linear ODE with constant coefficients, proof

Question: Let $
\sum\limits_{k = 0}^n {y^{\left( k \right)} a_k }  = 0
$ an homogeneous ODE, where $a_k$ are constants. How can I solve the equation when the roots are repeated?  One way, that I saw in wikipedia, is using the fact that if $ e^{cx} $ is a solution, then $ (x^r)(e^{cx}) $ is also too, How can I prove this? It´s difficult to me, to evaluate the sum, because i want to show that $
\sum\limits_{k = 0}^n {\left( {x^r e^{cx} } \right)^{\left( k \right)} a_k }  = 0
$ but I need to evaluate $
{\left( {x^r e^{cx} } \right)^{\left( k \right)} }
$ and I don´t know how to do it Dx
Comment: The idea is to perturb coefficients of your equation a little so as to make all the roots distinct, and then consider the limit as the amount of perturbation goes to zero.
Answer: I don't know if you have the background for this, but I do not think there is much choice if you allow the degree  $n \geq 3.$ It is not so bad for $n=2.$
Anyway, you introduce a bunch of variables $y_0 = y, y_1 = y', y_2 = y'',$ and so on. As your coefficients are constant, we may divide through by whatever $a_n$ might be to arrive at a revised constant coefficient equation with $a_n = 1.$
So the new system of equations is a linear system beginning with $$y_0' = y_1, \; y_1' = y_2, \ldots, \; y_{n-2}' = y_{n-1},$$ but finally
$$ y_{n-1}' = - b_0 y_0 - b_1 y_1 - \ldots - b_{n-1} y_{n-1},$$ where I have taken $b_j = a_j / a_n.$
We write a column vector $Y$ with entries $y_0, y_1, \ldots, y_{n-1}.$ that system now becomes
$$  Y' = B Y $$ with initial conditions written as
$$ Y(0) = Y_0$$
and the solution to the system is
$$ Y = e^{B x} Y_0$$
Note that $B$ has exactly the form of a companion matrix, see the square matrix at
http://en.wikipedia.org/wiki/Companion_matrix#Linear_recursive_sequences
The appearance of $x, x^2,\ldots$ comes from the Jordan normal form of $B,$ precisely when there are repeated eigenvalues (characteristic values) of $B,$ and when there are off-diagonal entries in the relevant Jordan block. If there are repeated roots but the Jordan normal form is diagonal anyway, then no polynomial terms appear. I do not know anything special about the Jordan normal form of a companion matrix, perhaps something precise can be said that need not hold for other types of coefficient matrix $B.$
This is half a semester of work if you have already had linear algebra. Anyway, see
http://en.wikipedia.org/wiki/Ordinary_differential_equation#Fundamental_systems_for_homogeneous_equations_with_constant_coefficients
Answer: An alternative to looking at the corresponding first order linear system is to consider factoring linear differential operators.
Given a homogeneous linear differential equation with constant coefficients, $a_ny^{(n)}+\cdots+a_1y'+a_0y=0$, we can define a linear differential operator $L=a_nD^n+\dots+a_1D+a_0$ where $D[y]=y'$, $D^2[y]=D[D[y]]=y''$, etc. and each constant stands for left multiplication by that constant: $a_k[y]=a_ky$.
So our differential equation becomes the operator equation: $L[y]=0$.
Example: $3y''+2y'+5y=0$ $\Longleftrightarrow$ $L=3D^2+2D+5$ and $L[y]=0$
Next, $D$ is a linear operator, so it commutes with constants: $D[ky]=kD[y]$. In addition, powers of $D$ commute with each other. This means we can factor each operator (like a polynomial).
Example: $y''-y=0$ gives us $L=D^2-1=(D-1)(D+1)=(D+1)(D-1)$ these are all equal because everything commutes.
Now to solve our equation we can focus on one factor at a time.
Example: Suppose $L=(D-1)(D-2)$. Then $L[e^t]=(D-1)(D-2)[e^t]=(D-2)(D-1)[e^t]=(D-2)[0]=0$ since $(D-1)[e^t]=e^t-e^t=0$ and $(D-2)[0]=0-2(0)=0$. Next, $L[e^{2t}]=(D-1)(D-2)[e^{2t}]=(D-1)[2e^{2t}-2e^{2t}]=(D-1)[0]=0$. This shows that both $y=e^t$ and $y=e^{2t}$ are solutions and hence $y=c_1e^t+c_2e^{2t}$ is the general solution (accepting some theory which tell us that 2 solutions are linearly independent and the space of solutions is 2-dimensional).
Let's focus on a single factor: $L=(D-r)^n$.
Consider $y=t^ke^{rt}$. $(D-r)[y]=(kt^{k-1}e^{rt}+t^kre^{rt})-r(t^ke^{rt})=kt^{k-1}e^{rt}$.
Next, $(D-r)^2[y]=(D-r)(D-r)[y]=(D-r)[kt^{k-1}e^{rt}]=k(k-1)t^{k-2}e^{rt}$. And in general $(D-r)^{k+1}[y]=k(k-1)\cdots 2\cdot 1\cdot 0 \cdot e^{rt}=0$.
This means that $(D-r)^n[t^ke^{rt}]=0$ when $n>k$.
Thus if the characteristic polynomial has a factor $(x-r)^n$ so that the corresponding linear differential operator $L$ has a factor $(D-r)^n$, we must have solutions $e^{rt}$, $te^{rt}$, $\dots$, $t^{n-1}e^{rt}$. These can be shown to be linearly independent by using a Wronskian determinant (this is fairly straight forward but difficult for a beginner). Solutions corresponding to distinct factors are independent because they are in distinct spaces of generalized eigenvectors.
Example: Suppose $y^{(5)}-4y^{(4)}+y'''+10y''-4y'-8y=0$. Then $L=D^5-4D^4+D^3+10D^2-4D-8=(D-2)^3(D+1)^2$. Solving $(D-2)^3[y]=0$, we get $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}$. Solving $(D+1)^2[y]=0$, we get $y=c_4e^{-t}+c_5te^{-t}$. The general solution is $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}+c_4e^{-t}+c_5te^{-t}$.
Similar methods allow one to deal with (repeated) irreducible quadratic factors.
Answer: I''l use two theorems to prove your point. One is:
THEOREM 1: If $y_1$, $y_2$, $y_3$,$\cdots$, $y_k$ are $k$ solutions of the
$$\phi(D)y = 0$$
where $\phi(D)$ is a polynomial in $D =\displaystyle \frac{d}{dx}$
then $$y = \sum_{i=i}^k c_i y_i$$
with each $c_i$ constant is a solution.
PROOF Given that $\phi(D)$ is a linear operator, if one has
$$\eqalign{
  & c_1\phi \left( D \right){y_1} &=& 0  \cr
  & c_2\phi \left( D \right){y_2} &=& 0  \cr
  &  \cdots  &=&  \cdots   \cr
  & c_k\phi \left( D \right){y_k} &=& 0 \cr} $$
then summing produces
$${c_1}\phi \left( D \right){y_1} + {c_2}\phi \left( D \right){y_2} +  \cdots  + {c_k}\phi \left( D \right){y_k} = 0$$
$$\phi \left( D \right)\left( {{c_1}{y_1} + {c_2}{y_2} +  \cdots  + {c_k}{y_k}} \right) = 0$$
So $$y = {{c_1}{y_1} + {c_2}{y_2} +  \cdots  + {c_k}{y_k}}$$ is a solution.
THEOREM 2
If $$(D-a)^n y =0$$ then $$y=\left(c_0+c_1 x+\cdots+c_n x^{n-1}\right)e^{ax}$$ is a solution.
PROOF
It is true that $$(D-a)(e^{ax} u) = e^{ax} Du$$ since
$$ae^{ax}u+e^{ax}u'-ae^{ax}u = e^{ax} u'$$
Let's show by induction that $$(D-a)^n(e^{ax}u) = e^{ax} D^nu$$
It is true for $n=1$. But $n=k$ implies $n=k+1$, since
$$(D-a)^{n+1}(e^{ax}u) =(D-a) (e^{ax} D^nu)$$
$$\eqalign{
  & {\left( {D - a} \right)^{n + 1}}\left( {{e^{ax}}u} \right) &=& D\left( {{e^{ax}}{D^n}u} \right) - a{e^{ax}}{D^n}u  \cr
  &  &=& a{e^{ax}}{D^n}u + {e^{ax}}{D^{n + 1}}u - a{e^{ax}}{D^n}u  \cr
  &  &=& {e^{ax}}{D^{n + 1}}u \cr} $$
Therefore it is true for all $n$.
Putting $y = e^{ax} u$ in our original DE produces.
$$\eqalign{
  & {\left( {D - a} \right)^n}y &=& 0  \cr
  & {\left( {D - a} \right)^n}\left( {{e^{ax}}u} \right) &=& 0  \cr
  & {e^{ax}}{D^n}u &=& 0  \cr
  & {D^n}u &=& 0 \cr} $$
If $$u = \sum\limits_{k = 1}^n {{c_k}{x^{k - 1}}} $$ then $D^{n}u =0$ and for having $n$ arbitrary constants, is a general solution. Thus
$$y = {e^{ax}}\sum\limits_{k = 1}^n {{c_k}{x^{k - 1}}} $$

Thus your expression can be put as $$y = \sum\limits_{k = 0}^n {{a_k}{y^{\left( k \right)}}}  = \sum\limits_{k = 0}^n {{a_k}{D^k}y}  = \phi \left( D \right)y$$
and you can use the theorem for each root by factoring $\phi(D)$.
Answer: While there are some other good answers to this question, I'll proceed along a different line.  Certainly, linear combinations of the solutions $f(x)=p_\lambda(x)e^{\lambda x}$ will work, where $\lambda$ is a root of the characteristic polynomial of the differential equation, $\chi(t) = \sum a_k t^k$, and $\deg p_\lambda$ is less than the multiplicity of $\lambda$ in $\chi$.  This we can check by hand, and the uniqueness theorem for solutions to linear differential equations says that the $\mathbb{C}$-vector space spanned by the possible $f(x)$'s  exhausts the solutions.
On the other hand, this offers no intuition as to how the solutions were found in the first place.  I'll show how to construct them, on the assumption that they represent holomorphic functions near $0$ (this shouldn't be too much of a leap of faith; in any case, it works).  With that assumption, let $f(x) = \sum c_n x^n$, and if $f$ solves our equation, we must have
$$ a_0 \sum c_n t^n + a_1 \sum n c_nt^{n-1} + \ldots + a_d \sum \frac{n!}{(n-d)!} c_n t^{n-d} = 0.$$
Equating coefficients, it follows that
$$a_0c_n + a_1(n+1)c_{n+1}+\ldots + a_d\frac{(n+d)!}{n!}c_{n+d}=0.$$
Next, we introduce $b_n = c_n n!$, such that
$$a_0 \frac{b_n}{n!} + a_1 \frac{b_{n+1}}{n!} + \ldots + a_d \frac{b_{n+d}}{n!} = 0.$$
Of course, this implies that $\sum a_k b_{n+k} =0$ for all $n$; that is, the $b_n$ satisfy a linear recurrence.  By the general theory of linear recurrences, we have $b_n = \sum p_\lambda(n) \lambda^n$, where $\chi(\lambda)=0$ with multiplicity greater than $\deg p_\lambda$.  In turn, we have $c_n = \sum p_\lambda(n) \lambda^n/n!$, so
$$f(x) = \sum c_n t^n = \sum_n \sum_\lambda p_\lambda(n)\frac{\lambda^n}{n!}x^n.$$
Here is where the powers of $x$ enter; rewrite the polynomials $p_\lambda$ with respect to the basis $1$, $n$, $n(n-1)$, etc., so the summand corresponding to $\lambda$ reads as $$\sum_n \left(s_0 \frac{\lambda^n}{n!} x^n + s_1n \frac{\lambda^n}{n!}x^n + \ldots s_k \frac{n!}{(n-k)!} \frac{\lambda^n}{n!} x^n\right).$$
Canceling, and recognizing the Taylor series for the exponential, gives us
$$s_0 e^{\lambda x} + s_1 (\lambda x) e^{\lambda x} + \ldots + s_k(\lambda x)^k e^{\lambda x}.$$
Now, we just sum the contributions over all $\lambda$ to recognize our solution in the form
$$ \sum_\lambda q_\lambda(x) e^{\lambda x}\;,$$
with $q_\lambda(x)$ a polynomial in $\mathbb{C}[x]$ of degree less than the multiplicity of $\lambda$ in $\chi$.  By our work, moreover, all holomorphic solutions to this differential equation must be of this form.
Note: with this, and some `backwards reasoning', one can transfer many results between the theory of differential equations with constant coefficients to the theory of linear recurrences.  To some extent, they are entirely analogous.


Lower version of Ubuntu?

Question: So I have a 702MB disk that I'll be using, and Ubuntu 12.10 is 753MB. Any way to get an older version of Ubuntu for my disk, then upgrade later? Thanks in advanced!
Answer: No need. Yes, the size of the iso exceeds that of the capacity of a standard CD. You can either create a live-dvd or live-usb with it. Yikes! The solution is to create a minimal cd. It downloads and installs all the missing packages after you've installed ubuntu. Follow this link to download it. You don't need to download an old distribution, you can download a minimal cd of 12.10.
You may want to see answers for this question
Answer: dont go for older version. just make bootable predrive using unetbootin. you will save you CD also

Lower version of Ubuntu?


Is the Born rule a fundamental postulate of quantum mechanics?

Question: Is the Born rule a fundamental postulate of quantum mechanics, or can it be inferred from unitary evolution?
Comment: As the page about postulates you linked to correctly says, the Born-like rules to calculate probabilities from state vectors and operators are among the general postulates of quantum mechanics. It doesn't mean that they can't be derived from some other assumptions. However, the other assumptions clearly have to be connected with the notion of "probability" in one way or another, so they will be either a special or generalized formulation of the Born rule, anyway. Saying that the evolution is unitary doesn't say anything about probabilities - it can't "replace" the Born rule.
Comment: @LubošMotl I felt that if experimental apparatus must obey the same laws as the system under observation, then Born rule must follow from unitary evolution in all situations. Please can you elaborate on this comment "However, the other assumptions clearly have to be connected with the notion of "probability" in one way or another, so they will be either a special or generalized formulation of the Born rule"
Comment: I gave a derivation of the Born rule in the last answer to the question http://physics.stackexchange.com/q/19500/
Comment: related or duplicate: http://physics.stackexchange.com/q/73329/
Answer: Strictly speaking, the Born rule cannot be derived from unitary evolution, furthermore, in some sense the Born rule and unitary evolution are mutually contradictory, as, in general, a definite outcome of measurement is impossible under unitary evolution - no measurement is ever final, as unitary evolution cannot produce irreversibility or turn a pure state into a mixture. However, in some cases, the Born rule can be derived from unitary evolution as an approximate result - see, e.g., the following outstanding work: http://arxiv.org/abs/1107.2138 (accepted for publication in Physics Reports). The authors show (based on a rigorously solvable model of measurements) that irreversibility of measurement process can emerge in the same way as irreversibility in statistical physics - the recurrence times become very long, infinite for all practical purposes, when the apparatus contains a very large number of particles. However, for a finite number of particles there are some violations of the Born rule (see, e.g., the above-mentioned work, p. 115).
Comment: Unfortunately the article is completely wrong. I know two of the authors and their works on perpetual machines and supposed violations of the second law of thermo.
Comment: @juanrga: Maybe you're right, and the article is indeed completely wrong, but until you offer some specific arguments, why should I believe you, rather than the authors and the referees of their published articles? You mentioned their articles on other topics, but I am not sure this is relevant.
Comment: Thank you, I will take a look at the article referred to see if there is any weight in their arguments. Probably they are wrong as juanrga says, as most papers in this field are.
Comment: @Prathyush: You may wish to start with their article http://arxiv.org/abs/quant-ph/0702135 , which is much shorter (see references to their journal articles there).
Comment: @akhmeteli Thank you I will look into it, Indeed since I haven't gone deeply into the article, I must not comment on its factual accuracy. May I ask what you thought about the article?
Comment: @Prathyush: For me it was an eye-opener: they show how the contradiction between final measurement and never-final unitary evolution is resolved - in the same way as the contradiction between irreversibility in thermodynamics and reversibility in mechanics (classic or quantum) is resolved: while there is no way the Poincare recurrence theorem can be circumvented, the recurrence time is mind-boggling. In their article the measurement occurs through interaction with a large paramagnetic system, and that ensures practical irreversibility (but there is no way around theoretical reversibility).
Comment: @akhmeteli Does it comment on the concept of memory?
Comment: @juanrga: I find their work on measurement theory to be some of the best in this field.
Comment: @Arnold Neumaier: Thank you, I am very glad we agree on that.
Comment: @Prathyush: Not  quite sure what kind of "memory" you have in mind, but they do discuss "robust registration".
Comment: @akhmeteli "the registration is a process taking the apparatus from its initially metastable state to one of its stable final states." thank you I will definitely read this paper. It looks interesting. Since it is not a well accepted theory I can't give you correct answer bonus but +1 is a subjective opinion ;)
Comment: @akhmeteli Prathyush: The authors work in the $\lambda^2t$ approx. --see (A.6)--, where $\Lambda \approx U$ and this is why they _believe_ that have derived the Born rule from unitarity, when they have _not_. A rigorous study shows that the Born rule follows from $\Lambda \neq U$. Their treatment of the large N limit also lacks rigour: there is not Hilbert space norm for (3.33) in the limit, and the spectral decomposition of (1.6) has to be made in an extended space, not in the Hilbert space as they believe...
Comment: @ArnoldNeumaier: I like their support for Ballentine ensemble interpretation of QM.
Comment: @juanrga: Your comment does not seem clear: I even failed to find letter $\Lambda$ in their large article.
Comment: @akhmeteli: $\Lambda$ is the usual symbol for denoting the non-unitary evolutor. You will not find the non-unitary evolutor $\Lambda$ in their paper because they misguidedly use $U$ by the reasons explained above.
Comment: @juanrga: I am trying to say that you're not giving enough details, so I cannot agree or disagree with your objections to the article. I am not able to read your thoughts and, with all due respect, cannot believe you on your word. They have published their articles, starting years ago. Is there any published critique? Your objections are not accessible, as far as I know, so I cannot discuss them. What you added in your last remark is not nearly enough and does not make your remark significantly clearer.
Comment: @akhmeteli: Being published is not synonym for being right. As stated above, they are working in a well-known approximation, for which the non-unitary evolutor looks as an unitary one. This is the source of their confusion. Moreover, their treatment of the large $N$ limit is without any basis (there is not Hilbert space norm). It is difficult to give details with limited-size comments. I understand that you are open to believe what you want. It is fine for me if you believe that their work is "_outstanding_".
Comment: @juanrga: I certainly agree that "being published is not synonymous to being right". Furthermore, I did not check their derivations. I admire their article because it provides unexpected (for me, at least) answers (to difficult questions) that seem to make sense. On the other hand, you'll probably agree that "not being published is not synonymous to being right", either. Your remarks on the large $N$ limit are not convincing either, as the authors mostly deal with finite $N$. So let us just agree that we cannot resolve our disagreement right now. Again, I cannot believe you on your word.
Comment: @akhmeteli: As stated above your attitude is fine for me, but I can count about 11 explicit instances of $N \rightarrow \infty$ on the paper, starting with (1.4), and there is more implicitly --e.g., when they write "_so that the spectrum can be treated as a continuum_"-- they are taking the same limit. As stated above their treatment of this limit is invalid and this has consequences for the physics.
Comment: @juanrga: It seems to me they use (1.4) for a qualitative discussion, while the phrase "can be treated as a continuum" is immediately followed by words "(except in sections 5.3 and 6)", where they do consider a discrete spectrum. So until you show where and how improper treatment of the limit of large $N$ leads the authors to wrong conclusions (and what these wrong conclusions are), I am not impressed.
Comment: @akhmeteli: Sorry, but I do not find any real argument in your comment obligating me to reconsider my position.
Comment: Neither do I find any real arguments in your large $N$ comments.
Answer: The use of the word "postulate" in the question may indicate an unexamined assumption that we must or should discuss this sort of thing using an imitation of the axiomatic approach to mathematics -- a style of physics that can be done well or badly and that dates back to the faux-Euclidean presentation of the Principia.  If we make that choice, then in my opinion Luboš Motl's comment says all that needs to be said. (Gleason's theorem and quantum Bayesianism (Caves 2001) might also be worth looking at.) However, the pseudo-axiomatic approach has limitations. For one thing, it's almost always too unwieldy to be usable for more than toy theories. (One of the only exceptions I know of is Fleuriot 2001.) Also, although mathematicians are happy to work with undefined primitive terms (as in Hilbert's saying about tables, chairs, and beer mugs), in physics, terms like "force" or "measurement" can have preexisting informal or operational definitions, so treating them as primitive notions can in fact be a kind of intellectual sloppiness that's masked by the superficial appearance of mathematical rigor.
So what can physical arguments say about the Born rule?
The Born rule refers to measurements and probability, both of which may be impossible to define rigorously. But our notion of probability always involves normalization. This suggests that we should only expect the Born rule to apply in the context of nonrelativistic quantum mechanics, where there is no particle annihilation or creation. Sure enough, the Schrödinger equation, which is nonrelativistic, conserves probability as defined by the Born rule, but the Klein-Gordon equation, which is relativistic, doesn't.
This also gives one justification for why the Born rule can't involve some other even power of the wavefunction -- probability wouldn't be conserved by the Schrödinger equation. Aaronson 2004 gives some other examples of things that go wrong if you try to change the Born rule by using an exponent other than 2.
The OP asks whether the Born rule follows from unitarity. It doesn't, since unitarity holds for both the Schrödinger equation and the Klein-Gordon equation, but the Born rule is valid only for the former.
Although photons are inherently relativistic, there are many situations, such as two-source interference, in which there is no photon creation or annihilation, and in such a situation we also expect to have normalized probabilities and to be able to use "particle talk" (Halvorson 2001). This is nice because for photons, unlike electrons, we have a classical field theory to compare with, so we can invoke the correspondence principle. For two-source interference, clearly the only way to recover the classical limit at large particle numbers is if the square of the "wavefunction" ($\mathbf{E}$ and $\mathbf{B}$ fields) is proportional to probability. (There is a huge literature on this topic of the photon "wavefunction". See Birula 2005 for a review. My only point here is to give a physical plausibility argument. Basically, the most naive version of this approach works fine if the wave is monochromatic and if your detector intercepts a part of the wave that's small enough to look like a plane wave.) Since the Born rule has to hold for the electromagnetic "wavefunction," and electromagnetic waves can interact with matter, it clearly has to hold for material particles as well, or else we wouldn't have a consistent notion of the probability that a photon "is" in a certain place and the probability that the photon would be detected in that place by a material detector.
The Born rule says that probability doesn't depend on the phase of an electron's complex wavefunction $\Psi$. We could ask why the Born rule couldn't depend on some real-valued function such as $\operatorname{\arg} \Psi$ or $\mathfrak{Re} \Psi$. There is a good physical reason for this. There is an uncertainty relation between phase $\phi$ and particle number $n$ (Carruthers 1968). For fermions, the uncertainty in $n$ in a given state is always small, so the uncertainty in phase is very large. This means that the phase of the electron wavefunction can't be observable (Peierls 1979).
I've seen the view expressed that the many-worlds interpretation (MWI) is unable to explain the Born rule, and that this is a problem for MWI. I disagree, since none of the arguments above depended in any way on the choice of an interpretation of quantum mechanics. In the Copenhagen interpretation (CI), the Born rule typically appears as a postulate, which refers to the undefined primitive notion of "measurement;" I don't consider this an explanation. We often visualize the MWI in terms of a bifurcation of the universe at the moment when a "measurement" takes place, but this discontinuity is really just a cartoon picture of the smooth process by which quantum-mechanical correlations spread out into the universe. In general, interpretations of quantum mechanics are explanations of the psychological experience of doing quantum-mechanical experiments. Since they're psychological explanations, not physical ones, we shouldn't expect them to explain a physical fact like the Born rule.
Aaronson, "Is Quantum Mechanics An Island In Theoryspace?," http://arxiv.org/abs/quant-ph/0401062
Bialynicki-Birula, "Photon wave function", 2005, http://arxiv.org/abs/quant-ph/0508202
Carruthers and Nieto, "Phase and Angle Variables in Quantum Mechanics", Rev Mod Phys 40 (1968) 411; copy available at http://www.scribd.com/doc/147614679/Phase-and-Angle-Variables-in-Quantum-Mechanics (may be illegal, or may fall under fair use, depending on your interpretation of your country's laws)
Caves, Fuchs, and Schack, "Quantum probabilities as Bayesian probabilities", 2001, http://arxiv.org/abs/quant-ph/0106133; see also Scientific American, June 2013
Fleuriot, A Combination of Geometry Theorem Proving and Nonstandard Analysis with Application to Newton's Principia, Springer, 2001
Halvorson and Clifton, "No place for particles in relativistic quantum theories?", 2001, http://philsci-archive.pitt.edu/195/
Peierls, Surprises in Theoretical Physics, section 1.3
Comment: @SebastianHenckel: This is not completely thought out and may be wrong. But suppose that the rule for electrons is not the Born rule but a rule saying that probability is $\propto|\Psi|^p$, where $p\ne 2$. If you scatter an EM wave off of an electron,  they interact through some wave equation such that the scattered part of $\Psi$ is proportional to the amplitude of the EM wave: amplitude is proportional to amplitude. But then the electron is acting like a detector, and $p\ne 2$ means that the probability of detection isn't proportional to the probability that the photon was there.
Comment: It took me a while to read this answer. you said "The OP asks whether the Born rule follows from unitarity. It doesn't, since unitarity holds for both the Schrödinger equation and the Klein-Gordon equation, but the Born rule is valid only for the former." Isn't Born rule applicable even in Relativistic Quantum mechanics(Any field theory in general), not in the sense of KG equation but the KG field. Also Would you comment on my recent answer on a related topic, http://physics.stackexchange.com/questions/76132/the-irreducible-observer
Comment: *Since the Born rule has to hold for the electromagnetic "wavefunction," and electromagnetic waves can interact with matter, it clearly has to hold for material particles as well, or else we wouldn't have a consistent notion of the probability that a photon "is" in a certain place and the probability that the photon would be detected in that place by a material detector.* Could you explain this in more detail?
Comment: I like this argument. The interaction between the photon and the electron however is quantum electrodynamics all the way through, and that's something I don't know much about. However, thanks for making a connection between electrons and waves I never thought about. The pure de Broglie argument always seemed very ad hoc, and this makes it somewhat more plausible.
Comment: @Prathyush: My relativistic field theory is pretty weak, so if you want a really coherent explanation of why the Born rule doesn't apply to the KG equation, you're probably better off posting that as a question and letting someone more competent answer. But basically I think the concept is that in relativistic QM, we have to give up on the idea of having eigenstates of position, so the whole Copenhagen-ish interpretation of a position measurement as projecting the wavefunction down to a delta function doesn't really work.
Answer: This is clearly a somewhat controversial topic, but Zurek claimed to derive the Born Rule from other postulates.
Answer: The idea of deriving the Born rule (and in fact the whole measurement postulate) from the usual unitary evolution of quantum systems is at the very heart of a realist interpretation of quantum theory. If the quantum state really describes a the true internal state of a system and measurement is just a certain kind of interaction, then there should be only one single law for the time evolution.
Quantum theory however is fundamentally non-local and separating systems is conceptually hard, which makes observer and experiment impossible to describe separately. There should be a system containing both parts however and which follows a simple law of time evolution. Of course, the obvious candidate for such a law is unitary evolution, simply because that is what we observe for systems that we isolate as good as possible.
It is usually argued that this route leads to the Everett interpretation of quantum theory, where observations are relative to the observer and realized by entangled states. There have been several attempts to derive the Born rule in this context, but all that seem valid require additional assumptions that are questionable (and may in fact be inconsistent with the realist approach or other fundamental assumptions).
The reason why there cannot be a derivation that just uses ordinary unitary evolution and results in the Born rule is not even unitarity but the linearity of the theory. Say there is an evolution that takes out input to the measurement output, and we decide to measure a|A>+b|B> in the basis {|A>,|B>}. Then independently from the environment the Born rule predicts that |A> and |B> are invariant under measurement. A superposition (|A>+B>)/sqrt(2) should end up in either |A> or |B> depending on a possible environment state if the Born rule applies. The linearity of the theory requires that the outcome is a superposition of |A> and |B> however (the phase may change though).
Everett's answer to this problem is that the superposition comes out, but with the outcomes entangled with the observer seeing either outcome. But this creates two observers that are unaware of their own amplitude. Because of the linearity their future evolution is independent from the branch amplitude, and it's therefore hard to argue that any aspects of their perceived reality would depend on the branch amplitude.
Interestingly approaches to fix this issue, like the use of decision theory, advanced branch counting, etc, in some form introduce a nonlinear element to the theory. Be it a measure of branch amplitude, a cutoff amplitude or amplitude discretization, a stability rule (envariance or quantum darwinism). There are also approaches that don't hide the nonlinearity in additional assumptions that may collide with the linear evolution. Those are explicit nonlinear variations of the Schroedinger equation that can in fact produce an evolution that allows the Born rule to emerge. Of course, this is not something that most theorists embrace, simply because the linearity of quantum theory is such an attractive feature.
But there's one more approach that I personally favor. The nonlinearity could be only subjective to an observer, caused by incomplete knowledge about the universe. An observer, i.e. a local mechanism realized within quantum theory, can only gather information by interacting with his environment. Certain information however is inaccessible dynamically, hidden outside the observer's light cone or just not available for direct interaction. Considering this, it can be shown that reconstructing the best possible state description an observer can come up with must follow a dynamic law that is not unitary all the time, but also contains sudden state jumps with random outcomes driven by incoming priorly unknown information from the environment. It can be shown that a photon from the environment with entirely unknown polarization can cause a subjective state jump that corresponds exactly to the Born rule. This is of course a bold claim. But please see http://arxiv.org/abs/1205.0293 for a proper derivation and discussion of the details. If you you would like to look at a more gently introduction to the idea you can also read the (less complete but more intuitive) blog I've set up for this: http://aquantumoftheory.wordpress.com
Comment: I don't know if environment is a necessary concept in the measurement problem, For example, will a photographic Plate work in perfect vacuum. Thought I don't have the opportunity of experiment with such a situation, I believe a photographic must work normally in vacuum where there is no environment or extraneous photons.
Comment: Even in your perfect vacuum you always have an interacting environment. And of course the environment may not be needed for the resolution of the measurement problem, but it might possibly be necessary, and so you cannot simply exclude it. It is at least a plausible source for randomness due to our lack of information about its state.
Comment: In some situations where you cannot remove it from the experimental setup you will have to include the environment in the theory. What do you mean even in perfect vacuum you have the interaction environment? The basic process in a photographic plate is a light sensitive chemical reaction right? So an environment wont play a role
Comment: The environment always plays a role in quantum theory. You cannot remove the quantum fields from space, no matter how perfect your vacuum is. There will always be interaction on some level, and ignoring that is surely not helpful for understanding the properties of quantum systems. You seem to be thinking is more or less classical terms with your photographic plate example.
Comment: Also, in order to see if your plate has been affected by light you have to look at it. So at the very latest then you will subject it to massive interaction with an unknown environment
Comment: Hmm yes indeed that is true, you have to look at it.
Answer: The Born rule is a fundamental postulate of quantum mechanics and therefore it cannot be derived from other postulates --precisely your first link emphasizes this--.
In particular the Born rule cannot be derived from unitary evolution because the rule is not unitary
$$A \rightarrow B_1$$
$$A \rightarrow B_2$$
$$A \rightarrow B_3$$
$$A \rightarrow \cdots$$
The Born rule can be obtained from non-unitary evolutions.
Comment: You are arguing that the same input state gives different output states, which is not unitary. That argument is false because you don't know that the input state is different for different outcomes, simply because you don't know the state of the unknown environment, by definition, that leads to the different outcomes. I'm not saying that your conclusion is wrong, but your argument certainly is.
Comment: This argument is actually not valid because it does not count in unknown states from the environment which could differ for different outcomes.
Comment: That is not true. Adding the environment and its equation of evolution gives an isolated system whose __exact__ evolution is non-unitary.
Comment: Either if you assume that the same initial environment state $A\otimes E$ or not $A\otimes E_1,A\otimes E_2,A\otimes E_3\dots$ the evolution of the composite isolated system continues being non-unitary. von Neuman understood this and introduced his non-unitary evolution postulate in orthodox QM.
Comment: That's not what you wrote in your answer however
Comment: Now you missed me entirely. I do not even know what do you mean by "answer": My answer to the OP? My first answer to your comment? My second answer?
Answer: It is independent, but it is not fundamental, as it applies only to highly idealized kinds of measurements. (Realistic measurements are governed by POVMs instead.)
In fact, the role of Born's rule in quantum mechanics is marginal (after the standard introduction and the derivation of the notion of expectation).
It is hardly ever used for the analysis of real problems, except to shed light on problems in the foundations of quantum mechanics.
Comment: "It is hardly ever used for the analysis of real problems [...]" This is wildly wrong in my experience.
Comment: One day I will learn about POVM's its been on my list of To Do's for a long time.
Comment: POVMs can be regarded as Born type measurements in a larger space, so you're back where you started.
Comment: @A.O.Tell: On the formal level, yes. But in this larger space, one never does any measurements that would deserve that name.
Comment: That statement would require an exact definition of what a measurement is and how it is applied to a subsystem. Also, it makes no practical difference. If you know how a Born style measurement works you understand how a POVM works.
Comment: @A.O.Tell: It is enough to know what is really measured. Measure the mass of the sun, the halflife of Technetium, or the width of a spectral line in the Balmer series, and try to express it in terms of the Born rule!
Comment: We're talking about quantum measurement here, which both of your example are not. So of course you cannot express them in terms of the Born rule. And that's totally missing the point.
Comment: @A.O.Tell: Each example mentioned is a quantum system, and each effect can be expressed in terms of quantum mechanics. Your answer only confirms my point that what you call ''quantum measurements'' covers only highly idealized kinds of measurements.
Comment: Quantum meausrements are the building blocks for more complex meauserements. Nobody would call a derived result like the halflife of an atom a quantum measurement. Also, this does not even fit a POVM scheme, so you're disagreeing with yourself. The OP clearly talked about quantum measurements in a way that is really hard to misunderstand. Why you're trying to take that to somewhere entirely unrelated is beyond me.
Comment: The OP didn't talk about measurements at all.
Comment: So you know an application of the Born rule outside of the context of quantum measurement?
Comment: The question was whether the Born rule is fundamental, not about its role in quantum measurement. It is not fundamental, except when you build the fundament on the textbook approach.
Answer: Yes it can be derived from unitarity of quantum evolution. This was shown by Lesovik in his paper, Derivation of the Born rule from the unitarity of quantum evolution.


Non-computable numbers in Physics

Question: It is known that there are real numbers that can't be calculated (non-computable numbers). Quite probably that some physical phenomena (it is possible still undetected) depend on this numbers. Whether means it, what we can't construct the theory of this phenomenon and-or (if probably to construct) to do predictions (i.e. to check up this theory in practice)?
Comment: such numbers typically require a recursively infinite number of operations in order to define. That is typically ok when each operation gives a correction to the value which is bounded by some polynomial or exponential bound. a candidate for this would behave as a number where the bounds of each progressive operation do not approach zero - There are many numbers like these in QFT when one tries to do perturbative expansions, However, they are also in principle computable in non-perturbative fashion, so they fail to be entirely non-computable in the strict, mathematical sense.
Answer: When you think of a physical parameter which is "uncomputable", what precisely do you mean? For us to know that it is uncomputable, it has to arise somehow, on theoretical grounds, from e.g. a computational process which is equivalent to the Halting Problem of theoretical computer science; so that we could not compute it from first principles. But any non-computable number can still be approximated, to arbitrary precision, by computable (even rational!) numbers: we've just stipulated that within the theory itself, those rational approximations are hard to come by and cannot be obtained systematically.
But there's another way of obtaining those approximations: experimentally! Ultimately what fixes constants is not theory, but observation. Even if you have a computable constant, if there's consistent experimental evidence against the theory, it's probably the theory which is wrong. Experiment then becomes a way of trying to fix the value of the uncomputable number. In order to prevent that, there would also have to be (at the least) theoretical grounds why any procedure for measuring it would also amount to solving an uncomputable problem: e.g. that any sequence of experimental set-ups, each revealing the mystery constant with higher precision, must either fail or would amount to a model of computations strictly stronger than Turing Machines.
Even if there was a theoretical justification for why the constant would be experimentally unmeasurable, that claim itself would have to be tested experimentally. What should we expect from experimentally unmeasurable parameters? Well, we don't mean that we can't measure it to any precision, just that there are real obstacles to measuring it to any substantial precision (like being stuck on "3.2" as an approximation to π, and unable to get a better result). If that were the case, attempts to measure to high precision should yield results which are routinely inconsistent with one another; and experiments to get high-precision values should be extraordinarily difficult to devise. (Though how you know that it's "hard enough", for a finite number of attempts to suspect the parameter might be uncomputable isn't an obviously answerable question.)
Edited to add: It's worth noting that — as Peter Shor noted in the comments — there may be other aspects of the physical theory which might prevent precision measurements  of any purportedly uncomputable constants. For instance, in any detector which is meant to make precision measurements of energy, the time you allow for this measurement becomes crucial to the precision of the measurement, by the uncertainty principle. Even if by some masterstroke you manage to supercede quantum theory and bypass the uncertainty principle (which would be quite an unlikely theoretical breakthrough!), the new superceding theory may involve new uncertainty principles of its own, which would limit the precision to which you could measure constants. In that case, the precise-but-unmeasurable value would be about as good as any number of finite-precision rational approximations.
Finally — and in a similar vein as the remarks above regarding limits on measurements of constants imposed by uncertainty relations — in the case that the real world did operate on the basis of a constant that was neither theoretically nor experimentally determinable, it is quite likely that the uncomputability of the constant would not have any discernable effect. After all: if its specific value played an important role in physics, we would be able to measure it by observing physical phenomena! If the value of a constant is so inaccessible that its value cannot be pinned down by experiment, or even just passive cosmological observation, it can't really be significant. It certainly doesn't seem like it would be a scientific theory. Not if the physics depends on the constant in any more-or-less continuous way, anyway...
In short, unless you're interested in diving into joint philosophical underpinnings of physical science and computability theory, I wouldn't be too concerned about the possibility your raise.
Comment: I want to rephrase this answer, although maybe I'm turning it into a different answer: Because of quantum mechanics and the uncertainty principle, the value of a physical constant can't be defined to more than 60 or 70 digits. And any finite-precision number by definition is computable.
Comment: @PeterShor: it depends on how "in principle" the answer should be, to be most useful to the OP. It's conceivable (but perhaps surprising) that parts of the cosmos might be described by Bohmian mechanics in "quantum non-equilibrium", in which case the uncertainty principle may not hold. Still, the fact that there might exist physical (as opposed to computational) barriers to arbitrary precision estimation of constants is a possibility which my answer didn't address; I'll add that in.
Answer: Noncomputable numbers are defined by infinite sequences of approximations. One such number is the number whose n-th binary digit is 1 if n is the concatenated ASCII code of a halting computer program and zero elsewhere. If you knew this real number, you solve the halting problem.
Almost all the real numbers that arise in physics are approximations, valid for finite time measurements. You have to idealize to infinite times in a physical model. But you can ask, for our known physical laws, assuming we can do measurements for an infinite amount of time, can we generate an uncomputable number?
There is a simple example. Every random number is non-computable. So if you measure the spin of a quantum system in a nontrivial superposition, and write down 1 in the n-th binary digit if you see spin up, and 0 if you see spin down, you get a noncomputable real in the limit. This means that if quantum mechanics is exact, one can easily generate numbers which are not output by any Turing machine, but these types of real numbers are output by a probabilistic computer, a probabilistic Turing machine.
I suppose though that you are interested in generating actual non-trivial non-random non-computable numbers. If you build a computer, then in the limit of infinite running time, you can write down the Halting number, but you write it peicemeal as each program halts. The question is then whether there is a more efficient way to write down the halting number than waiting for each program to halt, using random numbers, or using quantum mechanics. I think that this is not known.
The basic definitions of mathematics regarding computability basically coincide with the classical physics notion of system behavior, since you can simulate a classical physics system, and you can build a computer out of the motion of classical particles. Quantum mechanics is different computationally, as demonstrated most conclusively by Peter Shor, who showed that a quantum computer can solve the integer factoring problem, while a classical computer cannot. But for the purposes of decidability questions, not efficiency questions, quantum computers are equivalent to classical probabilistic computers. The probabilistic computer can output a sequence of random digits, and so can generate a non-computable real, as above.
Answer: This would most likely violate the Church Turing Thesis, namely that everything that is computable is computable by a turing machine. If we could measure this parameter, we could construct a machine that could measure it to arbitrary precision, we would have hypercomputation, which we do not believe exists.
Answer: There's a difference between noncomputable and limit computable real numbers. A limit computable real number is a real number which can be expressed as the limit of a computable sequence of real or rational numbers. Limit computable quantities abound in physics, but noncomputable ones violate the Church-Turing hypothesis which in essence states the universe can be simulated on a computer.

Non-computable numbers in Physics


Why don't you see multiple images of an object?

Question: Consider the ray model of light. Let's say an object such as a pencil is illuminated, and consider one point on that pencil. Since there could be many rays of light bouncing off the same point on the pencil, one ray could hit the left side of your eye and another ray from the same point could hit the right side of that same eye. Why don't you see that point in 2 different places then? Thanks!
Comment: The problem when I take off my glasses is that I *do* see multiple images (thought they lie close together and overlap so much that the total effects is simply on of blurriness).
Answer: I believe that the answer to this question involves multiple parts. I will try and hit all of them. Since you mentioned the ray model, I will assume you are relatively familiar with geometric optics.
First, we do see different images of the same object at times! Or rather, we see a blurry image rather than a sharp one. If you bring a pencil so close to your eyes that you cannot focus on it using your iris (more on that later), then you will see a blurry image. This happens for exactly the reason you mention. Rays from the same point of the object take different paths to your retina. If the path taken to your lens makes a large angle compared to the path that passes straight through the lens, then in general the rays will not recombine at a single point, but will have some spread. As the object moves closer to your eye, the angles increase, so the spread becomes larger and you see a blurry image.
Your iris is very important for creating sharp images. Generally, the distance between your lens and your retina is fixed. The distance between the object and your lens is not fixed, but we would like to be able to resolve detail for some range of object distances, rather than just a single one. So, the only parameter your body can control is the focal length. By constricting and relaxing, the iris changes the curvature of your lens. The change in curvature leads to a change in focal length. So, whatever object you are attempting to focus on, your iris constricts so that the object is beyond the focal length of your lens. This ensures that the rays will converge toward the retina and produce an image.
However, even with the object beyond the focal length you still get a blurry image if the rays make a large angle with the axis that is perpendicular to your eye/lens/retina (as discussed before) and this is one reason why you have a pupil. The pupil only allows rays that are approximately all parallel to each other to fall on your lens. It effectively acts as a collimator. So now, you have an object at or beyond the focal length and a set of approximately parallel rays that fall on your lens and are focused. This leads to a relatively sharp image on your retina (assuming that the object is far enough away that the pupil can do its job).
The final piece of the puzzle is your brain. Even though your iris, lens, and pupil do what they can to create a sharp image for your retina, it is still imperfect. There are still a number of aberrations in the image that falls on your retina. Your visual cortex and related support areas in the brain do all of the processing and reconstructing that leads to you perceiving a sharp image.
Answer: Ideally, all the rays coming from a point source intersect at the same point, to form an image. When that point is on your retina, the eye can "see" the object (eg: a specific point on the pencil). Note that to form a clear/distinct image all rays must intersect at a single point.
Similarly, when you want to estimate where an object lies, you trace all those rays (which hit your eyes) backwards, till they all intersect at some point. That point gives you the location of the object. Given that they all meet at the same point on the retina (since they form an image in your eye), just extending each one straight backwards would not make any light rays meet. So you won't be confused by imagining there are two copies of the object.
Instead of light rays hitting two sides of the same eye, consider light rays hitting two different eyes. Then they can (and do!) each estimate a slightly different direction for the position of the object, based on the different images seen by each eye. That difference comes in handy for depth perception.


How can I measure water/moisture content within a wall cavity?

Question: Related to improper placement of a vapour barrier within wall insulation, are there any tools or methods for measuring water/moisture content within a wall cavity?
Answer: I've used a moisture meter to do roughly this. My meter is the pinless kind. Some years ago, we had a contractor who did a less than perfect job on a bathroom remodel. (Ok, they left behind leaks in something like six separate places.) With the meter, I was able to tell where water was a problem, and also know when we had finally gotten all the leaks repaired.
I would not expect the actual meter reading to be highly accurate, since these meters are usually calibrated for wood. But the meter will indeed be able to tell the presence of moisture, and to differentiate between dry and wet. The problem is, these meters can only read to a certain depth, typically a half inch or so in wood. I'll look tomorrow to see, I might be able to test mine under similar conditions to see how deeply it can read.
Edit: I tested my moisture meter. I have a 12+ year old Wagner L609, an inexpensive pin-less meter that works nicely for wood samples. It is not tremendously accurate, but it does show moisture differences well between samples. Newer models might read more deeply than did mine.
The first test that I ran was to test a piece of 1/2 inch thick drywall, then move the flat of my hand up to it, until I saw a difference. By itself, the drywall reads at roughly 5-6%, my hand by itself reads off the top of the scale. When my hand approaches to within roughly 1/2inch, the meter sees a difference, jumping up by about 3-4%. It could also recognize the presence of my hand on the other side of a 3/4 inch thick telephone book.
Of course, this is a poor test for what you are doing. Suppose that moisture collects on the inside of a vapor barrier inside the wall, separated by over an inch from the interior wall surface? Were this to happen, the air space between the two surfaces will be maintained at a high humidity. In turn, this will significantly raise the moisture content of the interior wall surface.
I next did a test where I showed that drywall faced with a somewhat moister environment on one side would then easily register several percent high in moisture content.
This last test convinces me that you can indeed check for elevated moisture content of a wall cavity using a non-invasive and inexpensive moisture meter.
Comment: yes, please do test the depth if you have the time. Thanks!
Answer: You might be able to use a moisture meter like this one, but you'd probably have to push completely through the drywall into the wall cavity. I've never used a moisture meter but the writeup states the following:

The mini-Ligno S/D gives instant,
  accurate moisture readings between
  6-45%, indicated by a bright digital
  read out. Designed for inspectors,
  architects, contractors, painters and
  homeowners. This meter is a useful
  tool to measure moisture content in
  wood and soft building materials such
  as drywall, plaster and gypsum.
  Features include a green cap that
  serves as a handle for easy probe
  insertion and covers the pins during
  transportation. Longer pins stored
  inside the meter for measuring thicker
  material. To obtain readings the pins
  are pushed into the wood and the
  unique on-off switch between the pins
  turns the meter on.

I'm not certain it would work if you pushed it all the way through the drywall into the cavity. You might want to see if a local tool rental place has one you can rent before you sink the money into one for a one time reading.
Comment: Thanks. My concern with this kind of pushing through the interior wall surface (drywall in some places, wood panelling in others) to measure at the back of the wall cavity will change the behaviour of air & moisture movement (because now there are holes). Still it's likely better than just leaving it and waiting for the smell of mold or signs of water stains to indicate a problem.


Adjusting strategy for MTT

Question: I've been playing low stakes SNG's (10-50 participants) for a while and learned to beat the game to such a degree that I can play profitably. However, when I attempt playing in MTTs I rarely make it into the money. When the MTT have 500 participants or more it just seems hopeless. What kind of strategy adjustments are advisable compared to the SNG's that I usually play?
Answer: I think one of the most important aspects of becoming a good player in any kind of game is  knowing yourself. Everyone plays in a different way because they respond to other player's actions and to the cards they're dealt in a different way, because that's just how they are.
I suffer from the opposite problem of yours. I usually make into the money in 500~5000 players tournaments - which turns out to be really profitable - however, I often lose at small SNGs and am not-so-good not-so-bad at final tables. I found out that, although I'm certainly not a tight player, I tend to avoid risky hands and big decisions. I play what is called smallball poker, which is actually very common in MTTs, but not as much in small SNGs - because it doesn't work that well in it, specially the so called turbo ones, where the blinds go up real quickly and you have to make big decisions. So I had to adjust my strategy to play SNGs. I personally think smallball poker is a very good strategy for long, big MTTs, and it has been working pretty well for me, but that's because it fits my personality.
Players like Gus Hansen made their way to the top by playing very aggressively all through a lot of tournaments. There was this guy - which I can't remember the name - that came from limit poker and made a fortune playing very tightly on MTTs.
What I'm trying to say is: It doesn't really matter what strategy you use. Some usually work better than others in this or that type of game, but if it doesn't fit you, it might work sometimes, but it'll usually fail. Find out what's your kind of game and use a strategy that suits you the best - I listed some as suggestions.
Answer: The biggest difference in the 50 people sng and the 500 mtt is just that. Do not make any adjustments if you feel you are profitable. Remember though if you are itm 20% in this SNGs which is 1/5 times, you can now only expect to be itm 1/50 times due to the sheer volume of players. This means it takes you longer to approach that EV that you deserve. The other big difference is the amount of time these tournaments take. Make sure you feel fresh, because 5 hours in, it is easy to be less objective and pull off some silly moves that you would not otherwise.

Adjusting strategy for MTT


Cox rings of toric varieties over arbitrary fields

Question: The Cox ring of a toric variety X can be viewed as a generalisation of the homogeneous coordinate ring of projective n-space.  Over the complex numbers, the theory is outlined in The Homogeneous Coordinate Ring of a Toric Variety.
We associate to X a graded polynomial ring S and an ideal B.  The space $V=Spec(S)$ is isomorphic to $\mathbb{C}^n$ and has a subvariety Z cut out by the ideal B.  There is an action of a torus G on V such that V - Z is G-invariant with quotient isomorphic to X.  Furthermore, when X is smooth and complete, the automorphism group Aut(X) of the toric variety has a cover $\widetilde{Aut}(X)$ fitting into the exact sequence
$1 \to G \to \widetilde{Aut}(X) \to Aut(X) \to 1$.
The construction can be generalised to other varieties and base fields, often under the name of "universal torsors."  Unfortunately, this added generality lacks the wonderful combinatorial description in Cox's paper (for example, the Demazure root system).
Toric varieties over general base fields can be studied by considering the actions of Galois groups on the fan.  It should be fairly straghtforward to extend these ideas to the Cox ring, as well.  Rather then reinvent the wheel, I'd like to see if this has already been done.

Does there exist a reference?  A combinatorial description of Cox rings of toric varieties over arbitrary base fields, especially descriptions of analogs of $\widetilde{Aut}(X)$.
Answer: Parametrizations of toric varieties over any field: also available via google
Comment: The link to `sciencedirect.com` is broken, but the article can be found at [doi:10.1016/j.jalgebra.2006.08.016](https://doi.org/10.1016/j.jalgebra.2006.08.016) ([Zbl 1120.14043](https://zbmath.org/1120.14043)). The link to the PDF at A. Thoma's (old) website is now broken, and I couldn't find any snapshot on the Wayback Machine either, but it could perhaps be found at [CORE](https://core.ac.uk/download/pdf/82404918.pdf).


What are some approaches for avoiding rounded corners?

Question: Rounded corners require images (sprites), i.e. extra bandwidth and I'm working with slow Internet connections (think dialup at 56k or less). What are some approaches for designing so that rounded corners can be avoided? Are there any circumstances in which rounded corners are so important that I should consider taking a hit on page load time?
Comment: I did not vote the question down but it seems to be purely a coding/graphic design question. It's not really what UX is about in my mind.
Comment: If you're looking for rounded corners without CSS3 consider using a sprite for the corner images.
Comment: Yes, that's what I mean by "graphics". I'll edit the question.
Comment: Would appreciate comment on what the downvote is for.
Comment: Just let the people using old browsers miss the rounded borders. In general you've got to do quite a bit depending on rounded borders before you make a page look distinctly *bad* without them rendering.
Comment: @Gael The last part is ux concern. Can I get away without rounded corners? The answer appears to be yes.
Answer: In your situation I'm not sure what better practice there would be beyond not attempting to round the corners of elements. It looks sort of nice (if a bit over used) but even in modern full CSS 3 designs it's just a plus, not an integral part of design in most sites. Any technical work around you try to implement (sprites or images for button corners) is likely to result in added workload, added complexity/compatibility issues and added bandwidth usage. Don't forget that designing for pre-css3 browsers is as complex and incompatible enough as it is without trying to integrate css3 elements.
If you're introducing rounded borders on buttons or other elements to make them stand out as unique elements, consider more CSS2 friendly methods of doing so; borders, different colors (Don't overdo it though), white space, clear linking/physical proximity to related elements.
If for some explicit reason you need a rounded corner or you feel something is lost in the design without it, please post an image so that we know what the situation is. Generally rounded borders are only for making it look modern and nice (something I'm sure you're already sacrificing to adapt to low bandwidth condition) or making elements look distinct or "like buttons," but as mentioned there are pre-CSS ways of doing this as well.
Answer: Generally I consider rounded corners to be a nice to have but not necessary feature. However, having rounded corners vs not can change the character (or emotional appeal, perhaps?) of the design. I worked on a site where just about everything was rounded off, for a very bubbly, soft feeling overall. We switched from sprites to css3 rounded corners and the site felt very different in browsers that didn't support css3. The page felt so much snappier that we kept the change, but it was definitely a trade-off.


assign unique id to point features

Question: The MassGIS parcel standard (http://www.mass.gov/mgis/ParstndrdVer1_5_1.pdf) utilizes a concatenation of the whole number portion of x- and y-coordinates to create a unique id (LOC_ID) for features.  I'm considering doing the same for a point feature class.  I like the consistent methodology, but perhaps I'm overlooking something.  Is there a standard or best practice for creating unique ids for point features?
Answer: If you use that id for something like as a foreign key in relation to another table your whole database will get in big trouble if you have to move a point for some reason. Probably you then will have to keep the id even if it not describes the xy-coordinates any more.
As a unique key is often the best to have something not telling anything about the data, because most data might change.
Comment: +1 Never ever ever ever use ids that mean something or use business data in them. Someone at my work just got burnt because they used road names in the ID and lo and behold the road name had to be changed, now the ID is wrong.
Answer: Adding on from Nicklas answer and my comment.
I would say the most used convention and most recommended is just to use a auto-incrementing ID, eg start at 1 and just keep going. No logic and simple.
If you have a distributed system, or don't like auto-incrementing numbers, you could use a GUID.  Most databases will handle creating this kind of ID for you. However they are a pain for a user to enter manually, for searching etc, so just keep that in mind.
The other option is to use some kind of hash of the data but I would not recommend this.  It would mean that would you need to write an algorithm to do this for you, you can't always ensure uniqueness, they also tend to be a pain to enter for searching.
These are just my opinions, but from personal experience, trust me, never use business data in IDs.
Answer:
Follow up question: Is it best to create and maintain the unique id in the DBMS (SQL Server) or in the GIS software (ArcGIS)?

I'd strongly suggest checking for uniqueness inside the DBMS. That's one of the many strengths of DBMSs. It also allows you to access your data with different GIS software that probably wouldn't be aware of the unique constraints.
Answer: Our identification scheme was not chosen by me but is as follows: 2,3,4 character code which is the class of asset and a 6-digit sequential number (you would choose whatever digit count works for you).   A stored procedure creates these IDs and relies on a couple of non-geodatabase tables in the same SQL Server database.
I do keep a separate, sequential auto-increment ID.  I also keep a 13 character geohash field (for point features), but I would never use that as a key. The field is auto-populated (by a custom editor extension) whenever the feature is moved.
If your GIS data is to be used with any sort of asset management system, you'll want your IDs to be globally unique within your geodatabase (and perhaps unique within all geodatabases within your organization).  This will also make it a little easier to do geodatabase refactoring in the future.
Answer: Perrsonally I've used CRC calculations in the past to create similar values. Not too difficult to create and librarys/algorithms are available online.
The advantage there is that you can do features greater than points, whereas a concatenation would really have to be point features only (unless you want a really massive key).
And I think it's unlikely an end-user would be searching by this ID anyway, so I don't see that as a problem.
Having said that, I don't see many explicit benefits of assigning ID like that. I might use the method for change detection (because it's way mnore efficient to compare two CRC values than two sets of geometry) but even then - why use it as the primary ID?
Answer: ESRI's Arc Hydro Tools come with a toolbar that also installs a unique ID manager which runs in the background. The toolbar lets you assign unique IDs per feature class or per geodatabase. The ID manager per default only handles unique ID attributes called e.g. HydroID, which is part of the Arc Hydro data model. But it can be set up to handle other attributes, too. The tools come with a lot of documentation, so configuring the ID manager to your needs shouldn't be a problem.
The unique ID is, to my knowledge, always an integer. After assigning unique IDs once the manager takes care of assigning new unique IDs for every newly created feature whichs fits the configuration.
The unique ID manager could be useful for database backends which (AFAIK) don't support auto-incrementing numbers like the personal geodatabase.
Answer: "GUIDs can, of course, be generated using VB Script. But given the gradual deemphasis of VB Script by ESRI, we'll perform GUID generation in ArcMap using the ninth wonder of the world, Python. For those of you not already in the know, Python is God's gift to inveterate GIS hackers. My advice: Learn it! Live it! Love it!"
http://eaglemap.com/blog/bid/45555/How-to-Generate-GUIDs-in-ArcMap

assign unique id to point features


how to sample a certain sound on a song?

Question: I would like to find out one of the best approaches on how to sample a certain sound  in a song? Open-source/freeware software would be ideal but propriety software is welcomed as well. I'm looking to use these samples in live  DJ performances and not music production. Thank you
Answer: If you are wanting to separate out a single sound from within a performance, you will need to use filters to try and remove frequencies higher and lower than those in the piece you want, but you will still be left with some noises.
If you can find the sound you want in a part of the song with no other instruments playing, this will be your best bet. Them just record it into your sampler.
Answer: Recycle propeller heads is a good software that you can use to sample sounds. It allows you to cut audio according to different criteria such as bars, or you can place your own markers etc. unfortunately it is not free.
Ableton Live is also quite good to deal with sampling.
If you are looking for a free editor, there is none that comes to my mind now except audacity.
What are you using to launch your samples? A MPC, MPD, Maschine...?
Answer: There are two kinds of software, generally:
Programs for processing sound itself
Programs for putting sounds to good use
Nowadays, you can process digital sound in realtime. That means that DJing software could be used as both.
There is opensource DJing software such as Mixxx, and you should definitely checkout the demo version of Native Instruments' Traktor, which runs full for 30 minutes. There is also Virtual DJ, and others.
They all allow you to combine arbitrary samples with music tracks, without having to worry about timing and synchronization.


How do I prevent my child from freeing their hands in a SwaddleMe

Question: My daughter is currently in a SwaddleMe and has recently begun to wake herself up by getting her hand out and playing with her face This was always a slight problem but since she has started to seek her fist to chew on, it has increased exponentially. I have the SwaddleMe as tight as it can go and she can still break out of it. Is there anything more I can do to keep her hands in?
Comment: @Bobnix sorry, I missed that detail. I think that this might be a sign that she has outgrown the device. Any modifications to the functionality could affect the safety of the swaddle. On the other hand, I used to use a square cloth swaddle and tie the ends in a knot.
Comment: Does she wake herself up because she hits herself in the face? Or does she try to chew on her fist because she wakes?
Comment: Don't over tighten - it can be unsafe and can delay motor development, especially if swaddle is on for extended periods and large portions of each day
Comment: @RoryAlsop The former. She will make minor movements in a given time span, each moving her arm up. Once it gets to her face it is a constant movement from then on out.
Comment: @David I dont think I am over tightening it, as she can still manage to get out.
Comment: tight is not the same as secure. the first is related to the size of the enclosed swaddle, the second is related to the strength of the Velcro. If Velcro were replaced with zipper or buttons, the swaddle could be both looser and more secure
Comment: @David Ok, I see what happened, I was not clear. She gets her arm out without breaking the Velcro.
Answer: It sounds like she has developmentally outgrown the swaddle and is ready for a woombie or similar one piece jammy cacoon. It zips up and provides more room for movement while keeping baby in a comfy pacifying swaddle.
A similar effect can be obtained with a traditional square cloth swaddle by tying the ends in a knot around the chest.
Answer: I agree that your daughter may have outgrown the swaddle you're using. However, my own daughter is only a month old and can work her arms free of a SwaddleMe or a Halo SleepSack Swaddle given a few hours, unless the swaddle is drawn up very tightly.
I disagree with some of the comments; tight is not necessarily bad. You certainly don't want it loose, because then they just break out of it. Most parents do not swaddle tightly enough, leading to this problem. A "tight" swaddle has enough room for your child to breathe comfortably (it doesn't take much) but keeps their arms pinned. Contrary to initial fussing, babies feel comfortable when swaddled very firmly.
If your daughter is old enough to have conscious control over her arms (as in when you put her in the swaddle she will actively and intelligently wiggle her arms free relatively quickly) then she has probably outgrown the "arms-in" swaddle method in general. If not, then I would focus on tightening the swaddle around her shoulders; that will allow some movement and keep her abdomen free to breathe, while restricting her ability to actually get her hands out the top of the swaddle.
You may also try a swaddle with a different material. Most swaddles including the SwaddleMe brand are made with a fabric that will stretch relatively easily. While these fabrics are nice and soft to the touch, the stretching can limit your ability to get the swaddle as tight as you need. Complicating this with the SwaddleMe brand is that the hook side of the velcro is just a couple square-inch tabs, so there's more limited adjustment range.
By contrast the HALO SleepSack swaddles have long strips of the hook material that can securely adjust over a very wide range of tensions. However, the HALO brand aren't designed for use with a 5-point harness (and you should not attempt to use them when the child must be harnessed). We use both regularly, but generally prefer the SleepSacks when she's not in a carrier or car seat.
In either case, a swaddle made of a stretch fabric can stretch out over time. The SleepSack we got from the hospital is especially bad; it's "newborn" size but when she's wrapped firmly in it the velcro doesn't meet up. We have two other SleepSack swaddles, one exactly the same but issued from a different hospital and handed down to us, and one made of a microplush material that isn't near as stretchy. Both of them provide a much more secure swaddle with less stretch and more positive velcro contact, and we generally prefer these when putting her down for the night.
Comment: Active babies who fight swaddles WILL get the Halo swaddle fabric over their faces when swaddled arms in.  We stopped swaddling at 9 weeks due to this issue.


Why take Linguistics if you can cast Comprehend Languages?

Question: Given the nature of the spell Comprehend Languages in Pathfinder, is there any particular advantage to taking Linguistics for a character who can cast it? Especially given the chance of severe (or horribly entertaining, depending on your point of view) failure with Linguistics that doesn't seem to exist for Comprehend Languages.
Comment: The spell changed in *Pathfinder* from [*comprehend languages*](http://www.d20srd.org/srd/spells/comprehendLanguages.htm) in *Dungeons and Dragons 3.5*. In the *Pathfinder* antecedent the caster "must touch the creature" that is to be comprehended--which is much riskier and, potentially, socially awkward.
Comment: @GMJoe answered quite well, so I'll just add an additional comment:  Comprehend Languages / Tongues tends gives literal translations, but does not grant you understanding of idioms.  Linguistics has the important extra knowledge so you don't say 'fire it up' in gnomish to a gnome and he looks at you like you're crazy because why do you want to set his invention on fire?!
Answer: This has come up in one of the campaign I currently run. Comprehend languages, like most spells, has a limited duration and is therefore only useful as long as:

You don't encounter more than one piece of foreign-language text over the course of the day.
The foreign-languages you do encounter can be completely interpreted before the duration runs out.

Of course, players could prepare the spell more than once, but then they find themselves short on spell slots. Charged magic items can offset the disadvantage of spells without requiring slots, but they tend to have a worryingly large up-front cost for a non-guaranteed payoff.
The problem with translation spells is exacerbated by the fact that memories are rarely perfect: A wizard who read the riddle-poem of Llasynt the Long-winded probably won't remember it word for word, and if guessing the password for Llasynt's front door requires the third verb in the fourth insulting stanza, then he'd better have made some detailed notes...
On top of that, comprehend languages does what it says on the tin: It allows you to comprehend written or spoken languages, but doesn't allow you to speak or write those languages, making it useless if your cunning plan depends on asking a dryad where her bathroom is.
Linguistics, on the other hand is always on, is non-magical (and therefore works in situations where magic would be impossible, inadvisable or a social faux pas), and makes you a hit at parties. It also doesn't require you to touch the target for it to work, which again may allow you to avoid potential hazards, inconvenient distances, and more social faux pas. Plus, it grants you an additional language known with each rank, which is a nice way to avoid making mistakes with the skill entirely.
Comment: The added language is a huge benefit, not to be overlooked!
Answer: GMJoe has a great answer, but there's one major reason that comprehend languages is not a good solution that everyone else has missed.
From the spell:

You can understand the spoken words of creatures or read otherwise incomprehensible written messages. The ability to read does not necessarily impart insight into the material, merely its literal meaning. The spell enables you to understand or read an unknown language, not speak or write it.

Comprehend languages is a one-way spell.  It's useless if you need to actually communicate in the alternate language, rather than just understanding what someone says.  Tongues does provide that ability, but it's 3rd/4th level spell.
Answer: Our brave band of fearless heroes have braved the Chasm of Doom, crossed the Bridge of Tears and found their way to the Orc lair, from which they can make out some grunting in the fiendishly difficult greenskin tongue.
"What are they saying?", whispered Mik Half-finger.
His elven companion smiled and spent a few seconds incanting.
"They are saying…"
"Yes?", prompted Mik eagerly.
"They are saying that someone is casting magic behind this wall."
"Then shoot them with a magic missile!"
"I can't! I memorised Comprehend Languages this morning!"
"Ruuuuuuuuun!"

Comprehend Languages requires time (sound, spell components, etc) to cast and uses up a spell slot.
Comment: Yes the bit where you basically have to shout "Comprehend Languages!" before the spell is cast kinda nixes stealth.
Answer: From the PFSRD (with added emphasis):

You can decipher writing in an unfamiliar language or a message written in an incomplete or archaic form.

With Linguistics, you get the language, its history and past forms, and contextual knowledge to fill in blanks. You do not get this with Comprehend or Tongues. Such is the reward for spending skill points.
Additionally, Linguistics is used to create and detect forgeries. Either of those uses can be extremely useful.
Answer: And since the problem will arise, what about Linguistics vs permanent Tongues, or an equivalent item? Basically, the thing is that one of the often forgotten points of Linguistics  is that it is not useful only to speak/understand a language. It also represents how deeply you understand this language, etc. It can be seen as Knowledge: languages from this aspect.
For example, someone with Tongues/Comprehend Languages would be able to speak/understand French, sure. Ask him for the etymology of one word, and he is screwed, he just doesn't have the knowledge at all. On the same note, he probably has no idea how the grammar/conjugations works. It also mean that if he encounter some kind of slang, he won't be able to speak it, as he has a global understanding of the language, not of his workings/adaptations.
Answer: Linguistics also lets you create and detect forgeries and (in PFS) helps when deciphering encrypted messages.


How can I demonstrate that the same wireframe can be used for varied visual designs?

Question: I'm trying to demonstrate to someone that one wireframe or template can be used across multiple brand experiences (we want the UX to be similar, but the brand experience to be different). At some point in the past, I found an article or blog post that showed multiple examples of "one wireframe, two/three designs", but I cannot find it any more. Anyone know of any resources - either actual websites or articles/blog posts/portfolios that demonstrate this?
Answer: Try to look for pattern repositories, like ui-patterns.com or patternry
Another solution would be to show how different the finished product can be of a wireframe, based on, like, wireframeshowcase.
Personally, I like to define UI in patterns, that is, to define what is the problem the user is facing, and what is a corresponding solution, and provide a lot of examples to it, rather than provide a single mockup, so that the developers can tailor those to their needs. One of my such UX docs was for an open source mail archiver (it's like GMail, except it's made sure you can't delete mails so a security audit could be done), albeit it's in Hungarian, but it's here if you wish. But sometimes developers want clean, finished, processed ideas, that is, they basically want a full-scale design which they just "type in".
UX is about patterns, and patterns are about problems and solutions: there's a need (like, we want the user to understand which button to press if everything goes normal) and there's a corresponding solution (we highlight a single "primary action" button), and that's it.
These patterns have to be assembled together in order to become an application,for which the Wireframe is merely an illustration, perhaps a test if this combination is able to work together or not, before the "real" design starts, as there are things for which font choice and color choice don't matter that much, and we want to concentrate just on the interactions / navigation.
If you concentrate on patterns instead of widgets and full-page wireframes, that can be assembled a lot of different ways, while still maintaining a similar experience across a family of systems.
On the other side, I don't see real design alternative checking in the industry, but perhaps I've only seen a small part closely. When I was doing web design, I did show 4 alternatives to the customers as a minimum, but I've seen too many times that designers come up with a single mockup, they toss it around a bit, then come up with a single visual design based on the mockup, then toss it around, sometimes making it radically different, and that's it.
One exception is Apple, where it's known that there are design teams to compete with each other with prototypes.
Answer: If you're looking for a wireframe that's used in precisely the same way across different brands, then look no further. Either take one of the  beta-stage StackExchange sites, e.g. GraphicDesign or just create a wireframe of an SE site yourself, and then show the dozens of graduated sites that use the same HTML across completely different brands, changing only the CSS. You have about 40 of them in the footer of this page. I really can't think of a better example. Best do it for both the list page and the question page.
Answer: It's an impossible thing that: "we want the UX to be similar, but the brand experience to be different". It seems that you are not familiar with terms UX and brand experience. User Experience design consist of designing and managing all intellectual senses that transferred to user and visual design is only one portion of this whole. UX could not be different from brand experience.
But about re using some virtual elements as Aadaam mentioned, you could and in my opinion should use design patterns.


Which expressions can be used to close an email?

Question: At the end of written communication like emails and letters, it is customary to use a closing valediction or "complementary close". Which formal and informal expressions can be used to end emails?
Comment: I think the best inspirations in this area are in `URGENT MESSAGE FROM UNICEF REFUND DEPARTMENT` style scam mails. They tend to use clean, proper, formal English that is almost nowhere to be found elsewhere anymore.
Comment: Was this closed because it asks for "primarily opinion-based" answers, or because it asks for a **list** of expressions that can be used to close an email? I'm still a bit confused about the  circumstances under which a request for a list will render a question unacceptable on EL&U.
Comment: "proper" English?  Not sure how many of these you've seen (don't know how good your spam filters are ;-)), but quite a lot of them are horrible. ;D
Comment: I don't know about you, but I usually end my email with **sᴇɴᴅ** myself ☺.
Comment: "l8rs" is my preference
Answer: Putting formal salutations and complimentary closes into an e-mail tends to make them very formal compared to most e-mails.
In my experience working in the software industry, people who always put them into ordinary business e-mails come off as unnecessarily formal. Often foreigners have been taught in English class to do this, and the result is that I have been in meetings where foreign colleagues were ridiculed for e-mails constructed to a ceremonious level of formality.
I've just looked through my most recently sent business e-mails and the only complimentary close I ever use is "thanks", in about a quarter of my e-mails—mostly ones where I've asked the recipient to do something. The rest of them just end with a dash and my name. Looking through my received business e-mails, I see the same. Most e-mails have no complimentary close, and of the ones that do, nearly all of them have "thanks" or similar ("many thanks", "thank you", "thanks all", etc.). I got one "regards" in the last week as well (out of about 250 messages).
Your experience may vary, I suppose, depending on the culture of the company you work for.
A colleague of mine uses an e-mail template that begins every e-mail with "Heyas," and closes with "Humbly". He is a quirky fellow.
My personal recommendation (at least when corresponding via e-mail for business with Americans) is to err on the side of no complimentary close, or a very informal one. Using a formal complimentary close marks an e-mail as extremely serious and if used on an e-mail that is not extremely serious will make it seem out of place or even funny.
Comment: Then again, are those mostly company internal emails? When emailing a customer, or a person you don't know well, it *may*, depending on circumstances, be better to err on the side of using some complimentary close (e.g. "best regards", nothing more formal than that).
Comment: The software industry is not representative of all industries. Technical professions in general tend to strike a much less formal note (in all ways) than other industries. There is also a lot of variation among technical professions.
Comment: @Jonik: maybe I'm not formal enough, but I'd just end with "Regards"
Comment: By contrast, other industries will tend to be much more formal than the software industry. I would say the question is pitched at "normal" professions (i.e. the other ones).
Comment: @jae: "Regards" indeed sounds fine, and respectful enough, to me. :-) "Best regards" was just an example (one that I've seen used a lot).
Answer: For informal e-mails:

Best, or
  Regards, or
  Cheers,

For formal e-mails:

Kind regards, or
  Respectfully, or
  Sincerely,
Comment: I'm curious why you consider "Regards" as _informal_ and "Kind regards" as _formal_? If anything I would have said it was the other way round (or perhaps even have both in the _formal_ group)? (UK)
Answer: Business Email/Letter Closings:

Best regards,
Cordially,
Good wishes,
Many thanks,
Most sincerely,
Regards,
Sincerely,
Thank you,
With confidence.

Informal Email/Letter Closings:

Adios,
Blessings,
Cheerio,
Cheers,
God bless,
Gotta boogie,
Grace and peace,
Have fun,
Health and happiness,
Keep the faith,
Later Vader,
Later alligator,
Lots of love,
Onward and upward,
Over and out,
Peace, love & happiness,
Peace & blessings,
Rock on,
See ya,
Smell ya later,
SMILE,
TA TA,
Toodles,
Your friend,
Dream BIG.

Ref: http://languagelearningbase.com/257/how-to-end-emails
Answer: "Yours sincerely" and "Yours faithfully" now sound somewhat formal and I'd advise against them unless your email is otherwise official or formal in nature.
"Best regards" (or just "Br") is, in my experience, extremely common in business emails, and a safe choice for many situations. "Best wishes," is also a common alternative that falls into this category.
For informal situations or when messaging friends, I'd probably go with something lighter, such as:

Cheers,

or

Take care,

:-)
Comment: You ended your answer with a closing you didn't say anything about ;-)
Comment: BR (for best regards) and KR (for kind regards) are divisive - some people (I include myself) consider not actually writing out the phrase in dull to be a strange combination between formality and shortcut-taking.  If you want to be seen as respectful, take two seconds to spell it out; if you want to be seen as part of the in-group and no longer at the highest respect level, use a different closer.
Comment: @ijw, good point; I tend to agree. I've seen "Br," used quite often though, at least among businessy types here in the Nordics.
Answer: I'll tend to use "Best regards," for anything even semi-formal, including correspondence with people in a business context whom I don't know very well.
When I use to work at NEC, it was considered the way you must sign off your emails when dealing with any of the Tokyo managers or engineers, and so it kind of just stuck.
For less formal, it'll be "Thanks" if appropriate, "Take care," or occasionally "Peace and chocolate."
Never ever put it in your email signature. A colleague of mine has "Regards," in his signature and it grates every time I see it. It may save you some typing, but it comes off as disingenuous. You're better off with nothing if you can't be bothered.
Comment: +1 I always customize my complimentary close every time I write an e-mail
Answer: As a contractor I tend to just use:

Yours,

I like to think it implies their 'ownership' of me for that time.
The longer versions do feel too formal for e-mail, or too loaded with meaning.
Otherwise I use the ever popular:

Regards,
Comment: That's what I use as well.
Comment: I like "Yours," a lot.  Simple, neutral, just formal enough for many situations.
Answer: There seems to be a great uncertainty among business people on how to close - and address - emails. Looking at my emails, I have found that a lot of people simply have a template, which either has their name and position/contact details, or that preceded by "Regards"/"Kind Regards".
I personally dislike "regards", because it is obvious that no-one ever means it, and I also find it hilarious when I get a strongly worded email from a superior demanding some sort of action, which then ends in "Best Regards".
I have found that "cheers", is a very neutral ending, that is both informal and polite, and, in my experience, has been used a lot in business and informal contexts. Most of my close friends end their emails to me with "cheers", and at the moment, as I am interviewing for jobs, I have found that most recruiters end their emails to me with "cheers". (I am in New Zealand). One particular recruitment agent has been more creative than the others with her endings, and has used a different one with each correspondence. She has used

Cheers,
Best wishes,
Take care,
All the best,
Good luck,

I particularly like the last one in that list.
Comment: Somehow I am not thrilled with the idea of all emails from  my boss ending with "Good luck," :-o
Comment: Ah, NZ.  Well, that explains the "Cheers" (and your gravatar, I guess) ;-)
Answer: It seems to me that emails were originally designed to be written in memo style (hence From, To, Subject, etc.), for which it is standard to have no closing expression.
Of course, many emails are written more to resemble a letter, for which other conventions apply.
Answer: In the US military, we use Respectfully and Very Respectfully or even Hooah!. I use Respectfully when I am closing a formal letter to a peer. I also use Respectfully in any letter which might otherwise be considered informal, but in which I append my rank to my name, because the fact that I'm writing in my military capacity alone makes it formal in a sense, even if I'm just writing to inquire about some triviality.
I use Very Respectfully when closing any letter to a superior; when I need to convey the utmost formality and respect in my closing. These are military writing conventions, so I can not invent some flowery superlative closing to sound poetic and transcendent, because I would only appear unprofessional for not following the convention.
If I am writing an utterly casual letter in some military context, e.g., to inform my mother that I'm returning home in a week's time, I might say something corny like Hooah! Then again, this is really not important to mention; I would probably use Love, in that situation anyway.
In Thelemic correspondence, the convention is to open with 93 and to close with 93 93/93 as in:
93,
I will be attending the whatever-event on xxx date.
93, 93/93
Among my friends, I tend to use whatever closing comes to mind; I just pluck something out of my memory, like Hail Satan! or Praise Jesus! I sometimes open with a good old Latin Ave! or Salve.
There are my two cents. Of course, a long time has passed since that expression was coined (pun intended), so I suppose what I've written here would be closer to "my 50 cents."
Answer: For our corporate emails, amongst different departments, we use the "neutral":

Regards,
Answer: In emails within the organization I tend to use no salutation and no closes at all.
Answer: For formal email, just as in formal letter writing, the standard set of valedictions and advice apply.* Wikipedia and supporting sources cover this in great detail, so I recommend at least having a look:
http://en.wikipedia.org/wiki/Valediction
For informal email, creative use of the valediction can actually add warmth and intrigue to your original statement.
As general creative advice, abusing the valediction to include additional material or postscript can be greatly entertaining for the reader within the appropriate contexts. But obviously, selection of these scenarios is left to your own judgmental heuristics and creative tastes for writing.

*_In fact, I'm surprised that a backdoor editor improving the OP's question has been the only person to use this term thusfar.
Answer: Say nothing at the end.
Unlike in Oral speech, the end of text signal can be carried out by the software.
On the other hand, scrolling is tedious on mobile devices, so you may want to save on text lines.
Answer: I just use a dash:
-jeffamaphone
Answer: I would generally use the suitably neutral and meaningless "Regards," in a business email - sometimes changing to "Thanks," where appropriate (always keeping the comma in a very traditional manner).  On a couple of line reply I may drop any salutation.  I would keep to the formal "Yours sincerely/faithfully," in a business letter.
On personal emails I would vary widely from nothing to some elaborate blessing and anything inbetween depending on what I was saying.  My favourite in Christian/church contexts is "Your brother", either followed by my name or just a full stop.
Answer: I usually use Best, for informal messages.
Comment: ok, but best what?
Comment: @Ed Guinness: Just "best" is definitely "on my radar", even if it's not very common.
Answer: I usually end my mails with either "Thanks," or "Bye," depending on the context. For very formal mails I use just "Regards,". Only if you are writing official mails (e.g to customers) you should use a more complimentary form.
Answer: I have found a very interesting collection
of email closing. Hope you will like it. I usually end my emails with Regards or Happy hacking .
Comment: "Yabba dabba do" is quite ridiculous and amusing. Thanks for sharing.
Answer: 'Regards' means nothing. 'Kind regards' at least means something. But regards simply means 'thoughts'. Those thoughts could be as diverse as 'I love you' or 'you smell like sauerkraut'.
Cheers also smacks to me of informal pub language. Each to their own I guess. I smiled when I saw 'toodle pip' in the above list. I have resolved to use that more often.
Answer: Formally, in emails, most people find it acceptable to use
"Regards,
XYZ"
However, "Best Regards" would come next in accepatability & suitability for professional purposes.
Comment: Most people? I've rarely seen it in American emails. Is this most people in British English?
Comment: Interesting. Yes, I would agree it's certainly true in Britain. What would Americans write?
Comment: I tend to use the exact same closings that I would in a written letter- when it's at all formal.  If it's between friends, co-workers, or family a simple `-Jim` suffices. (An American)
Comment: @Urbycoz: 'Regards' is not uncommon but 'thanks' or some variation is much more common (in my experience). Most common of all nowadays is simply ones name: no phaticism, maybe a dash. That's in the business setting.
Comment: I'm in the US, and I've used 'Regards' for years in the business setting.
Comment: @Mitch: in my experience, this usage is common in both US as well as UK. Though I should have mentioned-"for official/business communication".