TITLE: Characteristic function is an identification function
QUESTION [0 upvotes]: Every characteristic function $\Phi_{\beta}^b: E^n_{\beta} \to e^{-n}_{\beta}$ is an identification function.

My book says the following:
This follows from the fact the the CW complex $X$ has $X^n$ a quotient space of $X^{n-1} \sqcup \coprod_{\beta}E^n_{\beta}$.
I am not sure I see how this follows. We know that $\Phi_{\beta} |S^{n-1}_{\beta} = \phi_{\beta}:S^{n-1}_{\beta} \to X^{n-1}$, but how do I use what the book gave me to prove it?
Note: $e^{n}_{\beta}$ is an open $n$-cell, $E$ is the unit ball, and $S$ is the unit sphere. The characteristic function is what I have defined above.

REPLY [0 votes]: I don't see an easy way to prove this using what the books says.  But if you know that CW complexes are Hausdorff, you can instead prove it as follows.  Let $Y$ be the quotient space of $E^n_\beta$ by the equivalence relation induced by $\Phi^b_\beta$, i.e. $x\sim y$ iff $\Phi^b_\beta(x)=\Phi^b_\beta(y)$.  Then $Y$ is compact since $E^n_\beta$ is, and $\Phi^b_\beta$ gives a continuous bijection $f:Y\to \overline{e^n_\beta}$.  Since $X$ is Hausdorff, so is $\overline{e^n_\beta}$, and so $f$ is automatically a homeomorphism (it is a continuous bijection from a compact space to a Hausdorff space).  But $\Phi^b_\beta$ is just $f$ composed with the quotient map $E^n_\beta\to Y$, so this means $\Phi^b_\beta$ is also a quotient map.


TITLE: If $X$ is Stein and $H^2(X,\mathbb{Z})\neq 0$ then there are no principal divisors?
QUESTION [2 upvotes]: There is a result that claims: If $X$ is a Stein manifold and $H^2(X,\mathbb{Z})=0$ then for any divisor $D$ there is a meromorphic function $f$ such that $(f)=D$.
What happens if $H^2(X,\mathbb{Z})\neq 0$ ? Is there any divisor that can not be of the form $(f)$ for some meromorphic function?
Thanks

REPLY [2 votes]: It is well known as the Cousin problem II.
If $M$ is a Stein manifold of dimension $n$, the exact sequence of sheaves
$$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$
yields a morphism of groups in cohomology (part of a long exact sequence) $$H^1(M,\mathcal O_M)\to H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)\to H^1(M,\mathcal O_M).$$
We have $H^1(M,\mathcal O_M)$ and $H^2(M,\mathcal O_M)$ both vanish thanks to Cartan's theorem B. Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the isomorphism $$c_1:Pic(M)\to H^2(M,\mathbb Z).$$
We then absolutely have $Pic(M)=0,$ thus any line bundle $\mathcal O(D)$ corresponding to the divisor $D$ is trivial.
Due to a classical result, a divisor $D$ is principal iff $\mathcal O (D)$ is trivial (A), we are done.
If $H^2(M, \mathbb Z)\neq 0$, we absolutely can find a non-principal divisor $D$ when $\mathcal O(D)$ is not trivial thanks to (A).


TITLE: Why are SDP solvers inherently inaccurate?
QUESTION [2 upvotes]: It's common to read in textbooks that semidefinite programming solvers are inherently inaccurate.

Are the authors referring to the general machine inaccuracy (things like $10^{-16}=0$) or a special case around SDPs only?
If it is a special case, does this problem occur in interior point methods for linear programming as well? Why does it happen?
Is there some way to avoid this or to increase the maximum precision?

Edit: As Prof. Borchers requested, in Moments, Positive Polynomials and Their Applications, by Jean Lasserre, he mentions, in the Algebraic Certificates of Convexity section, that:
"[...] in practice such a certificate is numerical and so can be obtained only up to a machine precision, because of numerical inaccuracies inherent to semidefinite programming solvers."
The same is written in Modern Optimization Modelling Techniques by Comminetti, Facchinei and Lasserre. I suspect that I have read it somewhere else as well.
Thank you!

REPLY [5 votes]: Both of the references are simply making the point that due to round-off errors, floating point numerical solutions to SDP are inherently inexact.
In the applications that Lasserre is interested in, we'd like to have exact certificates that (for example) a polynomial can be written as a sum of squares, but a numerical solution isn't really proof.  There has been work on rounding numerical solutions to produce exact (in rational form involving integer numerators and denominators that might require many digits of precision) solutions in such cases.
There's nothing special about SDP with respect to this- similar issues occur with linear programming and other optimization problems.
It happens that some weird things can occur in SDP (such as lack of strict duality) that don't occur in LP, but that's a separate issue from the fact that floating point arithmetic isn't exact.


TITLE: Does this quantity have a name in information theory?
QUESTION [4 upvotes]: The Kullback-Leibler divergence between two (discrete) probability distributions is defined as
$$
D_{KL}(P\|Q) = \sum_i p_i \log \frac{p_i}{q_i},
$$
where $p_i$ is the probability that $P$ assigns to the event $i$, and $q_i$ is the probability assigned by $Q$.
I know that the quantity $D_{KL}(P\|Q) + D_{KL}(Q\|P)$ (symmetrised Kullback-Leibler divergence) is sometimes used, because it is symmetric and thus behaves more like a distance between the two distributions.  But does anyone know of a case where the quantity
$$
 \sum_i (p_i-q_i) \log \frac{p_i}{q_i}
$$
is used, and whether it has a standard name?  I ask because it came up in some statistical mechanics work I'm doing and I want to know if it has an interpretation in terms of information theory, or any particularly interesting known properties.

REPLY [6 votes]: Since
\begin{align*}
   \sum_i (p_i - q_i) \log \frac{p_i}{q_i}
    &= \sum_i p_i \log \frac{p_i}{q_i} - \sum_i q_i \log \frac{p_i}{q_i}\\
    &= \sum_i p_i \log \frac{p_i}{q_i} + \sum_i q_i \log \left(\frac{p_i}{q_i}\right)^{-1}\\
    &= D_{KL}(P||Q) + D_{KL}(Q||P)
\end{align*}
the quantity in question is the symmetrised KL-divercence,
AB,


TITLE: Prove that the Diophantine equation has only one solution for $a,b,c$.
QUESTION [3 upvotes]: Prove that the Diophantine equation
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
has only one solution for positive integers $a,b,c$   with $a<b<c$

REPLY [5 votes]: there must be a number smaller than $3$ since there is going to be a number larger than $3$. So $a=2$. we now want $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$ with $b$ and $c$ larger than $2$. If we take $\frac{1}{4}$ we need to repeat it, so we need a number larger than $\frac{1}{4}$, this forces us to take $\frac{1}{3}$.

REPLY [3 votes]: Hint:
$$1\geq\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$$ so , $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{a}$$.
If $a>3$ the sum is less than $1$.
Check the cases $a=2$ , $a=1$.


TITLE: Question about covariant derivate on a manifold
QUESTION [0 upvotes]: Let $S$ be an n-dimensional manifold and $M$ be an m-dimensional submanifold of $S$. Let $[\xi^i]$ and $[u^a]$ be coordinate systems for $S$ and $M$, respectively, and let $\partial_i=\frac{\partial}{\partial \xi^i}$ and $\partial_a=\frac{\partial}{\partial u^a}$ Suppose also that $\nabla$ is an affine connection on $S$ and that $\Gamma_{ij}^{k}$  are the connection coefficients of $\nabla$ with respect to $[\xi^i]$ Now letting $X=X^a\partial_a$ and $Y=Y^a\partial_a$ be two vector fields in $M$ then we define $\nabla_{X_p}Y$ "the directional derivative of $Y$ along $X_p$" ,In general $\nabla_{X_p}Y$ is a tangent vector in $S$ but not in $M$.
Can someone give an example, I am finding it difficult to visualize why $\nabla_{X_p}Y$ is not tangent vector in a submanifold $M$.Thanks

REPLY [1 votes]: Consider a smooth curve $ u:(-1,1) \longrightarrow \mathbb{R}$ with $u_x, u_{xx} \neq0 $ on its domain. Then, $X= \frac{\partial}{\partial x} + u_x \frac{\partial}{\partial y}$ is tangent to the curve at every point. We have that $D_XX= u_{xx}\frac{\partial}{\partial y}$. Since $N =(u_x\frac{\partial}{\partial x}-\frac{\partial}{\partial y})$ is perpendicular to the curve at every point and
$$\langle D_XX,N \rangle= -u_{xx} \neq 0,$$
the normal component of $D_XX$ is not zero. This means $D_XX$ is not in the tangent space of the curve at any point.
Note that because $u_x$ is the slope of the tangent line on curve at every point, $y=u_x x $ is parallel to the tangent line, in turn, $(1,u_x) \in T_xM$. (One can also consider  parametrization  $ X :(-1,1) \longrightarrow \mathbb{R}^2$ by $X(x) =(x,u(x))$. Therefore $ d X (\frac{\partial}{\partial x}) = (1, u_x)$ is a basis for $T_xM$). And since $\langle N,X \rangle =0$,   $N\perp T_xM$.


TITLE: A bound on the derivative of a concave function via another concave function
QUESTION [2 upvotes]: Let $f:\mathbb{R}^+\to\mathbb{R}^+$ and $g:\mathbb{R}^+\to\mathbb{R}^+$ be two strictly concave, strictly increasing, twice differentiable functions, such that $f(x)=O(g(x))$ as $x\to\infty$, i.e. there exists $M>0$ and $x_0$ such that
$$f(x)\leq Mg(x)\qquad \forall x\geq x_0.$$
Is it true that $f'(x)=O(g'(x))$ as $x\to\infty$?
(this is an extensions of this question)

REPLY [0 votes]: No.  We first construct a function $h:[-1,\infty)\rightarrow\mathbb{R}$ as follows: For integers $n \in \{-1, 0, 1, 2, ...\}$ we want $h$ to be:
$$ h(n) = \frac{1}{(n+2)!} \quad \forall n \in \{-1, 0, 1, 2, ...\} $$
These values are positive and strictly decreasing.  So, we can interpolate them to get the full function $h(x)$ that is differentiable, positive, and satisfies $h'(x)<0$ for all $x \geq -1$.
Define $g:[-1,\infty)\rightarrow\mathbb{R}$ by $g(x) = \int_{-1}^{x} h(t)dt$.  Then $g'(x)=h(x)>0$ and $g''(x)=h'(x)<0$ for all $x\geq -1$. So $g$ is increasing, twice differentiable, and strictly concave.
Now define $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(x) = g(x-1)$.  Then
$$ f(x) = g(x-1) \leq g(x) \quad \forall x \in [0,\infty)$$
Furthermore, for integers $n \in \{1, 2, 3, ...\}$ we get:
$$ \frac{f'(n)}{g'(n)} = \frac{g'(n-1)}{g'(n)} = \frac{1/(n+1)!}{1/(n+2)!}\rightarrow \infty $$
and so the conjecture $f'(x) = O(g'(x))$ fails.


TITLE: Common insights into hypothetical complexity of graph problems
QUESTION [10 upvotes]: I came across two examples of hypothetical hardness of some graph problems. Hypothetical hardness means that refuting some conjecture would imply the NP-completeness of the respective graph problem. For instance, Barnette's conjecture states that every 3-connected cubic planar bipartite graph is Hamiltonian. Feder and Subi proved that refuting the conjecture would imply the NP-completeness of the Hamiltonian cycle problem on graphs in the class of the conjecture.
Tutte's 5-flow Conjecture states that every bridgeless graph has a nowhere-zero 5-flow. Kochol showed that if the conjecture is false, then the problem of determining whether a cubic graph admits a nowhere-zero 5-flow is NP-complete.

Are there common insights into the above conjectures that explain the hypothetical NP-completeness of the corresponding graph problems? Are there other examples of hypothetical complexity in the above sense?

P.S. This was posted on MathoverFlow without getting an answer.

REPLY [0 votes]: Are there common insights into the above conjectures that explain the hypothetical NP-completeness of the corresponding graph problems?

In my opinion there is a clear common insight in the opposite direction: if the conjectures are true then the corresponding problems are not NP-complete and turn out to be trivial in both cases (they switch from NPC to $O(1)$ ).
And  the common insight is that the natural problems, Hamiltonian cycle and nowhere zero flow in general graphs, are "strutured and powerful" enough to efficiently "simulate" the trace of a Turing machine (à la Cook-Levin) . Then you start adding more and more constraints until you get no "computational power" at all.
To me it's like adding more and more constraints on the transition graph of a Turing machine (or on the read/write tape device) until you get something trivial like "the transition graph doesn't contain a cycle".

Are there other examples of hypothetical complexity in the above sense?

As a (probably) "solved case" I can bring my experience related to the Rolling a Die over a Labeled Board problem.
A few years ago it was unknown if fully labeled boards can contain two distinct Hamiltonain cycles (uniquely-rollable conjecture was settled for all boards with side lengths at most 8). Domotor P. (user domotorp here) and I (independently) proved that such boards exist and the conjecture is false (... note that Joseph O'Rourke hasn't updated his page yet :-).
Then using that fact I was able to prove that rolling a die on fully labeled board with holes is NP-complete (the without holes case is still open); though this is an unpublished result.


TITLE: Coin Flip winning probability increase in sequence
QUESTION [1 upvotes]: Flipping a coin is an independent event, and has a chance of 50%. So Lets say that I flip a coin 10 times, and if the first 5 flips are heads.
 Then the probability of the next flip being a tails, is higher(so as to satisfy the 50% chance), How can we calculate the increase in probability of this happening.
I read an article about a roulette wheel. In a roulette wheel the reds and blacks have an equal chance of hitting.(ignoring the zero) just like a coin flip. Now the article said that if you place a bet on red, then you have a 52% chance of losing(as this includes the zero).
If you bet on red again, then you have only a 28% chance of losing in the second consecutive bet.
And the chances of a red coming keeps increasing each flip.

My Question is that, since these events are independent, then shouldnt the probability of me winning or losing be 50% everytime. May it be 10 consecutive blacks..
Also, if to satisfy the 50% over a large sample size, the probability of red hitting does increase, then how do we calculate it?

REPLY [1 votes]: The article you reference is perhaps a little misleading. The probability of losing if you bet on red stays 52% each individual time you roll, but the chance that you lose $\textit{every time}$ naturally decreases the more times you roll. In fact if you lose on a roll with probability $p$, then the probability that you have lost every time by the $n$-th roll is $p^n$. Note that this gives probabilities of losing as $.52^2\approx.27$ for the second roll and $.52^3\approx.14$ for the third roll and so on (I think probably the probability of losing should actually be $\frac{20}{38}\approx.526$ which gives numbers consistent with the table you gave).
The key point here is that if you have truly independent events, then the outcomes of previous events $\textit{do not affect}$ the likelihoods of future outcomes. In the case of coin flipping, the tendency of large samples to have approximately half heads and half tails just comes from having many repeated trials. The mistake that previous outcomes affect future outcomes is commonly called the gambler's fallacy.


TITLE: Why is the curl the limiting value of circulation density?
QUESTION [1 upvotes]: The circulation of vector field around some region is the line integral of the field over closed path surrounding that region so why do we need the curl at all if we have a quantity that represents the amount of circulation and why the curl is the circulation density and not the circulation?

REPLY [1 votes]: Given a vector field ${\bf F}$ in the plane ${\mathbb R}^2$ the circulation is a function $$\Phi_{\bf F}(\gamma):=\int_\gamma{\bf F}({\bf z})\cdot d{\bf z}$$
that produces a scalar value for each closed curve $\gamma\subset{\mathbb R}^2$, in particular for all boundary curves $\gamma:=\partial B$ of (reasonable) domains $B\subset{\mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $\Phi$. But this function is important: It measures the nonconservativity of the field ${\bf F}$. A conservative field ${\bf F}$ (meaning ${\bf F}=\nabla f$) has $\Phi_{\bf F}(\gamma)=0$ for all closed curves $\gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${\bf F}$ can be captured in a simple point-function $${\rm curl}\,{\bf F}:\quad{\mathbb R}^2\to{\mathbb R}, \qquad {\bf z}\mapsto{\rm curl}\,{\bf F}({\bf z})$$
in such a way that the circulation of ${\bf F}$ around the boundary $\partial B$ of a domain $B$ appears as integral of the density ${\rm curl}\,{\bf F}$ over $B$:
$$\Phi_{\bf F}(\partial B):=\int_B{\rm curl}\,{\bf F}({\bf z})\>{\rm d}({\bf z})\ ,$$
whereby ${\rm d}({\bf z})$ denotes the area element at ${\bf  z}=(x,y)$. This means that the nonconservativity of ${\bf F}$ is in a precise way localized to the points of the domain of ${\bf F}$.


TITLE: In proving A = B, A, B are sets, do you always have to show $\subseteq$ and $\supseteq$?
QUESTION [4 upvotes]: I am trying to show the DeMorgan's Law

$X \backslash \bigcup_{\alpha \in I} A_\alpha = \bigcap_{\alpha \in I}
(X \backslash A_\alpha)$

It seems I could directly approach this as follows:
$X \backslash \bigcup_{\alpha \in I} A_\alpha = X \bigcap (\bigcup_{\alpha \in I} A_\alpha)^c  = X \bigcap (\bigcap_{\alpha \in I} A_\alpha^c) = \bigcap_{\alpha \in I} X \backslash A_\alpha$
The last line follows from distributivity
But I thought in set theory proofs of the type Prove $A = B$, you have to show that $ A \subseteq B$ and $B \subseteq A$.
But in this case it seems I have directly proved that the two are equivalent without resorting to $A \subseteq B$ and $B \subseteq A$...
Can someone enlighten me as to whether my approach is correct? I am not very well versed in set theory proofs.

REPLY [1 votes]: A direct proof : For all $y$ we have $$y\in X\backslash \cup_{a\in I}A_a \iff (y\in X \land y\not \in \cup_{a\in A}A_a)\iff$$  $$\iff   (y\in X \land (\forall a\in A\;(y\not \in A_a)))  \iff (\forall a\in A\; (y\in X \land y\not \in A_a)) \iff $$ $$\iff (\forall a\in A (y\in X\backslash A_a))\iff y\in \cap_{a\in A}(X\backslash A_a).$$
In some problems it may be easier to show $A\subset B$ and $B\subset A$ to prove A=B. In many cases,one half is easier than the other, or different methods may  be more suitable for the two halves.


TITLE: Can this be solved by induction? (number of ways of cutting a rod into pieces)
QUESTION [5 upvotes]: I am reading an algorithm example.
The example is about Rod cutting.
The idea is that a steel rod can either be sold as it is, or be cut into integral pieces and sell the pieces. The goal is to find what is best. Sell a rod as a whole or in pieces (prices for rod lengths are assumed to be known).
In the presentation of the solution it says that a rod can be cut in $2^{n-1}$ ways.
I can see and verify that, for $n=1$, $n=2$, $n=3$, and $n=4$ but after that how can this be proved for any value of $n$?
By induction? I mean to verify it for $n=4$ I drew the various cuts. I didn't use some relation with $n=3$.
So although this problem seems to be proved by induction, it seems to me that it is proven some other way, but I am not sure on how.
How can we be sure that the rod can be cut in $2^{n-1}$ ways?

REPLY [0 votes]: Thomas Andrews has given a direct proof in his comment.  To put it in terms of induction, a one inch rod can be cut in  $1=2^{1-1}$ way (not at all).  If a length $k$ rod can be cut in $2^{k-1}$ ways, a length $k+1$ rod can be cut either by cutting off the right inch and cut the remaining in $2^{k-1}$ ways or by not cutting off the right inch and cut the remaining in $2^{k-1}$ ways, totaling $2^{(k+1)-1}$ ways.


TITLE: Can the matrix element of the momentum operator be found in the momentum basis?
QUESTION [0 upvotes]: In Shankar's Quantum Mechanics example 7.3.4, the problem is to find $\langle n'\rvert P\lvert n\rangle$ for the harmonic oscillator. The answer contains imaginary parts; you can derive such an answer by solving it in the $x$ basis (making $P= -\frac{i}{\hbar}\frac{\mathrm{d}}{\mathrm{d}x}$).
But I think there is another way to approach this, writing the $\psi$ function in the $p$ basis and simply integrate $\psi P \psi\,\mathrm{d}P$ from $-\infty$ to $+\infty$. By doing this I don't get any imaginary terms. Why not? Should this work?

REPLY [1 votes]: I don't see why you wouldn't get any imaginary terms in the matrix elements. Your suggestion of aproach is right and in this case the eigenstates $|n\rangle$ of the harmonic oscilator are represented by Hermite Functions in position representation, which happens to be Fourier transform's eigenfunctions as well. As the unitary map that changes form the position basis to the momentum basis is the Fourier transform $\mathcal{F}$, by writing the eigenstates in momentum representation, you will get the same functions but with a constant factor of difference, $(-i)^n$, which is the eigenvalue of $\psi_n$: $(\mathcal{F}\psi_n)(p)=(-i)^n\psi_n(p)$. So, when calculating the matrix elements, you will have (with $\hbar=1$):
\begin{align}\langle n'|P|n\rangle&=\int dp'\int dp\langle n'|p'\rangle\langle p'|P|p\rangle\langle p|n\rangle\\
&=\int dp'\int dp\langle n'|p'\rangle p\delta(p'-p)\langle p|n\rangle\\
&=\int dp\langle n'|p\rangle p\langle p|n\rangle\\
&=\int_{-\infty}^\infty i^{n'}\psi_{n'}(p)\,p\,(-i)^n\psi_n(p)dp\end{align}
which will be purely imaginary if $n+n'$ is odd. You can realize that this happens, because the Hamiltonian $H=X^2+P^2$ is symmetric between a change in position and momentum, so its eigenfunctions should be the same in both representantions.


TITLE: Breaking a stick of $n$ units into $n$ sticks
QUESTION [0 upvotes]: In how many ways can you break a stick of length $n> 1$ into $n$ pieces of unit length? At each step you can break the stick into two sticks of integer lengths.
I just want an insight into the problem. Thanks.

REPLY [4 votes]: You'll break the stick $n-1$ times, once each at each integer position $i$ with $1 \le i \le n-1$, and this can be done in any order.  So how many ways are there to arrange $n-1$ things in an order?


TITLE: Find the least square solution x
QUESTION [1 upvotes]: Find the least square solution $x$ for $Ax = b$ if
$$A =\left(\begin{array}{rrr} 2 & 0\\
1 & 3 \\
0 & 2\end{array}\right) \quad b =\left(\begin{array}{rrr} 7\\
0\\-1\end{array}\right) $$

My Solution:
I found $A^T A =\left(\begin{array}{rrr} 7 & -1\\
-1 & 8 \\
\end{array}\right) $
$$A^T b =\left(\begin{array}{rrr} 2\\
2\\
\end{array}\right) $$
$$\left(\begin{array}{rrr} 5 & -1\\
-1 & 5 \\
\end{array}\right) x = \left(\begin{array}{rrr} 2\\
2\\
\end{array}\right)  $$
But I can't solve this equation to find $x$ as there is no solution.
My solution after finding $x$ was to then find this:
$$Ax = \left(\begin{array}{rrr} 2 & 0\\
1 & 1 \\
0 & 2\end{array}\right)  \left(\begin{array}{rrr} x\\
x\\ \end{array}\right) = \left(\begin{array}{rrr} a\\
b\\c\end{array}\right)  $$
and then use $b-Ax^ = \sqrt{1+0+1} - \sqrt{a^2 + b^2 +c^2} = \textrm{final  answer}$
Where am I going wrong with my approach? I am stuck as I can't solve for $x$?

REPLY [5 votes]: Your first error is in computing $A^\top b$. We have
$$A^\top b = \begin{pmatrix} 2 & -1 & 0 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$$
So, we need to solve
$$\begin{pmatrix} 5 & -1\\
-1 & 5
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2\\
-2
\end{pmatrix}.$$
I don't know where you went wrong specifically here, but there is a unique solution, as $A^\top A$ is invertible (compute its determinant to see this). We can use the old shortcut to compute the inverse of this matrix:
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$$
In particular,
$$\begin{pmatrix} 5 & -1 \\ -1 & 5\end{pmatrix}^{-1} = \frac{1}{24}\begin{pmatrix} 5 & 1 \\ 1 & 5\end{pmatrix},$$
so
$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{24}\begin{pmatrix} 5 & 1\\1 & 5\end{pmatrix}\begin{pmatrix} 2\\-2\end{pmatrix} = \begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix}.$$
Therefore, the closest point to $(1, 0, -1)^\top$ in the columnspace of $A$ is
$$A\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} =\begin{pmatrix} 2 & 0\\-1 & 1 \\0 & 2\end{pmatrix}\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ -\frac{2}{3} \end{pmatrix}.$$


TITLE: An exercise about separable algebras
QUESTION [1 upvotes]: Let $R$ be a ring and let $A$ be a $R$-algebra. $A$ is separable over $R$ if and only if the localisation $A_P$ is separable over $R_P$ for all $P\in Spec(R)$.

REPLY [2 votes]: I assume that the ring $R$ is commutative (else I don't think  "separable algebra" makes sense.)
a) It is always true that
$A$ separable over $R$ implies  $A_P$ separable over $R_P$ for all $P\in Spec(R)$.
More generally separability is preserved under  base-change: if $A$ is separable over $R$ then, for any commutative $R$-algebra $S$, the $S$-algebra $S\otimes_R A$ is $S$-separable.
b) For the converse you need some supplementary hypothesis.
If for all $P\in Spec(R)$ the algebra $A_P$ is separable over $R_P$, then $A$ separable over $R$ under any of the following two assumptions:
i) $A$ is a finitely generated $R$-module.
ii) $A$ is commutative and a finitely generated $R$-algebra.


TITLE: Flows of vector fields and diffeomorphisms isotopic to the identity
QUESTION [22 upvotes]: Let $M$ be a compact manifold and $\varphi : M \longrightarrow M$ be a diffeomorphism which is isotopic to the identity. Does there exist a vector field $ X $ on $M$ such that $\varphi$ is the flow at time $1$ of $X$? If that is not always the case, where does the obstruction for such a $\varphi$ to be a flow lives in ?

REPLY [8 votes]: In fact, for every manifold $M$ with $\dim(M)\ge 2$, the group $Diff_c(M)$ of diffeomorphisms with compact support contains a smooth curve $c:\mathbb R\to Diff_c(M)$
(smooth in the sense that the associated mapping $\hat c: \mathbb R\times M\to M$ is smooth)
with $c(0)=Id$
such that the $\{c(t): t\ne 0\}$ are a free set of generators for a free subgroup of the diffeomorphism group such that no diffeomorphism in this free subgroup with the exception of the identity embeds into a flow.
This is proved in

Grabowski, J., Free subgroups of diffeomorphism groups, Fundam. Math. 131 (1988), 103–121.

The proof uses clever constructions with Nancy Kopell's diffeomorphisms on $S^1$ (those described in Andy Putman answer) which do not embed into a flow.
This free subgroup is contractible to the identity: slide the generators $c(t)$ to $c(0)$.
So the image of exponential mapping of $Diff_c(M)$ does not meet very many diffeomorphisms near the identity.


TITLE: Prove $(p \lor q) \land (\neg p \lor r) \to (q \lor r)$ is a tautology
QUESTION [0 upvotes]: I am looking for a way to prove that the statement,
$$(p \lor q) \land (\neg p \lor r) \to (q \lor r)$$
is a tautology without the help of the truth table, by using only laws and theorems like De Morgan's Law, Domination Law, etc. Also, I can't use the rules of inference. Please help, thank you.

REPLY [0 votes]: Start from
$$(p∨q) ∧ (¬p∨r) → (q∨r) $$
Apply $(p\rightarrow q)\Leftrightarrow(\neg p\vee q)$ have
$$(\neg p\wedge q)\vee(p\wedge r)\vee(q\vee r)$$

Notice $(\neg p\wedge q)$ and $(p\wedge r)$ are contradict to each other
That means one of them has to be true, therefore the whole statement will also be true
Which implies it's a tautology

Apply $(p\vee\neg p)\Leftrightarrow\top$ that
$$\top\vee (q\vee r)$$
Apply $(\top\vee p)\Leftrightarrow \top$ we have
$$\top\tag*{$\square$}$$


TITLE: How to find the limit $\lim_{x \to 0}\sin(x)^{2\sin(x)}$
QUESTION [3 upvotes]: I can't find the following limit:
$$\lim_{x \to 0}\sin(x)^{2\sin(x)}$$
My attempt :

Let $y =\sin(x)^{2\sin(x)}$. Then we take the natural logarithm of this expression:
$$y=\sin(x)^{2\sin(x)}$$
$$\ln(y) = 2\sin(x)\cdot \ln(\sin(x))$$

$$y=e^{2\sin(x)\cdot \ln(\sin(x))}$$
Then we put this in limit
$$\lim_{x \to 0}\sin(x)^{2\sin(x)}= \lim_{x \to 0}(e^{2\sin(x)\cdot \ln(\sin(x))})=e^{0\cdot (-\infty)}=e^0=1$$
I want to know whether it is correct.

REPLY [3 votes]: Let's focus on $\sin x\cdot\log\sin x$.  It is well-known that $\sin x$ is asymptotic to $x$ as $x\to0$ (see: Taylor series).  Thus we need only consider $$\lim_{x\to0^+}x\log x$$
(we approach $0$ from $+x$ because $\log$ of negative reals is not very nice to work with). If we substitute $u=\frac1x$:
$$ \lim_{x\to0^+}x\log x=\lim_{u\to\infty}\frac{-\log u}{u} $$
It should be plainly obvious that the latter expression approaches $0$ for large $u$, and if not one may apply the ever-useful L'Hopital's rule to simplify the limit to $\frac1u$.  Thus, we obtain that $$\lim_{x\to0}\sin^{2\sin x} x=\lim_{x\to0}e^{2\sin x\log \sin x}=e^0=1$$
Without this methodology, $0\cdot (-\infty)$ is indeterminate and you cannot assume it to be $0$.  For example, $\lim_{x\to\infty}x\cdot\frac1x$ becomes $\infty\cdot 0$ but evaluates to $1$.

REPLY [1 votes]: You can use that $\lim\limits_{x\to 0}\frac{\sin (x)}{x}=1$ and that $\lim\limits_{x\to 0}x\mathrm{ln}(x)= 0$ to show that $\lim\limits_{x\to 0}\sin(x)\mathrm{ln}(\sin(x))=0$ and then you can use the continuity of the exponential function to get the result.


TITLE: Direct proof that the product of odd integers is odd
QUESTION [0 upvotes]: Prove $P(x,y)$: If $x$ and $y$ are odd integers, then the product  $xy$ must also be odd.

I need a direct proof of this.
I know that $ x $ and $y$ both have to equal to $2n+1$ in order for them to be odd. But that's all I have. Any help will be appreciated.

REPLY [0 votes]: Hint $\, $ Multiplying by an odd $ $ preserves parity: $\ (1\!+\!2k)\,n = n + 2(kn)\,$ has same parity as $\,n.$


TITLE: Fourier Series Proof
QUESTION [1 upvotes]: I have concerns about this problem
Let $f_e(x)$ and $f_o(x)$ represent general continuous even and odd functions on $[-L,L]$.
Prove that $\int_{-L}^{L} f_e(x) dx$ = 2 $\int_{0}^{L} f_e(x) dx$
My attempt:
$\int_{-L}^{L} f_e(x) dx$ = $\int_{-L}^{L} \frac{1}{2}[ f(x)+f(-x)] dx$ =  $\frac{1}{2} \int_{-L}^{L} f(x) dx$ + $ \frac{1}{2} \int_{-L}^{L} f(-x) dx$. By the Fundamental Theorem Of Calculus, I would have
$\frac{1}{2} [F(L) -F(-L)] + \frac{1}{2} [F(-L) - F(L)]$. Is this correct?

REPLY [1 votes]: This has nothing to do with the Fourier series.
It holds for every even function $f$, that
$$\int_{-L}^{0} f(x) dx = \int_{-(-L)}^{0} f(-x) d(-x) =  \int_{0}^{L} f(x) dx.$$
The first step uses integration by substitution with $u(x)=-x$ and the second step the defining property of even functions $f(x)=f(-x)$.


TITLE: How can I simplify this statement?
QUESTION [0 upvotes]: $\text {I've simplified one statement down to}\\ (p \wedge \neg q) ∨ (r \vee \neg q)
\\ \text{And I thought I could use the distributive laws to simplify this, but the operator inside the parentheses aren't the same. }
\\ \text {How would I simplify this? }$
$
\text{It apparently simplfies down to }
\\(\neg q \vee r)
\\ \text { according to Matlab/WolframAlpha but I'm not entirely sure how with the rules that I've learned.}
$
I don't necessarily need the solution, just the next step to get me closer to the solution.
I'd appreciate any help.

REPLY [2 votes]: We have $$(p \wedge \neg q) ∨ (r \vee \neg q)$$The parenthesis surrounding $(r \vee \neg q)$ are not necessary here since we have an $\vee$ in the middle. So $$(p \wedge \neg q) ∨ r \vee \neg q$$
If $\neg q$ by itself is enough to make the statement true, then the $(p \wedge \neg q)$ is redundant.

REPLY [2 votes]: It's easy to see that this is simplification is equivalent to ignoring the first term of the $\vee$. But why can that be done?
If you examine the first term, you'll see that it's only true if both $p$ and $\neg q$ are true. But if you check the second term, having $\neg q$ to be true would also make $(r\vee\neg q)$ true, allowing you to ignore the first term altogether.


TITLE: Gordan's Theorem with $Ax=b$
QUESTION [2 upvotes]: Is there a version of Gordan's Theorem in which $Ax=0$ has been replaced by $Ax=b$? That is, I want a condition (possibly including conditions on $A$) for when $Ax=b$ has a solution $x \neq 0$, for $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}_+^m \setminus \{0\}$
Gordan's Theorem says that for all $A \in \mathbb{R}^{m \times n}$ we have
$
\text{either} \quad
\exists x \in \mathbb{R}_+^m\setminus{0} \centerdot Ax = 0,
\quad\text{or}\quad
\exists y\in\mathbb{R}^n\centerdot A^\top y > 0,
$$
In particular, I cannot apply Farkas' Lemma to my problem because I want a condition with non-zero solutions.

REPLY [4 votes]: All these results follow from the duality theory of linear programming.
Let $e$ be the vector in $R^n$ with all entries 1.
$Ax = b$, $x \ge 0$, $x \ne 0$ is solvable iff the problem P:
maximize $e^T x$ subject to $Ax = b$, $x \ge 0$
is unbounded.  This implies that the dual problem D:
minimize $b^T y$ subject to $A^T y \ge e$
is infeasible, which is equivalent to saying that $A^T y > 0$ is unsolvable.  However, it is possible for both problems P and D to be infeasible, so you don't have "if and only if".  What's special about $b=0$ is that in that case you know that the problem P is feasible, namely $x=0$ is a feasible solution.

REPLY [2 votes]: This certainly won't work when $A=0$ and $b\neq 0$.  In that case, $Ax=b$ has no solution and $A^{T}y>0$ has no solution.
Another counterexample to consider is $A=[1\;\; -1; -1\;\; 1]$ and $b=[2; 1]$.  In this case, $Ax=b$ has no solution and $A^{T}y>0$ has no solution.
There's a related theorem of the alternative that is easy to prove:  Either $Ax=b$, $x \geq 0$ has a solution (which would be nonzero if $A$ and $b$ are nonzero), or $A^{T}y \geq 0$, $b^{T}y<0$ has a solution, but not both.  It's easy to prove this using LP duality.