around those parts. The word can be of posh language or street language, either is fine because the main character is familiar with both sociolects.
Comment: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/116702/discussion-on-question-by-a-kvale-olden-version-of-psychopath).
Answer: As  user121863 said, in the 1850s the general understanding of mental disorders lacked nuance. So there weren't many "posh words", as you put it, and Hot Licks has them covered.
The following are synonyms for 'mad person'. I'm not sure if they would all be deemed slang:
hatchet-man from 1876
nasty man 1863-
cutter 1568–1734
frantic (noun!) 1574-1758
crack-brain c1570-1831
madbrain c1570-1616 + 1876
shatter-brain 1719 -
maniac 1763 -
dement 1857 -

berserker:  The adjective 'berserk' is still used to mean crazy, but
according to Etymonline, the noun 'berserker' (from the Old Norse
'berserkr', meaning "raging warrior of superhuman strength") is the
oldest form of the word in its revival in Modern English (1822).
'Berserk' appeared some twenty years later.

And perhaps your readers would believe these killers, from Old English) were still on the loose in Birmingham in the 1850s:
bane (from 'bana')
slaga - hence faederslaga (father-slayer)
queller (from 'cwellere') - hence boyqueller (boy-killer), which was even used in 1606!
morþ (morth)

Comment: You're missing the most obvious ones: *madman*, *maniac* and *lunatic*. (And since the comments have been moved to chat now, you should probably add them.)

Olden version of "psychopath"


How to use persistent and two(at least) casper-rw partitions on the same computer?

Question: I have a question about GRUB2,Ubuntu and initrd:

I need at kernel level(initrd) unknown for me command (with used grub2) to disable boot/detect/use specific partition of disk(by /dev/sdX# or by Label or by UUID).

For example, I have live-CD/USB of Ubuntu + persistent (With casper-rw Labeled partition) on my SSD. And I want to use another partition like USB-disk on the same computer with other persistent partition(also must be Labeled partition as casper-rw) I'll do this with kernel command "persistent persistent-path=/live-usb" where "/live-usb" is a dir where is the placed file named "casper-rw".
But boot priority is given to the partition on the SSD-disk and not USB-disk (partition with file). If I rename the partition label on the SSD-disk to something else than "casper-rw" - then persistent works well with the file partiton. So I need the ability to disable boot/detect/use of a specific partition on the SSD-disk at the boot time of Ubuntu.

Thanks.
Comment: Do you need to use a partition on the SSD instead of another FAT directory path?  Then you could use a variable in the persistent-path=$var set in grub to be the USB if present, default to SSD path.
Comment: I have tried many times and can only get one persistent partition to work. It is always the first `casper-rw` or `writable` partition the boot encounters, The only way I have found to get multiple persistence to work is by using `casper-rw` or `writable` **files** and by using `persistent-path`. YUMI manages to get multiple persistent files of unlimited size working, but only using BIOS boot. I managed to duplicate that method once, I recall the method uses GRUB for DOS with virtual partitions and persistent-path.
Answer: I think persistent-path works only with a casper-rw file, and not with a partition.
There are two alternatives with partitions that I can suggest.

Relabel the casper-rw partition that you don't want to use to something else, e.g. casper-rw0 (you know this alternative already).

If you use Ubuntu 20.04 LTS or newer, you can relabel the partition that you want to give priority to writable, and it should be selected (when detected) before the partition with the label casper-rw. This alternative may be more attractive.
Answer: Using Multiple casper-rw Partitions on the Same Computer?
One method to get multiple persistent partitions on the same computer, would be to put your Persistent partitions on separate USB drives and plug in the drive with the persistence you want to use, before booting.
This method does not seem have much advantage over using two persistent USB drives.


Filed earlier, published later

Question: US patent claiming Compound A or its salt (X salt included in laundry list) published after another US application filed as compound A's X salt and gets grant. They do not share common priority and inventors except applicant. Later patent can be challenged or can earlier patent be prior art for later patent?
Comment: Are you saying they do have one common inventor?
Comment: No common inventor but applicant is same
Answer: It is all about the filing dates. If application A is filed first then application A may be considered prior art for application B even if application B is filed before application A publishes. It doesn't matter whether or not application B was developed independently of patent A. This is true even if the application A never gets granted as a patent.
I am not a lawyer and this is not legal advice.

Filed earlier, published later


Duplicated object not showing when rendered

Question: So basically, duplicated items don't show in render. I've tried ticking all in filter, one by one, all, none and still doesn't work, I've also checked if if the visibility boxes are ticked, but still doesn't work. I am desperate, please help!
Comment: How are the objects duplicated? Can you share the relevant part of your scene on  [Blend-Exchange](https://blend-exchange.giantcowfilms.com/)?
Answer: Could you post some screenshots of the screen about the problems?
However, it may be enough to move the project to a new project file and see if the problem recurs, alternatively you can also make a copy of the object and delete the previous one, first I would tell you to try these simple procedures.
Alternatively, do the objects have applied textures? Does the project have lights? Are they correctly positioned?
Is the camera positioned correctly, does the field of view cover both positioned objects?
Add screenshots of your work screen to make a certain problem more understandable.
In the new project have you tried to import only what you see in the render? to import I mean through the '' append '' section from the blender library.
So to import only what you are interested in a project.
  However, you should also check the maximum view of both the camera and the viewport, or rather being in perspective the image maybe the camera does not get to physically see the most distant object. look at my screenshot
Comment: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/106011/discussion-on-answer-by-albizz-duplicated-object-not-showing-when-rendered).


Laminas\Mail\Protocol\Exception\RuntimeException "Authentication Failed"

Question: I'm order the product and proceed to checkout then show the error. I'm use the "zoho credentials" after that i remove the "zoho credentials". and use the gmail Own smtp but still error is coming. could be any suggestion please help.
Answer: You can review the SMTP configurations, especially, the authentication type for the SMTP server connection: NONE , PLAIN, or LOGIN and the verification type for the server security certificate: SSL or TLS
For Magento 2.4.6, it resolves my issues in this regard, https://experienceleague.adobe.com/docs/commerce-admin/systems/communications/email-communications.html?lang=en


What is the security and identity impact for current org due to new release winter 17

Question: Is there any security impact on current org due to new release winter17? I would like to see difference between old release and new release. Can anyone please suggest me?
Answer: If you have a Production Org with sandboxes then you will have the sandboxes updated prior to the introduction of the release in live meaning that it is possible to do a regression test around your current setup which will allow security and customisation to be tested with the new release.
The release notes also contain a specific section on Security and Identity which explains what has changed.
Ultimately this would have to be a Salesforce Org specific task as apart from general changes such as TLS1.0 being desupported and the Winter17 requirement for HTTPS for all logins it will depend on the features that you are using.
For example in Winter17 HTTPS is now required for all logins

Logging in to a Salesforce server instance, such as na1.salesforce.com, now requires an HTTPS secure connection. Previously, HTTPS was required for logins via login.salesforce.com and custom domain URLs. HTTPS is now required for all Salesforce logins. Replace HTTP with HTTPS any time you call yourInstance.salesforce.com. This feature is available in both Lightning Experience and Salesforce Classic.


Magento 2. RabbitMQ Ack or Reject message

Question: I'm wondering if it is possible to manually Ack or Reject message from the RabbitMQ within the Magento 2 consumer process? What is the default behavior?
Answer: Let's take first the case of vendor/magento/framework-message-queue/Consumer.php (this is the consumer instance used by most M2 OOTB consumers). Check the method process() and from it go to method getTransactionCallback().
Within this method, the class vendor/magento/framework-amqp/Queue.php is called when using acknowledge() and reject() (in their definition you will find as expected the basic_ack and basic_reject).
In the case of vendor/magento/framework-message-queue/BatchConsumer.php, if you go down the line, you will find usages of the vendor/magento/framework-message-queue/MessageStatusProcessor.php class with its acknowledgeMessages() and rejectMessages() which again point back to the vendor/magento/framework-amqp/Queue.php as described above.
So in conclusion, Magento already uses both mechanisms, depending on a specific logic. Check the methods mentioned above to get a grasp on when each is used.
Cheers!
Comment: It seems the reject method doesn't work, I'm trying to add `$queue->reject($message); return;` after vendor/magento/module-asynchronous-operations/Model/MassConsumerEnvelopeCallback.php:103, but messages are still ack-ed.
Comment: does the code reach your line at least? maybe you are on a different flow
Comment: yes, the code reaches normally, but reject doesn't seem to work
Comment: uhmm ok, that's weird, really weird. i mean i can't think of a different reason for implementing reject with basic_reject if not for rejecting messages.


OSX: Spotlight shows the content of locked notes

Question: When I search in the spotlight for a locked note, I can see the content of that looked note. I went ahead and rebuilt the spotlight index, but still showing the content. Am I missing something here or is it a bug? OSX-ELCapitan 10.11.5
Answer: You need to close your notes

I justed tested this in OS 10.11.5 and the contents of the note don't show, just the title.
I created a note and titled it for which it would be easy for Spotlight to find:  "Smattering test note"  (I have no other instance of "Smattering" on my system).  Here is what it looks like after I have locked and closed the note.

Doing a Spotlight search for "Smattering" yields this result:

It autocompletes the note title (ligher grey text) and clicking on it opens the Notes App.  Until I unlock the note with my password, the contents are not visible.
If I don't lock and close my notes, if I do a Spotlight search, the contents will be visible
Comment: Thank you a bunch! Here is the path in case some is having the same issue
/Users/USER/Library/Containers/com.apple.Notes/Data/Library/CoreData/ExternalRecords/HTMLNotes/40943718-ADA7-487A-8194-7FE94D6460DC/IMAPNote/_records/2/0
Comment: Thank you for the details. So I see what the issue is, I went ahead and renamed the existing locked note, but the spotlight is still showing the old version of the locked note that doesn't exist in notes anymore. It seems that I have duplicate of old notes somewhere and even if I lock it, it will show the other old version. Where can I find them?
Comment: When you do a spotlight search, look at the very bottom of the left column of sources.  Click on "Show All in Finder".  It will bring them up for you


server response xml or json objects android

Question: I have an application which uses restful web services, the service returns a json object.
I wanted to know , what would be better , returning a json or xml response?
Please help as i am new to the concept.
Comment: http://stackoverflow.com/questions/4862310/json-and-xml-comparison
Comment: http://stackoverflow.com/questions/2316966/should-i-use-xml-or-json-as-a-data-format-in-android-applications
Answer: The answer is that it doesn't really matter. They are both almost equivalent. They use slightly different API's so go with the one you feel more comfortable with.
Example of JSON parsing in Android you can find in this question.
Example of XML parsing in Android you can find in this question.
The only reason I can think of to use XML over JSON is when your webservice responses are huge. JSON usually requires the entire response to arrive before you can start parsing. XML easily supports a pull parser which can start parsing "on the fly" before all the data arrives - which can be much more efficient. But if your responses are small, you don't really need all that.
If your existing webservice already returns a JSON object, this could be reason enough to stick with JSON.


How do I draw CCG derivations in LaTeX?

Question: I have been trying to find an easy way to draw CCG derivations in Latex but nothing seems to work. I am working on a presentation, and I found this github repo but it for some reason when I try it it doesn't work. I am using Overleaf and I am a beginner with latex, but I really need to figure this out for a computational syntax course.
Comment: I did my PhD in an area of Combinatory Categorial Grammar. In the course of writing up, I adapted the `.sty` file you've linked to. It should work pretty well as it was designed not to have any LaTeX dependencies. Can you edit your question to give some idea of how you attempted to use it, and how it didn't work for you? It could be that you're using it incorrectly.
Answer: do you want something like what $\frac{\frac{the}{NP/N}\ \frac{dog}{N}}{NP}$ gets you?
If so, a few notes:

LaTeX has two main modes - regular mode and math mode. The thing I would do, if I am understanding you correctly, only works in math mode. You can use $ to toggle inline, \[ and \] do math mode on a separate line.
\frac{A}{B} gets you A over B. These can be nested.
\ (backslash space) gets you a space, but a regular space in math mode does not appear (it is used to show you are done with an escaped sequence (one set off with a \)
to make a backslash character, you I think need to do \backslash

How do I draw CCG derivations in LaTeX?


New tag for Simulation-nodes?

Question: Now that Simulation Nodes are here.  Should we create a new tag called simulation-nodes like we have for geometry-nodes?
Comment: Basically yes, but *"it's here"*, is currently not correct, because it has not reached a stable version yet *(15.5.2023)*. Establishing functions that are only available in alpha/beta versions as their own tags is therefore premature in my opinion.
Comment: I just wanted to get ahead of it they announced that it will be in the 3.6 version https://www.youtube.com/watch?v=4yq_2qsv1MI . Things related to Simulation-nodes has been showing up for awhile and getting this stuff tagged / labeled correctly will make it easier to search for things in the future... https://blender.stackexchange.com/search?q=simulation+nodes
Comment: From this point of view: Yes, Since it is (fairly) certain that *Simulation Nodes* will come sooner or later, it would be good to do some preliminary work here and tag the questions correctly already now (Even though they may not have much relevance in the long run at the moment). However, there was once a tag [tag:geometry-nodes-fields], which was also merged with [tag:geometry-nodes]. Strictly speaking, *Simulation Nodes* are simply *Geometry Nodes*. Currently, there are actually too many tags, and it would be better to clean up the mess and merge tags instead of creating more derivatives.
Comment: Good point. Given that question about it are bound to come pouring in soon now would be the time to create the tag (or synonym). Whether we actually want a dedicated [tag:simulation-nodes] tag or not is something else that needs discussion. As quellenform correctly points out it seems to me like just another type of Geometry Nodes that doesn't warrant a tag of its own. @RickT Do you think we'd benefit from a dedicated tag for simulation nodes? Do you see any situation we'd benefit from a tag specifically for simulations or would the old tag suffice?
Comment: @DuarteFarrajotaRamos Honestly, I go back and forth.  On one hand simulation-nodes is part of geometry nodes as @quellenform so `correctly` stated, on the other hand it really is it's own subgroup.  We currently have the `modifier` tag but we also have the `screw-modifier` tag  and the `bevel` tag (should be a `bevel-modifer` tag also) but we have no `weld-modifier` tag.  What are the requirements for a tag to be created.  Note this can quickly get out of hand and we start creating a tag for everything (Like putting post-it notes / labels on everything hahaha https://imgur.com/NWzeK6H)
Answer: I've been thinking a bit more about this and came to the conclusion that for the time being at least we do need the simulation-nodes tag.
It is true our tag system is a bit inconsistent at the time, we do have tags for some modifiers but not others.
The general (lack of) rule so far has been creating tags on a need-to basis. Users, some times newly arrived, create tags that lay around for a while, and we ended up adding wikis to them making them became more permanent. Some times in retrospect we come to the conclusion we don't need a certain tag and create a synonym after the fact.
I do think some specific modifier tags do have their value, they help us filter for more specific or substantial modifiers (like boolean or bevel), or distinguish between modifiers and their homonym edit mode operators.
As for the Simulation Nodes I'd vote for keeping it as dedicated tag not because they are distinct from regular geometry-nodes, but to help distinguish from legacy physics simulations, which I suspect will eventually be phased out in favor and the new system.
It may additionally be relevant in contrast with regular Geometry Nodes, since unlike the old Fields, (which became an integral part of the broader geometry nodes) physics simulations will always act as "box" inside a more general node tree, with well defined "gated" entry and exit points.
While input can be added at any point in the simulation box, output will only happen with a Simulation Output node. I haven't yet dug into the specifics, but I suspect this will open the door to trying to solve certain issues with the subset of nodes that can be used inside this simulation area, along with any additional limitations that may or may not apply.
For these reasons I'd argue to create a simulation-nodes and maintain tag to distinguish from the old physics simulation both for historical purposes and for any users for any reason stuck with older Blender versions.
Let me know if you have any arguments against it. If we all agree now would be a good time to create it, since questions about simulation are likely to start rising in volume in the near future.
Comment: OK, let's do this, I agree for now. I'm not quite convinced yet, since more tags don't make the tags less, but in this case a distinction might be handy for the future. And if the questions are tagged, they can always be retagged afterwards if the decision didn't turn out to be useful after all. I am so free and take care of it right away....
Comment: Yes, we can later remove the tag if we no longer see the need for it. It is easier to remove an unwanted tag, then it is to retroactively try to tag a large backlog of existing questions with an new tag.

New tag for Simulation-nodes?


Will applications for Ubuntu Touch have the same look as the Desktop version?

Question: For example, the File Manager and others softwares in development.
Answer: They should have similar look, but with more screen space and mouse+keyboard inputs it would be better to make slight changes to make better use of those.  Take the File Manager for example, on a desktop with more screen space it will show the Places list in a side panel, while on a phone the Places are accessed via a popup from a toolbar button.
Answer: I believe so. As they exist currently you can easily run all of these programs on your desktop and play around with them. The way they look on your desktop is how they appear on the phone.
These apps are being developed for the phone however and will therefore not be ideal for use in a desktop environment.


Can a NFA reach two final states at the same time?

Question: I'm studying nondeterministic finite automata, and I understand them in principle. Compared to deterministic FA, you can have more than one transition function with the same character starting from a single state and you can have "empty" transitions, and you can "choose" which one to follow based on the current string. So I came to an example which is confusing me.

So these two NFAs can recognize strings from the same alphabet. For example abaaaa.
What is confusing me is the first branch in both cases. For example in the first one something like this would happen:
Start, read a, go to 1, read b, go to 2, which is a final state, but the string isn't over. So what happens now? Does it go back to 0 and starts reading the final as? so 0, read a, go to 3, read a, go 3, read a, go 3, read a, go 3, finish?
Essentially, with NFA, can we follow two or more paths at the same time? (As if we "split" the string in two parts maybe?) Because otherwise I can't see how any string would be accepted in this case. ab would lead to state two, but the string isn't over, or a would lead to state 3, but then we would find b and the string would be rejected in that state.
In the second one how is the final state reached? We would still need to "go back" to 0.
Answer: The language accepted by an NFA is the set of all strings $s$ for which there exists an accept path in the NFA (meaning a path from the start state to an accept state) which is labelled by the letters of $s$ in order. You can't start over. You can't break the string into pieces and follow separate paths with separate pieces of the string. You have to start at the start state and look for paths along which you can read off each letter of the string one-at-a-time, and perhaps one of them will end with an accept state; if such a path exists, the string is accepted; if no such path exists, the string is not accepted.
So, for example, in your first automaton, let's consider a few examples.
First example: the string $aa$ is accepted, because there exists an accept path in that automaton labelled by $aa$, namely the path $0 \mapsto 3 \mapsto 3$. The fact that the path that starts $0 \mapsto 1$ cannot be continued is irrelevant.
Second example: the string $ab$ is accepted, because there exists an accept path in the automaton labelled by $ab$, namely the path $0 \mapsto 1 \mapsto 2$.
Third example: the string $abb$ is not accepted, because there does not exist any accept path in the automaton labelled by $abb$. You could start with $0 \mapsto 1 \mapsto 2$ labelled by the first two letters $ab$ of $abb$, but since you cannot continue with the third letter $b$, you're not going to get an accept path this way. Or you could start with $0 \mapsto 3$ labelled by the first letter $a$ of $abb$, but since you cannot continue with the second letter $b$, you're not going to get an accept path this way either. Since you have exhausted all of the possibilities for accept paths in the automaton labelled $abb$, and since you have demonstrated that none exists, it follows that $abb$ is not accepted.
Fourth example: consider the string $a$. You could start with the path $0 \mapsto 1$, and you're done with the string, but oops, you're not at an accept path; so there's no accept path for the string $a$ which starts with $0 \mapsto 1$. No worries, though, because instead you start with the path $0 \mapsto 3$, and the string is done and you're at an accept state, and so the string $a$ is accepted.
Comment: Understood... So in this case this NFA does not accept every possibile string from the language ab+aa*, right? But how would it work, because from this language, the "minimum" string would be aba, right? Because you must have ab from the left part and then at least an a because of the right part of the expression.
Comment: Oh no, I'm wrong, the language would be {ab, a, aa, aaa, aaaa...}. I see my mistake now, it makes sense, thank you very much.


Looking for intuition behind the Euler Product Formula and related Number Theory things

Question: I've been trying to dive into Number Theory to better understand prime numbers. I found Euler's Product Formula, $$\sum_{n\in\mathbb{N}}{\frac{1}{n^s}}=\prod_{\text{Prime }p}{\frac{1}{1-\frac{1}{p^s}}}$$ While looking into this, a number of things confused me. In one informal explanation, I don't get how they Add for n=1,2,3... on the very first page. In another book I have, I see them "expand the factors" by saying $$\frac{1}{1-\frac{1}{p^s}}=1+\frac{1}{p^s}+\frac{1}{(p^2)^s}+\frac{1}{(p^3)^s}+...$$ and I don't see where that comes from either. I'm also curious why we express the product as such an unwieldy fraction instead of simply writing $\frac{p^s}{p^s-1}$. I suppose my overarching question is... am I just blind and this isn't clicking with me, or am I in over my head and these steps involve other parts of number theory which aren't mentioned? I apologize if I've been too broad or not very specific, I'd be happy to clarify what I can or move this somewhere more appropriate.
Answer: Remember that $\dfrac{1}{1-x} = 1 + x + x^2 + ...$. If we let $x = 1/p$, we get a sequence of all inverse powers of $p$. Now, if we multiply these sequences over all $p$, we are generating every possible combination of various primes to various powers (distributive law) and by the Fundamental Theorem of Arithemtic we can know for sure that every inverse integer is expressed once and exactly once. Does that help?
Comment: It does, that makes a lot more sense. I feel like my intuition for number theory isn't built up very well. I've gone through elementary number theory, but I suppose I could use some better intuition on the stuff, any recommendations for material that might help with that?
Comment: @Vedvart1 I am just a sophomore in college taking my first number theory course, but that subject has always come naturally to me, plus I have read a lot of number theory over the years for competitions. I generally have found that most of number theory can be understood either by factoring things or using congruences. A good overview of Euler's work in number theory would provide a great background, or any expository work you could find in a place like Barnes and Nobles would also be a good resource.
Comment: I think I'll pour over Euler's work then, and try and look at things as no more than a bunch of numbers and basic algebra whenever possible. I'm still only a junior in high school and I've taught myself most of what I know, so it's entirely possible I might just need to let the basics simmer in the back of my mind for awhile. Thank you for the help though, I always love when this site and its community pull through!
Comment: Yeah I'd focus on linear congruences and the Chinese Remainder theorem before you get anywhere else. Those should provide a good background if you aren't yet familiar enough with congruences.


Holomorphic function in Taylor series of principal Logarithm

Question: Yesterday, I made a series expansion (Taylor's) of principal Logarithm i.e $Log: \mathbb C$ \ $(-\infty,0] \to \mathbb C$ at a point $z_0$, The radius of convergence comes out to be $|z_0|$, Now when i take $z_0=-1+i$ or anything near that my circle of convergence crosses negative axis where my function is not actually defined, which holomorphic function is represented by this series at that region?
Answer: On your given disc of convergence, your power series equals some branch of the complex logarithm that avoids that disc. It just so happens that this branch of log agrees with the principal Log away from the negative axis.
In your specific example, $z_0 = -1 + i$, the power series about that point would agree with the branch of logarithm given by
$$ \log{z} = \log{r} + i\theta: z = re^{i\theta}, -\pi/4 < \theta < 7 \pi/4$$
Of course, there are infinitely many branches that work, so long as the branch cut avoids the relevant disc of convergence.


Where on the web can I find a Stratum protocol specification document?

Question: I am writing my own mining pool server for Bitcoin and would like to work with a detailed and complete specification document for the Stratum protocol.  There are documents on the web that casually outline the protocol (https://braiins.com/stratum-v1/docs and https://en.bitcoin.it/wiki/Stratum_mining_protocol) but I am looking for something rigorous and complete and wasn't able to find it.  The reference documentation of BIP 41 and 40 doesn't exist (see Where is BIP 40 and 41?).  Something like this is good: https://github.com/aeternity/protocol/blob/master/STRATUM.md
Answer: The team at Braiins is working on Stratum v2 in collaboration with others. There have been a number of changes made since Stratum v1 was first specified.
There is a work in progress specification for Stratum v2 here.
Comment: @andrewz: I asked on Twitter and the Braiins link you included in your question is the closest thing that exists apparently https://twitter.com/michaelfolkson/status/1539217254709178368?s=20&t=QB7Sc9dmqtsd9-V7OXM-vQ
Comment: I need the specs for V1 as I'm working with miners from 2019


Magic Mouse 1 and Bluetooth 4.0

Question: I am considering adding a Bluetooth 4.0 dongle to my 2011 Mac (to enable Airdrop). Will my Magic Mouse 1 still work?
Answer: Yes, the Magic Mouse 1 works with standard Bluetooth 4.0.
Bluetooth is generally backwards compatible so that older devices keep working with new Bluetooth hosts (either built-in to computers or in dongle form), although newer devices generally are not guaranteed to work with older Bluetooth hosts.
The Magic Mouse 1 requires your Mac to be running at least Mac OS X 10.5.81, which you would already be running if your mouse is working before adding the dongle. Any 2011 Mac is capable of running 10.5.8 or later, so you will face no problems there.

Magic Mouse 1 and Bluetooth 4.0


Why doesn't amplitude increase when drive frequency is above resonance?

Question: Why doesn't amplitude increase when the frequency of external periodic force increases above the natural frequency of the vibrating object?
Comment: Try pushing a kid on a swing more often than the natural frequency.
Comment: The math behind a driven damped oscillator is e.g. reviewed here: http://physics.stackexchange.com/q/228279/2451
Answer: In one sense this is a consequence of the meaning of resonance. Resonance is identified by a maximum in amplitude. If amplitude increased above (or below) the resonant frequency then we have not found a maximum in amplitude.
Perhaps what you are really asking is, why is there a maximum in amplitude when driving frequency $f$ equals the natural frequency $f_0$? This was answered by What is the qualitative cause for a driven oscillator to have a max. amplitude during resonance?
At the simplest level this is because at the natural frequency $(f=f_0)$ the driving force is always in phase with the natural motion of the system. The driving force is then always adding energy to the system, which will increase indefinitely unless there is some form of damping (eg friction) which removes energy from the system at a faster rate as amplitude increases.
At all other frequencies the driving force is sometimes in phase and sometimes out of phase with the natural motion of the system. Sometimes it adds energy, sometimes it takes energy away.
However, this assumes that both the natural motion and the driving force are both sinusoidal. If the driving force is applied as an intermittent impulse or "kick" then sub-harmonic frequencies of $f_0/n$, where $n$ is an integer, will also result in resonance. For example, if the system oscillates once every second, giving it a "kick" in the right direction once every 2 or 3 or 4 etc seconds will also increase the amplitude of the system, provided that the energy imparted with each kick is not outweighed by the loss of energy in between. See Non-resonant but efficient frequencies.
Answer: A simple physical explanation follows directly from Newton's second law, $F = ma$.
Suppose you keep the amplitude $F_0$ of the external force $F = F_0 \cos \omega t$ constant, and increase the frequency $\omega$. The amount of time that the force is acting in one direction is inversely proportional to the frequency - it is a quarter of the vibration period, or $(1/4)(2 \pi / \omega) = \pi / ( 2\omega)$.
So as the frequency increases, the structure doesn't have enough time to move far, before the force reverses and accelerates it in the opposite direction.
If $\omega$ is much higher than the resonant frequency, the stiffness of the structure is not very important, because it never gets time to move far, and $Kx$ is always small compared with $F$.
There are situations where the amplitude of $F$ is not constant, but proportional to $\omega^2$. One example is the force acting on a rotating shaft which is not perfectly balanced - the radial force on a small element of material $dm$ of the rotating structure is $r \omega^2\, dm$.
In that situation, the amplitude of vibration is approximately constant for frequencies well above the resonant frequency. In fact the shaft is trying to rotate about its mass center, not its geometrical center, and the vibration amplitude is simply the distance between those two points.


Forward iMessages and SMS to e-mail?

Question: I want to forward all SMS and iMessages received on my iPhone to my e-mail address. I have tried IFTTT, but can't see a way to do it. Has anyone else managed to do this?
Answer:
you need to have your email address turned on for iMessage on your
  iPhone in order to enable Text Message Forwarding. If you don't, the
  numeric access code will not appear on your iPad or Mac during the
  setup process. As soon as you enable your email address for iMessage
  (you only need to do this on your iPhone) the numeric access codes
  appear as expected. Once you have text message forwarding setup you
  can disable your email address again in iMessage as it seems to only
  be necessary for the numeric access code setup step, not the actual
  text message forwarding itself.`
To enable your email in Messages, grab your iPhone and head to
  Settings > Messages > Send & Receive. There, you'll find all of the
  addresses at which you can be reached via iMessage. Enable your iCloud
  email address by tapping on it, then head back to your Mac or iPad and
  again request the pairing verification code from the Messages app.
  This time, you should see a box appear on your iPhone in which you can
  enter the code.

Source: http://www.tekrevue.com/tip/sms-relay-configuration-fix/
Comment: What about SMSes? (no iMessages here)

Forward iMessages and SMS to e-mail?


Where can I download the REAL MonoDevelop Windows binaries?

Question: The monodevelop.com website just takes me to this "Xamarin Studio" thing that has severe restrictions on the usage of the free version, and pricing that's almost as bad as Visual Studio on the paid versions! Is MonoDevelop for Windows dead? All I want to do is develop a GTK# application without installing Linux!
Answer: There's used to be a version up on Sourceforge, but 9 years after I wrote this answer it seems to have disappeared. Sounds reasonable enough...
If you really need version 4 you'll have to build it yourself. If you still have/find the sources that is.
Comment: Why don't they provide Windows binaries anymore?
Comment: Thanks! Actually found that .msi, but now I can't get it to install - it keeps demanding GTK# 2.12.9 or newer, and somehow the newer versions I have installed that came with Mono don't seem to satisfy it!
Comment: Installing the proper version of GTK# solved it! Wonder why it didn't like 2.12.20 or whatever I had before?
Comment: First link dead `Whoops, we can't find that page.`
Answer: You do not have to pay to use Xamarin Studio to develop GTK# applications. There are no limitations whatsoever (except your imagination!) on what you can create with Xamarin Studio for the desktop.
You only have to pay if you want to develop applications for Android (and even for Android apps there is a free Starter edition).
Comment: This is difficult information to find. I'd expect it on the [pricing](http://xamarin.com/faq#pricing) portion of the FAQ - or [MonoDevelop's FAQ](http://monodevelop.com/FAQ). It's not like the confusion [is new](http://phoronix.com/forums/showthread.php?77965-MonoDevelop-vs-Xamarin-Studio-IDEs). And [here's a Team Xamarin guy last year saying they're different](http://forums.xamarin.com/discussion/comment/5017/). I'm not saying Xamarin's doing a bad job being fair to the OSS community, but the good job could be communicated *much* more effectively.  ;^)
Comment: But is there a size limitation on GTK# applications as there is with Android apps? I don't want to come up against a paywall halfway through the development process...
Comment: Is it possible to run & debug in Mono using Xamarin Studio (with the intent of targeting Mono on Linux)?
Comment: I researched it not long time ago. Restrictions in Xamarin Studio are not allowing to link some essential libraries without exceeding limits. Unusable. Seems Mono is dead unless you build it yourselves which is not easy as most of people say. It would be available as an installer otherwise.
Comment: This doesn't answer the question of where he can download MonoDevelop.
Answer: Honestly, it is a real pain in the butt to build MonoDevelop from the source or even to install on some platforms.
I did a lot of research and worked out for hours but still could not manage to build MonoDevelop and at the end of the day Decided to use SharpDevelop 4.3 (http://www.icsharpcode.net/OpenSource/SD/) instead and I gotta tell everybody that SharpDevelop is easy peasy to install. There are other alternatives too like Webbo, QuickSharp and Xamarin Studio but I haven't tried all of them.
If it is that easy to build MonoDevelop they would have build by them self and put an exe or a msi file or even a zip file with all the binaries and dependencies to download.
They have a similar problem which is quite annoying when it comes to installing the latest version of MonoDevelop on Ububtu or PC BSD.
First thing, I reckon its Git's problem but every time I downloaded the MonoDevelop source from its Git repo, the zip file is smaller than the real size and always corrupted.
Anyway I tried a few times and managed to download the source.
Then I downloaded and installed all the prerequisites.
I could not even open the Solution file in Microsoft Visual C# 2010 x86 (express).
Then I decided to build using MonoDevelop 3.0.6 as it is suggested in http://monodevelop.com/Developers/Building_MonoDevelop
When I tried to install MonoDevelop 3.0.6 it keeps on saying that I need to install GTK# 2.12.9 or newer even if I have already installed GTK# 2.12.22 (the latest one).
At the end I was really frustrated and decided to use SharpDevelop 4.3 as an alternative to M$ Visual Studio and MoNo-Develop.


Graph Theory proving planarity

Question: I have these set's of graphs:

I used Euler's inequality, and the 4 color theorem which resulted in a inconclusive result. Using Kuratowski's theorem I was unable to create a K3 3 or K5 graph. So does this prove that the graphs are planar? How do we really know that there does not exist a K3 3 or K5? I did use planar embedding and was able to put it in a form where the edges do not overlap, but the method does not seem rigorous enough to prove planarity.
Comment: The 4-colour theorem can only tell you that a graph is non-planar if you can prove that it has no proper 4-colouring, which is likely to be quite difficult. (For example, it's not true that all 4-colourable graphs are planar, since there are non-planar bipartite graphs, such as $K_{3,3}$.)
Comment: The Kuratowski / Wagner / Robertson-Seymour theorems are about forbidden minor graphs. For an embedding of a graph onto a surface with an Euler characteristic of 2 (i.e. a 2D plane) you can demonstrate planarity by creating a planar embedding and non-planarity by finding either $K_{3,3}$ or $K_5$ forbidden minor graphs. For graph $G$ you can delete vertex $f$ and edge $cg$ and you have a $K_5$ minor graph which demonstrates non-planarity. For graph $H$ you can move $b$ inside the polygon $acgfe$ and $d$ inside the polygon $aefg$ and you have a planar embedding.
Answer: Here's an illustration of a subgraph of $G$ which is a subdivision of $K_5$, so Kuratowski's Theorem implies $G$ is non-planar:

Here's a planar drawing of $H$, demonstrating that it's planar:

If it contained a subgraph which is a subdivision of $K_5$ or $K_{3,3}$, this planar drawing would not be possible.
Comment: Nice pictures!..
Answer: Both of these graphs have spanning cycles. There is an algorithmic way to check if a graph with a spanning cycle is planar using the concept of a "conflict graph".
Pick a spanning cycle, and draw the graph with that spanning cycle as a circle, and all the edges off the spanning cycle as chords of that circle. For the first graph, I chose the cycle a, b, f, d, e, g, c. My graph drawing software (Sage) is awkward with vertex labels, and so here I've named them respectively 1, 2, 3, and so on.

The conflict graph is a new graph in which the vertices are the chords of this drawing of your graph. Two chords are considered to "conflict", and are thus adjacent in the conflict graph, if and only if they cross in the above drawing. The idea is this. Let G be your graph and C the spanning cycle. Any planar embedding of G must represent C as some sort of a closed curve - not necessarily a circle, but some closed curve which will divide the plane into an inside and an outside. If two chords conflict, then they necessarily have to be drawn on different sies of this curve (one outside and one inside) in order not to cross.
The conflict graph here looks like this:

The theorem is: a graph with a spanning cycle is planar if and only if its conflict graph is bipartite. This is obvious given what we said above: we need a 2-coloring for the conflict graph using the colors "inside" and "outside". If the conflict graph is non-bipartite, two chords will necessarily have to be drawn on the same side of the cycle. In this case, the conflict graph is obviously non-bipartite since it has an obvious odd cycle.
You can read more about conflict graphs in Douglas West's book, Introduction to Graph Theory.
Comment: An answer where I learned a new generally useful principle?  Excellent, +1.
Answer: Graph $G$ is not planar. The subgraph you get by deleting the edges $ab,$ $ad,$ $bd,$ $cg,$ $eg$ is homeomorphic to $K_{3,3};$ the vertices $a,b,d$ are connected to the vertices $e,f,g$ by the internally disjoint paths $ace,$ $af,$ $ag,$ $be,$ $bf,$ $bg,$ $de,$ $df,$ $dg.$
Graph $H$ is planar. Plot $a$ at $(0,0),$ $b$ at $(2,0),$ $c$ at $(1,1),$ $d$ at $(1,3),$ $e$ at $(1,4),$ $f$ at $(2,3),$ $g$ at $(1,2),$ and draw all the edges as straight line segments.
Answer: In the first graph, there is a subdivision or minor of $K_5$ formed by $a$, $\{b,c\}$, $d$, $\{e,f\}$, $g$.
Therefore it is not planar.
The second graph has a planar embedding. This is the definition of a graph being planar, so I have no idea what you mean by whether it is not rigorous.

Graph Theory proving planarity


Can an H-1B holder work for a foreign institute? Does this affect the future application for permanent residency?

Question: I am an H1-B holder working at a university. I was recently invited to join a research project in Europe as an external investigator (I will receive a small amount of salary). I consulted with the immigration office in my university, the staff told me that this is a grey area, and this could affect my future application for permanent residency.
I wonder if anyone in this community had similar experiences? Any advice will be greatly suggested.
Comment: Your situation is odd enough that you should pay for a consultation  with an immigration attorney.
Comment: This being not academia specific, I think you'll get better answers at Expats.SE
Answer: Experience 1
I paid for a consultation with an attorney once over an immigration issue (seeking to obtain permanent residency).  The immigration official handling the case had basically said that if the proper documents were not all submitted exactly as prescribed that the whole process could be aborted and would have to be started again from zero.  The woman did not even detail what exactly she was expecting, leaving questions about why the documents submitted had not been adequate.  That's when I went to the attorney.
The attorney gave excellent advice, but it was of a more pragmatic/social nature than a legal nature.  He basically said two things:

The official was out of place and not following standard procedures to make such threats and provide no documentation to support what was being required; and

It would do no good to address the official's errors, but that, instead, one should approach the officials with a humble, asking-for-help, attitude, and let them decide how best they could help.  (He also warned against informing the official of having gone to an attorney.)

It boils down to this: officials are human; and many humans like the sense of power they derive by their position.  It is best to meekly submit, and not to challenge this.
Experience 2
On another occasion I called multiple officials trying to ascertain the validity of an advance parole document to return to the home country.  I ended up calling all the way to Washington, D.C. (headquarters) to learn that once in hand, the document could be used regardless of the reason--even though, technically, it is not supposed to be issued during the initial residence period (prior to permanent residency) for returning to one's native country.
This experience taught me that the officials themselves can sometimes seem at odds with each other in the interpretation of the legal codes.
Conclusion
Putting these two pieces of wisdom together, I advise you to seek your answers from more than one authority/official, if possible, and always in the attitude of a humble petitioner, letting them help you.


If I already have windows as separate partition, is there a way to run this with VirtualBox?

Question: I already have windows he installed on my computer in a separate partition. I would like to be able to use this with VirtualBox, but I'm unsure how to do this. Since my computer came with windows installed, I do not have a windows instillation disc. Is there a way to create an .iso file that contains my windows installation? There are a few programs under windows but I really would like to use, but they do not have a linux equivalent.
Answer: Yes its possible, to run windows (at real partition in your hd) in native and at virtualbox, but (we always have "but") its more problems than honey.

we have chance to catastrophic errors (boot active host linux as guest)
windows detect a new hardware (now its virtualbox hardware)
maybe you need re-register windows
and for sure need reinstall lot of drivers

I try it. Sure you have solution for each above points, but you get more problems than solutions. Its a lot easy install a new windows and re-install windows app you need.
(my 5 cents)
Answer: I am not sure if it is possible to virtualize a real partition using vbox or vmware but I am pretty sure there is a program installed in your windows partition that will make recovery discs for your notebook. Every notebook, that has windows preinstalled, also has an application to create recovery discs. This usually needs two DVDs and should be found under start menu > programs
You could use that to create a vbox image to run when in ubuntu.. Btw did you try to run those windows apps using wine?
Let us know what notebook do you have? manufacturer, model etc. also what programs do you want to run that you can't find equivalent in ubuntu?
Comment: I'm running an Acer Aspire One AOA-150. I have a digital pen (http://goo.gl/a3mP) that I can't get to run under linux. I can see in the logs that ubuntu registers that it is plugged in, but it doesn't recognize it. It seems that the software is Windows only. I have tried running it with Wine, mono, and crossover but so far no luck.
Answer: There's a guide on VirtualBox site for this topic.
I'd stick with dual boot option in your case - most likely the pre-installed Windows system you've got has lots of hardware specific programs which will be a real headache in virtualized environment. Besides that you might get issues with performance and/or USB device support.
Comment: Does answer partially, because the guide refer to a new virtual hard drive creation, not the direct use of the real windows partition.
Comment: For that specific case there's another piece of information[1]. Please note, that it is better to create a second hardware profile (as author suggests) to avoid problems when booting natively (not into VirtualBox) later again. [1] http://blarts.wordpress.com/2007/12/06/how-to-run-virtualbox-using-a-physical-partition-using-ubuntu-feisty-fawn/


Spherical completions and flatness

Question: Let $k$ be a non-Archimedean field. Does there exist a spherical completion $K$ of $k$ such that for any $k$-Banach space $X$, the natural map $X \to X \widehat{\otimes}K$ is an isometric embedding? If so I might also ask if it is possible that an exact admissible sequence $X \to Y \to Z$ of bounded morphisms of $k$-Banach spaces leads to an exact admissible sequence $X \widehat{\otimes}K \to Y\widehat{\otimes}K \to Z\widehat{\otimes}K$ (or even conversely also)?  Note: if we only demanded that $K$ is non-trivially valued then such a fact appears in Berkovich's book.
Comment: Just to be sure: you do not assume that the field $k$ is complete, do you?
Comment: Thinking a little further, I do not see how you could use anything from Berkovich's book if your base field is not complete. Then your statements would be classical I believe. Could you clarify this point please? (And maybe also provide a more precise reference to Berkovich.)
Comment: Dear Jerome, Yes, my base field is complete. I forgot to mention that.
Comment: I was talking about Proposition 2.1.2 from Berkovich's book. He shows that you can pass to a situation where the field is non-trivially valued. I am looking for a similar way to pass to the situation where the field is spherically complete.
Comment: Dear Oren, OK, thanks. I have other questions then :) I have to say I am a little puzzled by what you asked, since I thought that what you want holds for arbitrary extensions $K/k$. For example, the second part should follow from Gruson's article "Théorie de Fredholm $p$-adique" (http://www.numdam.org/numdam-bin/fitem?id=BSMF_1966__94__67_0). Am I missing something here? Do you have examples of extensions where you statements fail?
Answer: Let me try to give an answer. In fact, I think that the statements you want always hold, i.e. for any complete valued field extension $K/k$. Which makes wonder whether I have missed something obvious... Anyway, writing things up clearly should help us start a discussion, so I will give it a try.
As regards the first question about the isometric embedding $X \to X\hat{\otimes}_k K$, it can be deduced from lemma  3.1 (with $A=k$, $B=K$ and $C=X$) from my paper "Les espaces de Berkovich sont angéliques" in Bulletin de la SMF 141 (2), p. 267-297 (2013) (http://smf4.emath.fr/Publications/Bulletin/141/html/smf_bull_141_267-297.php). It eventually relies on arguments from BGR about the existence of $\alpha$-cartesian bases for finite-dimensional spaces over complete valued fields.
As I said in the comments above, the second question about the admissible exact sequence is dealt with by Gruson in "Théorie de Fredholm p-adique" Bulletin de la SMF 94, p. 67-95 (1966) (www.numdam.org/numdam-bin/fitem?id=BSMF_1966_94_67_0).
Comment: Dear Jerome, thanks very much. I still will probably want to speak about it with you more, but because it will be slow, I approved this answer in the meantime!

Spherical completions and flatness