Question:
Polyploidy is the multiplication of number of chromosomal sets from 2n to 3n (triploidy), 4n (tetraploidy) and so on. It is quite common in plants, for example many crops like wheat or Brassica forms. It seems to be rarer in animals but still it is present among some amphibian species like Xenopus.
As I know in mammals polyploidy is lethal (I don't mean tissue - limited polyploidy). I understand that triploidy is harmful due to stronger influence of maternal or paternal epigenetic traits that cause abnormal development of placenta, but why there is no tetraploid mammals?

Great question, and one about which there has historically been a lot of speculation, and there is currently a lot of misinformation. I will first address the two answers given by other users, which are both incorrect but have been historically suggested by scientists. Then I will try to explain the current understanding (which is not simple or complete). My answer is derived directly from the literature, and in particular from Mable (2004), which in turn is part of the 2004 special issue of the Biological Journal of the Linnean Society tackling the subject.
The 'sex' answer...
In 1925 HJ Muller addressed this question in a famous paper, "Why polyploidy is rarer in animals than in plants" (Muller, 1925). Muller briefly described the phenomenon that polyploidy was frequently observed in plants, but rarely in animals. The explanation, he said, was simple (and is approximate to that described in Matthew Piziak's answer):

animals usually have two sexes which are differentiated by means of a process involving the diploid mechanism of segregation and combination whereas plants-at least the higher plants-are usually hermaphroditic.

Muller then elaborated with three explanations of the mechanism:

He assumed that triploidy was usually the intermediate step in chromosome duplication. This would cause problems, because if most animals' sex was determined by the ratios of chromosomes (as in Drosophila), triploidy would lead to sterility.
In the rare cases when a tetraploid was accidentally created, it would have to breed with diploids, and this would result in a (presumably sterile) triploid.
If, by chance, two tetraploids were to arise and mate, they would be at a disadvantage because, he said, they would be randomly allocated sex chromosomes and this would lead to a higher proportion of non-viable offspring, and thus the polyploid line would be outcompeted by the diploid.

Unfortunately, whilst the first two points are valid facts about polyploids, the third point is incorrect. A major flaw with Muller's explanation is that it only applies to animals with chromosomal ratio-based sex determination, which we have since discovered is actually relatively few animals. In 1925 there was comparatively little systematic study of life, so we really didn't know what proportion of plant or animal taxa showed polyploidy. Muller's answer doesn't explain why most animals, e.g. those with Y-dominant sex determination, exhibit relatively little polyploidy. Another line of evidence disproving Muller's answer is that, in fact, polyploidy is very common among dioecious plants (those with separate male and female plants; e.g. Westergaard, 1958), while Muller's theory predicts that prevalence in this group should be as low as in animals.
The 'complexity' answer...
Another answer with some historical clout is the one given by Daniel Standage in his answer, and has been given by various scientists over the years (e.g. Stebbins, 1950). This answer states that animals are more complex than plants, so complex that their molecular machinery is much more finely balanced and is disturbed by having multiple genome copies.
This answer has been soundly rejected (e.g. by Orr, 1990) on the basis of two key facts. Firstly, whilst polyploidy is unusual in animals, it does occur. Various animals with hermaphroditic or parthenogenetic modes of reproduction frequently show polyploidy. There are also examples of Mammalian polyploidy (e.g. Gallardo et al., 2004). In addition, polyploidy can be artificially induced in a wide range of animal species, with no deleterious effects (in fact it often causes something akin to hybrid vigour; Jackson, 1976).
It's also worth noting here that since the 1960s Susumo Ohno (e.g. Ohno et al. 1968; Ohno 1970; Ohno 1999) has been proposing that vertebrate evolution involved multiple whole-genome duplication events (in addition to smaller duplications). There is now significant evidence to support this idea, reviewed in Furlong & Holland (2004). If true, it further highlights that animals being more complex (itself a large, and in my view false, assumption) does not preclude polyploidy.
The modern synthesis...
And so to the present day. As reviewed in Mable (2004), it is now thought that:

Polyploidy is an important evolutionary mechanism which was and is probably responsible for a great deal of biological diversity.
Polyploidy arises easily in both animals and plants, but reproductive strategies might prevent it from propagating in certain circumstances, rather than any reduction in fitness resulting from the genome duplication.
Polyploidy may be more prevalent in animals than previously expected, and the imbalance in data arises from the fact that cytogenetics (i.e. chromosome counting) of large populations of wild specimens is a very common practise in botany, and very uncommon in zoology.

In addition, there are now several new suspected factors involved in ploidy which are currently being investigated:

Polyploidy is more common in species from high latitudes (temperate climates) and high altitudes (Soltis & Soltis, 1999). Polyploidy frequently occurs by the production of unreduced gametes (through meiotic non-disjunction), and it has been shown that unreduced gametes are produced with higher frequency in response to environmental fluctuations. This predicts that polyploidy should be more likely to occur in the first place in fluctuating environments (which are more common at higher latitudes and altitudes).
Triploid individuals, the most likely initial result of a genome duplication event, in animals and plants often die before reaching sexual maturity, or have low fertility. However, if triploid individuals do reproduce, there is a chance of even-ploid (fertile) individuals resulting. This probability is increased if the species produces large numbers of both male and female gametes, or has some mechanism of bypassing the triploid individual stage. This may largely explain why many species with 'alternative' sexual modes (apomictic, automictic, unisexual, or gynogenetic) show polyploidy, as they can keep replicating tetraploids, thus increasing the chance that eventually a sexual encounter with another tetraploid will create a new polyploid line. In this way, non-sexual species may be a crucial evolutionary intermediate in generating sexual polyploid species. Species with external fertilisation are more likely to establish polyploid lines - a greater proportion of gametes are involved in fertilisation events and therefore two tetraploid gametes are more likely to meet.
Finally, polyploidy is more likely to occur in species with assortative mixing. That is, when a tetraploid gamete is formed, if the genome duplication somehow affects the individual so as to make it more likely that it will be fertilised by another tetraploid, then it is more likely that a polyploid line will be established. Thus it may be partly down to evolutionary chance as to how easily a species' reproductive traits are affected. For example in plants, tetraploids often have larger flowers or other organs, and thus are preferentially attractive to pollinators. In frogs, genome duplication leads to changes in the vocal apparatus which can lead to immediate reproductive isolation of polyploids.

References

Furlong, R.F. & Holland, P.W.H. (2004) Polyploidy in vertebrate ancestry: Ohno and beyond. Biological Journal of the Linnean Society. 82 (4), 425–430.
Gallardo, M.H., Kausel, G., Jiménez, A., Bacquet, C., González, C., Figueroa, J., Köhler, N. & Ojeda, R. (2004) Whole-genome duplications in South American desert rodents (Octodontidae). Biological Journal of the Linnean Society. 82 (4), 443–451.
Jackson, R.C. (1976) Evolution and Systematic Significance of Polyploidy. Annual Review of Ecology and Systematics. 7209–234.
Mable, B.K. (2004) ‘Why polyploidy is rarer in animals than in plants’: myths and mechanisms. Biological Journal of the Linnean Society. 82 (4), 453–466.
Muller, H.J. (1925) Why Polyploidy is Rarer in Animals Than in Plants. The American Naturalist. 59 (663), 346–353.
Ohno, S. (1970) Evolution by gene duplication.
Ohno, S. (1999) Gene duplication and the uniqueness of vertebrate genomes circa 1970–1999. Seminars in Cell & Developmental Biology. 10 (5), 517–522.
Ohno, S., Wolf, U. & Atkin, N.B. (1968) EVOLUTION FROM FISH TO MAMMALS BY GENE DUPLICATION. Hereditas. 59 (1), 169–187.
Orr, H.A. (1990) ‘Why Polyploidy is Rarer in Animals Than in Plants’ Revisited. The American Naturalist. 136 (6), 759–770.
Soltis, D.E. & Soltis, P.S. (1999) Polyploidy: recurrent formation and genome evolution. Trends in Ecology & Evolution. 14 (9), 348–352.
Stebbins, C.L. (1950) Variation and evolution in plants.
Westergaard, M. (1958) The Mechanism of Sex Determination in Dioecious Flowering Plants. In: Advances in Genetics. Academic Press. pp. 217–281.

(I'll come back and add links to the references later)


Question:
Is mathematics one big tautology? Let me put the question in clearer terms:
Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work!
If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic?
I'm trying to figure out the true nature of what's going on here.

Disclaimer: different people view this differently. I side with Lakatos: Logic is a tool. Proofs are a way to verify one's intuition (and in many cases to improve one's intuition) and it is a tool to check the consistency of theories in a process of refining the axioms. The fact that every proof boils down to a tautology is true but irrelevant to mathematics.
Here is an isomorphic question to the question you posed: A painting is just blobs of paint of different colour on canvas. So, are we to deduce from this fact that the art of painting is reduced to just placing paint on canvas? Technically, the answer is yes. But the painter does much more than that. In fact, it is clear that while the painter must possess quite a large amount of skill in placing paint on canvas, this skill is the least relevant (while absolutely necessary) for the creative process of painting.
So it is with mathematics. Being able to prove is essential, but is the least relevant skill for doing mathematics. In mathematics we don't deduce things from axioms. Rather we try to capture a certain idea by introducing axioms, check which theorems follow from the axioms and compare these results against the idea we are trying to capture. If the results agree we are happy. If the results disagree, we change the axioms. The ideas we try to capture transcend the deductive system. The deductive system is there to help us find consequences from the axioms, but it does not tell us how to gauge the validity of results against the idea we try to capture, nor how to adjust the axioms.
This is my personal point of view of what mathematics is (or at least what a sizable portion of it is). It is very close to what physics is. Physics is not just some theories about matter and its interactions with stuff. Rather it is trying to model reality. So does mathematics, it's just not entirely clear which reality it is trying to model.


Question:
Despite the fact that oxygen is much more electronegative than carbon, the bond in $\ce{CO}$ presents a weak dipole moment.
This observation can easily be explained using the concept of "dative bond", that is, one bond is formed with two electrons from oxygen, producing a polarization $\ce{O\bond{->}C}$ which equilibrates the expected polarization $\ce{C->O}$.
I would like to know if the molecular orbital model could be used to explain this phenomenon.
Here's the diagram for $\ce{CO}$:

Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can only be explained by molecular orbital theory. A complete analysis of this can be found in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.)
Responsible for the dipole moment is the highest occupied molecular orbital, a $\pmb{\sigma}$ orbital, which has its largest coefficient at the carbon atom. In first order approximation, this orbital can be considered the lone pair of carbon. All other valence orbitals are more strongly polarised towards the oxygen. The orbital that can in first order approximation be considered as the oxygen lone pair has almost only s character and therefore contributes only little to the dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}
I have reproduced the MO scheme of carbon monoxide for you below. Please note, that the blue/orange coloured orbitals are virtual (unoccupied) orbitals, which should be taken with a grain of salt.

There are two possible decomposition schemes to explain the bonding, both of them involve donor-acceptor interactions. The term "dative bonding" should be avoided here, it is better to use it only for bonds, that consist purely of donor-acceptor interactions, as for example in $\ce{H3N\bond{->}BH3}$.
Below, the two decomposition schemes are reproduced from figure 6 (b & c) in the linked paper. Please note, that this decomposition does not include hybridised orbitals.

The left decomposition is a better description, since it retains the $C_{\infty{}v}$ symmetry of the molecule. We can see a donor-acceptor $\sigma$ bond and two electrons sharing $\pi$ bonds.
In the right configuration we assume an electron sharing $\sigma$ bond, an electron sharing $\pi$ bond and a donor-acceptor $\pi$ bond.
It is very important to understand, that the concept of a dative bond, that you are trying to employ here is only right by coincidence. The reason that the dipole moment is oriented towards the carbon is only to find in the weakly bonding HOMO, the lone pair of carbon.


Question:
Evaluate the following integral
$$
\tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\Bbb dx
$$

My Attempt:
Letting $x=\frac{\pi}{2}-x$ and using the property that
$$
\int_{0}^{a}f(x)\,\Bbb dx = \int_{0}^{a}f(a-x)\,\Bbb dx
$$
we obtain
$$
\tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\Bbb dx
$$
Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.

Here is an approach.
We give some preliminary results.

 The poly-Hurwitz zeta function

The poly-Hurwitz zeta function may  initially be defined  by the series
$$
\begin{align}
  \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)},
  \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1
\end{align}
$$
This special function is a natural extension of the Hurwitz zeta function initially defined as
$$
\zeta(s,a)=\sum_{n=0}^{\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \Re s>1, \tag2
$$
which is a natural extension itself of the Riemann zeta function initially defined as
$$
\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, \quad \Re s>1. \tag3
$$
The poly-Hurwitz function appears in different places with different notations, one may find it here:
 [Masri, p. 2 and p. 15  (2004)], [Murty, p. 17 (2006)], [Sinha, p. 45 (2002)]. In this answer we are dealing with a simplified version of a general poly-Hurwitz function.
The series in $(1)$ converges absolutely for $\displaystyle \Re s>0$. Moreover, the convergence of the series is uniform on every
half-plane $$\displaystyle H_{\delta}=\left\{s \in \mathbb{C}, \Re s \geq  \delta \right\}, \, \delta \in \mathbb{R},\, \delta>0,$$ therefore the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot \mid a,b)$
is analytic on the half-plane $\displaystyle \Re s>0$.
Let $a$, $b$ and $s$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re s >0$. One may observe that
$$
\begin{align}
  \zeta(s\mid a,b) & =  \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}\\
 & = \sum_{n=1}^{+\infty} \frac{(n+b)+(a-b)}{(n+a)^{s+1}(n+b)}\\
 & = \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}}+(a-b)\sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s+1}(n+b)} \tag4
\end{align}
$$
giving the functional identity
$$
\begin{align}
 \zeta(s \mid a,b) = \zeta(s+1,a+1) +(a-b) \zeta(s+1 \mid a,b) \tag5
\end{align}
$$
where $\displaystyle \zeta(\cdot,\cdot)$ is the standard Hurwitz zeta function.
From $(5)$, we obtain by induction, for $n=1,2,3,\ldots $,
$$
\begin{align}
 \zeta(s\mid a,b) = \sum_{k=1}^{n}(a-b)^{k-1}\zeta(s+k,a+1) +(a-b)^n\zeta(s+n \mid a,b). \tag6
\end{align}
$$
We use $(6)$ to extend $\displaystyle \zeta(\cdot \mid a,b)$ to a meromorphic function on each open set $\Re s>-n $, $n\geq 1$. Since the Hurwitz zeta function is analytic on the whole complex plane except for a simple pole at $1$ with residue $1$, then from $(6)$ the poly-Hurwitz zeta function $\displaystyle \zeta(\cdot\mid a,b)$ is analytic on the whole complex plane except for a simple pole at $0$ with residue $1$.

 The poly-Stieltjes constants

In 1885 Stieltjes has found  that the Laurent series expansion around $1$ of the Riemann zeta function
$$
\zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag7
$$  is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align}
\gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right).
\end{align} \tag8
$$ Euler was the first to define a constant of this form (1734)
$$
\begin{align}
\gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots.
\end{align}
$$ The constants $\displaystyle \gamma_k$ are called the Stieltjes constants and due to the fact that $\displaystyle \gamma_0=\gamma$   they are sometimes called the generalized Euler's constants.
Similarly, Wilton (1927) and Berndt (1972) established that the  Laurent series expansion in the  neighbourhood of $1$ of the Hurwitz zeta function
$$
\begin{align} \zeta(1+s,a)
= \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag{9}
\end{align}
$$ is such that the coefficients of the regular part of the expansion are given by
$$
\begin{align}
\gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0,
\end{align} \tag{10}
$$
with $\displaystyle \gamma_{0}(a)=-\psi(a)=-\Gamma'(a)/\Gamma(a)$. The coefficients $\gamma_k(a)$ are called the generalized Stieltjes constants.
We have seen from $(6)$ that the poly-Hurwitz zeta function admits a Laurent series expansion around $0$. Let's denote by $\displaystyle\gamma_k(a,b)$ the coefficients of the regular part of $\displaystyle \zeta(\cdot\mid a,b)$ around $0$. I will call these coefficients the poly-Stieltjes constants.
Do we have an analog of $(10)$ for $\displaystyle\gamma_k(a,b)$?
The following result is new.

Theorem 1. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider
  The poly-Hurwitz zeta function $$
\begin{align}
   \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)},
  \quad \Re s>0. \tag{11}
\end{align}
$$
  Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$,
  $$
\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag{12}
$$ where the poly-Stieltjes constants $\displaystyle \gamma_k(a,b)$ are given by $$
\begin{align}
\gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)
\end{align} \tag{13}
$$ with $$ \gamma_{0}(a,b)=-\psi(b+1)=-\Gamma'(b+1)/\Gamma(b+1). \tag{14}$$

Proof. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$.
We first assume $\Re s>0$. Observing that, for each $n \geq 1$,
$$
\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}<\infty
$$ and that
$$
\sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| =
\sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}<\infty,$$
we obtain
$$
\begin{align}
&\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right)  s^k \\\\
&= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\
&=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\
&=\zeta(s \mid a,b)-\frac1{s}
\end{align}
$$  as desired. Then, using $(6)$, we extend the preceding identity by analytic continuation  to all $s \neq 0$. To prove $(14)$, we start from a standard series representation of the digamma function (see Abram. & Steg. p. 258 6.3.16):
$$
\begin{align}
-\psi(b+1) &= \gamma - \sum_{n=1}^{\infty} \left( \frac1n - \frac1{b+n}  \right) \\
&=\lim_{N\to+\infty}\left(\gamma - \sum_{n=1}^N\left( \frac1n - \frac1{b+n}  \right)\right)\\
&=\lim_{N\to+\infty}\left(\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)-\left(\sum_{n=1}^N\frac1n-\ln N-\gamma \right)\right)\\
&=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{b+n} -\ln N\right)\\\\
&=\gamma_0(a,b)
\end{align}
$$ using $(13)$.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

One of the consequences of Theorem 1 is the new possibility to express some series in terms of the poly-Stieltjes constants.

Theorem 2. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.
Then
  $$
\begin{align}
  (b-a)\sum_{n=1}^{+\infty} \frac{\log (n+c)}{(n+a)(n+b)}=\gamma_1(c,a)-\gamma_1(c,b), \tag{15}
\end{align}
$$ similarly
  $$
\begin{align}
 \sum_{n=1}^{+\infty} \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)=\gamma_1(a,b)-\gamma_1(c,b), \tag{16}
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).
$$

Proof. Let $a,b,c$ be complex numbers such that $\Re a >-1, \, \Re b >-1, \, \Re c >-1$.
We have
$$
 (b-a)\frac{\log (n+c)}{(n+a)(n+b)}=\frac{\log (n+c)}{n+a}-\frac{\log (n+c)}{n+b}
$$ giving, for $N\geq1$,
$$
\begin{align}
  (b-a)&\sum_{n=1}^N \frac{\log (n+c)}{(n+a)(n+b)}=\\\\
&  \left(\sum_{n=1}^N\frac{\log (n+c)}{n+a}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{17}
\end{align}
$$
letting $N \to \infty$  and  using $(13)$ gives $(15)$.
We have, for $N\geq1$,
$$
\begin{align}
  &\sum_{n=1}^N \frac1{n+b}\left({\log (n+a)-\log (n+c)}\right)\\\\
&=  \left(\sum_{n=1}^N\frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right)-\left(\sum_{n=1}^N\frac{\log (n+c)}{n+b}-\frac{\log^2 \!N}2\right) \tag{18}
\end{align}
$$
letting $N \to \infty$ and  using $(13)$ gives $(16)$.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

 Juantheron's integral

Let’s first give a numerical evaluation of Juantheron’s integral.
I would like to thank Jonathan Borwein and David H. Bailey who obtained the result below to $1000$ digits, in just 3.9 seconds run time, using David's new MPFUN-MPFR software, along with the tanh-sinh quadrature program included with the MPFUN-MPFR  package.
They also tried the integral with David's MPFUN-Fort package, which has a completely different underlying multiprecision system, and they obtained the same result below.
Finally, they computed the integral with Mathematica $11.0$;  it agreed with the result below, although it required about 10 times longer to run.

Proposition 1. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!&\frac1{(1+x^2)(1+\tan x)}\mathrm dx\\\\=
0.&59738180945180348461311323509087376430643859042555\\
&67307703207161550311033249824121789098990404474443\\
&73300942847961727020952797366230453350097928752529\\
&62099371263365268445580755896768905606293308536674\\
&89639352215352393870280616186538538722285601087082\\
&81730013060929540132583577240799018025603130403772\\
&83596189879605956759516344861849456740112012597646\\
&30195536341071109827787231788650530475635336662512\\
&50757672078320586388500276160658476344052492489409\\
&64026178233152015087197531148322444147655936720008 \tag{19}\\
&40650450631581050321100329502169853063154902765446\\
&58804861176982696627707544105655815406116180984371\\
&54148587721902800400109013880620460529382772599713\\
&06874977209651994186527207589425408866256042399213\\
&80515694164361264997143539392018681691584285790381\\
&65536517701019826846772718498479534803417547866296\\
&23842162877309354675086691711521468623807334908897\\
&71491673168051054009130049879837629516862688171756\\
&13790927986073268994254629238035029442300668334396\\
&901581838911515359223628586133156893962372426055\cdots
\end{align}
$$

David H. Bailey confirmed that Mathematica $11.0$, in spite of the great numerical precision, could not find a closed-form of the integral.
The next result proves that the OP integral admits a closed form in terms of the poly-Stieltjes constants.

Proposition 2. We have
  $$
\begin{align}  \int_{0}^{\Large\frac{\pi}2}\!\!\frac1{(1+x^2)(1+\tan x)}\mathrm dx
&=\frac{(e^2+1)^2}{2(e^4+1)}\arctan \! \frac{\pi}{2}-\frac{e^4-1}{4(e^4+1)}\log\left(1+\frac{\pi^2}{4}\right)\\\\
&+\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{20}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$

Proof. We  proceed in three steps.
Step 1. One may write
$$
\require{cancel}
\begin{align}
&\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\mathrm dx \\
&=\int_{0}^{\Large\frac{\pi}{2}}\frac{\cos x}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&=\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{(\cos x+\sin x)+(\cos x-\sin x)}{(1+x^2)(\cos x+\sin x)}\mathrm dx\\
&=\frac12\int_{0}^{\Large\frac{\pi}{2}}\!\frac{1}{1+x^2}\mathrm dx+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)}\frac{(\cos x-\sin x)}{(\cos x+\sin x)}\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{1+x^2}\tan (x-\pi/4)\:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_{-\Large\frac{\pi}{4}}^{\Large\frac{\pi}4}\frac{1}{1+(x+\pi/4)^2}\tan x \:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}+\frac12\int_0^{\Large\frac{\pi}4}\left(\frac1{1+(x+\pi/4)^2}-\frac1{1+(x-\pi/4)^2}\right)\tan x \:\mathrm dx\\
&=\frac12 \arctan\! \frac{\pi}{2}-\frac{\Im}2 \!\int_{0}^{\Large\frac{\pi}{4}}\!\!\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx \tag{21}
\end{align}
$$
Let’s evaluate the latter integral.
Step 2. One may recall that the tangent function, as a meromorphic function, can be expressed as an infinite sum of rational functions:
$$
\tan x = \sum_{n=0}^{+\infty} \frac{2x}{\pi^2 (n+1/2)^2-x^2}, \quad x \neq \pm \pi/2, \pm 3\pi/2,\pm 5\pi/2,\ldots. \tag{22}
$$
We have the inequality
$$
\sup_{x \in [0,\pi/4]}\left|\frac{2x}{\pi^2 (n+1/2)^2-x^2}\right|\leq \frac1{(n+1/2)^2}, \quad n=0,1,2,\ldots, \tag{23}
$$ the convergence in $(22)$ is then uniform on  $[0,\pi/4]$. Thus, plugging $(22)$ into $(21)$, we are allowed to integrate $(21)$ termwise.
Each term, via a partial fraction decomposition, is evaluated to obtain
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}4}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right)\frac{2x}{\pi^2 (n+1/2)^2-x^2}\:\mathrm dx\\
&=\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&+\frac1{\pi}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&+\frac1{\pi}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)
\end{align} \tag{24}
$$
where for the sake of convenience we have set $\tau:=\pi/4+i$.
Step 3. We sum $(24)$ from $n=0$ to $\infty$ obtaining
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&=\sum_{n=0}^{\infty}\frac{2\tau}{\pi^2 (n+1/2)^2-\tau^2}\log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)\\
&+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2+\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right)\\
&+\frac1{\pi}\sum_{n=0}^{\infty}\frac1{(n+1/2-\tau/\pi)}\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right) \right), \tag{25}
\end{align}
$$ then, singling out the first terms in the two last series and using Theorem $2$ $(16)$, we get
$$
\begin{align}
\int_{0}^{\Large\frac{\pi}{4}}\!&\left(\!\frac1{x+\pi/4+i}+\frac1{x-\pi/4-i}\!\right) \tan x \:\mathrm dx\\
&=\tan \tau \log \left( \frac{4\tau-\pi}{4\tau+\pi}\right)
+\frac{4\pi}{\pi^2 -4\tau^2}\log 3
\\&+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 +\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 +\frac{\tau}{\pi}\!\right)\tag{26}\\ &+\frac1{\pi}\gamma_1\!\left(\!\frac34,\frac12 -\frac{\tau}{\pi}\!\right) -\frac1{\pi}\gamma_1\!\left(\!\frac14,\frac12 -\frac{\tau}{\pi}\!\right)
\end{align} $$ and the substitution $\tau=\pi/4+i$ gives the desired result.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

Achille Hui's conjecture is true.
Achille Hui has announced in the comments that the OP integral is equal to
$$
\begin{align}
\frac{\arctan(\frac{\pi}{2}) - t\log\sqrt{1+\frac{\pi^2}{4}}}{1+t^2}+\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) } \tag{27}
\end{align}
$$ with $\displaystyle t := \tanh(1)$.
The first term in $(27)$, with a little algebra is seen  to be equal to the sum of the first two terms on the right hand side of $(20)$.
We establish the veracity of the conjecture using Proposition $2$ and using the next result.

Proposition 3. We have
  $$
\begin{align}  &\frac{\pi^2}4 \sum_{n=0}^{\infty}\frac{ (2n+1)\left(\log\left(n+\frac34\right)-\log\left(n+\frac14\right)\right) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }\\\\
&=\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)} \\\\
&+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac34 +\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac34 +\frac{i}{\pi}\!\right)\tag{28}\\\\ &+\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac34,\frac14 -\frac{i}{\pi}\!\right) -\frac{\Im}{2\pi}\gamma_1\!\left(\!\frac14,\frac14 -\frac{i}{\pi}\!\right)
\end{align}
$$ with the poly-Stieltjes constant
  $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).$$

Proof. Observe that the first term of the series on the left hand side of  $(28)$, given by $n=0$, is just equal to
$$
\frac{64 \pi^2\log 3}{(\pi^2+16)(9\pi^2+16)}.
$$ By a partial fraction decomposition, one may check that
$$
\begin{align}
\frac{\pi^2}4 &\frac{ (2n+1) }{ \left(1+\pi^2\left(n+\frac14\right)^2\right)\left(1+\pi^2\left(n+\frac34\right)^2\right) }
=\frac{\Im}{2\pi}\left(\!\frac1{n+\frac34+\frac{i}{\pi}}-\frac1{n+\frac14+\frac{i}{\pi}}\!\right) \tag{29}
\end{align}
$$
then, multiplying $(29)$ by $\left(\log\!\left(n+\frac34\right)-\log\!\left(n+\frac14\right)\right)$ and summing from $n=1$ to $\infty$ we get, using Theorem $2$ $(16)$, the result $(28)$.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$


Question:
In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.
Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

I would like to prove the Quadratic Formula in a cleaner way.  Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.
Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)
We want to solve the equation
$$ax^2+bx+c=0,$$
where $a \ne 0$.  The usual argument starts by dividing by $a$.  That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.
Instead, multiply both sides by $4a$.  We obtain the equivalent equation
$$4a^2x^2 +4abx+4ac=0.\tag{1}$$
Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
$$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
So our equation can be rewritten as
$$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$
or equivalently
$$(2ax+b)^2=b^2-4ac. \tag{3}$$
Now it's all over. We find that
$$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$
and therefore
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$
No fractions until the very end!
Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest  strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.
Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots.  Ideally, each type of proof should be presented, since each connects to an important family of ideas.  And a twice proved theorem is twice as true.


Question:
I'm working on a simple web app that allows the user to tune his/her guitar. I'm a real beginner in signal processing, so please don't judge me too harshly if my question is inappropriate.
So, I managed to get the fundamental frequency using an FFT algorithm and at this point the application is somehow functional. However, there is room for improvement. Right now, I send raw PCM to the FFT algorithm, but I was thinking that maybe there are some pre/post algorithms/filters that may improve the detection. Can you suggest any?
My main problem is that when it detects a certain frequency, it shows that frequency for 1-2sec and then jumps to other random frequencies and comes back again and so on, even if the sound is continuous.
I'm also interested in any other type of optimization if anyone has experience with such things.

I'm guessing the other frequencies it gets are harmonics of the fundamental?  Like you're playing 100 Hz and it picks out 200 Hz or 300 Hz instead?  First, you should limit your search space to the frequencies that a guitar is likely to be.  Find the highest fundamental you're likely to need and limit to that.
Autocorrelation will work better than FFT at finding the fundamental, if the fundamental is lower in amplitude than the harmonics (or missing altogether, but that's not an issue with guitar):

You can also try weighting the lower frequencies to emphasize the fundamental and minimize harmonics, or use a peak-picking algorithm like this and then just choose the lowest in frequency.
Also, you should be windowing your signal before applying the FFT.  You just multiply it by a window function, which tapers off the beginning and end of the waveform to make the frequency spectrum cleaner.  Then you get tall narrow spikes for frequency components instead of broad ones.
You can also use interpolation to get a more accurate peak.  Take the log of the spectrum, then fit a parabola to the peak and the two neighboring points, and find the parabola's true peak.  You might not need this much accuracy, though.
Here is my example Python code for all of this.


Question:
If I have a square invertible matrix and I take its determinant, and I find that $\det(A) \approx 0$, does this imply that the matrix is poorly conditioned?
Is the converse also true? Does an ill-conditioned matrix have a nearly zero determinant?
Here is something I tried in Octave:
a = rand(4,4);
det(a) %0.008
cond(a)%125
a(:,4) = 1*a(:,1) + 2*a(:,2) = 0.000000001*ones(4,1);
det(a)%1.8E-11
cond(a)%3.46E10

It's the largeness of the condition number $\kappa(\mathbf A)$ that measures the nearness to singularity, not the tininess of the determinant.
For instance, the diagonal matrix $10^{-50} \mathbf I$ has tiny determinant, but is well-conditioned.
On the flip side, consider the following family of square upper triangular matrices, due to Alexander Ostrowski (and also studied by Jim Wilkinson):
$$\mathbf U=\begin{pmatrix}1&2&\cdots&2\\&1&\ddots&\vdots\\&&\ddots&2\\&&&1\end{pmatrix}$$
The determinant of the $n\times n$ matrix $\mathbf U$ is always $1$, but the ratio of the largest to the smallest singular value (i.e. the 2-norm condition number $\kappa_2(\mathbf U)=\dfrac{\sigma_1}{\sigma_n}$) was shown by Ostrowski to be equal to $\cot^2\dfrac{\pi}{4n}$, which can be seen to increase for increasing $n$.


Question:
I noticed that after stirring, a bubble in the centre of my mug of tea changed the speed it was rotating at periodically. Speeding up, then slowing down, then speeding up again, etc. Almost like when a ballerina pulls in her arms to increase her speed.
Tea after stirring
Edit:
I've repeated this with room temperature water to try and rule out any temperature-related effects and the same effect is present.

Just looking at the video, it appears that the shape of the surface is varying quasi-periodically, as if the liquid is moving outward (and upward) toward the cup walls, then moving inward and rising in the center of the cup.  This can be expected, if in the beginning the shape is not a perfect equilibrium shape (e.g., like a parabolic surface in a rotating cup).  But when the the liquid moves toward the center, the rotation necessarily speeds up due to conservation of angular momentum; and when it moves outward the opposite happens.
A crude analogy:  If you rolled a marble in a large wok with smooth spherical curvature, in such a way that it looped near the center/bottom, then out near the edge, you would see that its angular velocity increases when it approaches the center/bottom, and decreases when it recedes from the center/bottom.  You can think of the volume of liquid doing the same thing as the surface shape changes from a shallow curve to a deep curve.


Question: What's the difference between a microcontroller and a microprocessor?

A microprocessor generally does not have RAM, ROM and IO pins. It usually uses its pins as a bus to interface to peripherals such as RAM, ROM, Serial ports, Digital and Analog IO. It is expandable at the board level due to this.
A microcontroller is 'all in one', the processor, ram, IO all on the one chip, as such you cannot (say) increase the amount of RAM available or the number of IO ports. The controlling bus is internal and not available to the board designer.
This means that a microprocessor is generally capable of being built into bigger general purpose applications than a microcontroller. The microcontroller is usually used for more dedicated applications.
All of these are very general statements. There are chips available that blur the boundaries.


Question:
There was a reason why I constantly failed physics at school and university, and that reason was, apart from the fact I was immensely lazy, that I mentally refused to "believe" more advanced stuff until I understand the fundamentals (which I, eventually, never did).
As such, one of the most fundamental things in physics that I still don't understand (a year after dropping out from the university) is the concept of field. No one cared to explain what a field actually is, they just used to throw in a bunch of formulas and everyone was content. The school textbook definition for a field (electromagnetic in this particular case, but they were similar), as I remember it, goes like:

An electromagnetic field is a special kind of substance by which charged moving particles or physical bodies with a magnetic moment interact.

A special kind of substance, are they for real? This sounds like the authors themselves didn't quite understand what a field is so they decided to throw in a bunch of buzzwords to make it sounds right. I'm fine with the second half but a special kind of substance really bugs me, so I'd like to focus on that.
Is a field material?
Apparently, it isn't. It doesn't consist of particles like my laptop or even the light.
If it isn't material, is it real or is it just a concept that helps to explain our observations? While this is prone to speculations, I think we can agree that in scope of this discussion particles actually do exist and laws of physics don't (the latter are nothing but human ideas so I suspect Universe doesn't "know" a thing about them, at least if we're talking raw matter and don't take it on metalevel where human knowledge, being a part of the Universe, makes the Universe contain laws of physics). Any laws are only a product of human thinking while the stars are likely to exist without us homo sapiens messing around. Or am I wrong here too? I hope you already see why I hate physics.
Is a field not material but still real?
Can something "not touchable" by definition be considered part of our Universe by physicists? I used to imagine that a "snapshot" of our Universe in time would contain information about each particle and its position, and this would've been enough to "deseralize" it but I guess my programmer metaphors are largely off the track. (Oh, and I know that the uncertainty principle makes such (de)serialization impossible — I only mean that I thought the Universe can be "defined" as the set of all material objects in it). Is such assumption false?
At this point, if fields indeed are not material but are part of the Universe, I don't really see how they are different from the whole Hindu pantheon except for perhaps a more geeky flavor.
When I talked about this with the teacher who helped me to prepare for the exams (which I did pass, by the way, it was before I dropped out), she said to me that, if I wanted hardcore definitions,

a field is a function that returns a value for a point in space.

Now this finally makes a hell lot of sense to me but I still don't understand how mathematical functions can be a part of the Universe and shape the reality.

I'm going to go with a programmer metaphor for you.

The mathematics (including "A field is a function that returns a value for a point in space") are the interface: they define for you exactly what you can expect from this object.
The "what is it, really, when you get right down to it" is the implementation. Formally you don't care how it is implemented.
In the case of fields they are not matter (and I consider "substance" an unfortunate word to use in a definition, even though I am hard pressed to offer a better one) but they are part of the universe and they are part of physics.
What they are is the aggregate effect of the exchange of virtual particles governed by a quantum field theory (in the case of E&M) or the effect of the curvature of space-time (in the case of gravity, and stay tuned to learn how this can be made to get along with quantum mechanics at the very small scale...).
Alas I can't define how these things work unless you simply accept that fields do what the interface says and then study hard for a few years.

Now, it is very easy to get hung up on this "Is it real or not" thing, and most people do for at least a while, but please just put it aside. When you peer really hard into the depth of the theory, it turns out that it is hard to say for sure that stuff is "stuff". It is tempting to suggest that having a non-zero value of mass defines "stuffness", but then how do you deal with the photo-electric effect (which makes a pretty good argument that light comes in packets that have enough "stuffness" to bounce electrons around)? All the properties that you associate with stuff are actually explainable in terms of electro-magnetic fields and mass (which in GR is described by a component of a tensor field!). And round and round we go.


Question:
I'm trying to understand algorithm complexity, and a lot of algorithms are classified as polynomial. I couldn't find an exact definition anywhere. I assume it is the complexity that is not exponential.
Do linear/constant/quadratic complexities count as polynomial? An answer in simple English will be appreciated :)

First, consider a Turing machine as a model (you can use other models too as long as they are Turing equivalent) of the algorithm at hand. When you provide an input of size $n$, then you can think of the computation as a sequence of the machine's configuration after each step, i.e., $c_0, c_1, \ldots$ . Hopefully, the computation is finite, so there is some $t$ such $c_0, c_1, \ldots, c_t$. Then $t$ is the running time of the given algorithm for an input of size $n$.
An algorithm is polynomial (has polynomial running time) if for some $k,C>0$, its running time on inputs of size $n$ is at most $Cn^k$. Equivalently, an algorithm is polynomial if for some $k>0$, its running time on inputs of size $n$ is $O(n^k)$. This includes linear, quadratic, cubic and more. On the other hand, algorithms with exponential running times are not polynomial.
There are things in between - for example, the best known algorithm for factoring runs in time $O(\exp(Cn^{1/3} \log^{2/3} n))$ for some constant $C > 0$; such a running time is known as sub-exponential. Other algorithms could run in time $O(\exp(A\log^C n))$ for some $A > 0$ and $C > 1$, and these are known as quasi-polynomial. Such an algorithm has very recently been claimed for discrete log over small characteristics.


Question:
As enrichment analysis a usual step is to infer the pathways enriched in a list of genes. However I can't find a discussion about which database is better. Two of the most popular (in my particular environment) are Reactome and KEGG (Maybe because there are tools using them in Bioconductor).
KEGG requires a subscription for ftp access, and for my research I would need to download huge amounts of KGML files I am now leaning towards Reactome
Which is the one with more genes associated to pathways ?
Which is more completely annotated ?
Is there any paper comparing them ?

One big downside of KEGG is the licensing issue. One big advantage of Reactome are various crosslinks to other databases and data.
ad 1, This depends on which pathway, they are both primary databases. Sometimes other databases that for instance combine data of primary databases have better annotation of pathways (there is an example in the review paper bellow)
ad 3, There is very extensive relatively new (2015) review on this topic focused on human pathways: Comparison of human cell signaling pathway databases—evolution, drawbacks and challenges. However I could not find there which one is more complete ...


Question:
So I've got a decent head for what problems I work with are best one in serial, and which can be managed in parallel. But right now, I don't have much of an idea of what's best handled by CPU-based computation, and what should be offloaded to a GPU.
I know its a basic question, but much of my searching gets caught in people clearly advocating for one or the other without really justifying why, or somewhat vague rules of thumb. Looking for a more useful response here.

GPU hardware has two particular strengths: raw compute (FLOPs) and memory bandwidth.  Most difficult computational problems fall into one of these two categories.  For example, dense linear algebra (A * B = C or Solve[Ax = y] or Diagonalize[A], etc) falls somewhere on the compute/memory bandwidth spectrum depending on system size.  Fast Fourier transforms (FFT) also fit this mold with high aggregate bandwidth needs.  As do other transformations, grid/mesh-based algorithms, Monte Carlo, etc.  If you look at the NVIDIA SDK code examples, you can get a feel for the sorts of problems that are most commonly addressed.
I think the more instructive answer is to the question `What kinds of problems are GPUs really bad at?'  Most problems that don't fall into this category can be made to run on the GPU, though some take more effort than others.
Problems that don't map well are generally too small or too unpredictable.  Very small problems lack the parallelism needed to use all the threads on the GPU and/or could fit into a low-level cache on the CPU, substantially boosting CPU performance.  Unpredictable problems have too many meaningful branches, which can prevent data from efficiently streaming from GPU memory to the cores or reduce parallelism by breaking the SIMD paradigm (see 'divergent warps').  Examples of these kinds of problems include:

Most graph algorithms (too unpredictable, especially in memory-space)
Sparse linear algebra (but this is bad on the CPU too)
Small signal processing problems (FFTs smaller than 1000 points, for example)
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Question: In my field of research the specification of experimental errors is commonly accepted and publications which fail to provide them are highly criticized. At the same time I often find that results of numerical computations are provided without any account of numerical errors, even though (or maybe because) often questionable numerical methods are at work. I am talking about errors which result from discretization and finite precision of numerical computations etc. Sure, these error estimates are not always easy to obtain, such as in the case of hydro-dynamical equations but often it seems to result from laziness while I believe that the specification of numerical error estimates should be standard just as much as they are for experimental results. Hence my question: Are there resources which discuss in some detail the treatment of numerical errors or propose scientific standards for the specification of numerical errors which result from typical approximations such as discretization?

Your question is asking about model Verification. You can find numerous resources on methods and standards by searching for Verification and Validation (Roache 1997, 2002, 2004, Oberkampf & Trucano 2002, Salari & Knupp 2000, Babuska & Oden 2004), as well as the broader topic of Uncertainty Quantification. Rather than elaborate on methods, I would like to highlight a community that took a firm stand on the issue.
In 1986, Roache, Ghia, and White established the Journal of Fluids Engineering Editorial Policy Statement on the Control of Numerical Accuracy which opens with

A professional problem exists in the computational fluid dynamics community and also in the broader area of computational physics. Namely, there is a need for higher standards on the control of numerical accuracy.
[...] The problem is certainly not unique to the JFE and came into even sharper focus at the 1980-81 AFOSRHTTM-Stanford Conference on Complex Turbulent Flows. It was a conclusion of that conference's Evaluation Committee that, in most of the submissions to that conference, it was impossible to evaluate and compare the accuracy of different turbulence models, since one could not distinguish physical modeling errors from numerical errors related to the algorithm and grid. This is especially the case for first-order accurate methods and hybrid methods.

They conclude with very direct guidelines:

The Journal of Fluids Engineering will not accept for publication any paper reporting the numerical solution of a fluids engineering problem that fails to address the task of systematic truncation error testing and accuracy estimation.
[...] we must make it clear that a single calculation in a fixed grid will not be acceptable, since it is impossible to infer an accuracy estimate from such a calculation. Also, the editors will not consider a reasonable agreement with experimental data to be sufficient proof of accuracy, especially if any adjustable parameters are involved, as in turbulence modeling.

The current version contains a comprehensive set of criteria and represents a standard that, in my opinion, other fields should aspire to match. It is shameful that even today, awareness about the importance of model verification is absent in so many fields.


Question:
I don't understand why the Halting Problem is so often used to dismiss the possibility of determining whether a program halts. The Wikipedia article correctly explains that a deterministic machine with finite memory will either halt or repeat a previous state. You can use the algorithm which detects whether a linked list loops to implement the Halting Function with space complexity of O(1).
It seems to me that the Halting Problem proof is nothing more than a so-called "paradox," a self-referencing contradiction in the same way as the Liar's paradox. The only conclusion it makes is that the Halting Function is susceptible to such malformed questions.
So, excluding paradoxical programs, the Halting Function is decidable. So why do we hold it as evidence of the contrary?
4 years later: When I wrote this, I had just watched this video (update: the video has been taken down). A programmer gets some programs, must determine which ones terminate, and the video goes on to explain that it's impossible. I was frustrated, because I knew that given some arbitrary programs, there was some possibility the protagonist could prove whether they terminated. I knew many real-life algorithms had already been formally proven to terminate. The concept of generality was lost somehow. It's the difference between saying "some programs cannot be proven to terminate," and, "no program can be proven to terminate." The failure to make this distinction, by every single reference I found online, was how I came to the title for this question. For this reason, I really appreciate the answer that redefines the halting function as ternary instead of boolean.

Because a lot of really practical problems are the halting problem in disguise. A solution to them solves the halting problem.
You want a compiler that finds the fastest possible machine code for a given program? Actually the halting problem.
You have JavaScript, with some variables at a high security level and some at a low security level. You want to make sure that an attacker can't get at the high security information. This is also just the halting problem.
You have a parser for your programming language. You change it, but you want to make sure it still parses all the programs it used to. Actually the halting problem.
You have an anti-virus program, and you want to see if it ever executes a malicious instruction. Actually just the halting problem.
As for the Wikipedia example, yes, you could model a modern computer as a finite-state machine. But there's two problems with this.

Every computer would be a different automaton, depending on the exact number of bits of RAM. So this isn't useful for examining a particular piece of code, since the automaton is dependent on the machine on which it can run.

You'd need $2^n$ states if you have n bits of RAM. So for your modern 8GB computer, that's $2^{32000000000}$. This is a number so big that wolfram alpha doesn't even know how to interpret it. When I do $2^{10^9}$ it says that it has $300000000$ decimal digits. This is clearly much too large to store in a normal computer.

The Halting problem lets us reason about the relative difficulty of algorithms. It lets us know that, there are some algorithms that don't exist, that sometimes, all we can do is guess at a problem, and never know if we've solved it.
If we didn't have the halting problem, we would still be searching for Hilbert's magical algorithm which inputs theorems and outputs whether they're true or not. Now we know we can stop looking, and we can put our efforts into finding heuristics and second-best methods for solving these problems.
UPDATE: Just to address a couple of issues raised in the comments.
@Tyler Fleming Cloutier: The "nonsensical" problem arises in the proof that the halting problem is undecidable, but what's at the core of undecidability is really having an infinite search space. You're searching for an object with a given property, and if one doesn't exist, there's no way to know when you're done.
The difficulty of a problem can be related to the number of quantifiers it has. Trying to show that there exists ($\exists$) an object with an arbitrary property, you have to search until you find one. If none exists, there's no way (in general) to know this. Proving that all objects ($\forall$) have a property is hard, but you can search for an object without the property to disprove it. The more alternations there are between forall and exists, the harder a problem is.
For more on this, see the Arithmetic Hierarchy. Anything above $\Sigma^0_0=\Pi^0_0$ is undecidable, though level 1 is semi-decidable.
It's also possible to show that there are undecidable problems without using a nonsensical paradox like the Halting problem or Liar’s paradox. A Turing Machine can be encoded using a string of bits, i.e. an integer. But a problem can be encoded as a language, i.e. a subset of the integers. It's known that there is no bijection between the set of integers and the set of all subsets of the integers. So there must be some problems (languages) which don't have an associated Turing machine (algorithm).
@Brent: yes, this admits that this is decidable for modern computers. But it's decidable for a specific machine. If you add a USB drive with disk space, or the ability to store on a network, or anything else, then the machine has changed and the result doesn't still hold.
It also has to be said that there are going to be many times where the algorithm says "this code will halt" because the code will fail and run out of memory, and that adding a single extra bit of memory would cause the code to succeed and give a different result.
The thing is, Turing machines don't have an infinite amount of memory. There's never a time where an infinite amount of symbols are written to the tape. Instead, a Turing machine has "unbounded" memory, meaning that you can keep getting more sources of memory when you need it. Computers are like this. You can add RAM, or USB sticks, or hard drives, or network storage. Yes, you run out of memory when you run out of atoms in the universe. But having unlimited memory is a much more useful model.


Question:
I'm using a 5 V / 2 A voltage regulator (L78S05) without a heatsink. I'm testing the circuit with a microcontroller (PIC18FXXXX), a few LEDs and a 1 mA piezzo buzzer. The input voltage is aprox. 24 VDC. After running for a minute, the voltage regulator starts to overheat, meaning it burns my finger if I keep it there for more than a second. Within a few minutes it starts to smell like it's burnt. Is this a normal behavior for this regulator? What could cause it to heat that much?

Other components used in this circuit:
L1: BNX002-01 EMI filter
R2: Varistor
F1: Fuse 0154004.DR

Summary:  YOU NEED A HEATSINK NOW !!!!! :-)
[and having a series resistor as well wouldn't hurt :-) ]

Well asked question Your question is asked well - much better than usual.
The circuit diagram and references are appreciated.
This makes it much easier to give a good answer first time.
Hopefully this is one ... :-)
It makes sense (alas): The behavior is entirely expected.
You are thermally overloading the regulator.
You need to add a heat sink if you want to use it in this manner.
You would benefit greatly from a proper understanding of what is happening.
Power = Volts x Current.
For a linear regulator Power total = Power in load + Power in regulator.
Regulator Vdrop = Vin - Vload
Here Vdrop in regulator = 24-5 = 19V.
Here Power in = 24V x Iload
Power in load = 5V x Iload
Power in regulator = (24V-5V) x Iload.
For 100 mA of load current the regulator will dissipate
Vdrop x Iload (24-5) x 0.1 A = 19 x 0.1 = 1.9 Watt.
How Hot?: Page 2 of the data sheet says that the thermal resistance from junction to ambient (= air) is 50 degrees C per Watt. This means that for every Watt you dissipate you get 50 degrees C rise. At 100 mA you would have about 2 Watts dissipation or about 2 x 50 = 100°C rise. Water would boil happily on the IC.
The hottest most people can hold onto long term is 55°C. Yours is hotter than that. You didn't mention it boiling water (wet finger sizzle test). Let's assume you have ~~ 80°C case temperature. Let's assume 20°C air temperature (because its easy - a few degrees either way makes little difference.
Trise = Tcase-Tambient = 80°C - 20°C = 60°C.
Dissipation = Trise/Rth = 60/50 ~= 1.2 Watt.
At 19v drop 1.2 W = 1.2/19 A = 0.0632 A or about 60 mA.
ie if you are drawing about 50 mA you will get a case temperature of 70°C - 80°C degrees range.
You need a heatsink.
Fixing It: The data sheet page 2 says Rthj-case = thermal resistance from junction to case is 5C/W = 10% of junction to air.
If you use a say 10 C/W heatsink then total Rth will be R_jc + Rc_amb  (add junction to case to case to air).
= 5+10 = 15°C/Watt.
For 50 mA you will get 0.050A x 19V = 0.95W or a rise of 15°C/Watt x 0.95 ~= 14°C rise.
Even with say 20°C rise and a 25V ambient you will get 20+25 = 45°C heatsink temperature.
The heatsink will be hot but you will be able to hold it without (too much) pain.
Beating the heat:
As above, heat dissipation in a linear regulator in this situation is 1.9 Watt per 100 mA or 19 Watt at 1A. That's a lot of heat. At 1A, to keep temperature under the temperature of boiling water (100°C) when ambient temperature was 25°C you'd need an overall thermal resistance of no more than (100°C-25°C)/19 Watt = 3.9°C/W. As the junction to case Rthjc is already greater than 3.9 at 5°C/W you cannot keep the junction under 100°C in these conditions. Junction to case alone at 19V and 1A will add 19V x 1A x 5°C/W = 95°C rise. While the IC is rated to allow temperatures as high as 150°C, this is not good for reliability and should be avoided if at all possible. Just as an exercise, to JUST get it under 150°C in the above case the external heatsink would need to be (150-95)°C/19W  = 2.9°C/W. That's attainable but is a larger heatsink than you'd hope to use. An alternative is to reduce the energy dissipated and thus the temperature rise.
The ways of reducing heat dissipation in the regulator are:
(1) Use a switching regulator such as the NatSemi simple switchers series. A  performance switching regulator with even only 70% efficiency will reduce the heat dissipation dramatically as only 2 Watt is dissipated in the regulator!.
ie Energy in = 7.1 Watts. Energy out = 70% = 5 Watts. Current at 5 Watts at 5V = 1A.
Another option is a pre-made drop-in replacement for a 3 terminal regulator.
The following image and link are from the part referred to in a comment by Jay Kominek. OKI-78SR 1.5A, 5V drop in switching regulator replacement for an LM7805. 7V - 36V in.

At 36 Volts in, 5V out, 1.5A efficiency is 80%. As Pout = 5V x 1.5A = 7.5W = 80%, the power dissipated in the regulator is 20%/80% x 7.5W = 1.9 Watts. Very tolerable. No heatsink required and can provide 1.5A out at 85 degrees C. [[Errata: Just noticed the curve below is at 3.3V. The 5V part manages 85% at 1.5A so is better than the above.]]

(2) Reduce the voltage
(3) Reduce the current
(4) Dissipate some energy external to the regulator.
Option 1 is the best technically. If this is not acceptable and if 2 & 3 are fixed then option 4 is needed.
The easiest and (probably best) external dissipation system is a resistor. A series power resistor which drops from 24V to a voltage that the regulator will accept at max current will do the job well. Note that you will want a filter capacitor at the input to the regulator due to the resistance making the supply high impedance. Say about 0.33uF, more won't hurt. A 1 uF ceramic should do. Even a larger cap such as a 10 uF to 100 uF aluminum electrolytic should be good.
Assume Vin = 24 V. Vregulator in min = 8V (headroom / dropout. Check data sheet. Selected reg says 8V at <1A.) Iin = 1 A.
Required drop at 1A = 24 - 8 = 16V. Say 15V to be "safe".
R = V/I = 15/1 = 15 ohms.
Power = I2*R = 1 x 15 = 15 Watts.
A 20 Watt resistor would be marginal.
A 25W + resistor would be better.
Here's a 25W 15R resistor priced at $3.30/1 in stock lead free with datasheet here. Note that this also needs a heat sink!!! You CAN buy free air rated resistors up to 100's of Watts. What you use is your choice but this would work well. Note that it is rated at 25 Watt commercial or 20 Watt military so at 15W it is "doing well". Another option is a suitable length of properly rated resistance wire mounted appropriately. Odds are a resistor manufacturer already does this better than you do.
With this arrangement:
Total power = 24W
Resistor power = 15 Watt
Load power = 5 Watt
Regulator power = 3 Watt
Regulator junction rise will be 5°C/W x 3 = 15°C above the case. You will need to provide a heatsink to keep regulator and heatsink happy but that is now "just a matter of engineering".

Heatsink examples:
21 degrees °C (or °K) per Watt

7.8°C/W

Digikey - many heatsink examples including this 5.3 C/W heatsink

2.5°C/W

0.48°C/W!!!
119mm wide x 300mm long x 65 mm tall.
1 foot long x 4.7" wide x 2.6" tall

Good article on heatsink selection
Forced convection heatsink thermal resistance

Reducing linear regulator dissipation with a series input resistor:
As noted above, using a series resistor to drop voltage prior to  a linear regulator can greatly reduce dissipation in the regulator. While cooling a regulator usually requires heatsinks, air-cooled resistors can be obtained cheaply that are able to dissipate 10 or more Watts without needing a heatsink. It is not usually a good idea to solve high input voltage  problems in this manner but it can have its place.
In the example below an LM317 5V output 1A supply operated from 12V. Adding a resistor can more than halve the power dissipation in the LM317 under worst case conditions by adding a cheap air cooled wire mounted series input resistor.
The LM317 needs 2 to 2.5V headroom at lower currents or say 2.75V under extreme load and temperature conditions. (See Fig 3 in the datasheet,  - copied below).
LM317 headroom or dropout voltage

Rin has to be sized such that it does not drop excessive voltage when V_12V is at its minimum, Vdropout is worst case for the conditions and the series diode drop and output voltage are allowed for.
Voltage across resistor must always be less than =

Minimum Vin

less Maximum Vdiode drop

less Worst case dropout relevant to situation

less output voltage

So  Rin <= (v_12 - Vd - 2.75 - 5)/Imax.
For 12V minimum Vin, and say 0.8V diode drop and say 1 amp out that's
(12-0.8-2.75-5)/1
= 3.45/1
= 3R45
= say 3R3.
Power in R = I^2R = 3.3W so a 5W part would be marginally acceptable and 10W would be better.
Dissipation in the LM317 falls from > 6 Watt to < 3 Watt.
An excellent example of a suitable wire lead mounted air-cooled resistor would be a member of this nicely specified Yageo family of wire-wound resistors with members rated from 2W to 40W air cooled. A 10 Watt units is in stock at Digikey at $US0.63/1.

Resistor ambient temperature ratings and temperature rise:
Nice to have are these two graphs from the datasheet above which allow real world results to be estimated.
The left hand graph shows that a 10 Watt resistor operated at 3W3 = 33% of its rate Wattage has an allowable ambient temperature of up to 150 C (actually about 180 C if you plot the operating point in the graph but the manufacturer says 150 C max is allowed.
The second graph shows that temperature rise for a 10 W resistor operated at 3W3 will be about 100°C above ambient. A 5 W resistor from the same family would be operating at 66% of rating and have a temperature rise of 140°C above ambient. (A 40 W would have about 75°C rise but 2 x 10 W = under 50°C and 10 x 2 W only about 25°C !!!.
The decreasing temperature rise with an increasing number of resistors with the same combined Wattage rating in each case is presumably related to "Square cubed law" action as there is less cooling surface area per volume as size increases.

________________________________________
Added August 2015 - Case study:
Somebody asked the reasonable question:

Isn't a more likely explanation the relatively high capacitive load (220 µF)? E.g. causing the regulator to become unstable, oscillations causing a lot of heat dissipated in the regulator. In the datasheet, all of the circuits for normal operation only have a 100 nF capacitor on the output.

I answered in comments, but they MAY be deleted in due course and this is a worthwhile addition to the subject, so here are the comments edited into the answer.
In some cases oscillation and instability of the regulator certainly is an issue but, in this case and many like it, the most likely reason is excess dissipation.
The 78xxx family are very old and predate both the modern low dropout regulators and the series powered (LM317 style) ones. The 78xxx family are essentially unconditionally stable with respect to Cout. They in fact need none for proper operation and the 0.1uF often shown is to provide a reservoir to provide extra surge or spike handling.
In some of the related data sheets they actually say that Cout can be "increased without limit" but I do not see such a note here - but also (as I'd expect) there is no note suggesting instability at high Cout. In fig 33 on page 31 of the datasheet they show the use of a reverse diode to "protect against "high capacitance loads" - i.e., capacitors with high enough energy to cause damage if discharged into the output - i.e., far more than 0.1 uF.
Dissipation: At 24 Vin and 5 Vout the regulator dissipates 19 mW per mA. Rthja is 50 C/W for the TO220 package so you'd get ABOUT 1°C rise per mA of current.
So with say 1 Watt dissipation in 20 C ambient air the case would be at about 65°C (and could be more depending how the case is oriented and located). 65°C is somewhat above the lower limit of "burn my finger" temperature.
At 19 mW/mA it would take 50 mA to dissipate 1 Watt. The actual load in the example given is unknown - he shows an indicator LED at about 8 or 9 mA (if red) plus a load of the regulator internal current used (under 10 mA)  +  "PIC18FXXXX), a few LEDs ..." That total could reach or exceed 50 mA depending on the PIC circuit, or MAY be much less. |
Overall given regulator family, differential voltage, actual cooling uncertainty, Tambient uncertainty, C/W typical figure and more it seems like sheer dissipation is a reasonable reason for what he sees in this case - and for what many people using linear regulators will experience in similar cases. There is a chance that it's instability for reasons less obvious, and such should never be rejected without good reason, but I'd start on dissipation.
In this case a series input resistor (say 5W rated with air cooling) would move much of the dissipation into a component better suited to deal with it.
And/or a modest heatsink should work marvels.


Question: In some literature, I have read that a regression with multiple explanatory variables, if in different units, needed to be standardized.  (Standardizing consists in subtracting the mean and dividing by the standard deviation.)  In which other cases do I need to standardize my data?  Are there cases in which I should only center my data (i.e., without dividing by standard deviation)?

In regression, it is often recommended to center the variables so that the predictors have mean $0$. This makes it easier to interpret the intercept term as the expected value of $Y_i$ when the predictor values are set to their means. Otherwise, the intercept is interpreted as the expected value of $Y_i$ when the predictors are set to 0, which may not be a realistic or interpretable situation (e.g. what if the predictors were height and weight?). Another practical reason for scaling in regression is when one variable has a very large scale, e.g. if you were using population size of a country as a predictor. In that case, the regression coefficients may be on a very small order of magnitude (e.g. $10^{-6}$) which can be a little annoying when you're reading computer output, so you may convert the variable to, for example, population size in millions. The convention that you standardize predictions primarily exists so that the units of the regression coefficients are the same.
As @gung alludes to and @MånsT shows explicitly (+1 to both, btw), centering/scaling does not affect your statistical inference in regression models - the estimates are adjusted appropriately and the $p$-values will be the same.
Other situations where centering and/or scaling may be useful:

when you're trying to sum or average variables that are on different scales, perhaps to create a composite score of some kind. Without scaling, it may be the case that one variable has a larger impact on the sum due purely to its scale, which may be undesirable.

To simplify calculations and notation. For example, the sample covariance matrix of a matrix of values centered by their sample means is simply $X'X$. Similarly, if a univariate random variable $X$ has been mean centered, then ${\rm var}(X) = E(X^2)$ and the variance can be estimated from a sample by looking at the sample mean of the squares of the observed values.

Related to aforementioned, PCA can only be interpreted as the singular value decomposition of a data matrix when the columns have first been centered by their means.

Note that scaling is not necessary in the last two bullet points I mentioned and centering may not be necessary in the first bullet I mentioned, so the two do not need to go hand and hand at all times.


Question:
The division of the chromatic scale in $7$ natural notes (white keys in a piano) and $5$ accidental ones (black) seems a bit arbitrary to me.
Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with $12$ notes? (apparently there's a musical scale called Swara that does just that)
I've asked several musician friends, but they lack the math skills to give me a valid answer. "Notes are like that because they are like that."
I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this.
Mathematically, is there any difference between white and black notes, or do we make the distinction for historical reasons only?

The first thing you have to understand is that notes are not uniquely defined.  Everything depends on what tuning you use.  I'll assume we're talking about equal temperament here.  In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave.  Why twelve?
At the end of the day, what we want out of our musical frequencies are nice ratios of small integers.  For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$.  As you can see, this is not a fifth; however, it is quite close.
Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$.  Again, this is not a perfect fourth, but is quite close.
And so on.  What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music.
Here's how this coincidence works.  You get the white keys from $C$ using (part of) the circle of fifths.  Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$.  Then go down a fifth to get $F$.  These are the "neighbors" of $C$ in the circle of fifths.  You get the black keys from here using the rest of the circle of fifths.  After you've gone up a "perfect" perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$.  This happens to be rather close to $2^7 : 1$, or seven octaves!  And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves.  In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$.  Said another way,
$$\log_2 3 \approx \frac{19}{12}$$
happens to be a good rational approximation, and this is the main basis of equal temperament.  (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.)
It is a fundamental fact of mathematics that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have "perfect" perfect fifths regardless of how many notes you use.  However, you can write down good rational approximations by looking at the continued fraction of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes.
Of course, you can use other types of temperament, such as well temperament; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse.  In particular, if you don't use equal temperament then different keys sound different.  This is a major reason many Western composers composed in different keys; during their time, this actually made a difference.  As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning.

Edit:  I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers.  This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation.
You probably know that sound is a wave.  More precisely, sound is a longitudinal wave carried by air molecules.  You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$.  Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it overtones and has a Fourier series
$$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$
where the coefficients $a_n, b_n$ determine the timbre of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well.  So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$.
If you play two notes of frequencies $f, f'$ together, then the resulting sound corresponds to what you get when you add their Fourier series.  Now it's not hard to see that if $\frac{f}{f'}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones.  Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough.

Edit:  You should probably check out David Benson's "Music: A Mathematical Offering", the book Rahul Narain recommended in the comments for the full story.  There was a lot I didn't know, and I'm only in the introduction!


Question: I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.

These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by
$$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$
On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients,  we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.


Question:
An Amazon EC2 compute cluster costs about \$800-\$1000 (depending on duty cycle) per physical CPU core over the course of 3 years. In our last round of hardware acquisition, my lab picked up 48 cores worth of hardware very similar to that of Amazon's clusters for about ~$300 a core.
Am I missing something here? Are there any situations in which it makes economic sense to build a cluster in the cloud for high CPU tasks such as molecular dynamics simulations? Or am I always better off just building and baby-sitting the dang machine myself?
(I should mention that my lab doesn't pay for electricity in our server room (at least not directly), but even with this benefit Amazon still seems extremely expensive).

The main advantage, in my opinion, of using Cloud-based resources is flexibility, i.e. if you have a fluctuating workload, you only pay for what you need.
If this is not the case in your application, i.e. you know you will have a quantifiable and constant workload, then you're probably better-off building your own cluster. In the Cloud, you pay for flexibility, and if you don't need flexibility, you would be paying for something you don't need.
If your workload is flexible but somewhat intense and relies on certain hardware features (see aeismail's answer), you may want to try sharing a cluster with other people in your university to amortize the idle cycles. My old university runs such a shared cluster with a "Shareholder Model" in which every group is guaranteed a share of the computing power proportional to their investment in the hardware and idle cycles can be used by anyone. The only difficulty is centralizing the cluster administration.


Question:
When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.
What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.
Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.

As I understand this, there are basically two effects at work here.
When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $\mathrm{d}$-orbitals and putting one in an $\mathrm{s}$-orbital increases as you fill the $\mathrm{d}$-orbitals.
Additionally, pairing electrons in one orbital (so, adding the second $\mathrm{s}$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.
I'm assuming that the effect isn't strong enough to avert fluorine having a $\mathrm{2s^2}$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $\mathrm{s}$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $\mathrm{5d}$ instead of the $\mathrm{4f}$ orbitals.
Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $\mathrm{6s^2}$ occupation, but is for gold - more $\mathrm{d}$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $\mathrm{s}$-orbital.


Question:
An exam for high school students had the following problem:
Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle?

Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square.
The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?

Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!):

Consequently, as several others have noted,
$$
\frac{\text{perimeter of the circle}}{\text{perimeter of the square}}
  = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2}
 < 1.
$$

For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius:

Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off
$$
x + 1 = 2y\quad\text{(or $x = 2y - 1$).}
$$
Invoking the Pythagorean theorem and substituting the preceding line,
\begin{align*}
0 &= x^{2} + y^{2} - 1 \\
  &= (2y - 1)^{2} + y^{2} - 1 \\
  &= 5y^{2} - 4y \\
  &= y(5y - 4).
\end{align*}
Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.


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