Question:
I recently began experimenting with gnuplot and I quickly made an interesting discovery. I plotted all of the prime numbers beneath 1 million in polar coordinates such that for every prime $p$, $(r,\theta) = (p,p)$. I was not expecting anything in particular, I was simply trying it out. The results are fascinating.
When looking at the primes beneath 30000, a spiral pattern can be seen:

For comparison, here is the same graph with the multiples of 3 and 7 superimposed on it. Primes are in yellow, multiples of 3 and 7 in green and red respectively.

What is really interesting to me, though, is the behavior when the range is increased. Multiples of a given number appear to spiral out in the same pattern into infinity, but the primes begin to form rays in groups of 3 or 4. See below:

Compared to multiples of 3 and 7 again:

Now, I must admit that I am very much a novice mathematician with little experience beyond trigonometry. I am just going into Calculus and Discrete Mathematics this upcoming fall.
I know that there is something called the Prime Number theorem - are these patterns related to it? Are these rays the same phenomenon as the diagonal lines found in Ulam Spirals?
EDIT:
In response to Greg Martin's explanation, I decided to add a couple more graphs. To see why they are relevant, read his answer.
$(r,\theta)=(n,n), n \in  \mathbb{N}$

What we're seeing is arithmetic progressions (not prime-producing polynomials) of primes, combined with a classical phenomenon about rational approximations.
When the integers (or any subset of them) are represented by the polar points $(n,n)$, of course integers that are close to a multiple of $2\pi$ apart from each other will lie close to the same central ray. Figuring out when integers are close to a multiple of $2\pi$ apart is a perfect job for continued fractions. The continued fraction of $2\pi$ is $\langle 6; 3,1,1,7,2,146,\dots\rangle$, giving the convergents
$$
\left\{6,\frac{19}{3},\frac{25}{4},\frac{44}{7},\frac{333}{53},\frac{710}{113},\frac{103993}{16551},\dots\right\},
$$
which are the rational approximations of $2\pi$ that will dominate the picture on different scales.
For example, if you plot the polar points $(n,n)$ for $1\le n\le 25000$, you will notice the points aligning themselves into $44$ spirals: jumping ahead from $n$ to $n+44$ is almost the same as going around the circle $7$ times (note the convergent $\frac{44}7$ showing up); moving from $n$ to $n+1$ jumps ahead $7$ spirals. Each spiral corresponds to an arithmetic progression $a\pmod{44}$;  going from one spiral to the next one counterclockwise corresponding to changing the arithmetic progression from $a\pmod{44}$ to $a+19\pmod{44}$ (note that $19\equiv7^{-1}\pmod{44}$).
If instead you plot only the primes $(p,p)$, you will get reasonable representation in the $\phi(44)=20$ spirals corresponding to arithmetic progressions $a\pmod{44}$ where $\gcd(a,44)=1$, and no primes in the other $24$ spirals. That's what we're seeing in the top two pictures.
As the scale moves farther out, these particular spirals become more tightly wound and harder to see (go from the 1st picture to the 5th, then the 3rd, then the 4th), and the next convergent takes over. In this case, the convergent $\frac{710}{113}$ is an extremely good rational approximation to $2\pi$ (as we know from the large partial quotient $146$). Therefore the integer points $(n,n)$ will group themselves into $710$ spirals, but these spirals are so close to straight lines at the beginning that they almost don't look like spirals, and persist for a large interval of possible scales. Each ray thus corresponds to an arithmetic progression $a\pmod{710}$.
When we plot only prime points $(p,p)$ (the 4th picture is best here), we will only see the $\phi(710)=280$ arithmetic progressions $a\pmod{710}$ where $\gcd(a,710)=1$. The fact that the visible rays are mostly grouped in fours is a consequence of the fact that $5\mid710$ and so every fifth ray doesn't contain primes. Really, though, we are seeing four out of every ten rays rather than four out of every five; the arithmetic progressions $a\pmod{710}$ with $a$ even have no primes at all and are thus invisible. There are four exceptional groups containing only three rays instead of four; these correspond to the four arithmetic progressions $a\pmod{710}$ where $a$ is a multiple of $71$ but not a multiple of $2$ or $5$.


Question:
I recently read a post from R-Bloggers, that linked to this blog post from John Myles White about a new language called Julia. Julia takes advantage of a just-in-time compiler that gives it wicked fast run times and puts it on the same order of magnitude of speed as C/C++ (the same order, not equally fast). Furthermore, it uses the orthodox looping mechanisms that those of us who started programming on traditional languages are familiar with, instead of R's apply statements and vector operations.
R is not going away by any means, even with such awesome timings from Julia. It has extensive support in industry, and numerous wonderful packages to do just about anything.
My interests are Bayesian in nature, where vectorizing is often not possible. Certainly serial tasks must be done using loops and involve heavy computation at each iteration. R can be very slow at these serial looping tasks, and C/++ is not a walk in the park to write. Julia seems like a great alternative to writing in C/++, but it's in its infancy, and lacks a lot of the functionality I love about R. It would only make sense to learn Julia as a computational statistics workbench if it garners enough support from the statistics community and people start writing useful packages for it.
My questions follow:

What features does Julia need to have in order to have the allure that made R the de facto language of statistics?
What are the advantages and disadvantages of learning Julia to do computationally-heavy tasks, versus learning a low-level language like C/++?

I think the key will be whether or not libraries start being developed for Julia. It's all well and good to see toy examples (even if they are complicated toys) showing that Julia blows R out of the water at tasks R is bad at.
But poorly done loops and hand coded algorithms are not why many of the people I know who use R use R. They use it because for nearly any statistical task under the sun, someone has written R code for it. R is both a programming language and a statistics package - at present Julia is only the former.
I think its possible to get there, but there are much more established languages (Python) that still struggle with being usable statistical toolkits.


Question:
The accepted range for the wavelengths of light that the human eye can detect is roughly between 400nm and 700nm.  Is it a co-incidence that these wavelengths are identical to those in the  Photosynthetically Active Radiation (PAR) range (the wavelength of light used for normal photosynthesis)?
Alternatively is there something special about photons with those energy levels that is leading to stabilising selection in multiple species as diverse as humans and plants?

Good question.
If you look at the spectral energy distribution in the accepted answer here, we see that photons with wavelengths less than ~300 nm are absorbed by species such as ozone. Much beyond 750 infrared radiation is largely absorbed by species such as water and carbon dioxide. Therefore the vast majority of solar photons reaching the surface have wavelengths that lie between these two extremes.
Therefore, I would suggest that surface organisms will have adapted to use these wavelengths of light whether it be used in photoreceptors or in photosynthesis since these are the wavelengths available; i.e., organisms have adapted to use these wavelengths of light, rather than these wavelengths being special per se (although in the specific case of photosynthesis there is a photon energy sweet spot).
For example this study suggests that some fungi might actually be able to utilize ionizing radiation in metabolism. This suggests that hypothetical organisms on a world bathed in ionizing radiation may evolve mechanisms to utilize this energy.


Question:
I found this math "problem" on the internet, and I'm wondering if it has an answer:

Question: If you choose an answer to this question at random, what is the probability that you will be correct?
a. $25\%$
b. $50\%$
c. $0\%$
d. $25\%$

Does this question have a correct answer?

No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth.
It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement

This statement is false.

is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.

Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%.
Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question

Is the correct answer to this question "yes"?

would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless.
One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on

True or false?

The Great Pumpkin exists.
Both of these statements are false.

where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed.
On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful.
In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.


Question:
I need to numerically evaluate the integral below:
$$\int_0^\infty \mathrm{sinc}'(xr) r \sqrt{E(r)} dr$$
where $E(r) = r^4 (\lambda\sqrt{\kappa^2+r^2})^{-\nu-5/2} K_{-\nu-5/2}(\lambda\sqrt{\kappa^2+r^2})$, $x \in \mathbb{R}_+$ and $\lambda, \kappa, \nu >0$. Here $K$ is the modified Bessel function of the second kind. In my particular case I have $\lambda = 0.00313$, $\kappa = 0.00825$ and $\nu = 0.33$.
I am using MATLAB, and I have tried the built-in functions integral and quadgk, which gives me a lot of errors (see below). I have naturally tried numerous other things as well, such as integrating by parts, and summing integrals from $kx\pi$ to $(k+1)x\pi$.
So, do you have any suggestions as to which method I should try next?
UPDATE (added questions)
I read the paper @Pedro linked to, and I don't think it was too hard to understand. However, I have a few questions:

Would it be okay to use $x^k$ as the basis-elements $\psi_k$, in the univariate Levin method described?
Could I instead just use a Filon method, since the frequency of the oscillations is fixed?

Example code
>> integral(@(r) sin(x*r).*sqrt(E(r)),0,Inf)
Warning: Reached the limit on the maximum number of intervals in use. Approximate
bound on error is   1.6e+07. The integral may not exist, or it may be difficult to
approximate numerically to the requested accuracy.
> In funfun\private\integralCalc>iterateScalarValued at 372
In funfun\private\integralCalc>vadapt at 133
In funfun\private\integralCalc at 84
In integral at 89
ans =
3.3197e+06

I've written my own integrator, quadcc, which copes substantially better than the Matlab integrators with singularities, and provides a more reliable error estimate.
To use it for your problem, I did the following:
>> lambda = 0.00313; kappa = 0.00825; nu = 0.33;
>> x = 10;
>> E = @(r) r.^4.*(lambda*sqrt(kappa^2 + r.^2)).^(-nu-5/2) .* besselk(-nu-5/2,lambda*sqrt(kappa^2 + r.^2));
>> sincp = @(x) cos(x)./x - sin(x)./x.^2;
>> f = @(r) sincp(x*r) .* r .* sqrt( E(r) );

The function f is now your integrand. Note that I've just assigned any old value to x.
In order to integrate on an infinite domain, I apply a substitution of variables:
>> g = @(x) f ( tan ( pi / 2 * x ) ) .* ( 1 + tan ( pi * x / 2 ).^2 ) * pi / 2;

i.e. integrating g from 0 to 1 should be the same as integrating f from 0 to $\infty$. Different transforms may produce different quality results: Mathematically all transforms should give the same result, but different transforms may produce smoother, or more easily integrable gs.
I then call my own integrator, quadcc, which can deal with the NaNs on both ends:
>> [ int , err , npoints ] = quadcc( g , 0 , 1 , 1e-6 )
int =
  -1.9552e+06
err =
   1.6933e+07
npoints =
       20761

Note that the error estimate is huge, i.e. quadcc doesn't have much confidence in the results. Looking at the function, though, this is not surprising as it oscillates at values three orders of magnitude above the actual integral. Again, using a different interval transform may produce better results.
You may also want to look at more specific methods such as this. It's a bit more involved, but definitely the right method for this type of problem.


Question:
According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained.
So, can an atom have more than 8 valence electrons?
If this is not possible, why can't an atom have more than 8 valence electrons?

2017-10-27 Update
[NOTE: My earlier notation-focused answer, unchanged, is below this update.]
Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.
However, the same bonding mechanisms that enable the formation of greater-than-8 valence shells also enable alternative structural interpretations of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. Manishearth's excellent answer explores this issue in much greater detail than I do here.
Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:
(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or
(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.
While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.
The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic ("astronomical” might be a more apt word).
A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggest that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.
Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.
Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.
That is why my overall answer to this question is simply "yes".
Question: Why are octets special?
The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?
In a nutshell, the reason is that for any energy level above the special case of the $n=1$ shell (helium), the "closed shell" orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.
It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.
The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus avoid the larger energy penalties imposed by Pauli exclusion.
Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.
A final note
I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.
(A hint: If you have ever watched people create two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)
Original 2014-ish Answer (Unchanged)
This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.
I call it donation notation, and it dates back to my high school days when none of the chemistry texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.
The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.
Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indicating that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)
Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.
Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, some of the pairs in the central octet can remain unattached. This results in less oxidized anions such as sulfites and nitrites.
Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.
2014-04-04 Update
Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.
The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.
The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.
The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.
Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."
Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four-line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.
Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologs, one of the consequences of which is that they have nearly identical physical chemistry properties.
Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:


Question:
My friend's 3-year-old daughter asked "Why are there circles there?"

It had either rained the night before or frost had thawed. What explains the circles?
Follow-up question: Ideally, are these really circles or some kind of superellipse?

Both thawing and evaporation involve heat exchange between the stone tile, the water sitting atop the stone tile, any water that's been absorbed by the stone tile, and the air around. The basic reason that the center and the edges of the tile evaporate differently is that the gaps between the tiles change the way that heat is exchanged there. However the details of how that works are a little more involved than I can get into at the moment, and would be lost on a three-year-old anyway.
A good way to explain this phenomenon to a three-year-old would be to bake a batch of brownies in a square pan, and watch how the brownies get done from the outside of the pan inwards.  Even after you have finished them you can still tell the difference between the super-crispy corner brownies, the medium-crispy edge brownies, and the gooey middle-of-the-pan brownies.  The three-year-old would probably ask you to repeat this explanation many times.
I think the shapes are not exactly circles, superellipses, or any other simple mathematical object --- there's too much real life in the way --- but they do become more circular as the remaining puddle gets further from the edges.
A related explanation.


Question:
Paper is an extremely flexible material, at least when it is in sheet form. It will deform significantly according to the pressure applied and it is easy to fold.
Therefore, it's extremely counterintuitive that a sheet of paper could cut through human skin and probably through stiffer/harder materials, since when the skin applies a pressure on the paper, one would expect it to fold or bend. Yet it is easy to have a severe cut from paper, through both the epidermis and the dermis. How is that possible? Certainly the width of the sheet of paper plays a big role: the smaller it is, the sharper it is, but also the more flexible it becomes and the less it should sustain an applied pressure without folding up!
I can think of other materials such as thin plastic films and aluminium foils. My intuition tells me the plastic foil would not cut through skin but the aluminium foil would, although I am not sure since I did not try the experiment. If this hold true, what determines whether a material would be able to cut through skin? A hair for example, which is flexible and thinner than a paper sheet, is unable to cut through the skin. What makes paper stand out? What is so different that makes it a good cutter?
Maybe it has to do with its microscopic properties and that it contains many fibers, but I highly doubt it because the aluminium foil does not contain these and yet would probably also cut.

Paper, especially when freshly cut, might appear to have smooth edges, but in reality, its edges are serrated (i.e. having a jagged edge), making it more like a saw than a smooth blade. This enables the paper to tear through the skin fairly easily. The jagged edges greatly reduce contact area,  and causes the pressure applied to be rather high. Thus, the skin can be easily punctured, and as the paper moves in a transverse direction, the jagged edge will tear the skin open.
Paper may bend easily, but it's very resistant to lateral compression (along its surface). Try squeezing a few sheets of paper in a direction parallel to its surface (preferably by placing them flat on a table and attempting to "compress" it laterally), and you will see what I mean. This is analogous to cutting skin with a metal saw versus a rubber one. The paper is more like a metal one in this case. Paper is rather stiff in short lengths, such as a single piece of paper jutting out from a stack (which is what causes cuts a lot of the time). Most of the time, holding a single large piece of paper and pressing it against your skin won't do much more than bend the paper, but holding it such that only a small length is exposed will make it much harder to bend. The normal force from your skin and the downward force form what is known as a torque couple. There is a certain threshold torque before the paper gives way and bends instead. A shorter length of paper will have a shorter lever arm, which greatly increases the tolerance of the misalignment of the two forces. Holding the paper at a longer length away decreases this threshold (i.e. you have to press down much more precisely over the contact point for the paper to not bend). This is also an important factor in determining whether the paper presses into your skin or simply bends.
Paper is made of cellulose short fibers/pulp, which are attached to each other through hydrogen bonding and possibly a finishing layer. When paper is bent or folded, fibers at the folding line separate and detach, making the paper much weaker. Even if we unfold the folded paper, those detached fibers do not re-attach to each other as before, so the folding line remains as a mechanically weak region and decreasing its stiffness. This is why freshly made, unfolded paper is also more likely to cause cuts.
Lastly, whether a piece of paper cuts skin easily, of course depends on its stiffness. This is why office paper is much more likely to cut you than toilet paper. The paper's density (mass per unit area), also known as grammage, has a direct influence on its stiffness.


Question:
I have this problem from University Physics with Modern Physics (13th Edition):

The inside of an oven is at a temperature of 200 °C (392 °F). You can put your hand in the oven without injury as long as you don't touch anything. But since the air inside the oven is also at 200 °C, why isn't your hand burned just the same?

What I understood from this problem is that my hand won't be as hot as the air temperature, but then my first conjecture was: It’s the nature of the air (i.e., a gas) that its molecules are more disperse than those of a solid.
Is my reasoning right? Or what thermodynamics concepts do I need to understand better to tackle this problem?

There are two points relevant for the discussion: air itself carries a very small amount of thermal energy and it is a very poor thermal conductor.
For the first point, I think it is interesting to consider the product $\text{density} \times \text{specific heat}$, that is the amount of energy per unit volume that can be transferred for every $\text{K}$ of temperature difference. As of order of magnitudes, the specific heat is roughly comparable, but the density of air is $10^3$ times smaller than the density of a common metal; this means that for a given volume there are much less "molecules" of air that can store thermal energy than in a solid metal, and hence air has much less thermal energy and it is not enough to cause you a dangerous rise of the temperature.
The rate at which energy is transferred to your hand, that is the flow of heat from the other objects (air included) to your hand. In the same amount of time and exposed surface, touching air or a solid object causes you get a very different amount of energy transferred to you. The relevant quantity to consider is thermal conductivity, that is the energy transferred per unit time, surface and temperature difference. I added this to give more visibility to his comment; my original answer follows.
Air is a very poor conductor of heat, the reason being the fact that the molecules are less concentrated and less interacting with each other, as you conjectured (this is not very precise, but in general situations this way of thinking works). On the opposite, solids are in general better conductors: this is the reason why you should not touch anything inside the oven. Considering order of magnitudes, according to Wikipedia, air has a thermal conductivity $ \lesssim 10^{-1} \ \text{W/(m K)} $, whereas for metals is higher at least of two orders of magnitude.
I really thank Zephyr and Chemical Engineer for the insight that they brought to my original answer, that was much poorer but got an unexpected fame.


Question:
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?

OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9
(EDIT: ...which is actually the proof that I read in Aigner & Ziegler).
When $0 < x < \pi/2$ we have $0<\sin x < x < \tan x$ and thus
$$\frac{1}{\tan^2 x} < \frac{1}{x^2} < \frac{1}{\sin^2 x}.$$
Note that $1/\tan^2 x = 1/\sin^2 x - 1$.
Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum
the inequality over the (inner) "gridpoints" $x_k=(\pi/2) \cdot (k/2^n)$:
$$\sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k} - \sum_{k=1}^{2^n-1} 1 < \sum_{k=1}^{2^n-1} \frac{1}{x_k^2} < \sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k}.$$
Denoting the sum on the right-hand side by $S_n$, we can write this as
$$S_n - (2^n - 1) < \sum_{k=1}^{2^n-1} \left( \frac{2 \cdot 2^n}{\pi} \right)^2 \frac{1}{k^2} < S_n.$$
Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with,
$$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$
Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short,
$$S_n = 4 S_{n-1} + 2.$$
Since $S_1=2$, the solution of this recurrence is
$$S_n = \frac{2(4^n-1)}{3}.$$
(For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.)
We now have
$$ \frac{2(4^n-1)}{3} - (2^n-1) \leq  \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2}  \leq \frac{2(4^n-1)}{3}.$$
Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!


Question:
Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?
Is there a good reason for this or is it simply "because it works".

I reproduce a blog post I wrote some time ago:
We tend to not use higher derivative theories. It turns out that there
is a very good reason for this, but that reason is rarely discussed in
textbooks. We will take, for concreteness, $L(q,\dot q, \ddot
  q)$, a Lagrangian which depends on the 2nd derivative in an
essential manner. Inessential dependences are terms such as $q\ddot q$
which may be partially integrated to give ${\dot q}^2$. Mathematically,
this is expressed through the necessity of being able to invert the
expression $$P_2 = \frac{\partial L\left(q,\dot q, \ddot
  q\right)}{\partial \ddot q},$$ and get a closed form for $\ddot q
  (q, \dot q, P_2)$. Note that usually we also require a
similar statement for $\dot q (q, p)$, and failure in this
respect is a sign of having a constrained system, possibly with gauge
degrees of freedom.
In any case, the non-degeneracy leads to the Euler-Lagrange equations in
the usual manner: $$\frac{\partial L}{\partial q} -
  \frac{d}{dt}\frac{\partial L}{\partial \dot q} +
  \frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot q} = 0.$$ This is then
fourth order in $t$, and so require four initial conditions, such as
$q$, $\dot q$, $\ddot q$, $q^{(3)}$. This is twice as many as usual, and
so we can get a new pair of conjugate variables when we move into a
Hamiltonian formalism. We follow the steps of Ostrogradski, and choose
our canonical variables as $Q_1 = q$, $Q_2 = \dot q$, which leads to
\begin{align} P_1 &= \frac{\partial L}{\partial \dot q} -
  \frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \\ P_2 &=
  \frac{\partial L}{\partial \ddot q}. \end{align} Note that the
non-degeneracy allows $\ddot q$ to be expressed in terms of $Q_1$, $Q_2$
and $P_2$ through the second equation, and the first one is only
necessary to define $q^{(3)}$.
We can then proceed in the usual fashion, and find the Hamiltonian
through a Legendre transform: \begin{align} H &= \sum_i P_i \dot{Q}_i -
  L \\ &= P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1,
  Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time
derivative of the Hamiltonian to find that it is time independent if the
Lagrangian does not depend on time explicitly, and thus can be
identified as the energy of the system.
However, we now have a problem: $H$ has only a linear dependence on
$P_1$, and so can be arbitrarily negative. In an interacting system this
means that we can excite positive energy modes by transferring energy
from the negative energy modes, and in doing so we would increase the
entropy — there would simply be more particles, and so a need to
put them somewhere. Thus such a system could never reach equilibrium,
exploding instantly in an orgy of particle creation. This problem is in
fact completely general, and applies to even higher derivatives in a
similar fashion.


Question:
What are the key differences between VCF versions 4.1 and 4.2?
 It looks like v4.3 contains a changelog (specs available here) but earlier specifications do not.
This biostar post points out one difference: the introduction of Number=R for fields with one value per allele including REF — can anyone enumerate the other changes between these two versions?

This is easy to check, you can download both specs in .tex format and do diff.
Changes to the v4.2 compared to v4.1:

Information field format: adding source and version as recommended fields.
INFO field can have one value for each possible allele (code R).
For all of the ##INFO, ##FORMAT, ##FILTER, and ##ALT metainformation, extra fields can be included after the default fields.
Alternate base (ALT) can include *: missing due to a upstream deletion.
Quality scores, a sentence removed: High QUAL scores indicate high confidence calls. Although traditionally people use integer phred scores, this field is permitted to be a floating point to enable higher resolution for low confidence calls if desired.
Examples changed a bit.


Question: What is your best guess how Google Image Search works? I can upload a photo and can search for similar images. What algorithm does it use to identify similar images?

I don't know which algorithm Google uses. But, since you wanted a best guess, let me give some ideas on how a similar system could be constructed.
The whole field dealing with search-image-base-by-image is called Content Based Image Retrieval (CBIR). The idea is somehow to construct an image representation (not necessarily understandable by humans) that contains the information about image content.
Two basic approaches exist:

retrieval using low-level (local) features: color, texture, shape at specific parts of images (an image is a collection of descriptors of local features)
semantic approaches where an image is, in some way, represented as a collection of objects and their relations

Low-level local approach is very well researched. The best current approach extracts local features (there's a choice of feature extraction algorithm involved here) and uses their local descriptors (again, choice of descriptors) to compare the images.
In newer works, the local descriptors are clustered first and then clusters are treated as visual words -- the technique is then very similar to Google document search, but using visual words instead of letter-words.
You can think of visual words as equivalents to word roots in language: for example, the words: work, working, worked all belong to the same word root.
One of the drawbacks of these kinds of methods is that they usually under-perform on low-texture images.
I've already given and seen a lot of answers detailing these approaches, so I'll just provide links to those answers:

CBIR: 1, 2
feature extraction/description: 1, 2, 3, 4

Semantic approaches are typically based on hierarchical representations of the whole image. These approaches have not yet been perfected, especially for the general image types. There is some success in applying these kind of techniques to specific image domains.
As I am currently in the middle of research of these approaches, I can not make any conclusions. Now, that said, I explained a general idea behind these techniques in this answer.
Once again, shortly: the general idea is to represent an image with a tree-shaped structure, where leaves contain the image details and objects can be found in the nodes closer to the root of such trees. Then, somehow, you compare the sub-trees to identify the objects contained in different images.
Here are some references for different tree representations. I did not read all of them, and some of them use this kind of representations for segmentation instead of CBIR, but still, here they are:

binary partition trees and mention of min/max trees: P. Salembier, M.H.F Wilkinson: Connected Operators
binary partition trees: V. Vilaplana, F. Marques, P. Selembier: Binary Partition Trees for Object Detection
tree of shapes (component tree): P. Monasse, F. Guichard: Fast Computation of Contrast-Invariant Image Representation, C. Ballester, V. Castellis, P. Monasse: The Tree of Shapes of an Image
monotonic trees: Y. Song, A. Zhang: Analyzing scenery images by monotonic tree
edit: further digging shows that tree of shapes and monotonic tree are equivalent, except processing the image in 4-/8- (tree of shapes) or 6-connectivity (monotonic)
extrema-watershed tree: A. Vichik, R. Keshet, D. Malah: Self-dual morphology on tree semilattices and applications
constrained connectivity, alpha-trees, ultrametric waterseads: P. Soille, L. Najman: On morphological hierarchical representations for image processing and spatial data clustering


Question:
USB specifies 4 pins:
1.   VBUS   +5V
2.   D-     Data-
3.   D+     Data+
4.   GND    Ground

Why is this not 3?  Could the Data and Power not share a common ground?  Am I correct in understanding that D- is the ground for D+?

No, D- is not ground. Data is sent over a differential line, which means that D- is a mirror image of D+, so both Data lines carry the signal. The receiver subtracts D- from D+. If some noise signal would be picked up by both wires, the subtraction will cancel it.

So differential signalling helps suppressing noise. So does the type of wiring, namely twisted pair. If the wires ran just parallel they would form a (narrow) loop which could pick up magnetic interference. But thanks to the twists the orientation of the wires with respect to the field changes continuously. An induced current will be cancelled by a current with the opposite sign half a twist further.
Suppose you have a disturbance working vertically on the twisted wire. You could regard each half twist as a small loop picking up the disturbance. Then it's easy to see that the next tiny loop sees the opposite field (upside down, so to speak), so that cancels the first field. This happens for each pair of half twists.
A similar balancing effect occurs for capacitance to ground. In a straight pair one conductor shows a higher capacitance to ground than the other, while in a twisted pair each wire will show the same capacitance.

edit
Cables with several twisted pairs like cat5 have a different twist length for each pair to minimize crosstalk.


Question:
Everyone knows computing speed has drastically increased since their invention, and it looks set to continue. But one thing is puzzling me: if you ran an electrical current through a material today, it would travel at the same speed as if you did it with the same material 50 years ago.
With that in mind, how is it computers have become faster? What main area of processor design is it that has given these incredible speed increases?
I thought maybe it could be one or more of the following:

Smaller processors (less distance for the current to travel, but it just seems to me like you'd only be able to make marginal gains here).
Better materials

if you ran an electrical current through a material today, it would travel at the same speed as if you did it with the same material 50 years ago.
With that in mind, how is it computers have become faster? What main area of processor design is it that has given these incredible speed increases?

You get erroneous conclusions because your initial hypothesis is wrong: you think that CPU speed is equivalent to the speed of the electrons in the CPU.
In fact, the CPU is some synchronous digital logic. The limit for its speed is that the output of a logical equation shall be stable within one clock period. With the logic implemented with transistors, the limit is mainly linked to the time required to make transistors switch. By reducing their channel size, we are able to make them switch faster. This is the main reason for improvement in max frequency of CPUs for 50 years. Today, we also modify the shape of the transistors to increase their switching speed, but, as far as I know, only Intel, Global Foundries and TSMC are able to create FinFETs today.
Yet, there are some other ways to improve the maximum clock speed of a CPU: if you split your logical equation into several smaller ones, you can make each step faster, and have a higher clock speed. You also need more clock periods to perform the same action, but, using pipelining techniques, you can make the rate of instructions per second follow your clock rate.
Today, the speed of electrons has become a limit: at 10GHz, an electric signal can't be propagated on more than 3cm. This is roughly the size of current processors. To avoid this issue, you may have several independent synchronous domains in your chip, reducing the constraints on signal propagation. But this is only one limiting factor, amongst transistor switching speed, heat dissipation, EMC, and probably others (but I'm not in the silicon foundry industry).


Question:
I've studied convex optimization pretty carefully, but don't feel that I have yet "grokked" the dual problem.  Here are some questions I would like to understand more deeply/clearly/simply:

How would somebody think of the dual problem?  What thought process would lead someone to consider the dual problem and to recognize that it's valuable/interesting?
In the case of a convex optimization problem, is there any obvious reason to expect that strong duality should (usually) hold?
It often happens that the dual of the dual problem is the primal problem.  However, this seems like a complete surprise to me.  Is there any intuitive reason to expect that this should happen?
Does the use of the word "dual" or "duality" in optimization have anything to do with the dual space in linear algebra?  Or are they just different concepts that go by the same name.  What about the use of the word "dual" in projective geometry — is there a connection there?
You can define the dual problem and prove theorems about strong duality without ever mentioning the Fenchel conjugate.  For example, Boyd and Vandenberghe prove a strong duality theorem without mentioning the Fenchel conjugate in their proof.  And yet, people often talk as if the Fenchel conjugate is somehow the "essence" of duality, and make it sound as if the whole theory of duality is based on the Fenchel conjugate.  Why is the Fenchel conjugate considered to have such fundamental importance?

Note: I will now describe my current level of understanding of the intuition behind the dual problem.  Please tell me if you think I might be missing any basic insights.
I have read the excellent notes about convex optimization by Guilherme Freitas, and in particular the part about "penalty intuition".  When we are trying to solve
\begin{align*}
\text{minimize} &\quad f(x) \\
\text{such that} & \quad h(x) \leq 0
\end{align*}
one might try to eliminate the constraints by introducing a penalty when constraints are violated.  This gives us the new unconstrained problem
\begin{equation}
\text{minimize} \quad f(x) + \langle \lambda ,h(x) \rangle
\end{equation}
where $\lambda \geq 0$. It's not hard to see that for a given $\lambda \geq 0$, the optimal value of this unconstrained problem is less than or equal to the optimal value for the constrained problem.  This gives us a new problem — find $\lambda$ so that the optimal value for the unconstrained problem is as large as possible.  That is one way to imagine how somebody might have thought of the dual problem.  Is this the best intuition for where the dual problem comes from?
Another viewpoint: the KKT conditions can be derived using what Freitas calls the "geometric intuition".  Then, if we knew the value of the multipliers $\lambda$, it would be (often) much easier to find $x$.  So, a new problem is to find $\lambda$.  And if we can somehow recognize that $\lambda$ is a maximizer for the dual problem, then this suggests that we might try solving the dual problem.
Please explain or give references to any intuition that you think I might find interesting, even if it's not directly related to what I asked.

Here's what's really going on with the dual problem.  (This is my attempt to answer my own question, over a year after originally asking it.)
(A very nice presentation of this material is given in Ekeland and Temam.  These ideas are also in Rockafellar.)
Let $V$ be a finite dimensional normed vector space over $\mathbb R$.  (Working in an inner product space or just in $\mathbb R^N$ risks concealing the fundamental role that the dual space plays in duality in convex optimization.)
The basic idea behind duality in convex analysis is to think of a convex set in terms of its supporting hyperplanes.  (A closed
convex set $\Omega$ can be "recovered" from its supporting hyperplanes
by taking the intersection of all closed half spaces containing $\Omega$.
The set of all supporting hyperplanes to $\Omega$ is sort of a
"dual representation" of $\Omega$.)
For a convex function $f$ (whose epigraph is a convex set), this strategy leads
us to think about $f$ in terms of affine functions $\langle m^*, x \rangle - \alpha$
which are majorized by $f$.  (Here $m^* \in V^*$ and we are using the notation $\langle m^*, x \rangle = m^*(x)$.)
For a given slope $m^* \in V^*$, we only need to consider the "best" choice of $\alpha$ -- the other affine minorants with slope $m^*$ can be disregarded.
\begin{align*}
& f(x) \geq \langle m^*,x \rangle - \alpha \quad \forall x \in V \\
\iff & \alpha \geq  \langle m^*, x \rangle - f(x) \quad \forall x \in V \\
\iff & \alpha \geq \sup_{x \in V} \quad \langle m^*, x \rangle - f(x)
\end{align*}
so the best choice of $\alpha$ is
\begin{equation}
f^*(m^*) = \sup_{x \in V} \quad \langle m^*, x \rangle - f(x).
\end{equation}
If this supremum is finite, then
$\langle m^*,x \rangle - f^*(m^*)$ is the best affine minorant of
$f$ with slope $m^*$.
If $f^*(m^*) = \infty$, then there is no affine minorant of $f$ with slope $m^*$.
The function $f^*$ is called the "conjugate" of $f$.  The definition and basic facts about $f^*$ are all highly intuitive.  For example, if $f$ is a proper closed convex function then $f$ can be recovered from $f^*$, because any closed convex set (in this case the epigraph of $f$) is the intersection of all the closed half spaces containing it.  (I still think the fact that the "inversion formula" $f = f^{**}$ is so simple is a surprising and mathematically beautiful fact, but not hard to derive or prove with this intuition.)
Because $f^*$ is defined on the dual space, we see already the fundamental role played by the dual space in duality in convex optimization.
Given an optimization problem, we don't obtain a dual problem until we specify how to perturb the optimization problem.  This is why equivalent formulations of an optimization problem can lead to different dual problems.  By reformulating it we have in fact specified a different way to perturb it.
As is typical in math, the ideas become clear when we work at an appropriate level of generality.  Assume that our optimization problem is
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,0).
\end{equation*}
Here $\phi:X \times Y \to \bar{\mathbb R}$ is convex.  Standard convex optimization problems can be written in this form with an appropriate choice of $\phi$.  The perturbed problems are
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,y)
\end{equation*}
for nonzero values of $y \in Y$.
Let $h(y) = \inf_x \phi(x,y)$.  Our optimization problem is simply to evaluate $h(0)$.
From our knowledge of conjugate functions, we know that
\begin{equation*}
h(0) \geq h^{**}(0)
\end{equation*}
and that typically we have equality.  For example, if $h$ is subdifferentiable at $0$ (which is typical for a convex function) then $h(0) = h^{**}(0)$.
The dual problem is simply to evaluate $h^{**}(0)$.
In other words, the dual problem is:
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - h^*(y^*).
\end{equation*}
We see again the fundamental role that the dual space plays here.
It is enlightening to express the dual problem in terms of $\phi$.  It's easy to show that the dual problem is
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - \phi^*(0,y^*).
\end{equation*}
So the primal problem is
\begin{equation*}
\operatorname*{minimize}_{x \in X} \quad \phi(x,0)
\end{equation*}
and the dual problem (slightly restated) is
\begin{equation*}
\operatorname*{minimize}_{y^* \in Y^*} \quad \phi^*(0,y^*).
\end{equation*}
The similarity between these two problems is mathematically beautiful, and we can see that if we perturb the dual problem in the obvious way, then the dual of the dual problem will be the primal problem (assuming $\phi = \phi^{**}$).  The natural isomorphism between $V$ and $V^{**}$ is of fundamental importance here.
The key facts about the dual problem -- strong duality, the optimality conditions, and the sensitivity interpretation of the optimal dual variables -- all become intuitively clear and even "obvious" from this viewpoint.
An optimization problem in the form
\begin{align*}
\operatorname*{minimize}_x & \quad f(x) \\
\text{subject to} & \quad g(x) \leq 0,
\end{align*}
can be perturbed as follows:
\begin{align*}
\operatorname*{minimize}_x & \quad f(x) \\
\text{subject to} & \quad g(x) + y \leq 0.
\end{align*}
This perturbed problem has the form given above with
\begin{equation*}
\phi(x,y) =
\begin{cases}
f(x) \quad \text{if } g(x) + y \leq 0 \\
\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}
To find the dual problem, we need to evaluate $-\phi^*(0,y^*)$, which is a relatively straightforward calculation.
\begin{align*}
-\phi^*(0,y^*) &= -\sup_{g(x) + y \leq 0} \quad \langle y^*,y \rangle - f(x) \\
&= -\sup_{\substack{ x \\ q \geq 0 }} \quad \langle y^*, -g(x) - q \rangle - f(x) \\
&= \inf_{\substack{ x \\ q \geq 0 }} \quad f(x) + \langle y^*, g(x) \rangle + \langle y^*, q \rangle.
\end{align*}
We can minimize first with respect to $q$, and we will get $-\infty$ unless $\langle y^*, q \rangle \geq 0$ for all $q \geq 0$.  In other words, we will get $-\infty$ unless $y^* \geq 0$.
The dual function is
\begin{equation*}
-\phi^*(0,y^*) =
\begin{cases}
\inf_x \quad f(x) + \langle y^*, g(x) \rangle \quad \text{if } y^* \geq 0 \\
-\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}
This is the expected result.


Question:
There was some discussion on this question

What are some reasons to connect capacitors in series?
What are some reasons to connect capacitors in series?

which I don't see as being conclusively resolved:

"turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics."
"No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator."
'Kind of like "you can't make a BJT from two diodes"'
"it is a process that a tinkerer cannot do"

So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?  Does it not survive the same voltages?  What happens to the reverse-biased cap when a large voltage is placed across the combination?  Are there practical limitations other than physical size?  Does it matter which polarity is on the outside?
I don't see what the difference is, but a lot of people seem to think there is one.
Summary:
As posted in one of the comments, there's a sort of electrochemical diode going on:

The film is permeable to free electrons but substantially impermeable to ions, provided the temperature of the cell is not high. When the metal underlying the film is at a negative potential, free electrons are available in this electrode and the current flows through the film of the cell. With the polarity reversed, the electrolyte is subjected to the negative potential, but as there are only ions and no free electrons in the electrolyte the current is blocked. — The Electrolytic Capacitor by Alexander M. Georgiev

Normally a capacitor cannot be reverse-biased for long, or large currents will flow and "destroy the center layer of dielectric material via electrochemical reduction":

An electrolytic can withstand a reverse bias for a short period, but will conduct significant current and not act as a very good capacitor. — Wikipedia: Electrolytic capacitor

However, when you have two back-to-back, the forward-biased capacitor prevents a prolonged DC current from flowing.
Works for tantalums, too:

For circuit positions when reverse voltage excursions are unavoidable,
  two similar capacitors in series connected “back to back” ... will create a
  non-polar capacitor function ... This works because almost all the circuit voltage
  is dropped across the forward biased capacitor, so that the reverse
  biased device sees only a negligible voltage.

Solid Tantalum Capacitors Frequently Asked Questions (FAQs):

The oxide dielectric construction that is used in tantalum capacitors
  has a basic rectified property which blocks current flow in one
  direction and at the same time offers a low resistance path in the
  opposite direction.

Summary:

Yes "polarised" aluminum "wet electrolytic" capacitors can legitimately be connected "back-to-back" (ie in series with opposing polarities) to form a non-polar capacitor.
C1 + C2 are always equal in capacitance and voltage rating
Ceffective =  = C1/2 = C2/2
Veffective = vrating of C1 & C2.
See "Mechanism" at end for how this (probably) works.

It is universally assumed that the two capacitors have identical capacitance when this is done.
 The resulting capacitor  with half the capacitance of each individual capacitor.
 eg if two x 10 uF capacitors are placed in series the resulting capacitance will be 5 uF.
I conclude that the resulting capacitor will have the same voltage rating as the individual capacitors. (I may be wrong).
I have seen this method used on many occasions over many years and, more importanttly have seen the method described in application notes from a number of capacitor manufacturers. See at end for one such reference.
Understanding how the individual capacitors become correctly charged requires either faith in the capacitor manufacturers statements ("act as if they had been bypassed by diodes" or additional complexity BUT understanding how the arrangement  works once initiated is easier.
 Imagine two back-to-back caps with Cl fully charged and Cr fully discharged.
 If a current is now passed though the series arrangement such that Cl then discharges to zero charge then the reversed polarity of Cr will cause it to be charged to full voltage.   Attempts to apply additional current and to further discharge Cl so it assumes incorrect polarity would lead to Cr being charge above its rated voltage. ie it could be attempted BUT would be outside spec for both devices.
Given the above, the specific questions can be answered:

What are some reasons to connect capacitors in series?

Can create a bipolar cap from 2 x polar caps.
 OR can double rated voltage as long as care is taken to balance voltage distribution. Paralleld resistors are sometimes used to help achieve balance.

"turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics."

This can be done with oridinary electrolytics.

"No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator."

Works OK if ratings are not exceeded.

'Kind of like "you can't make a BJT from two diodes"'

Reason for comparison is noted but is not a valid one. Each half capacitor is still subject to same rules and demands as when standing alone.

"it is a process that a tinkerer cannot do"

Tinkerer can - entirely legitimate.

So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?

It coild be but the manufacturers usually make a manufacturing change so that there are two Anode foils BUT the result is the same.

Does it not survive the same voltages?

Voltage rating is that of a single cap.

What happens to the reverse-biased cap when a large voltage is placed across the combination?

Under normal operation there is NO reverse biased cap.  Each cap handles a full cycle of AC whole effectively seeing half a cycle. See my explanation above.

Are there practical limitations other than physical size?

No obvious limitation that i can think of.

Does it matter which polarity is on the outside?

No. Draw a picture of what each cap sees in isolation without reference to what is "outside it. Now change their order in the circuit. What they see is identical.

I don't see what the difference is, but a lot of people seem to think there is one.

You are correct. Functionally from a "black box" point of view they are the same.

MANUFACTURER'S EXAMPLE:
In this document Application Guide, Aluminum Electrolytic Capacitors bY Cornell Dubilier, a competent and respected capacitor manufacturer it says (on age 2.183 & 2.184)

If two, same-value, aluminum electrolytic capacitors
are connected in series, back-to-back with the positive
terminals or the negative terminals connected, the
resulting single capacitor is a non-polar capacitor with
half the capacitance.
The two capacitors rectify the
applied voltage and act as if they had been bypassed
by diodes.
When voltage is applied, the correct-polarity capacitor gets the full voltage.
In non-polar aluminum electrolytic capacitors and motor-start aluminum electrolytic capacitors a second anode foil substitutes for the cathode foil to achieve a non-polar capacitor in a single case.

Of relevance to understanding the overall action is this comment from page  2.183.

While it may appear that the capacitance is between
the two foils, actually the capacitance is between the
anode foil and the electrolyte.
The positive plate is the
anode foil;
the dielectric is the insulating aluminum
oxide on the anode foil;
the true negative plate is the
conductive, liquid electrolyte, and the cathode foil
merely connects to the electrolyte.
This construction delivers colossal capacitance
because etching the foils can increase surface area
more than 100 times and the aluminum-oxide dielectric is less than a micrometer thick.  Thus the resulting
capacitor has very large plate area and the plates are
awfully close together.

ADDED:
I intuitively feel as Olin does that it should be necessary to provide a means of maintaining correct polarity. In practice it seems that the capacitors do a good job of accommodating the startup "boundary condition". Cornell Dubiliers "acts like a diode" needs better understanding.

MECHANISM:
I think the following describes how the system works.
As I described above, once one capacitor is fully charged at one extreme of the AC waveform and the other fully discharged then the system will operate correctly, with charge being passed into the outside "plate" of one cap, across from inside plate of that cap to the other cap and "out the other end". ie a body of charge transfers to and from between the two capacitors and allows net charge flow to and from through the dual cap. No problem so far.
A correctly biased capacitor has very low leakage.
 A reverse biased capacitor has higher leakage and possibly much higher.
 At startup one cap is reverse biased on each half cycle and leakage current flows.
 The charge flow is such as to drive the capacitors towards the properly balanced condition.
 This is the "diode action" referred to - not formal rectification per say but leakage under incorrect operating bias.
 After a number of cycles balance will be achieved. The "leakier" the cap is in the reverse direction the quicker balance will be achieved.
 Any imperfections or inequalities will be compensated for by this self adjusting mechanism.
 Very neat.


Question:
This is a general question that was asked indirectly multiple times in here, but it lacks a single authoritative answer. It would be great to have a detailed answer to this for the reference.
Accuracy, the proportion of correct classifications among all classifications, is very simple and very "intuitive" measure, yet it may be a poor measure for imbalanced data. Why does our intuition misguide us here and are there any other problems with this measure?

Most of the other answers focus on the example of unbalanced classes. Yes, this is important. However, I argue that accuracy is problematic even with balanced classes.
Frank Harrell has written about this on his blog: Classification vs. Prediction and Damage Caused by Classification Accuracy and Other Discontinuous Improper Accuracy Scoring Rules.
Essentially, his argument is that the statistical component of your exercise ends when you output a probability for each class of your new sample. Mapping these predicted probabilities $(\hat{p}, 1-\hat{p})$ to a 0-1 classification, by choosing a threshold beyond which you classify a new observation as 1 vs. 0 is not part of the statistics any more. It is part of the decision component. And here, you need the probabilistic output of your model - but also considerations like:

What are the consequences of deciding to treat a new observation as class 1 vs. 0? Do I then send out a cheap marketing mail to all 1s? Or do I apply an invasive cancer treatment with big side effects?
What are the consequences of treating a "true" 0 as 1, and vice versa? Will I tick off a customer? Subject someone to unnecessary medical treatment?
Are my "classes" truly discrete? Or is there actually a continuum (e.g., blood pressure), where clinical thresholds are in reality just cognitive shortcuts? If so, how far beyond a threshold is the case I'm "classifying" right now?
Or does a low-but-positive probability to be class 1 actually mean "get more data", "run another test"?

Depending on the consequences of your decision, you will use a different threshold to make the decision. If the action is invasive surgery, you will require a much higher probability for your classification of the patient as suffering from something than if the action is to recommend two aspirin. Or you might even have three different decisions although there are only two classes (sick vs. healthy): "go home and don't worry" vs. "run another test because the one we have is inconclusive" vs. "operate immediately".
The correct way of assessing predicted probabilities $(\hat{p}, 1-\hat{p})$ is not to compare them to a threshold, map them to $(0,1)$ based on the threshold and then assess the transformed $(0,1)$ classification. Instead, one should use proper scoring-rules. These are loss functions that map predicted probabilities and corresponding observed outcomes to loss values, which are minimized in expectation by the true probabilities $(p,1-p)$. The idea is that we take the average over the scoring rule evaluated on multiple (best: many) observed outcomes and the corresponding predicted class membership probabilities, as an estimate of the expectation of the scoring rule.
Note that "proper" here has a precisely defined meaning - there are improper scoring rules as well as proper scoring rules and finally strictly proper scoring rules. Scoring rules as such are loss functions of predictive densities and outcomes. Proper scoring rules are scoring rules that are minimized in expectation if the predictive density is the true density. Strictly proper scoring rules are scoring rules that are only minimized in expectation if the predictive density is the true density.
As Frank Harrell notes, accuracy is an improper scoring rule. (More precisely, accuracy is not even a scoring rule at all: see my answer to Is accuracy an improper scoring rule in a binary classification setting?) This can be seen, e.g., if we have no predictors at all and just a flip of an unfair coin with probabilities $(0.6,0.4)$. Accuracy is maximized if we classify everything as the first class and completely ignore the 40% probability that any outcome might be in the second class. (Here we see that accuracy is problematic even for balanced classes.) Proper scoring-rules will prefer a $(0.6,0.4)$ prediction to the $(1,0)$ one in expectation. In particular, accuracy is discontinuous in the threshold: moving the threshold a tiny little bit may make one (or multiple) predictions change classes and change the entire accuracy by a discrete amount. This makes little sense.
More information can be found at Frank's two blog posts linked to above, as well as in Chapter 10 of Frank Harrell's Regression Modeling Strategies.
(This is shamelessly cribbed from an earlier answer of mine.)

EDIT. My answer to Example when using accuracy as an outcome measure will lead to a wrong conclusion gives a hopefully illustrative example where maximizing accuracy can lead to wrong decisions even for balanced classes.


Question: In machine learning, people talk about objective function, cost function, loss function. Are they just different names of the same thing? When to use them? If they are not always refer to the same thing, what are the differences?

These are not very strict terms and they are highly related. However:

Loss function is usually a function defined on a data point, prediction and label, and measures the penalty. For example:

Square loss: $l(f(x_i|\theta),y_i) = \left (f(x_i|\theta)-y_i \right )^2$, used in linear regression
Hinge loss: $l(f(x_i|\theta), y_i) = \max(0, 1-f(x_i|\theta)y_i)$, used in SVM
0/1 loss: $l(f(x_i|\theta), y_i) = 1 \iff f(x_i|\theta) \neq y_i$, used in theoretical analysis and definition of accuracy

Cost function is usually more general. It might be a sum of loss functions over your training set plus some model complexity penalty (regularization). For example:

Mean Squared Error: $MSE(\theta) = \frac{1}{N} \sum_{i=1}^N \left (f(x_i|\theta)-y_i \right )^2$
SVM cost function: $SVM(\theta) = \|\theta\|^2 + C \sum_{i=1}^N \xi_i$ (there are additional constraints connecting $\xi_i$ with $C$ and with training set)

Objective function is the most general term for any function that you optimize during training. For example, a probability of generating training set in maximum likelihood approach is a well defined objective function, but it is not a loss function nor cost function (however you could define an equivalent cost function). For example:

MLE is a type of objective function (which you maximize)
Divergence between classes can be an objective function but it is barely a cost function, unless you define something artificial, like 1-Divergence, and name it a cost

Long story short, I would say that:
A loss function is a part of a cost function which is a type of an objective function.
All that being said, thse terms are far from strict, and depending on context, research group, background, can shift and be used in a different meaning. With the main (only?) common thing being "loss" and "cost" functions being something that want wants to minimise, and objective function being something one wants to optimise (which can be both maximisation or minimisation).


Question:
I am new to signal processing and especially to FFT, hence I am not sure if I am doing the correct thing here and I am a bit confused with the result.
I have a discrete real function (measurement data) and want to set up a low pass filter on that. The tool of choice is Python with the numpy package. I follow this procedure:

compute the fft of my function
cut off high frequencies
perform the inverse fft

Here is the code that I am using:
import numpy as np
sampling_length = 15.0*60.0 # measured every 15 minutes
Fs = 1.0/sampling_length
ls = range(len(data)) # data contains the function
freq = np.fft.fftfreq(len(data), d = sampling_length)
fft = np.fft.fft(data)
x = freq[:len(data)/2]
for i in range(len(x)):
if x[i] > 0.005: # cut off all frequencies higher than 0.005
    fft[i] = 0.0
    fft[len(data)/2 + i] = 0.0
inverse = np.fft.ifft(fft)

Is this the correct procedure? The result inverse contains complex values, that confuses me.

The fact that the result is complex is to be expected. I want to point out a couple things:
You are applying a brick-wall frequency-domain filter to the data, attempting to zero out all FFT outputs that correspond to a frequency greater than 0.005 Hz, then inverse-transforming to get a time-domain signal again. In order for the result to be real, then the input to the inverse FFT must be conjugate symmetric. This means that for a length-$N$ FFT,
$$
X[k] = X^*[N-k], k = 1, 2, \ldots , \frac{N}{2} - 1\;\;\;\;\;\;\;(N\;\; even)
$$
$$
X[k] = X^*[N-k], k = 1, 2, \ldots , \lfloor \frac{N}{2} \rfloor\;\;\;\;\;\;\;\;(N\;\; odd)
$$

Note that for $N$ even, $X[0]$ and $X[\frac{N}{2}]$ are not equal in general, but they are both real. For odd $N$, $X[0]$ must be real.

I see that you attempted to do something like this in your code above, but it is not quite correct. If you enforce the above condition on the signal that you pass to the inverse FFT, then you should get a real signal out.
My second point is more of a philosophical one: what you're doing will work, in that it will suppress the frequency-domain content that you don't want. However, this is not typically the way a lowpass filter would be implemented in practice. As I mentioned before, what you're doing is essentially applying a filter that has a brick-wall (i.e. perfectly rectangular) magnitude response. The impulse response of such a filter has a $sinc(x)$ shape. Since multiplication in the frequency domain is equivalent to (in the case of using the DFT, circular) convolution in the time domain, this operation is equivalent to convolving the time domain signal with a $sinc$ function.
Why is this a problem? Recall what the $sinc$ function looks like in the time domain (below image shamelessly borrowed from Wikipedia):

The $sinc$ function has very broad support in the time domain; it decays very slowly as you move in time away from its main lobe. For many applications, this is not a desirable property; when you convolve a signal with a $sinc$, the effects of the slowly-decaying sidelobes will often be apparent in the time-domain form of the filtered output signal. This sort of effect is often referred to as ringing. If you know what you're doing, there are some instances where this type of filtering might be appropriate, but in the general case, it's not what you want.
There are more practical means of applying lowpass filters, both in the time and frequency domains. Finite impulse response and infinite impulse response filters can be applied directly using their difference equation representation. Or, if your filter has a sufficiently-long impulse response, you can often obtain performance benefits using fast convolution techniques based on the FFT (applying the filter by multiplying in the frequency domain instead of convolution in the time domain), like the overlap-save and overlap-add methods.


Question:
Every once in a while, my eighth-inch audio jack will slip loose and I'll seemingly lose only the voice part of a track -- leaving somewhat of a "karaoke" version. What I would guess about how audio plugs work suggests that I'd be making this up; however, I've asked and others tell me they've experienced this as well.
What causes this stripped vocals from audio when a 1/8" audio jack is partially unplugged?

When the plug starts to slip out of the jack, very often it's the ground contact (sleeve) that breaks its connection first, leaving the two "hot" leads (left and right, tip and ring) still connected.
With the ground open like this, both earpieces still get a signal, but now it's the "difference" signal between the left and right channels; any signal that is in-phase in both channels cancels out.
Recording engineers tend to place the lead vocal signal right in the middle of the stereo image, so that's just one example of an in-phase signal that disappears when you're listening to the difference signal.


Question:
I was soldering a very thin wire today, and when I had one end firmly soldered, I accidentally bumped the wire diagonally with my tweezers. What I'd expect to happen is that the wire oscillates for a little while in one axis, then stops. However, what actually occurred is quite different and much more interesting! I recorded it in real-time;  (sorry for the poor macro focus), and recorded it again at 480FPS and imported it into Tracker video analysis;
As you can see, the rotational motion fully reverses!
Here are some still frames from Tracker:
The wire begins to rotate clockwise after being excited:

The wire begins to oscillate in one axis:

And, mindbogglingly, begins to rotate counterclockwise!

(clearer views in the videos above)
The X and Y axis motion plotted by tracker raises even more questions:

As you can see, the X axis motion simply stops, then restarts!
What's going on?
My first thought was that (because this wire wasn't originally straight) there was some sort of unusual standing wave set up, but this occurs even with a completely straight wire.
I'm absolutely sure there's something about two-axis simple harmonic motion that I'm missing, but I just cannot think of what is causing this. I've seen many other "home-experiment" questions on this site, so I thought this would an acceptable question; I hope it's not breaking any rules.
EDIT:
Okay, I've got some more data! I've set up a little solenoid plunger system that produces no torque or two-axis motion, and it's very repeatable. Here:
What I've noticed is that I can get almost any wire (even with a 90-degree bend!) to exhibit single-axis motion with this setup, with no spinning or deviation; and if I try enough, the same thing can happen with the tweezers. It seems like if I slide the tweezers slightly when exciting the wire, I can reliably produce this odd motion. I don't know what that indicates.
EDIT2:
Okay, seems like with the plunger-solenoid I still can get this circular motion even with a straight wire.
EDIT3:
Okay, so I wanted to test @sammy's suggestion once and for all. I assume that changing the moment of inertia to torsion of the wire would affect his theory, so I soldered a small piece of wire perpendicularly to the end of the main wire:

Then I recorded the motion;

And then I took off the perpendicular wire, and re-recorded the data:

And then I did it again (got noisy data first time):

EDIT N:
The final test!
Floris's hypothesis requires that the resonant frequency of a wire in each cardinal direction be different. To measure this, I used my solenoid setup that did not cause rotation, as above. I put a straight piece of wire between a light source and a light-dependent resistor and connected it to an oscilloscope;

The signal was very faint (42 millivolts), but my scope was able to pull it out of the noise. I have determined this:
In the +x direction, the resonant frequency of a just-straightened straight sample wire (unknown cycle frequency) is 51.81hz,+/-1hz;

In the +y direction, the resonant frequency of a sample wire is 60.60hz,+/-1hz;

So there's definitely a significant difference (~10 percent!) between the cardinal directions. Good enough proof for me.
EDIT N+1:
Actually, since my light detector above produces two pulses per sine wave, the actual vibration frequency is f/2; so the actual frequencies are 25.5 hz, and 30hz, which agrees roughly with @floris's data.

Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes1.
You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, is exactly what you are showing.
The first signal you show - which seems to "die" then come back to life, is exactly what you expect to see when you have two oscillations of slightly different frequency superposed; in fact, from the time to the first minimum we can estimate the approximate difference in frequency: it takes 19 oscillations to reach a minimum, and since the two waves started out in phase, that means they will be in phase again after about 38 oscillations, for a 2.5% difference in frequency.
Update
Here is the output of my little simulation. It took me a bit of time to tweak things, but with frequencies of 27 Hz and 27.7 Hz respectively and after adjusting the angle of excitation a little bit, and adding significant damping I was able to generate the following plots:

which looks a lot like the output of your tracker.
Your wire is describing a Lissajous figure. Very cool experiment - well done capturing so much detail! Here is an animation that I made, using a frequency difference of 0.5 Hz and a small amount of damping, and that shows how the rotation changes from clockwise to counterclockwise:

For your reference, here is the Python code I used to generate the first pair of curves. Not the prettiest code... I scale things twice. You can probably figure out how to reduce the number of variables needed to generate the same curve - in the end it's a linear superposition of two oscillations, observed at a certain angle to their principal axes.
import numpy as np
import matplotlib.pyplot as plt
from math import pi, sin, cos

f1 = 27.7
f2 = 27
theta = 25*pi/180.

# different amplitudes of excitation
A1 = 2.0
A2 = 1.0

t = np.linspace(0,1,400)

#damping factor
k = 1.6

# raw oscillation along principal axes:
a1 = A1*np.cos(2*pi*f1*t)*np.exp(-k*t)
a2 = A2*np.cos(2*pi*f2*t)*np.exp(-k*t)

# rotate the axes of detection
y1 = cos(theta)*a1 - sin(theta)*a2
y2 = sin(theta)*a1 + cos(theta)*a2

plt.figure()
plt.subplot(2,1,1)
plt.plot(t,-20*y2) # needed additional scale factor
plt.xlabel('t')
plt.ylabel('x')

plt.subplot(2,1,2)
plt.plot(t,-50*y1) # and a second scale factor
plt.xlabel('t')
plt.ylabel('y')
plt.show()

1. The frequency of a rigid beam is proportional to $\sqrt{\frac{EI}{A\rho}}$, where $E$ is Young's modulus, $I$ is the second moment of area, $A$ is the cross sectional area and $\rho$ is the density (see section 4.2 of "The vibration of continuous structures"). For an elliptical cross section with semimajor axis $a$ and $b$, the second moment of area is proportional to $a^3 b$ (for vibration along axis $a$). The ratio of resonant frequencies along the two directions will be $\sqrt{\frac{a^3b}{ab^3}} = \frac{a}{b}$. From this it follows that a 30 gage wire (0.254 mm) with a 2.5% difference in resonant frequency needs the perpendicular measurements of diameter to be different by just 6 µm to give the effect you observed. Given the cost of a thickness gage with 1 µm resolution, this is really a very (cost) effective way to determine whether a wire is truly round.


Question:
Mini USB connectors were standardized as part of USB 2.0 in 2000. In 2007, the USB Implemeters Forum standardized Micro USB connectors, deprecating Mini USB connectors four months later.
Why? What are the advantages of Micro USB over Mini USB that made USB-IF rip out an existing standard and replace it with another one that's basically the same thing?

Added mid 2022:
A lightly edited version of a comment by @LittleWhole
In 2022 the world is moving towards the far more robust and convenient USB-C connector. While there are still issues with USB-C (including even mechanical incompatibilities), things are slowly being addressed (i.e. USB4 standard on the protocol side) and I have only ever encountered one USB-C cable that wouldn't plug into a USB-C receptacle in my life. Adoption of USB-C is definitely picking up the pace - not just in consumer electronics, but a motor controller for my school's robotics club has even adopted USB-C
_____________________
A major flaw:
A major factor in abandoning mini-USB is that it was fatally flawed mechanically. Most people who have used a mini-USB device which requires many insertions will have experienced poor reliability after a significant but not vast number of uses.
The original mini-USB had an extremely poor insertion lifetime - about 1000 insertions total claimed. That's about once a day for 3 years. Or 3 times a day for one year. Or ...  For some people that order of reliability may be acceptable and the problems may go unnoticed. For others it becomes a major issue. A photographer using a flash card reader may expend that lifetime in well under a year.
The original mini-USB connector had sides which sloped as at present but they were reasonably straight. (Much the same as the sides on a micro-A connector). These are now so rare that I couldn't find an image using a web search. This image is diagrammatic only but shows the basic shape with sloped but straight sides.

Efforts were made to address the low lifetime issues while maintaining backwards compatibility and the current "kinked sides" design was produced. Both plug and socket were changed but the sockets ("receptacle") will still accept the old straight sided plugs. This is the shape that we are all so used to that the old shape is largely forgotten.

Unfortunately, this alteration "only sort of worked".  Insertion lifetime was increased to about 5,000 cycles. This sounds high enough in theory but in practice the design was still walking wounded with respect to mechanical reliability. 5,000 cycles is a very poor rating in the connector industry. While most users will not achieve that many insertion cycles, the actual reliability in heavy use is poor.
The micro-USB connector was designed with these past failings in mind and has a rated lifetime of about 10,000 insertion cycles. This despite its apparent frailty and what may appear to be a less robust design. [This still seems woefully low to me. Time will tell].
Latching Unlike mini USB, Micro USB has a passive latching mechanism which increases retention force but which allows removal without active user action (apart from pulling). [Latching seems liable to reduce the plug  "working" in the receptacle and may increase reliability].
Size matters:
The micro and mini USB connectors are of similar width. But the micro connector is much thinner (smaller vertical dimension). Some product designs were not able to accommodate the height of the mini receptacle and the new thinner receptacle will encourage and allow thinner products. A mini-USB socket would have been too tall for thin design. By way of example - a number of Motorola's "Razr" cellphones used micro-USB receptacles, thus allowing  the designs to be thinner than would have been possible with a Mini-USB receptacle.

Specific Razr models which use MICRO-USB include RAZR2 V8, RAZR2 V9, RAZR2 V9m,
RAZR2 V9x, DROID RAZR, RAZR MAXX & RAZR VE20.

Wikipedia on USB - see "durability".
Connector manufacturer Molex's micro USB page
They say:

Micro-USB technology was developed by the USB Implementers Forum, Inc. (USB-IF), an independent nonprofit group that advances USB technology. Molex's Micro-USB connectors offer advantages of smaller size and increased durability compared with the Mini-USB. Micro-USB connectors allow manufacturers to push the limits of thinner and lighter mobile devices with sleeker designs and greater portability.

Micro-USB replaces a majority of Mini-USB plugs and receptacles currently in use. The specification of the Micro-USB supports the current USB On-The-Go (OTG) supplement and provides total mobile interconnectivity by enabling portable devices to communicate directly with each other without the need for a host computer.
... Other key features of the product include high durability of over 10,000 insertion cycles, and a passive latching mechanism that provides higher extraction forces without sacrificing the USB's ease-of-use when synchronizing and charging portable devices.
All change:
Once all can change, all tend to. A significant driver to a common USB connector is the new USB charging standard which is being adopted by all cellphone makers. (Or all who wish to survive). The standard relates primarily to the electrical standards required to allow universal charging and chargers but a common mechanical connection system using the various micro-USB components is part of the standard. Whereas in the past it only really mattered that your 'whizzygig' could plug into its supplied power supply, it is now required that any whizzygig's power supply will fit any other device. A common plug and socket system is a necessary minimum for this to happen. While adapters can be used this is an undesirable approach. As USB charging becomes widely accepted not only for cellphones but for xxxpods, xxxpads, pda's and stuff in general, the drive for a common connector accelerates. The exception may be  manufacturers whose names begin with A who consider themselves large enough and safe enough to actively pursue interconnect incompatibility in their products.
Once a new standard is widely adopted and attains 'critical mass" the economics of scale tend to drive the market very rapidly to the new standard. It becomes increasingly less cost effective to manufacture and stock and handle parts which have a diminishing market share and which are incompatible with new facilities.
I may add some more references to this if it appears there is interest - or ask Mr Gargoyle.
Large list of cellphones that use micro-USB receptacle
_______________________________
_______________________________
A few more images allowing comparisons of a range of aspects including thickness, area of panel, overall volume (all being important independently of the others to some for various reasons) and retention means.
Large Google image samples each linked to a web page
and more
Useful discussion & brief history Note: they say (and, as Bailey S also notes)

Why Micro types offer better durability?
Accomplished by moving leaf-spring from the PCB receptacle to plug, the most-stressed part is now on the cable side of the connection.
Inexpensive cable bears most wear instead of the µUSB device.

Maybe useful:
USB CONNECTOR GUIDE — GUIDE TO USB CABLES
USB connections compared
What is Micro USB vs Mini USB


Question:
I was drinking a glass of milk the other day and that got me thinking that no other animal to my knowledge drinks milk past their infant stages. One could argue that cats might but it isn't good for them to do.
Are humans the only animal that are able to drink milk as adults and not have it cause issues?
Of course, I know some people do have lactose intolerance too.

Good observation!
Gene coding for the lactase
Gene LCT
Mammals have a gene (called LCT C/T-13910) coding for the lactase enzyme, a protein able to digest lactose. Lactose is a disaccharide sugar found in milk.
Expression of LCT
In mammals, the gene LCT is normally expressed (see gene expression) only early in development, when the baby feeds on his/her mother's milk. Some human lineages have evolved the ability to express LCT all life long, allowing them to drink milk and digest lactose at any age.
Today, the inability to digest lactose at all ages in humans is called lactose intolerance.
Evolution of lactose tolerance in human
Three independent mutations
Tishkoff et al. 2007 found that the ability to express LCT at an old age has evolved at least three times independently. Indeed, they found three different SNPs (stands for Single Nucleotide Polymorphism; it is a common type of mutation), two of them having high prevalence in Africa (and people of African descent) and one having high prevalence in Europe (and people of European descent). The three SNPs are G/C-14010, T/G-13915 and C/G-13907.
Pastoralist populations
Lactose tolerance is much more common in people descending from pastoralist populations than in people descending from non-pastoralist populations, suggesting a strong selection for lactose tolerance Durham 1991.
Selective sweep
On top of that, Tishkoff et al. 2007 focusing on the locus 14010 (one of the three SNP's mentioned above) showed that there is a clear selective sweep (which is a signature of past and present selection) around this locus.
They estimated the age of the allele allowing lactose tolerance at this locus (allele C is derived, the ancestral being G; see nucleotide) at around 3,000 to 7,000 years (with a 95% confidence interval ranging from 1,200 to 23,200 years) and a selection coefficient of 0.04 - 0.097 (with a 95% confidence interval ranging from 0.01 to 0.15).
I recommend reading Tishkoff et al. 2007. It is a classic, is short and is relatively easy to read, even for someone with only basic knowledge in evolutionary biology.
Are humans the only animal that is able to drink milk as adults?
I don't really know... but I would think so, yes!
Drink vs digest thoroughly
As @anongoodnurse rightly said in his/her answer

"Drink" and "digest thoroughly" are two different things

Pets
According to many dog health websites (such this one for example) claim that there is also variance among dogs where some dogs are lactose tolerant and others are lactose intolerant. I could not find any paper on the underlying genetics of lactose intolerance in dogs or other pets. It is not impossible our pets have also been under selection to be able to digest lactose as we humans could have given milk to them. It is also possible that pets do not actually produce any lactase at adult age but rather that some pets are just able to deal with having indigestible lactose in their guts! But then again, "Drink" and "digest thoroughly" are two different things.
Tits and robins in 20th century England
A funny and famous case is the case of blue tits and robins in the 20th century, in England. At that time, in England, the milkman was bringing the milk at home in the morning and would leave glass bottles with a simple aluminum cap in front of people's home. At some point, blue tits and robins learnt that by pecking through the aluminum they can get access to the milk. See this (non-peer-reviewed) article that tells the story.
Somewhat related
There are already a number of good posts on milk digestion in humans on Biology.SE. Consider having a look at:

What inactivates pepsin in infants? and Seriously, do humans produce rennin?
Are there any non-mammalian species known that lactate?
Can an adult without genetic lactase persistence still develop a tolerance for dairy foods?


Question: Why do some assemblers like SOAPdenovo2 or Velvet require an odd-length k-mer size for the construction of de Bruijn graph, while some other assemblers like ABySS are fine with even-length k-mers?

From the manual of Velvet:

it must be an odd number, to avoid palindromes. If you put in an even
  number, Velvet will just decrement it and proceed.

the palindromes in biology are defined as reverse complementary sequences. The problem of palindromes is explained in this review:

Palindromes induce paths that fold back on themselves. At least one
  assembler avoids these elegantly; Velvet requires K, the length of a
  K-mer, to be odd. An odd-size K-mer cannot match its reverse
  complement.

It is possible to construct graph with palindromes, but then the interpretation will be harder. Allowing only graphs of odd k-mers is just an elegant way to avoid writing a code for interpretation of a more complicated graph.