Need Atmel 89C5131 microcontroler library for Altium designer v14.3.9

Question: I want to design a small electronic circuit using Altium designer v14.3.9, and I need the microcontroller Atmel's 89C5131 (the symbol, not for PCB), but couldn't find it in any of my libraries. Does any one know where can I get it?
Answer: I assume you are looking for a schematic symbol, and perhaps footprint.  If so, you should just make the required symbols yourself - then you will know that it matches the component you are using (and, for the schematic symbol, that it has the pins arranged (and named) in a way that will suit your use.)
I don't trust other people's libraries (even Altium's - I once found an IC symbol that bore no relation to the maker's datasheet!).


Transparent HTTP Proxy (Windows)

Question: I have SquidNT set up on my home network for HTTP caching and it's working great.  I want to use it as a transparent proxy server, though.  Googling tells me to use Softperfect Bandwidth Manager to do the port redirection stuff that I would normally use iptables for, but it costs $99 after 30 days.  I'm wondering if there is a FLOSS (or even freeware) app I can use to forward traffic outbound on port 80 to localhost:3128 on Windows XP?
Answer: Why don't you just setup squid cache on the target proxy machine? It works as a transparent proxy right out of the box (by sending x-forwarded-for headers) You can disable caching as you wish inside the config file.
Comment: We have Squid running.  We need a port forwarder to actually do the traffic capturing :)
Answer: You could try redirection server rinetd service available here http://www.boutell.com/rinetd/ . But am not sure it might do the stuff. You could give it a shot.
Answer: I'm doing the same thing on my server. I found Routix NetCom which does exactly that. The free version will handle the first 5 rules only, which should be enough for your needs.

Transparent HTTP Proxy (Windows)


Stopping desktopcouch and gwibber-service?

Question: powertop shows desktopcouch and gwibber-service are not-inconsiderable power-users. I don't use gwibber or Ubuntu One, so can I safely kill these processes? If so, how could I stop them starting up, especially if my laptop is on battery power?
Answer: If you don't use them, you could uninstall them both. DesktopCouch isn't used for much yet. I just checked my database, and it's only really used for Gwibber and Evolution's contacts (and an experimental build of Caffeine).
Comment: Tomboy doesn't use couch, when you connect it to Ubuntu One it uses the Snowy API that upstream tomboy uses.
Comment: DesktopCouch is also used for Tomboy notes.
Comment: This is what I've ended up doing.
Answer: Open the System menu, go to Preferences and select Startup Applications. You can scroll through the list and deselect the two "Ubuntu One" entries. You might also want to turn off the Bluetooth service if you don't use that.
Comment: Yep, I already had all of those turned off. Thanks anyway.
Answer: To disable gwibber-service, instead of removing the package you could just disable the autostart service:
echo "Hidden=true" | sudo tee -a /etc/xdg/autostart/gwibber.desktop

source: http://geekvigarista.com/os/linux/ubuntu-linux/removendo-o-gwibber-e-desabilitando-o-gwibber-service-no-ubuntu
Answer: I installed Ubuntu 10.04 on an x86-64 laptop and disabled Ubuntu One already from the "Startup applications". Yet I still have desktop couch processes running:

1886 ?        S      0:00 /bin/sh -e /usr/bin/couchdb -n -a \"/etc/couchdb/default.ini\" -a \"/etc/xdg/desktop-couch/compulsory-auth.ini\" -a \"/home/wirawan/.config/desktop-co
 1914 ?        S      0:00  \_ /bin/sh -e /usr/bin/couchdb -n -a \"/etc/couchdb/default.ini\" -a \"/etc/xdg/desktop-couch/compulsory-auth.ini\" -a \"/home/wirawan/.config/deskto
 1915 ?        Sl     6:38      \_ /usr/lib/erlang/erts-5.7.4/bin/beam.smp -Bd -K true -- -root /usr/lib/erlang -progname erl -- -home /home/wirawan -- -noshell -noinput -smp au
 1937 ?        Ss     0:01          \_ heart -pid 1915 -ht 11
 1790 ?        S      7:20 /usr/bin/python /usr/lib/desktopcouch/desktopcouch-service
 1976 ?        SN     8:00  \_ /usr/bin/python /usr/lib/desktopcouch/desktopcouch-service

Anyone having idea to disable this completely without uninstalling desktop couch stuff?
Comment: It looks like gwibber service is not disabled in my system. I think this one causes couchdb service to be invoked. How to disable that too?
Comment: (comment deleted. It's the same as above. Sorry for typing the same thing twice.)


DataPlotly - displaying percentages over bar graph (QGIS)

Question: Is there a way to display percentages above a bar plot? So its easier for viewers to determine how much of an environment is one type (another visual aid). We aren't supposed to show pie graphs. Alternatively, is there a way to display this in the legend(if the other option is not possible)? Otherwise I will go and manually do it for all the graphs. Also is there a way to colour the bar graph in accordance of the feature colour it already has?
Comment: Give a look to this https://gitlab.com/GIS-projects/qgis-geometry-generator-examples/-/tree/master/QML-files%2Fgraph_charts_with_labels
Answer: Give a look to this example with Geometry Generator created by Michel Stuyts.
You can also download a QML file that works with point layers.

Obviously, you can reuse the code to adapt to your geometry and playing a bit with the expression you can show percentage instead of value.
Answer: From DataPlotly 3.5 a new feature has been introduced (https://github.com/ghtmtt/DataPlotly/pull/203).

You can choose to display as text the additional hover label in the plot advanced configuration. You can also add an expression rather than a pure layer field, so you can use the aggregation functions of QGIS like <code>"my_field"/sum("my_field")*100</code> to get the desired result


Inequality regarding diameter, maximum order and number of vertices.

Question: Suppose I have a connected graph on $n$ vertices with maximum degree $x$. What is the minimum value of the Diameter $D$?
Comment: You might wish to look into [Moore graphs](http://en.wikipedia.org/wiki/Moore_graph).
Answer: I'm coming late to this discussion, and perhaps you already know the answer, but anywyay, here it is: Let $n$ be the number of vertices of the graph, $D \geq 2$ its diameter, and $\Delta \geq 2$ the maximum degree. If $\Delta=2$, then $D \geq (n-1)/2$. For $\Delta > 2$ we have that $$D \geq \log_{\Delta-1}[(n-1)(\Delta-2)+\Delta] - \log_{\Delta-1}\Delta.$$ This comes by solving for $D$ in the Moore bound (see "Moore graphs and beyond: A survey in the Degree/Diameter Problem", by Mirka Miller and Josef Siran, Elec. J. of Combinatorics, Dynamic Survey 14).
Answer: If $G$ is k-regular, then the diameter is $log_{k}(n)$. So with a maximum degree, we get a lower bound as to the diameter using the same approximation: $diam(G) \geq log_{x}(n)$.


Bird flying in an open or a closed cage

Question: Assume that you are holding a cage containing a bird.  Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage?
Does it make a difference whether the cage is completely closed or it has rods to let air pass?
Comment: Are you asking if the weight of the cage + bird is reduced if the bird is hovering in the middle of the cage?
Comment: @DImension10 Abhimanyu PS: When OP is just asking for a conceptional qualitative (as opposed to a quantitative) understanding of a physics phenomenon, we usually don't tag it as 'homework' (unless there are other indications that it is indeed homework). That said, I'm pretty sure it is a duplicate. Ah yes, here we go: http://physics.stackexchange.com/q/12756/2451
Comment: @Qmechanic: I tagged it as such, because it seemed to be pretty common to get this as homework, even if it is   qualitative.
Comment: M. Tarun, see https://www.youtube.com/watch?v=lVeP6oqH-Qo
Answer: If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. Since no additional external force is acting on the cage-bird-system when the bird is flying as compared to when it's not, the acceleration on the cage can be no different. The effect due to the flying bird only concerns the internal forces and since action=reaction, they cancel.
However, if the cage would not be closed, some of the 'wind' due to the bird could escape the cage and would become an external force, making the cage-bird-system lighter.
Comment: Maybe this will make it more believable: If you sit on a skateboard and you have a hair dryer with you, it will cause you to accelerate. But the hair dryer is not just expelling air, it's also sucking air in. Still, you accelerate.
Comment: Well, of course air from the environment would go back into the cage, but it does not (substantially) alter its weight. Careful, you are confusing weight (the force pulling something down which may alter) with mass (an invariant).
Comment: Wouldn't air from the environment replace the vacuum created by the 'escaping wind' thus the keeping the mass of the air in the cage constant and not make the bird-cage-system ligter?
Answer: To answer the second question first, a completely closed cage is a system which comprises the exterior cage, the air inside and the bird. So what all forces do we need to balance now? Gravity on air, bird and the cage. OK, So can the bird apply an additional force if it sits or swings or flap around the walls of the cage? And the answer might be against intuition but it is NO! Let's imagine if it was a 'yes'. If the bird could apply a force on it's own closed system (or internal force) it could accelerate its own system as long it desires. Now this isn't right, is it? Otherwise space ships could just propel themselves by their travelers constantly kicking on the inside!The air inside the cage does two things it expands and contracts (changes pressure) when the bird moves around or flaps its wings. So when the bird pushes some air towards the bottom of the cage the cage gets a force (a small impulse) which is immediately redrawn by another force now, on the top of the cage, by the elastic air that obeyed Newton's third law. In fact the bird itself will feel this recoil (that's why a bird can fly) and if the flap was strong enough it will hit the cage's top and does the same as above. So all you'll feel is an up and down movement of the cage! This is just a rearrangement of masses inside the cage, but no net forces are acting on the center of mass.Now, for an open system external forces comes in to action. To put it simply, the bird will draw some air to the inside and this new air will apply new force that you will feel as a pull (since the air could now escape through the openings, the case above is not valid!)Buoyant force has nothing much to do here except that it is what slightly reduces our weight (or the bird's) due to pure gravity by pushing us upwards since we are displacing the medium (air) we are in (Archimedes, eureka moment!) Thus the bird need to apply a force slightly less than it's gravitational force by earth in order to fly, but as I said no big deal here. Now all this would get even more interesting by imagining a balloon instead of a bird. Think about it and tell me what you think.
Answer: the weight of the cage will not change when a bird does anything inside it because it becomes an isolated system and all the force applied by the bird will be counted as internal force...................however when the cage is open air is free to come in and go thus affecting the weight of the cage ........in the second case momentum is not conserved....
Answer: Some of the above discussions are incorrect; bird accelerating changes weight of cage/bird for bird in isolated cage sitting on a weight scale due to acceleration of center of mass. Assume closed cage.  System is cage and bird.  External forces are gravity (mass bird + mass cage)x g down and force of constraint holding cage up against gravity (e.g. cord holding cage off ground or weight on a scale with cage/bird on the scale on the ground).  The force of constraint is the "weight" of the cage/bird system.  Key is (basic physics) "Sum of external forces = mass of center of mass x acceleration of center of mass"; internal forces do not matter.  As long as bird is not accelerating (e.g. stationary or hovering bird), then (mass bird + mass cage)x g down =  force of constraint holding cage up and weight (force of constraint) is not changed.  If bird is accelerating up or down the center of mass is accelerating up or down and there must be a net external force; this net external force is due to a change in force of constraint (increases if bird accelerating up so weight of system increases).  See Halliday and Resnick/ Physics (best physics book ever written).
Comment: This doesn't add much to existing answers.
Answer: Whether the cage is enclosed or open, makes no difference. So long as the bird is not close to the top or bottom of the cage (when "ground effect" would come into play) the wings circulate air around the bird/cage. There is no net wind force either up or down - just recirculation.  The bird doesn't need the ground to "push against"! It just needs to create a local pressure differential. Therefore, so long as ground effect is negated, the cage will get lighter when the bird flies. (the bird itself will have a cyclic "weight" according to whether its wings are going down or up)
Comment: In a closed cage, what happens to the momentum of the molecules that your bird pushes down in order to stay midair?
Comment: Probably lost as heat due to the recirculation - obviously the energy has to go somewhere.
Comment: I mean the momentum, not the energy. The momentum must be conserved. It will be transferred to the bottom of the cage (via other molecules), and thus increase its weight by exactly the same as the bird weighs. So a closed cage will not decrease in weight because the bird is flying. You answer is only correct for the open cage, where the ait can escape.

Bird flying in an open or a closed cage


Uniqueness of quasiclassical consistent histories

Question: The current zeitgeist here is on interpretations of quantum mechanics, so let me add my own two cents here. As you may know, consistent histories is an alternative interpretation proposed in a series of papers 1990-1994 by Gell-Mann and Hartle. In a shocking article, Dowker and Kent proved there are many distinct consistent histories for most quantum systems. Even if we constrain ourselves to quasiclassical histories, i.e. those histories which are close to classical, being quasiclassical now does not guarantee being quasiclassical in the future, or in the past. This is disturbing. Do histories which are quasiclassical for all times exist? If they do, are they unique? Is the nightmare scenario of two everywhere quasiclassical consistent histories which are mutually incompatible possible?
Comment: Much of the "current zeitgeist" is to close a lot of vague philosophical "questions". Just sayin'.
Comment: @dmckee, I know the consistent histories literature very little, but my understanding is that the question of whether quasiclassical histories exist for all times and whether they are unique are detailed mathematical questions, they are not philosophical. One can ask how Useful the histories approach is, as one can of any math, but it is more Mathematics than Philosophy. I suspect these are "Hilbert"-style questions, too difficult to be Answered here except insofar as they have already been addressed in the literature, but I recommend this should not be [closed] for that. This shows effort.
Comment: @Peter: Indeed, I have nothing against this question aside from the only possible answers being literature references. But I *do* want to combat the idea that throwing in interpretations of quantum mechanics automatically qualifies questions to stay open. Must of them seem to be ill-informed bunkum.
Answer: Dear Billy, it is not quite clear why you add the emotional words such as "shocking" and "disturbing". It is a manifestation of the essence of quantum mechanics that the classical limit or the spectrum of allowed questions is not God-given. The classical limit occurs in situations that depend on the state of the system as well as its Hamiltonian. The process of decoherence also depends on these two things.
So it shouldn't be surprising that properties that behave classically generally cease to behave classically in the future (or ceased/started in the past). Consider an analogy in statistical physics and thermodynamics. The thermodynamics description works if there are many particles etc. But you may produce lots of new particles e.g. by the LHC proton-proton collisions. Thermodynamics is not applicable to the initial state of 2 particles - too few - but it is useful to describe the quark-gluon plasma and other things that may be formed.
The case of decoherence and consistency of histories is analogous. These things are simply not eternal and God-given. Classical limits only work in some situations, the situations are described by some inequalities involving values of parameters, and because parameters evolve with time, so does the applicability of the classical limit - and so does the consistency of various histories.
It's really a feature of quantum mechanics, not a vice.
Concerning the non-uniqueness, the situations in which the questions are non-unique are unrealistic, but yes, in principle, quantum mechanics allows one to ask different types of questions with different alternative answers that can't be reduced to the same fine-grained set of alternatives. Why do you think it's bad? Again, the right questions depend on the Hamiltonian and the state. They're not God-given. And classical physics isn't a "skeleton" of quantum mechanics that quantum mechanics is only allowed to "slightly disturb". The world is quantum mechanical and quantum mechanics is a qualitatively different description than classical physics, and classical physics is a limit. So classical limit is not a skeleton of quantum physics you may rely upon; it is the sweat on the skin, a derived product that only works on the boundaries of the parameter spaces.
I recommend you a paper by Zurek

http://arxiv.org/abs/quant-ph/0105127

that addresses and summarizes many issues. Search for "Dowker" in the paper above. The paper has 1,000+ citations. You will see that there's really no problem in understanding classicality etc.


Why does my bass buzz so much when plugged in

Question: I plug my bass in and it starts buzzing right from the start, it plays, and sounds great, but buzzes...I tried different cords, and even different amps, same results..I plugged another of my basses in, and no buzz!!!
Comment: Can you give more details? Is the buzzing loud or quiet? Is it 60 cycle hum or something more distressing? Does it change when you touch the strings, wiggle the cable, turn the knobs, rotate your body left to right? Does it go away if you select both pickups together?
Comment: Need much more detail.
Comment: It's likely a dirty or corroded jack. I've been told that TV tuner cleaner will help but I have no idea where such a substance can be obtained.
Answer: I believe you've isolated the problem to the instrument itself. Has it always been this way? It would help to at least know the instrument and pickup configuration. If it is a single coil pickup (or two pickups and you're only using one of them) you will get a hum or buzz like @luser droog mentioned in his comment. This type of hum can be fairly loud but not as much as a live instrument cable unplugged from an amp. If it is that loud then you need to check the internal electronics for dirt and loose connections as well as the jack like @Don Hosek said. If the jack is just dirty some alcohol on a Q-tip might do the trick. If it's corroded then the corrosion needs to be scraped off or the jack replaced.
You can also pick up interference from lights and other electric and electronic devices if the instrument isn't properly shielded on the inside. Turn the instrument in different directions and move around the room. If the intensity of the buzz varies adding foil tape inside the cavity might help.
Comment: @S.Imp cheap cables and using a speaker cable instead of an instrument cable can lead to problems like you say but I don't think that's the case here since OP said he tried different cables and that the sound with another bass was clean.
Comment: A cheap cable can also be the cause. a lot of times, people confuse a speaker cable for an instrument cable.


Has Russian to Spanish transliteration changed much over the centuries?

Question: I want to know which norms might have governed the spelling of some Russian names which were written down in Spanish around around 1750 - 1850. A number of formal standards exist today, such as ISO 9, but how would it have been done a couple of centuries ago?
Comment: @A.M.Bittlingmayer: I think, the most sensible place to ask for transliteration is the stackexchange dedicated to the target language. This is where you have a large proportion of native speakers and readers that have seen transliterations of different styles. Also the governing authorities of the target language tend to issue transcription rules (the main exception is CHina, where the source language country dictates transcription).
Comment: Maybe it is a good idea to ask for this on [spanish.se]
Comment: @jknappen Would we ask a question about Russian transliteration (into English, implicitly) on English SE or Russian SE?
Answer: Yes, although I'm afraid to say that then, as today, healthy chaos reigned.  Transliteration conventions depend as much on other aspects of the context - the writer's familiarity with Russian, the intended audience, the keyboard or press available - as on the year.
So which conventions do correlate with year?  These days, we can find the English convention 'sh' being used more, and also the letter 'k' in places where 'c' would work, and 'y' in 'Krylov'.  We also see 'y' used in ways that correspond to Latin American pronunciations.
For example, Pushkin vs Puchkin:

Some aspects are still open questions - hence Aseev/Aseyev/Aseiev, Yershov/Ershov - and it is likely that the precise balance has shifted over time.
You can select from Rusos del siglo XVII and search variants of their names in the Google Books n-gram viewer for Spanish books from 1800 on.
If we go back a few centuries further, then 'x' had yet another value, and English was even less dominant.
Transliteration aside, in the past names were more frequently simply translated, hence Luis Adolfo Pedro de Sayn-Wittgenstein, José Stalin etc.


Invariants of the maximal unipotent subgroup of GL(n) acting on the space of n by n matrices

Question: Let $G=GL(n,\mathbb{C})$ and let $U\subset G$ be a maximal unipotent subgroup. (For example,assume that U is the set of upper triangular matrices with ones in the diagonal.) Now let $X=M_{n}(\mathbb{C})$ be the space of $n$ by $n$ complex matrices and consider the usual action of $U$ on $X$ given by left multiplication. Is it possible to describe the space $\mathbb{C}[X]^{U}$ explicitly using generators and relations? In other words, is there an explicit realization of the categorial quotient of $X$ by $U$ as an affine algebraic variety?
Comment: Just for clarification: is $\mathbb{C}[X]$ the polynomial ring in $n^2$ variables? (It can be the group algebra as well...) And $\mathbb{C}[X]^U$ is the invariants under $U$?
Comment: @WhatsUp Yes, it is the polynomial algebra.
Comment: When you ask about the "categorical quotient", in what category, precisely, are you working?  The categorical quotient in the category of schemes is not an affine scheme (just consider the case when $n$ equals $2$).  You might want to look up "Matsushima's criterion" (extended to positive characteristic by Richardson).
Comment: @Jason Sorry I was imprecise. What I'm looking for is a description of $\operatorname{Spec} \mathbb{C}[X]^{U}$
Comment: @RaulGomez. "What I'm looking for is a description of $\text{Spec} \ \mathbb{C}[X]^U$."  Are you looking for a description as a subvariety $Z$ of an affine space?  I gave that below: up to the $\mathbb{G}_m$-factor $Z^n$, $Z$ is just the affine cone over the flag variety.
Comment: @JasonStarr You are right. Sorry it took me this long to digest your answer.
Answer: Here is a coordinate-free description of the affine hull of the quotient of $G=\text{Aut}_k(V)$ by the right action of the unipotent radical $U$ of a Borel subgroup $B$.  The group $B$ is the stabilizer of a flag of $k$-linear subspaces, $$\{0\} = F^0 \subsetneqq F^1 \subsetneqq \dots \subsetneqq F^r \subsetneqq F^n = V.$$  For $r=1,\dots,n-1$, denote by $H^r$ the affine $k$-space with underlying $k$-vector space $\text{Hom}_k(\bigwedge^r F^r,\bigwedge^r V)$.  Denote by $Z^r\subset H^r$ the affine cone over the Grassmannian $\text{Grass}(r,V)$, i.e., the homogeneous, Zariski closed subset parameterizing all linear maps of the form $\bigwedge^r T^r$ for a $k$-linear morphism $$T^r:F^r \to V.$$ The defining ideal of $Z^r$ in the (polynomial) coordinate ring $k[H^r]$ is generated by explicit Plücker quadratic relations, cf. Griffiths-Harris (for instance).  Denote by $Z^n$ the open complement of the origin in the $1$-dimensional affine space $\text{Hom}_k(\bigwedge^n F^n, \bigwedge^n V) =\text{End}_k(\bigwedge^n V)$.
Inside $Z^1 \times_k Z^2 \times_k \dots \times_k Z^{n-1}$, let $Z'$ be the affine cone over the flag variety $\text{Flag}(1,2,\dots,n-1;V)$.  This is cut out by explicit equations similar to the Plücker relations.  For instance, for every $r=2,\dots,n-1$, one relation on $(z^1,\dots,z^{n-1})$ is that $z^1\wedge z^r$ equals $0$ as an element in $\text{Hom}_k(\bigwedge^1F^1\otimes_k \bigwedge^r F^r,\bigwedge^{r+1}V)$.  I do realize that you want an explicit list of relations, and I am not quite giving these.  However, you can quickly find the explicit relations by searching for "Cox ring" and "flag variety".
Anyway, the affine hull of the quotient $G/U$ of the right action of $U$ on $G$ is $Z = Z'\times Z^n$.  The quotient morphism is $$q: G \to Z, \ \ q(T) = \left( \bigwedge^1(T|_{F^1}), \bigwedge^2(T|_{F^2}),\dots, \bigwedge^{n-1}(T|_{F^{n-1}}),\bigwedge^n(T|_{F^n}) \right).$$
Although $Z$ is affine, the morphism $q$ is not surjective.  In particular, an element $(z^1,\dots,z^{n-1},z^n)$ is in the image if and only if $z^r$ is nonzero for every $r=1,\dots,n-1$.  This is the point I was making in my comment above.
Edit. I checked a textbook on the combinatorics of these kinds of commutative rings.
Combinatorial Commutative Algebra
Ezra Miller, Bernd Sturmfels
Graduate Texts in Mathematics (Book 227), 420 pages.
Springer.
ISBN-10: 0387237070
On p. 275, Definition 14.2, they define the ring $k[Z']$ to be the Plücker algebra.  It is generated by the linear bases for the dual vector spaces of $H^1,\dots,H^{n-1}$, which total $2^n-2$ variables.  The ideal of relations is generated in degree $2$.  Most of Chapter 14 of that book is devoted to describing these quadratic relations.
Comment: I just have a small question regarding your answer. It seems  to me that you should be able to extend the morphism $q$ that you are defining above to the whole of $\operatorname{End}_{k}(V)$. Is this extended map then surjective?
Comment: @RaulGomez: "Is this extended map then surjective?"  No, the extended map is not surjective.  For instance, the element $(q^1,q^2,\dots,q^{n-1},q^n) = (0,0,\dots, 0,1)$ is not in the image.
Answer: Questions of this type can be approached from the direction of algebraic geometry (in a suitable generality) or from the direction of invariant theory involving algebraic group actions.   I'd point especially to Chapter 3 in the 1997 monograph by Frank D. Grosshans, Algebraic homogeneous spaces and invariant theory, Lect. Notes in Math. 1673, Springer.   He works in some generality throughout, taking $k$ to be any algebraically closed field of any characteristic and using "traditional" algebraic geometry language.  In particular, he looks at the case of general linear groups in $\S13$ of that chapter and describes (for either left or right action) the rings of invariants for $U$ on suitable algebras of polynomials.
Grosshans builds on older foundations for reductive group actions on varieties and resulting homogeneous spaces, going back to Hilbert, Mumford, and others.  His results rely a lot on combinatorics and are often more comprehensive than what you ask for.   On the other hand, he is fairly straightforward and careful about the details.  Though he works in arbitrary characteristic, his approach is not too different from that of the Russian school (Vinberg, Popov, Panyushev, ...) who typically work over an algebraically closed field of characteristic 0.  Along the way, Grosshans does provide a fairly detailed concrete picture of the quotients obtained, starting with the generators of rings of invariants.


Is there a bijective function $f: \mathbb{R}\to\mathbb{R}$ that is discontinuous?

Question: Is there a bijective function that is discontinuous?
Comment: Take a continuous bijection and exchange the images of two random points.
Comment: It would be more interesting to ask for a everywhere discontinuous bijection
Answer: Discontinuous everywhere: $f(x)=x$ if $x$ rational, $x+1$ if $x$ irrational.
Comment: To make this an even more fascinating example: If we use $-x$ and $1-x$ instead of $x$ and  $x+1$, then $f$ is discontinuous everywhere and $f\circ f$ is the identity.
Comment: @HagenvonEitzen Nice variation, since it gives a compositional inverse pair of maps with each everywhere discontinuous.
Comment: @HagenvonEitzen Does this work if we make $f(x)=-x$ if $x$ algebraic and $1-x$ if $x$ transcendental?
Answer: $$f(x) =\left\lbrace \begin{array}{ll}
x & \text{ if } x\not\in\lbrace 0,1\rbrace \\
1 & \text{ if } x=0 \\
0 & \text{ if } x=1
\end{array}  \right.$$
for example.
Answer: Yes. In fact, one can construct a maximally disconnected function in the sense that it is discontinuous on every open interval $(a, b)$.
This is constructed by having every open interval $(a, b)$ map to $\mathbb{R}$. An example of this is Conway's base 13 function
Another example is the indicator function for rational numbers.
Define a function
$$
f(x) = \begin{cases} x \in \mathbb{Q} & 1 \\x \notin \mathbb{Q} & 0\end{cases}
$$
This function is discontinuous everywhere because the rationals are dense in the reals, and so are the irrationals.
Comment: Also if in the first one every open interval maps to $\mathbb{R},$ it won't be a bijection, at least if these maps are each onto.
Comment: @CarlMummert The version you have would be continuous at $0.$ But that could probably easily be adjusted for.
Comment: Is the indicator function a bijection?
Comment: No, but $g(x) = (-1)^{f(x)}\cdot x$ is a discontinuous bijection, the identity on the irrationals and $-x$ on the rationals


What is the health bar in Halo games?

Question: In most Halo games, there's usually a "shield capacity bar" which depletes with damage. When this bar is fully depleted, there's a secondary bar that starts to deplete. What is this bar? What does it represent? Is it the physical health of Master Chief? Is it the strength of the armor? In either case, how does it regenerate?
Comment: @TimmyJim I was looking for a general lore based answer. I didn't see any tags for lore. Something like what Rapitor posted. It's not really specific to any one game, but it might differ between armor models.
Comment: If I recall, this would vary from game to game in the series. Halo CE for example had an actual health bar that you needed medkits for to restore, while most of the other games did not have medkits at all. My point being though - are you asking for a specific Halo game or all of them? It's not very clear. I see you tagged "halo" which  doesn't have any guidance, so it's not clear. The Master chief collection has multiple games too.
Answer: Short answer: it's the health of master chief and the armor is healing him(in some games) with sci-fi biofoam.
Long answer: This question kind of depends on the game, CE, Reach, 5, and Infinite have a seperate health display while 2, 3, and 4 do not.
In Halo CE, your health does not regenerate on its own and you need to collect medkits throughout the game. At this point, the Mjolnir armor system is MkV.
In Halo 2,3, and 4, the Mjolnir armor system is MkVI, which has Biofoam injectors built into the suit. Capable of keeping the Spartan alive continuously.
In Halo Reach, being a prequel, we're back in MkV... but your health regenerates partially, and needs a  med kit for full recovery. Lore wise is explained as the MkV armor Noble Team uses is modified... game reason is its just more "modern".
In Halo 5, you have a visible health bar regenerates fully on its own using MkVI Gen 2. We're in MKVI so we got the biofoam injectors...
Halo Infinite is a bit of a wild one, since you are able to choose different Mjolnir Systems as part of the customization, but in "actuality"... Master Chief is in MKVI Gen 3 armor while Spartan IVs (your own MP character) is suited with MKVII, both of which equipped with the same internal biofoam and operating systems.

What is the health bar in Halo games?


Is it possible for a long carbon chain to have both acidic and basic groups?

Question: I was reading about a zwitter ion and I had a thought that is the same possible for an acidic and basic groups. Like a very long carbon chain with a basic group and an acidic group on either sides, I am guessing it'd be called amphoteric but amphoteric compounds themselves do not contain acidic and basic group, so is it possible to synthesis one? And if it is done already what is it called?
Answer: There is no reason why basic and acidic groups could not coexist being attached to different carbons of the carbon chain, when they can coexist on the same carbon of aminoacids $\ce{H2N-CHR-COOH}$, occuring free or condensed in proteins in living cells.
These aminoacids exchange protons via $\ce{H+(aq)}$, forming zwitterions $\ce{H3N^{+}-CHR-COO-}$, neutralizing themselves.
For longer chains, consider e.g. $\ce{H2N-(CH2)_n-COOH}$ with an acidic and a basic group at the ends (simple example is GABA, $\gamma$-aminobutyric acid, a natural neurotransmitter where $n=3$). similarly as above, in water solution they would exist as zwittterions. In strongly acidic/basic solutions, only amino/carboxy group would have charge, respectively.
For $n=5$, it is epsilon-aminokaproic acid with an acidic and a basic group. Its cyclical lactam with amidic bond $\ce{-CO-NH-}$ is used as the monomer for the original polyamide Silon $\ce{[-NH-(CH2)_5-CO-]_n}$, with more simple production, compared to the Nylon-66. It was produced in the post-WW-II Czechoslovakia. For the curiosity, it was invented by famous Czech organic chemist Otto Wichterle, who invented also soft contact lenses.
Analogically, the above occurs for other acidic and basic groups, e.g.: $\ce{R1-NH-R2-SO3H ->[H2O]R1-NH2^{+}-R2-SO3-}$.
Comment: Can you cite a reference for Wichterle as well as inventor? If Wikipedia is to be believed both IG Farben and Toray first developed nylon 6 in 1938. Wichterle is not mentioned
Comment: @Andrew https://en.wikipedia.org/wiki/Otto_Wichterle // https://en.wikipedia.org/wiki/Nylon_6 // https://en.wikipedia.org/wiki/Silon // (CZ2EN)  https://cs-m-wikipedia-org.translate.goog/wiki/Silon?_x_tr_sl=cs&_x_tr_tl=en&_x_tr_hl=cs&_x_tr_pto=wapp
Comment: those pages give a date of 1940 for Wichterle's nylon work. Farben's patent was filed in 1938: https://patents.google.com/patent/US2241321A/en
Comment: It could be parallel work. He worked on it secretly to hide it from nacists. Perhaps he improved the synthesis. I assume he was not in position and circumstances to get it patented.


Justification for using geometric weights in linear regression

Question: In practical application, I have witnessed often the following practice. One observes a pair  $(x_t, y_t)$ over time. Under the assumption that they are linearly related, we regress one against the other using geometric weights rather than uniform ones, i.e., the OLS minimizes
$$\sum_{t=0}^\infty k^{t} (y_{T-t}- a x_{T-t}-b)^2$$
for some $k\in (0,1)$. This is very intuitive: we weight less observations far in the past. Compared to a "boxcar" weighting scheme, it has also the advantage of producing estimates that are changing smoothly over time, because observations do not fall abruptly off the observation window. However, I wonder if there is a probabilistic model underlying the relationship between $x_t$ and $y_t$ that justifies this choice.
Comment: I doubt there is a formal derivation, hence the quotes around poor man's version of adaptive parameters.
Comment: Just the other day someone somewhere on one of the related StackExchange sites was commenting on this scheme as "poor man's Kalman filter".  If I manage to unearth the link I will add it here.
Comment: Thanks. I would like to see how this can be reframed as a Kalman filter.
Answer: "Linearly related" usually means
$$y_t = a x_t + b + \varepsilon_t$$
for constant $a$, $b$ and iid random errors $\varepsilon_t$, $t=0,1,\ldots,T$.  One reason one would make an exponentially weighted OLS estimate is the suspicion that $a$ and $b$ might themselves be (slowly) varying with time, too.  Thus we really think the correct model is
$$y_t = \alpha(t) x_t + \beta(t) + \varepsilon_t$$
for unknown functions $\alpha(t)$ and $\beta(t)$ which vary slowly (if at all) over time and we're interested in estimating their current values, $a = \alpha_T$ and $b = \beta_T$.  Let's assume these functions are smooth, so we can apply Taylor's Theorem.  This asserts that
$$\alpha(t) = \alpha(T) + \alpha'(t_{\alpha,t})(t-T)$$
for some $t_{\alpha,t}, 0 \le t_{\alpha,t} \lt T$, and similarly for $\beta(t)$.  We think of $a$ and $b$ as being the most recent values, $\alpha_T$ and $\beta_T$, respectively.  Use this to re-express the residuals:
$$y_t - (a x_t + b) = \alpha'(t_{\alpha,t})(t-T)x_t + \beta'(t_{\beta,t})(t-T) + \varepsilon_t\text{.}$$
Now a lot of hand-waving needs to occur.  We will consider the entire right hand side to be random.  Its variance is that of $\varepsilon_t$ plus $x_t^2(t-T)^2$ times the variance of $\alpha'(t_{\alpha,t})$ plus $(t-T)^2$ times the variance of $\beta'(t_{\beta,t})$.  Those two variances are completely unknown, but (abracadabra) let's think of them as resulting from some kind of (stochastic) process in which possibly systematic (not random, but still unknown) "errors" or "variations" are accumulated from one time to the other.  This would suggest an exponential change in those variances over time.  Now just simplify the explicit (but essentially useless) expression for the right hand side, and absorb the quadratic terms $(t-T)^2$ into the exponential (since we're waving our hands so wildly about anyway), to obtain
$$y_t - (a x_t + b) = \delta_t$$
with the variance of $\delta_t$ equal to $\exp(\kappa(t-T))$ for some constant $\kappa$.  Ignoring possible temporal correlations among the $\delta_t$ and assuming they have Normal distributions gives a log likelihood for the data proportional to
$$\sum_{t=0}^{T} k^{-t} (y_{T-t}- a x_{T-t}-b)^2$$
(plus an irrelevant constant depending only on $k$) with $k = \exp{\kappa}$.  The exponentially weighted OLS procedure therefore maximizes the likelihood, assuming we know the value of $k$ (kind of like a profile likelihood procedure).
Although this entire derivation clearly is fanciful, it does show how, and approximately to what degree, the exponential weighting attempts to cope with possible changes in the linear parameters over time.  It relates the parameter $k$ to the temporal rate of change of those parameters.
Comment: I agree on the hand-waving part... I am fine with simplifying assumptions on the time-varying form of the regression parameters, as long as they are clearly stated. Of course feel free to reference existing literature.
Comment: @whuber - I would say that the exponentially weighted regression is a very crude approximation *for the particular model you have described*.  But it could well be an exact solution to a different model.  For the model you describe it would be much better to include the heteroscedastic component due to variation in $\alpha(t)$ (or assume that it has no variation, and you are dealing with random intercept).  You are making it seem as though geometric weighting is always sub-optimal, which is isn't.  It depends on your prior information.
Comment: @prob I agree, but I haven't been able to find a model that exactly justifies this approach, so I had to settle for pointing out some of the the things such a model might entail.  I notice your reply doesn't make any progress in this direction, either ;-).
Comment: @whuber - and where do I make an approximation in my equation for it not to be exact?
Comment: @probability You provide no justification: You simply announce the result I had already posted.  In other words, you observe that when OLS minimizes such an expression it's really doing weighted least squares.  OK, but isn't that perfectly obvious?  What justifies this choice of weights?  Where do they come from?
Comment: @whuber - I do say what justifies the weights - your prior information that the noise or error in the observations becomes large as you move backward in time from $T$.  How is this not any more justified than your claim that "the coefficients vary with time"?  Your solution talks about variability in the co-efficients, and mine talks about the variability in the noise level.  I can't see how your approach can be justified any more than mine - you make no explanation of what justifies the choice of $k$ in your example either (and neither do I).
Comment: @Prob the question is, why?  What conceivable conditions would have to apply for the noise to increase backwards in time?  My analysis, non-rigorous as it is, nevertheless indicates the kinds of assumptions and gross simplification that may be needed.
Comment: @whuber - if you apply the same argument to your own line of reasoning, you will find the same response "because my prior information indicates this is the case".  This is the same answer as you would give for "slowly varying" co-efficients.  One more concrete example I can give is if you are dealing with data which is of varying quality due to continuous improvement in the data collection process, thus leading you to expect less noise as time progresses.  Another is when the measurement process improved over time (you upgrade your measuring equipment each year)
Comment: @prob But why a geometric sequence of weights?  Where does that come from?
Comment: @whuber - but why a choice of time varying covariates? where does that come from?  My answer to this question is the same as yours
Answer: I think in that you actually mean $k^{t}$ as your weight, or that $k>1$.  If $0<k<1$ and we take $k^{-t}$ as the weight then $k^{-\infty}=\infty$.  So this actually weights the present observation the least.  For example, if we take $k=0.5$ then $k^{0}=1,\;k^{-1}=2,\;k^{-2}=4,\dots,k^{-20}\approx 10^{6}$, and so on.
This is just stating something that you know about the how the variance changes with each observation (it gets bigger as you mover further backward in time from time $T$):
$$(y_{T-t}|x_{T-t},a,b,k,s) \sim Normal(ax_{T-t}+b,s^{2}k^{-t})$$
Denoting $Y\equiv\{y_{T},y_{T-1},\dots,y_{1}\}$ and $X\equiv\{x_{T},x_{T-1},\dots,x_{1}\}$ we have a joint log-likelihood of:
$$\log\left[p(Y|X,a,b,k,s)\right]=-\frac{1}{2}\left(T\log(2\pi s^{2} k^{-t})+\sum_{t=0}^{T-1}\frac{(y_{T-t}-ax_{T-t}-b)^{2}}{s^{2}k^{-t}}\right)$$
So in order to get the maximum likelihood estimates of $a$ and $b$ you have the following objective function:
$$\sum_{t=0}^{T-1}k^{t}(y_{T-t}-ax_{T-t}-b)^{2}$$
Which is the one you seek


Why don't free electrons escape from a conductor?

Question: The thermal velocity of the free electron in a metallic conductor varies from $10^5\ \mathrm{m/s}$ to $10^6\ \mathrm{m/s}$. In spite of high velocity, free electrons fail to escape from the metallic surface. Why is that?
Comment: They do escape if the temperature is high enough; it's called [thermionic emission](https://en.wikipedia.org/wiki/Thermionic_emission) and it has quite a few industrial uses.
Comment: It is not just the magnitude of the velocity (which you have emphasized), but also its direction.  If the electron has a high velocity in the direction of the conductor, it will never "escape" from the conductor.  To escape, the velocity must have a component perpendicular to the conductor!
Comment: Is that velocity from $\sqrt{k_bT/2m}$?  That is assuming a classical gas which occurs at a density of electrons much less then that in a solid.
Comment: Do they not also escape if you happen to provide them with an "easier" path to ground than what's offered by the conductor?
Answer: Electrons are bound to the metal by the attraction of the nuclei.   After screening of the nuclei by other electrons in the metal, there is a net electric field creating a potential barrier for the electrons to escape.
This potential barrier is called the work function and is defined with respect to the Fermi energy of the electrons.  The work function is usually around a couple electron volts, while the fermi energy is usually around $10\,\mathrm{eV}$.  This means there is around a $12\,\mathrm{eV}$ potential barrier for the electrons to over come before they can escape.
On a temperature scale, the fermi energy corresponds to $T=E/k_b\approx10\,000\,\mathrm{K}$
This means at room temperature, and up to a couple $1000\,\mathrm{K}$, the "free" electron gas is a degenerate fermi gas, and the electrons rarely have energy greater then $10\,\mathrm{eV}$.
The relevant velocity in degenerate electron systems is the fermi velocity or the velocity of the fastest electron in the degenerate electron gas:
$$v_f=\sqrt{E_f/2m}=\sqrt{10\,\mathrm{eV} ~/~ (2\times 0.5\,\mathrm{MeV}/c^2)}\approx 10^{-3}c,$$
which is even faster than the velocity you quoted.
Comment: Okay fine, but in this context, why free electron is not emitted in case of photo-electric emission ??
Answer: The electrons in the conductor which are not free are also travelling at high speed but they are bound to particular atoms. It requires energy to remove them. The 'free' electrons in the conductor are not really free. They are not bound to individual atoms but they are shared by and bound to a large number of atoms which form a microscopic crystal called a 'grain'. It still requires energy to remove them from a grain, but not as much as to remove electrons attached to individual atoms.
Grains are very close together so it requires very little energy for an electron to overcome the energy barrier and 'hop' from one grain to another along the conductor. There are usually no adjacent grains outside of the conductor, so electrons do not leap out of it unless, for example, another conductor is placed in close contact with it, or the electric field at the surface of the conductor is strong enough to overcome the force keeping electrons inside the grains, or a collision with a photon gives the electron more energy. The minimum potential difference for this to happen is called the Work Function.
Answer: If you remove $n$ electrons off the metal, the metal becomes $+ne$ positively charged. An electron which tries to leave the surface will be pulled back by the attractive forces. In other words, an electron does not have sufficient energy to over come the potential barrier.
You can supply energy to the electrons by heating or by shining light on the metal. This will give enough energy for the electrons to overcome the potential barrier. The phenomenons are known as thermionic emission and photoelectric effect respectively.
Comment: Well, this doesn't really explain it. If the metal is hot enough, it may [start emitting electrons](https://en.wikipedia.org/wiki/Thermionic_emission).
Comment: I think that is included in the last sentence. If thermionic emission has to be mentioned, then why not photoelectric effect? I think the last statement clearly establishes the fact there is a potential barrier which an electron has to overcome and it generally does not have sufficient energy to do so.
Comment: Anyway, added a note about it.
Answer: I can't provide the exact numerical data just yet, but I can give a basic explanation. Simply put, the electrons in a metal lattice form a bounded cloud,free to move inside the lattice. This cloud of free electrons forms metallic bonds with the positively charged nuclei.These bonds hold a metal together. Now, based on spectroscopic analysis, the energy of the electrons is simply not enough to overcome the potential barrier-leaving the metal surface.
Answer: If a electron were able to just escape the conductor, then it would be pulled back by the positive charge left behind (An image charge potential, one could say). Due to this it would be pulled back and will not be able to escape until it is able to obtain enough energy to become completely free.
Electrons can attain energy by the photoelectric effect, or by heating the conductor (Thermionic emissions).Two example of Thermionic emissions is the Child-Langmuir Law and The Richardson Law. (This PDF gives you a short intro to both ).
Also, note that if a photoelectric effect were continuously illuminated, and were not connected to a circuit, then after some time the photoelectric emission would stop since the positive charges left behind would accumulate and increase the potential barrier required to escape.


Is it safe to watch solar eclipse's reflection in water?

Question: There will be a solar eclipse soon at my area... naturally I want to watch this with my son.
Is it safe for us to watch the solar eclipse's reflection in the water? There is a swimming pool near my house and I plan to watch it there.. is it safe to do so?
I remember vividly doing this with my dad when I was little, but some website says we are not supposed to watch it in a bucket of water..
So is it ok to watch it in the water?
Comment: No. There will be a very brief period where is could potnentially be safe, but in order to watch for that moment you would need to watch the reflection while it is damagingly bright. There are better low-tech methods available. But good on you for planning this event with your son! I was in my mid-20s when I brought my dad along to see our first eclipse together. It was special even then. :)
Comment: @pela Thanks for the CD suggestion.. I have a few CDs here that fit the description on NASA site.. I will use it for quick viewing as a  last resort if I could not find proper solar filters.
Comment: Also, I think even in a pool the image of the Sun will be so distorted that it will kind of ruin the experience. I have watched Solar eclipses through a CD without damaging my eyes, but I won't recommend this method to others, as I don't really know about the potential hazards. [NASA](http://eclipse.gsfc.nasa.gov/SEhelp/safety2.html) seems to think it's okay, though, as long as the CD isn't transparent, but other sites discourages this.
Answer: If it's not safe to look at the sun's reflection in the water normally then it's no safer during an eclipse because even if 1% of the sun is showing this is more than enough light to seriously damage your eyes. The best way to view the sun, if you haven't ordered your eclipse glasses yet, is to poke a hole in a piece of paper and project an image of the sun onto another, like so:

There are instructions on how to do this here on the JPL website.
Comment: @RosdiKasim Yeah that will work, you'll get two images instead of one but  that doesn't matter. If you can get them to focus on the paper well then you should see a great image.
Comment: I just want to add that people are so focused on observing the eclipsing disk that they miss the best part of the experience. If in a region of totality, you want to observe your environment on the ground. The temperature drops about 10 degrees as a shadow races across the ground toward you at a thousand miles an hour. All of the animals and plants act as if it were night time (it gets quiet, crickets chirp, plants recede, etc.) It is cool to see the corona though!
Comment: Thanks.. after pondering about it a little bit I think I will go for the binocular option.. something like this will do right? http://www.lelong.com.my/powerful-16x52-high-definition-monocular-travel-telescope-binocular-justeprix-I1710520-2007-01-Sale-I.htm it is just 10 bucks.. I will tape this to a tripod and project it to a paper on the ground.. should make for much better image right?


How does a circular polarizer affect the "stars" from strong lights at night?

Question: When taking night photos at narrower apertures strong light sources such as streetlights demonstrate a star effect based on the number of blades in the lens' aperture. What effect would a polarizer rotated to its most effective orientation have on these "stars"?
Answer: Unless you put a polarizing sheet on the lamps, no effect (other than the global ND effect inherent in these filters), except if there is a reflection on the ground from them, then the reflection will be attenuated.
The polarizer removes the reflections from non-metallic surfaces and the stars are direct light from emitting sources.
Since the stars appear due to diffraction caused by narrow aperture slits and they have a discrete amount of angles, causing double that amount star lines (even number of blades overlap), I don't see how the polariser could affect it.
One thing is theory, another is practice. I set up an ikea lamp in my vision lab to make the star effect and then took an image with and without polariser:

Indeed, theory is shown to be true. We only get an ND effect. Rotating the polarizer had no effect at all.
If we put a polarizer behind the aperture, like this:

Then we get this (1. without filter 0.5x exposure, 2. with pol, 3. with pol rotated 90d):

The small differences I will blame the fingerprints on the filter that I tried to wipe off, but there is still some left :)
Comment: added practical example.
Comment: No star came up with freelensing so instead I cut a piece out of the polsheet and mounted it inside. It doesnt affect the stars neither. Btw, the best part of my vision lab is the white chair on top of the rack! :)
Comment: The light from the lamps are direct, but the stars are the result of the light's interaction with the aperture blades that occurs behind a front mounted polarizer.
Comment: and how does polarity affect the angle of the aperture slit, in turn affecting its diffraction?
Comment: Since the aperture slits are more or less evenly distributed around a 360 degree circle, if we had an eight blade diaphragm we could rotate the polarizer until some of the slits would be parallel to the direction of polarization and some would be perpendicular. I'm wondering what, if any, effect this might have on the shape of the "star".
Comment: Now, if only we could try that with a lens that uses drop-in filters behind the aperture...
Comment: I might be able to simulate that by freelensing and putting a polarising sheet in between.


Is it possible to install Geoserver on Zentyal?

Question: We want to setup a small WebGIS environment based on Geoserver/PostGIS or even the OpenGeoSuite if possible.
Our Server runs on Zentyal. Is it possible to install  Geoserver and PostGIS on Zentyal and if so is there any documentation available?
Or what would be an easy to operate platform so that we can create our WebGIS (we do not have much experience in running web servers though).
Answer: As long the application you want to install does not change Zentyal created configfiles, it is very well possible to install the application. Depending on the version of Zentyal you use, applications that can be installed on the base OS (Ubuntu 10.04 for Zentyal 2.x and Ubuntu 12.04 for Zentyal 3.x) can also be installed on Zentyal.


Computing Hartley transform of zero-padded signal from transform of original signal?

Question: If I have the discrete Hartley transform of a signal of length 2n, how can I use it to efficiently obtain the discrete Hartley transform of the same signal padded to length 2n+m for some m > 0?
Answer: Since the convolution theorem applies to the Hartley transform as well as the DFT, you may be able to use the same technique as can be used for FFT interpolation or spectrum resampling.  Use the Hartley transform of the implied rectangular window as an interpolation kernel, and interpolate to a larger N. Perhaps window the interpolation kernel for a trade-off between accuracy and computational efficiency.
Depending on the size and efficiency of the interpolation needed to meet your requirements, it may be more efficient to do an inverse and forward transform instead.


Programmatically add Search Box and Search Results webparts in SP2013

Question: I have created a custom Search Results page and need to add the Search Box webpart and Search Results webpart to the page programmatically. Has anyone ever accomplished this before?
Answer: Use the Web Part Manager class, either SPWebPartManager or SPLimitedWebPartManager. There are a ton of blog posts that explain how to do this in PowerShell and as part of Feature Deployment.
Comment: I've used this class to accomplish adding CEWP and LVWP/CQWP's on the page. I think it was a little different to add Search webparts. I did figure out how to do this. I ended up referencing the classes "Microsoft.Office.Server.Search" and "Microsoft.Office.Server.Search.WebControls" then was able to access the Search Results and Search Box as webparts and add them successfully to the page
Comment: Are you missing a feature dependency? If you browse to the site can you see the web parts in the gallery?
Comment: I wasn't missing a feature dependency. Yes, I can see the web parts in the gallery. I will post a snippet of the code later
Comment: A gogle-doesn't-hurt-like answer, very helpful.
Answer: Here is a snippet of the code I used
<code>using Microsoft.SharePoint;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Web.UI.WebControls.WebParts;
using Microsoft.Office.Server.Search;
using Microsoft.Office.Server.Search.WebControls;

     private static void SetupSearch( SPSite site)
     {
        try
        {
            MyUtil.createPage(site, "/", "Minimal.Search.aspx" , "SearchResults.aspx" , "Search Results" );
            SearchBoxScriptWebPart searchBox = new SearchBoxScriptWebPart ();
            ResultScriptWebPart resultsParts = new ResultScriptWebPart ();
            site.RootWeb.AddWebPartToPage( "Pages", "SearchResults.aspx" , "RightColumn", 0, true , searchBox, PartChromeType.None);
            site.RootWeb.AddWebPartToPage( "Pages", "SearchResults.aspx" , "RightColumn", 1, false , resultsParts, PartChromeType.None);

            foreach ( SPWeb oWeb in site.AllWebs)
            {
                oWeb.AllProperties[ "SRCH_SB_SET_SITE"] = "{\"Inherit\":false,\"ResultsPageAddress\":\"~sitecollection/Pages/SearchResults.aspx\",\"ShowNavigation\":false}" ;
                oWeb.Update();
            }
        }
        catch ( Exception ex)
        {
            Logging.LogError( "Error: MyLoggingError for Search " + ex.Message);
        }

     }
</code>


Ubuntu 12.04. display stuck on 640x480

Question: Good day. i have a problem. i have been running ubuntu 12.04 on my asus n76v. with nvidia geforce gt 650m 2gb graphics card for two weeks. yesterday while trying to get my accelerated graphics to work i installed the experimental drivers 310 from synaptic. after which my display went from 1280x whatever it is, to 640x480. i followed Desktop does not show when I installed nvidia drivers!.  and   Cannot install Nvidia drivers. and Ubuntu 12.04 Nvidia drivers. to no avail. then i did a fresh install. it worked until i did the updates. after the updates it went back to 640x480. not sure if i am missing something, or just screwed, i just know it worked for two weeks with no problems and now i cant fix the resolution.
Answer: Nvidia cards use their own display settings when you install the proprietary drivers. Try changing the resolution in the Nvidia Settings Manager by opening the Unity dash and typing in "nvidia". The Settings Manager should be the first thing that comes up.
From there, go to "X Server Display Settings" and change the resolution to your screen's native resolution, then click "Apply" to test it. Your screen will flicker and bounce a little as everything adjusts, but if all goes well, you should have your desktop at the proper resolution. Make sure to confirm that setting in the dialog that pops up, so that it doesn't revert automatically.
Finally, click "Write to configuration" to make the changes permanent. It will likely prompt you for your password, since writing to the video config requires sudo.
Comment: @JavierRivera - Not display manager as in X, but display manager as in settings manager. IE - Instead of using Settings -> Display to change your resolution, you use the Nvidia Settings Manager. Until 12.10, Settings->Display didn't even pick up resolutions when the Nvidia proprietary drivers are active.
Comment: Any sources on: "Nvidia cards use their own display manager". As far as I know this isn't even remotely true.
Comment: Yes, that makes more sense :).

Ubuntu 12.04. display stuck on 640x480


Is there any possible to create master node in one aws region and worker node in another region?

Question: Is there any possibility to create cluster with eks in one region (eg. ap-south-1) and join worker node from another region (eg. ap-north-east-2). If yes, can anyone tell me how to do this ?
Answer: This is not possible, even with self-managed node groups, because the node in the distant region is unable to authenticate to join the cluster.
This authentication is performed by a component called aws-iam-authenticator, which is one of the Kubernetes master components that EKS runs for you. When a node tries to join the cluster, aws-iam-authenticator tries to map the IAM role that the node authenticates with to a Kubernetes username, system:node:<nodename>. <nodename> must precisely match the name of the node, which is its private EC2 DNS name. In order to generate this username, aws-iam-authenticator looks up the private DNS name of the node using the EC2 API, but it does so in the same region as the EKS cluster, so it will fail for a node that is trying to join from another region.
It would be possible to patch this behaviour in aws-iam-authenticator, but since AWS runs this component for you as part of the EKS master components, you can't run a patched version.
You can of course achieve a multi-region K8s cluster by deploying your own cluster without using EKS. However, there are risks with distant nodes. If the two regions lose network connectivity with each other, or latency increases beyond a certain amount, then control plane operations involving the nodes in the distant region will fail and the master will see those nodes as lost. Among other things, this means that pods that were running on the distant nodes may be rescheduled elsewhere despite still actually running in the distant region. Note that connectivity could be lost or impaired between the regions even while both regions still had connectivity to other parts of the Internet.
You should deploy another cluster in the second region. There are various solutions under SIG Multicluster for federating the clusters allowing them to be managed together.
Answer: No. It's because you have no power to change anything on master node.
Amazon EKS is fully managed Kubernetes service, you would need to make your own cluster on EC2 instances.
Comment: But I'm not going to change the master node , master node will be in the same region , I need to join worker node from the another region
Comment: You are not able to manually add new nodes so you cannot choose where they be located at. You can scale up the cluster but this will be done automatically.
Comment: Thankyou@ShantyMan


What can fiberglass insulation NOT touch?

Question: I have some rolls of 15" R-13 insulation that I am planning on installing in my floor joists to help sound proof my basement ceiling. Is there anything that the insulation can't touch? I know that the lighting has to be rated to touch the insulation but what about pipes/wiring?
Comment: Generally, nothing.
Comment: And lighting that is not IC rated needs at least a 3" gap.
Answer: This is pretty late for the op, but for anyone else that comes across this threat.
Resilient channel is extremely easy to short circuit, and there are numerous examples for installs gone wrong with sheetrock falling.  You want hat channel with the appropriate amount of clips to actually hold the weight of the sheetrock.
Just a home theatre enthusiast here, but I have used their products in our first house.  Check out the soundproofing company website for general guidelines.

Pink or yellow fluffy will keep the the joist bays from becoming a drum.
Using 5/8 type x sheetrock will give weight.
"Hat channel" with the basic clips is all that's needed to break the connection.
Double sheetrock with green glue works as a dampener

I like the idea of rockwool instead of fluffy for moisture in our situation, but in a dry space it does virtually nothing better than pink/yellown fluffy insulation to stop the drum situation.
Hope this helps.
Answer: It doesn't matter to piping.  As a rule, piping likes insulation.  Although for cold piping, condensation may be an issue; insulation can hold that water and cause mold and mildew.
It's an issue with electrical, because some types of wire need to be able to cool, and they can't do that covered in insulation, and this has caused house fires.
I am hoping some of the electrical-code experts who know more about this area of practice will chime in.
Comment: There is no code as far as I know involving electrical wiring and insulation.  The whole point of electrical code is to make sure the proper wire is used to be ran through insulation - there is an assumption there.  There is however rated electrical devices (pot lights) for insulation and others that have to have a minimum distance.
Comment: The issue is when the insulation comes after the wiring:  insulation is added to existing wire that may not be rated for insulation, e.g. knob and tube.
Comment: Yep but that isn't code once he makes changes to that space.
Comment: You're saying adding insulation counts as renovation, which obliges him to bring the electrical to current code.  Seems like that would be widely overlooked, but ok - there's his answer.  He needs to make sure all electrical in the area is to current codes, i.e. wiring compatible with insulation.
Answer: Typically you want the basement insulation installed with the paper (vapor barrier) facing the heated part of the home. Moisture from inside air can condense and become trapped in the insulation during cold winter months leading to mold, especially at rim joists or skirts (paper facing outside).
However, if you will be spending a little more time in the basement, you will want to do this differently. You will most likely want to use a foam board toward the house, then turn the insulation over so that the paper is facing down toward the basement. But if you are going to finish the basement and use it as a proper living space (and finish the ceiling with drywall), then you should not insulate the ceiling at all. You'll need to insulate the walls of the basement, and buy some soundproofing foam for the ceiling.

With all that in mind, cables (wires) should be exposed if possible (keep the wires cool). But pipes are more efficient when insulated (less heat loss).
Oh, and technically, hydrofluoric acid will react with (fiber) glass.
Comment: I think if you have open vats of hydrofluoric acid in your home, soundproofing  is the LEAST of your worries. That stuff eats almost EVERYTHING.
Comment: I would like to mention that the only reason I'm installing insulation in the ceiling is for sound proofing. I don't have the resources to use legitimate sound proofing materials.
Comment: I believe that you can find egg crate foam pretty cheaply... cheaper than replacing wood rot anyway. I hate shopping for people, but here's a pretty good deal: http://www.combinationtherapymattress.com/p-2573-val-med-convoluted-foam-hospital-bed-pad.html
Answer: Have you considered mounting sheetrock on the ceiling using resilient channel? Perhaps in addition to whatever insulation you put in the ceiling. The resilient channel will hold the sheetrock off of the joists. Since the sheetrock doesn't touch the wood, sound doesn't transmit directly from the framing to the sheetrock. Be sure to talk to somebody and/or watch some videos. You want to screw the sheetrock to the channel between the joists. Otherwise, you'll end up driving screws clear through the channel, into the joist, and acoustically coupling your sheetrock to your joists, defeating the purpose of the channel.


Need help setting up lighting for a render like this

Question: If anybody can help me with lighting and settings to get something like this image

Thanks
Comment: Welcome to Blender.SE, Could you specify what render engine you are using? Looks like it could works with a single light source (maybe two...) rightly placed over the subject. What's going wrong? Could you show a picture of your best result? It would be easier to guide you from there.
Answer: Basic setup
First, make the world background black, then add a single light because in your image there is no light whatsoever in the shadows.
Position/Orientation
The position of the light will depend on the type you will use, e.g., for a sun lamp the position doesn't matter. Approximate orientation can be determined by examining the shadows and highlights on the chin and hair regions. See in this picture how the shadow around the chin (relatively smoothly curving surface) is left of the center of his chin and the highlights in his hair extend to the right of the midline through his face.

Example 1
In this case, I used a single area lamp with a size of 1. You can see that with the more finely detailed model due to texturing, the lighting looks better than in the next example because of the finer shadows.

Example 2
In this case, I used a sun lamp, but of course, you could also use a spot, point, area, or even a mesh light.

Extras

You can adjust the lamp size to change the how blurred or scattered the shadows are.
If you want any ambient light to show on your object (which is not in your picture), you can change the strength of the world background based on if a ray is coming from the camera.

If you want more photorealistic light behavior, get Filmic.

The models that I showed in the demos are available on Blend Swap here and here. You can download the rendering setup for the skull project that I edited here:
Answer: Keeping in mind that light travels in a straight line. If you study the shadows there are clues as to how to recreate the lighting.
In the reference image the shadows are black = One light source. The angle of the shadows reveal the light is slightly up and to the back of the subject. The shadow edges are hard which means the light source is either a spot light or small.
EDIT: A little more on light source relative to shadows.There are two major factors in the light source that effect the resulting shadow. It's size and distance from the subject. Light sources that are smaller and/or further away yield a shadow with a "harder" edge. Conversely, light sources that are larger and/or closer to the subject yield "softer" shadow edges.
Comment: If you feel Its solved please mark my early answer as solved... smiles


What's this drum notation symbol (X in the top gap)

Question: What's the drum notation symbol that I've circled in the attached image?
Answer: That looks like an open hat to me
Comment: I agree.  It looks like the hi-hat is noted in the top space, with "+" for closed and "o" for open.
Answer: Drum notation seems fairly whimsical. Certainly the 'X' means a cymbal of some sort. The notation I use, which is not universal (there probably isn't such a thing!) is hi-hat (stick) where a G note is above the stave; ride cymbal where A is (above), and crash where B is above that. Obviously all treble clef.
The closest I can get is cross stick, but that's 'X' on the D line.
The 'o' and '+' indicate open and closed hi-hat respectively, so logic says that the 'X' in the E space must be hi-hat in this incarnation.
It seems that everyone has their own drum notation - some don't even use a stave - but with a bit of intuition they're all playable - with guesswork thrown in...
Other references - bass drum is generally in F space at the bottom, and the snare on middle B line, although here, it's probably the dot on the C space. For me, the spaces are all for toms, which makes it an easy read.
Comment: Thx. I'm personally not a drummer myself. I was just transcribing the above drum pattern and the specific drum beat that I mentioned wasn't on any chart that I saw when I first tried to find it. The closest I found was the snare. I did see it here - https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcR3i2IJhFzA5WmL4q964MtSlF0Kq-XizuaTQ1IqgJRpZrM4mMpSxA to be honest but I think I got confused because of the + and O above the chart.
Answer: Looks like a cymbal note to me. That's the cymbal note notation I see in Musescore. Perhaps the clearest tell that the X-headed note is for cymbals is the word "splash" (for splash cymbal) just before one such note.
Answer: That's a rim shot on the first tom. I'm a drummer I do drum set and marching band snare.
Comment: How could the "o" or "+" symbols apply to a rimshot? These are usually used for open and close on a hi-hat.
Answer: Very late answer, but I have seen this before in some basic rock exercises and it meant to hit the stick on the rim of the high tom.

What’s this drum notation symbol (X in the top gap)


Can I walk faster?

Question: I'm playing my way through Bioshock (and then Bioshock 2) to prepare for Bioshock Infinite. I'm doing okay apart from I've got to a part where I have no weapons apart from a wrench and a camera. I want to shock the enemy using plasmids and then run up and whack him with my wrench, but I seem to move very slowly (this is all the time, not just in this area). Can I up my move speed at all or is there a button for sprint?
Answer: Research the Thuggish Splicers to obtain the SportBoost physical tonics, which increases movement speed and also lets you swing your wrench faster. You can use both versions of this tonic at the same time.
Comment: In Bioshock 2, the first speed boost tonic is acquired easily early on, but for the second you need to do a lot of camera research on spider splicers (the ones that climb on the roof a lot and carry hooks) and they only appear in one or two places in the game, so make sure you get it. I missed out.
Answer: I assume you are playing PC version.
Firstly check if you didn't press walk(should be CTRL) or crouch (C) buttons. So far I checked in forums thats most common problem. Otherwise it will be natural game speed.
While playing Bioshock 1 I don't remember having sprint, but in 2nd part you will be able to unlock charge with drill :D
Answer: There's no way to move faster without tonics. I suggest taking cover, and shocking enemies at close range.

Can I walk faster?


What's the order of the Final Fantasy games?

Question: There are a lot of Final Fantasy video games out there, and I don't know which one to play.  What order should they be played in?
Comment: Presumably, first to last.
Comment: If I'm not mistaken, Final Fantasy is like Mario or Zelda; there is no order in which you should play them.
Comment: First way: start from the first one, and work your way up to the latest (Keep in mind a couple of them are MMOs), this order may help build an appreciation for how each iteration builds on and (tries to) improve on the mechanics and story of its predecessors. Second way: find one you really like to start with to get you hooked, and then branch out into the rest of the series, as they are all largely unrelated, the order doesn't matter.
Answer: It doesn't really matter.
Despite their numbered iterations, most of the Final Fantasy games are in no way related to each other. There's no overarching plot between the various entries, no shared setting, nothing that implies any particular order that they should be played in, plot-wise. A few exceptions do exist, however.

FFIV: The After Years - Sequel to FFIV
X-2 - Direct sequel to X
Final Fantasy XII: Revenant Wings - Sequel to XII
XIII-2 - Direct sequel to XIII
Lightning Returns: Final Fantasy XIII - Sequel to XIII-2
Final Fantasy Tactics A2 - Sequel to Final Fantasy Tactics Advance
Dirge of Cerberus - Sequel to VII
Before Crisis - Prequel to VII
Crisis Core - Prequel to VII
Comment: There's a *bit* of a shared setting in [Ivalice](http://finalfantasy.wikia.com/wiki/Ivalice_Alliance). But, yeah, the main Roman numeral games are largely unconnected save for a few recurring names (Cid) and creatures (Moogles, Chocobos, etc)
Comment: Don't forget about Lightning's Return: XIII
Comment: @Wondercricket Whoops, had that, but forgot to put it back in when I reformatted. Thanks.


What's the 26" tire that is the same width as 700×35c

Question: I have a 26" front wheel and 700c rear wheel. I put 35c on my rear wheels and I'd like to put the same tire witdh Infront. What 26" tire is the same as 700×35c?
Comment: Your 26" wheel should also have a metric value, maybe on your old tyre. Those seem to be more precise.
Answer: 35mm is 1.378". That's the same as 1.375 for our purposes. So you'd get a 26x1.375 tyre.
But tyre sizing isn't that precise, and anyway is affected by tread and the width of the rim you put it on. To get a similar feel, the same make and model in 26x1.375 will be closest, assuming the same rim width.
Even then, the smaller wheel diameter and different frame will mean a different sensation as you pass over rough bits.
There's no real need to put the same tyre front and rear; in your case you may want a bigger tyre at the front to offset the smaller wheel.
Note also the existence of rare 26" wheels with fractional rather than decimal widths. These aren't compatible with tyres giving their widths in decimal
Comment: @Criggie I haven't gone that extreme but ran 28 front and 35 rear by choice when I had a child seat fitted.  And for commuting bikes I regard 28-35mm as interchangeable.  OTOH my MTB has 2.3" at the front and 2.1" at the back
Comment: Concur - I have one bike with a 28mm front tyre and a 2" rear tyre.


Is selling or trading my World of Warcraft account a violation of the Terms of Service agreement?

Question: Would I be in violation of the Terms of Service agreement if I were to buy, sell, trade, or give away my World of Warcraft account?
Answer: According to the terms of use, yes

Ownership/Selling of the Account or Virtual Items.
Blizzard does not recognize the transfer of WoW Accounts or Blizzard Accounts (each an "Account"). You may not purchase, sell, gift or trade any Account, or offer to purchase, sell, gift or trade any Account, and any such attempt shall be null and void. Blizzard owns, has licensed, or otherwise has rights to all of the content that appears in the Game. You agree that you have no right or title in or to any such content, including without limitation the virtual goods or currency appearing or originating in the Game, or any other attributes associated with the Account or stored on the Service. Blizzard does not recognize any purported transfers of virtual property executed outside of the Game, or the purported sale, gift or trade in the "real world" of anything that appears or originates in the Game. Accordingly, you may not sell in-game items or currency for "real" money, or exchange those items or currency for value outside of the Game.
Comment: This paragraph apparently also bans the sale or trade of virtual property, which would include gold.
Comment: Yes, buying or selling gold is also against the terms of use.
Comment: Yet, apparently you can use it as payment for services, and it's taxable income.
Answer: Blizz will ban any account found to be sold or traded for financial gain. They will also ban the account of anyone related to the transaction. So, if  you sell your account and start a new one, they will ban your new account as well.


Which filament is best for printing parts that will endue a constant load for a long period of time?

Question: I thought the answer was ABS, but I read about how some people used it to print parts that were used in plumbing, and they failed when put under constant load, and some said that PETG is better for this application. I want to print bases for my table legs, would PETG be a better material to use?
Comment: source for ABS failing? it's quite important to know the source to get further details. Maybe (likely) it was ABS+
Comment: Table base means high compressive strength. Even if they are for a piano, any material will do. PLA will work just fine.
Answer: PETG creeps more than ABS/ASA, so I think you should still pick those. Be sure to print them hot and in a closed chamber, but you are loading in compression, so you won't have big issues.
See these tests for creep:

The conclusion is that ASA deforms in the beginning and then it stops. PETG in tensile keeps stretching, but in compression seems fine.
I think ABS is still the best, especially if you use a pure ABS, not an ABS+. Rather than an ABS+ use ASA.
Since it's not critical and if it breaks it's no big deal (likely it will get thinner with time), PETG is an alternative you can try.
Comment: PETG creeps but PET doesn't.
Comment: It creeps but not much
Answer: For table leg stoppers and things like sandwich signs I use TPU, never had a problem with them yet. They're flexible and strong. Which means you can design them to insert easily and lock themselves in place. They're forgiving if your table leg is not quite evenly prepared at the bottom.
The only problem I can think of which would also apply to any other material is if they have to deal with a sharp edge.


Is there a community chat room?

Question: Is there any way to chat with moderators or other members of the community?
Comment: I'm taking the `[featured]` tag off this because the chat links are now quite prominently displayed in a few different places on the main page.  Thought it was time for some cleanup. :)
Answer: Yes, there is. Join The Frying Pan today!
All are welcome to stop by.
Comment: I don't have the necessary rep to edit this, but note that this has now been migrated to chat.SE
Answer: cooking.stackexchange now has a chat home on chat.stackexchange; the "chat" option is now at the top of every Q&A page.
You aren't limited to 1 room - feel free to use chat for all (sensible) cooking chat-related purposes.

Is there a community chat room?


How can I move a MySQL database to a new localhost?

Question: I have a localhost (XAMPP) on an old pc. I need the MySQL database on my new mac with a localhost (MAMP). Issue is, pc doesn't have internet connection. Any way I can move the database over? Detailed instructions are highly welcomed as I am by no means a server expert. Thanks
Comment: Use a USB stick?
Answer: You should be able to just use a USB Stick as suggested in the comments above.  Just make sure to remove/modify the permissions on the DB file(s) first to avoid having issues with not being able to access the DB once it is moved over.  That is a painful, painful, painful lesson if something goes wrong... experience talking.

Backup MySQL
Consequently, use the import link on the new PC
Comment: I assumed XAMP = Windows MAMP = Mac
Comment: I've not seen it referred to as MAMP =D  As a SQA tester I've used it on both Mac and Windows so was just wanting to clarify and was pretty sure that apache friends just called it XAMP for each os(win/mac/linux)
Answer: XAMP should have phpmyadmin installed. If so, I would use it to create a backup / dump of the data base. Put that on your usb, move it to the mac, and again use phpmyadmin to first create a data base, then import your data.
You could also use mysqldump (at least on Linux from command line. Not sure about XAMP on windows):
mysqldump -u username -p database | gzip > database.sql.gz  ( the pipe gzip is optional)
so say your db name is db_important and your db username is phil
mysqldump -u phil -p db_important | gzip > db_important.sql.gz
( when you run this you will be asked for your db password)
Or wihout gzip:
mysqldump -u phil -p db_important > db_important.sql
phpmyadmin should be easiest and safest route for you if your unfamiliar with mysql command line
these links may help:
back up with php admin
how to use phpmyadmin to back up and restore


Loudspeaker detected yet won't connect

Question: Laptop Dell E5430
Running Ubuntu 16.04
Speaker Unotec Bullet sound
recognized by laptop in bluetooth but not through jack
Recognized in bluetooth but won't connect anyway
Any advices ?
Answer: ok, I got it working by tricking around.
Had to enter the connection menu in bluetooth.
Then, quitted it.
Now, went in the bluetooth thing (B symbol) in the taskbar,
told it to connect, then it worked.
Put music on, switched the output (sound menu) to the device.
Shazam !!

Loudspeaker detected yet won't connect


ASP.NET JavaScript Callbacks Without Full PostBacks?

Question: I'm about to start a fairly Ajax heavy feature in my company's application. What I need to do is make an Ajax callback every few minutes a user has been on the page.

I don't need to do any DOM updates before, after, or during the callbacks.
I don't need any information from the page, just from a site cookie which should always be sent with requests anyway, and an ID value.

What I'm curious to find out, is if there is any clean and simple way to make a JavaScript Ajax callback to an ASP.NET page without posting back the rest of the information on the page. I'd like to not have to do this if it is possible.
I really just want to be able to call a single method on the page, nothing else.
Also, I'm restricted to ASP.NET 2.0 so I can't use any of the new 3.5 framework ASP AJAX features, although I can use the ASP AJAX extensions for the 2.0 framework.
UPDATE
I've decided to accept DanP's answer as it seems to be exactly what I'm looking for. Our site already uses jQuery for some things so I'll probably use jQuery for making requests since in my experience it seems to perform much better than ASP's AJAX framework does.
What do you think would be the best method of transferring data to the IHttpHandler? Should I add variables to the query string or POST the data I need to send?
The only thing I think I have to send is a single ID, but I can't decide what the best method is to send the ID and have the IHttpHandler handle it. I'd like to come up with a solution that would prevent a person with basic computer skills from accidentally or intentionally accessing the page directly or repeating requests. Is this possible?
Answer: If you don't want to create a blank page, you could call a IHttpHandler (ashx) file:
public class RSSHandler : IHttpHandler
    {
        public void ProcessRequest (HttpContext context)
        {
            context.Response.ContentType = "text/xml";

            string sXml = BuildXMLString(); //not showing this function,
            //but it creates the XML string
            context.Response.Write( sXml );
        }

        public bool IsReusable
        {
            get { return true; }
        }

    }

Answer: You should use ASP.Net Callbacks which were introduced in Asp.Net 2.0. Here is an article that should get you set to go:
Implementing Client Callbacks Programmatically Without Postbacks in ASP.NET Web Pages
Edit: Also look at this:
ICallback & JSON Based JavaScript Serialization
Answer:
What do you think would be the best method of transferring data to the IHttpHandler? Should I added variables to the query string or POST the data I need to send? The only thing I think I have to send is a single ID, but I can't decide what the best method is to send the ID and have the IHttpHandler handle it. I'd like to come up with a solution that would prevent a person with basic computer skills from accidentally or intentionally accessing the page directly

Considering the callback is buried in the client code, it would take someone with equal determination to get either the querystring or the POST request. IE, if they have firebug, your equally screwed.
So, in that case, do whatever is easiest to you (Hint: I'd just use the querystring).
To handle repeating requests/direct access, I'd generate a key that is sent with each request. Perhaps a hash of the current time (Fuzzy, I'd go down to minutes, but not seconds due to network latency) + the client IP.
Then in the HTTPHandler, perform the same hash, and only run if they match.
Answer: You are not just restricted to ASP.NET AJAX but can use any 3rd party library like jQuery, YUI etc to do the same thing. You can then just make a request to a blank page on your site which should return the headers that contain the cookies.
Answer: My vote is with the HTTPHandler suggestion as well. I utilize this often. Because it does not invoke an instance of the page class, it is very lightweight.
All of the ASP.NET AJAX framework tricks actually instantiate and create the entire page again on the backend per call, so they are huge resource hogs.
Hence, my typical style of XmlHttpRequest back to a HttpHandler.
Answer: Since you are using only ASP.NET 2.0 I would recommend AjaxPro will which create the .ashx file.  All you have to do is to pull the AjaxPro.dll into your web site.  I developed an entire application with AjaxPro and found it worked very well.  It uses serialization to pass objects back and forth.
This is just a sample on how to simply use it.
namespace MyDemo
{
  public class Default
  {
    protected void Page_Load(object sender, EventArgs e)
    {
      AjaxPro.Utility.RegisterTypeForAjax(typeof(Default));
    }

    [AjaxPro.AjaxMethod]
    public DateTime GetServerTime()
    {
      return DateTime.Now;
    }
  }
}

To call it via JavaScript it is as simple as
function getServerTime()
{
  MyDemo._Default.GetServerTime(getServerTime_callback);  // asynchronous call
}

// This method will be called after the method has been executed
// and the result has been sent to the client.

function getServerTime_callback(res)
{
  alert(res.value);
}

EDIT
You also have to add

To the config.  AjaxPro also works well side by side with APS.NET Ajax and you can pass C# objects from Client to Sever if the class is marked as [Serializable]
Answer: Just to offer a different perspective, you could also use a PageMethod on your main page.  Dave Ward has a nice post that illustrates this.  Essentially you use jQuery ajax post to call the method, as illustrated in Dave's post:
$.ajax({
    type: "POST",
    url: "Default.aspx/GetFeedburnerItems",
    // Pass the "Count" parameter, via JSON object.
    data: "{'Count':'7'}",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    success: function(msg) {
      BuildTable(msg.d);
    }
  });

No need for Asp.Net Ajax extensions at all.
Answer: You can also use WebMethods which are built into the asp.net ajax library. You simply create a static method on the page's codebehind and call that from your Ajax.
There's a pretty basic example of how to do it here
Answer: You should use a web service (.asmx).  With Microsoft's ASP.NET AJAX you can even auto-generate the stubs.