TITLE: Does the series $\sum_{n=2}^\infty \frac{(-1)^n}{n \ln(n)}$ converge or diverge?
QUESTION [0 upvotes]: I'm a student, solving convergence problems today.
I can show that the series $\sum_{n=2}^\infty \frac{1}{n \ln{n}}$ diverges by integral test ($F(x) = \ln{(\ln{x})}, F'(x) = f(x) =  \frac{1}{x \ln{x}}$)
But I don't have any idea for the convergence of $\sum_{n=2}^\infty \frac{(-1)^n}{n \ln(n)}$
any help would be appreciated.

REPLY [3 votes]: The alternating series test immediately implies that the series converges.

REPLY [0 votes]: Yes, this does converge. In alternating series', as long as the absolute value of the numbers get closer to zero, it converges. 1/ln(n)n always get closer to 0 as n gets closer to infinity.


text:
\section{Sparse Separable NMF}
\label{sec:ssnmf}

We explained in the previous section why separable NMF does not allow for the identification of ``interior vertices'', as they are nonnegative linear  combinations of other vertices.
However, if we assume a certain column-wise sparsity on the coefficient matrix $H$, they may become identifiable.
For instance, the vertex~$W_4$ of \cref{fig:intvertex} can be expressed as a combination of the three exterior vertices ($W_1$, $W_2$, and $W_3$), but not as a combination of any two of these vertices.
Moreover, some data points cannot be explained using only pairs of exterior vertices, while they can be if we also select the interior vertex~$W_4$.

\subsection{Problem Statement and Complexity}
\label{subsec:pbm}

Let us denote $\|x\|_0$ the number of non-zero entries of the vector $x$.

\begin{definition}\label{def:sparsesep}
A matrix $M$ is $k$-sparse $r$-separable if there exists a subset of $r$ columns of $M$, indexed by $\mathcal{J}$, and a nonnegative matrix $H \geq 0$ with $\|H(:,j)\|_0 \leq k$ for all $j$ such that $M = M(:,\mathcal{J}) H$.
\end{definition}
\Cref{def:sparsesep} corresponds to \cref{def:separability} with the additional constraint that $H$ has $k$-sparse columns, that is, columns with at most $k$ non-zero entries.

A natural assumption to ensure that we can identify $W$ (that is, find the set $\mathcal{J}$), is that the columns of $W$ are not $k$-sparse combinations of any other columns of $W$; see \cref{sec:analysis} for the details.

\begin{problem}[SSNMF]
Given a $k$-sparse $r$-separable matrix $M$, find $W = M(:,\mathcal{J})$ with $|\mathcal{J}| = r$ and a column-wise $k$-sparse matrix $H \geq 0$  such that $M = W H$.
\end{problem}

As opposed to separable NMF, given $\mathcal{J}$, computing $H$ is not straightforward. It requires to solve the following $\ell_0$-constrained optimization problem
\begin{equation}\label{eq:sparseproj}
    H^* = \argmin_{H \geq 0} f(M - M(:,\mathcal{J}) H) \text{ such that }  \|H(:,j)\|_0 \leq k \text{ for all } j.
\end{equation}
Because of the combinatorial nature of the $\ell_0$-``norm'', this $k$-sparse projection is a difficult subproblem with $\binom{r}{k}$ possible solutions, which is known to be NP-hard~\cite{natarajan1995sparse}.

In particular, a brute-force approach could tackle this problem by solving $\mathcal{O}(r^k)$ NNLS problems.
However, this combinatorial subproblem can be solved exactly and at a reasonable cost by dedicated branch-and-bound algorithms, such as  \arbo{}~\cite{nadisicexact}, given that $r$ is sufficiently small, which is typically the case in practice.
Even when $k$ is fixed, the following result shows that no provably correct algorithm exists for solving SSNMF in polynomial time (unless P=NP):
\begin{theorem} \label{th:nphardssnmf}
SSNMF is NP-complete for any fixed $k \geq 2$.
\end{theorem}

\begin{proof}
The proof is given in Appendix~\ref{sec:nphard}.
Note that the case $k = 1$ is trivial since each data point is a multiple of a column of $W$.
\end{proof}

However, in Section \ref{sec:analysis}, we show that under a reasonable assumption, SSNMF can be solved in polynomial time when $k$ is fixed.

\subsection{Related Work} \label{sec:relwork}

To the best of our knowledge, the only work presenting an approach to tackle SSNMF is the one by Sun and Xin (2011)~\cite{sun2011underdetermined} --- and it does so only partially.
It studies the blind source separation of nonnegative data in the underdetermined case.
The problem tackled is equivalent to NMF in the case $m < r$.
The
assumptions
used in this work are similar to ours, that is, separability and sparsity.

However, the setup considered is less general than SSNMF because the sparsity assumption (on each column of $H$) is limited to $k = m - 1$, while the only case considered theoretically is the case $r = m+1$ with only one interior vertex.

The proposed algorithm first extracts the exterior vertices using the method LP-BSS from~\cite{naanaa2005blind}, and then identifies the interior vertex using a brute-force geometric method.
More precisely,
they select an interior point, and check whether at least two of the $m-1$ hyperplanes generated by this vertex with $m-2$ of the extracted exterior vertices contain other data points.

If it is the case, then they conclude that the selected point is an interior vertex, otherwise they select another interior point.
For example, when $m = 3$,  this method consists in constructing the  segments between the selected interior point and all the exterior vertices.  If two of these segments contain at least one data point, then the method stops and the selected interior point is chosen as the interior vertex. Looking at Figure~\ref{fig:intvertex}, the only interior point for which two segments joining this point and an exterior vertex   contain data points is $W_4$.
Note that, to be guaranteed to work,
this method requires at least two hyperplanes containing the interior vertex and $m-2$ exterior vertices to contain data points. This will not be a requirement in our method.

config: arxiv


TITLE: Sum of three dice is eleven
QUESTION [8 upvotes]: the sample space of one toss of three dice is:
$\Omega = \left \{ (1,1,1), ..., (6,6,6) \right \}$
so there are $6^3 = 216$ possible outcomes.
What is the probability to obtain an outcome where the sum of its three components is equal to 11?
I've considered the possible value can assume dice without a particular position and then I have considered the permutations to include every position:
$(6,4,1), 3! = 6 \\ (6,3,2), 3! = 6 \\ (5,5,1), \frac{3!}{2!} = 3 \\ (5,4,2), 3! = 6 \\ (5,3,3), \frac{3!}{2!} = 3 \\ (4,4,3), \frac{3!}{2!} = 3$
so I've summed up obtaining $27$ and the probability would be $\frac{27}{216}$
Now, consider if I have to do this same passages for sums from 3 to 18, it is very exhausting.
So, my question is: Is there any "faster" way to do that?

REPLY [6 votes]: For the favorable cases:  Since the dice can assume integer values between $1$ and $6$, we wish to find the number of solutions in the positive integers of the equation
$$x_1 + x_2 + x_3 = 11 \tag{1}$$
subject to the restrictions that $x_k \leq 6$ for $1 \leq k \le 3$.
A particular solution of equation 1 corresponds to the placement of two addition signs in the ten spaces between successive ones in a row of eleven ones.  For instance,
$$1 1 1 1 1 + 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 5$, $x_2 = 2$, and $x_3 = 4$.  Hence, the number of solutions of equation 1 in the positive integers is the number of ways we can select two of the ten spaces between successive ones in a row of eleven ones in which to place addition signs, which is
$$\binom{10}{2}$$
More generally, the equation
$$x_1 + x_2 + \cdots + x_k = n$$
has
$$\binom{n - 1}{k - 1}$$
solutions in the positive integers since we must choose which $k - 1$ of the $n - 1$ spaces between successive ones in a row of $n$ ones will be filled with addition signs.
However, we have counted solutions in which one of the variables exceeds $6$.  Notice that it is not possible for two of the variables to exceed $6$ simultaneously since $2 \cdot 7 = 14 > 11$.
Suppose $x_1 > 6$.  Let $x_1' = x_1 - 6$. Then $x_1'$ is a positive integer.  Substituting $x_1' + 6$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 6 + x_2 + x_3 & = 11\\
x_1' + x_2 + x_3 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with
$$\binom{5 - 1}{3 - 1} = \binom{4}{2}$$
solutions.  By symmetry, there are an equal number of solutions in which $x_2 > 6$ or $x_3 > 6$.  Hence, the number of solutions we must exclude is
$$\binom{3}{1}\binom{4}{2}$$
Hence, the number of permissible solutions is
$$\binom{10}{2} - \binom{3}{1}\binom{4}{2}$$
Addendum:  You asked about the sums from $3$ to $18$.  If $3 \leq n \leq 8$, then the number of solutions of the equation
$$x_1 + x_2 + x_3 = n \tag{3}$$
in the positive integers with $x_k \leq 6$ for $1 \leq k \leq 3$ is
$$\binom{n - 1}{3 - 1} = \binom{n - 1}{2}$$
since it is not possible for one of the variables to exceed $6$.
Notice also that by symmetry, the number of solutions with sum $3$ (all ones) is equal to the number of solutions with sum $18$ (all sixes), the number of solutions with sum $4$ (two ones and a two) is equal to the number of solutions with sum $17$ (two fives and a six), and so forth.  Hence, knowing the number of solutions for $3 \leq n \leq 8$ also tells us the number of solutions for $13 \leq n \leq 18$.
The following argument shows that the number of solutions with sum $n$ is equal to the number of solutions with sum $21 - n$.  If $y_k = 7 - x_k$, then $1 \leq x_k \leq 6  \implies 1 \leq y_k \leq 6$.  Moreover, substituting $7 - y_k$ for $x_k$ for $1 \leq k \leq 3$ in equation 3 yields
\begin{align*}
7 - y_1 + 7 - y_2 + 7 - y_3 & = n\\
21 - y_1 - y_2 - y_3 & = n\\
-21 + y_1 + y_2 + y_3 & = -n\\
y_1 + y_2 + y_3 & = 21 - n
\end{align*}
By an argument similar to that given above for $n = 11$, the number of solutions of the equation
$$x_1 + x_2 + x_3 = 9$$
in positive integers not exceeding six is
$$\binom{8}{2} - \binom{3}{1}\binom{2}{2} = \binom{8}{2} - \binom{3}{1}$$
By symmetry, this is also the number of solutions for $n = 12$.
Also, by symmetry, the number of solutions for $n = 10$ is equal to the number of solutions for $n = 11$.


TITLE: Sufficient and necessary conditions for a given operator $T$ to be $T \in \mathcal{B}$, $T \in \mathcal{K}$ or $T \in \mathcal{F}$
QUESTION [3 upvotes]: I was rewiewing for an upcoming exam and found this problem. It is quite basic and straightforward but still causing me some confusion mainly since I would have to justify my answer.
Let a linear operator $T$ on Banach space $\ell^1$ be given with respect to the normal baiss $(e_n)_ {n=1}^\infty$ s.t
$Te_n := \alpha_n (e_n +e_{n+1} + ... + e_{2n})$  for all  $n \in \mathbb{N}$.
Here $e_1 = (1,0,0,...)$, $e_2 = (0,1,0,0,...)$ etc.
Give sufficient and necessary conditions for the sequence $(\alpha_n)_{n=1}^\infty$ ($\alpha_n \in \mathbb{C}$) so that
a) $T \in \mathcal{B}(\ell^1)$
b) $T \in \mathcal{K}(\ell^1)$
c) $T \in \mathcal{F}(\ell^1)$
The a) part seems quite clear, I just need to check when $T$ is bounded i.e. when $||T|| < \infty$. As result I got $\sup_j |\alpha_j| < \infty$.
The other two parts, however, are a bit more confusing.

REPLY [0 votes]: a) If $T$ is bounded, then the sequence $\left(\left\lVert Te_n\right\rVert_1\right)_{n\geqslant 1}$ should be bounded. Since $\left\lVert Te_n\right\rVert_1=(n+1)\left\lvert \alpha_n\right\rvert$, the sequence $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ should be bounded.
Conversely, if $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ is bounded, then $\sup_{n\geqslant 1}\left\lVert Te_n\right\rVert_1$ is finite which is sufficient to ensure that $T$ is bounded.
b) Assume that $T$ is compact. Then the set $\left\{Te_n,n\geqslant 1\right\}$ should have a compact closure hence
$$\lim_{N\to+\infty}\sup_{n\geqslant 1}\sum_{i\geqslant N}\left\lvert \left(Te_n\right)_i\right\rvert=0,$$
where $\left(v\right)_i$ denotes the $i$-th coordinate of $v\in\ell^1$. Since
$$
\sup_{n\geqslant 1}\sum_{i\geqslant N}\left\lvert \left(Te_n\right)_i\right\rvert\geqslant \sum_{i\geqslant N}\left\lvert \left(Te_N\right)_i\right\rvert =(N+1)\left\lvert \alpha_N\right\rvert
$$
we should have that $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ converges to $0$.
Conversely, if $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ converges to $0$, define the linear operator $T_N$ in the following way: if $1\leqslant n\leqslant N$, then $T_N(e_n)=T(e_n)$ and if $n\gt N$, then $T_N(e_n)=0$. The operator $T_N$ has a finite rank and the operator norm of $T-T_N$ can be bounded by
$\sup_{i\geqslant N}(i+1)\left\lvert \alpha_i\right\rvert $.
c) If $a_n=0$ for $n$ large enough, then $T$ has a finite rank. If  $a_n\neq 0$ for infinitely many $n$'s, let $n_i$ be integers such that $a_{n_i}\neq 0$ and $n_{i+1}\gt 3n_i$. Then $\left\{Te_{n_i}\right\}$ is linearly independent and infinite.


TITLE: How do first integrals help you solve differential equations?
QUESTION [1 upvotes]: I am reading about Euler-Lagrange equations and this particular section is a little unclear. Consider the differential equation
$$\begin{bmatrix}
\dot{x}\\
\dot{y}
\end{bmatrix} = \begin{bmatrix}
-7x^{\frac{6}{7}}y^{\frac{1}{7}}\\
7x^{\frac{1}{7}}y^{\frac{6}{7}}
\end{bmatrix}\quad (x(0),y(0)) = (x_{0},y_{0}), \quad x_{0},y_{0} \neq 0.$$
How do I solve this using the fact that $H(x,y) = x^{\frac{2}{7}} + y^{\frac{2}{7}}$ is a first integral of the above differential equation?
If we let $u = (x,y)$, $u(0) = (x_{0},y_{0})$ be a solution to the above differential equation, then $H(u)$ is a constant. In particular, $H(u) = H(u(0))$ for all $t$ since a first integral is constant along the solutions of the differential equation.
After this, I am a confused. I know that the level sets of $H(x,y)$ looks like an asteroid. But how does this help me? Is each level set of $H$ the orbit of a solution to the differential equation? How do I know that there aren't any extraneous points in the level set that are not part of $u$?
If the above is correct, I don't understand how to write down the general solution from this level curve.

REPLY [1 votes]: The level sets of a first integral are not necessarily orbits, but the orbits are pieces of these level sets.  If $L$ is a level set and $E$ the set of equilibrium points,  the connected components of $L \backslash E$ should be orbits.
EDIT: If you can parametrize a level set (say as $x = X(s), y=Y(s)$), then substituting $x = X(s(t)), y = Y(s(t))$ in to the differential equation should give you a single first-order differential equation for $s(t)$.  In your example
I would try $x = c\; \cos(s)^7$, $y = c\;\sin(s)^7$.


TITLE: Can you mod under the root?
QUESTION [0 upvotes]: I've been using the following fact for a long time, and as I have never seen it fail, I've started to believe it. However, I have never actually seen a proof. For $m, n, p ∈ \Bbb Z, p$ prime:

$m ≡ n \mod p \Rightarrow \sqrt{m} ≡ \sqrt{n} \mod p$

This usually comes up when doing something like applying the quadratic formula in a finite field.

REPLY [0 votes]: I guess the reason no-one talks about even though it is true (if $p$ is prime)
[If $a,b\in \mathbb N;$ and $a^2 = n; b^2 = m$ then if $n\equiv m \pmod p$ so $p|m^2 - n^2=(a-b)(a+b)$ so either $a-b=1$ and $a+b=p$ so $\sqrt{n}=a\equiv -b=-\sqrt m \pmod p$ or $a-b=-p$ and $a+b=-1$ and $\sqrt n = a \equiv b=\sqrt m \pmod p$]
Is that it is naive to think about $n \pmod p$ as the integer $n$.  If he integer $n$ happens to be a perfect square that is pretty irrelevant as the integer $m = n+kp$ wont be.  And we don't care at all about $n$ as an integer; we only care about $n$ as a representative of an equivalence class, so $n$ and $n+kp$ are considered to be the exact same thing.
Which makes the only reasonable meaning for $\sqrt n$ is that $\sqrt n\equiv a\pmod p$ if $a^2 \equiv n\pmod p$ and we have some criterion for selecting $a$ for from all the other representatives whose square are equivalent to $n$ (possibly in the lowest integer representative between $0$ and $p-1$....).
But then the observation is trivial and near meaningless.  If $m \equiv n\pmod p$ then $m$ and $n$ are the same thing.  So if $\sqrt{n} \equiv a$ so $a^2 \equiv n$ then... $a^2 \equiv m$ and $\sqrt{m} \equiv a$.
However...  To put it in more sophisticated terms:

If $p$ is prime the if $a^2 \equiv b^2 \equiv n\pmod p$ then $a\equiv \pm b \pmod p$.  But only half of the residue class $1,...., p-1$ will have solutions to $x^2\equiv a\pmod p$ and for those classes there will be exactly $2$ solutions.

If $p$ is not prime $x^2 \equiv n$ may have more than two solutions.  Or may have none at all.


TITLE: Find $\lim\limits_{x\to0}\frac 1x(x^{-\sin x}-(\sin x)^{-x})$
QUESTION [7 upvotes]: The question is to evaluate this limit:$$\lim_{x\to0}\frac{\big(\frac{1}{x}\big)^{\sin x}-\big(\frac{1}{\sin x}\big)^x}{x}$$
I tried using l'Hospital's rule, taking the logarithm, doing some manipulations using known limits, but without success.

REPLY [2 votes]: I deleted my previous answer as there were (stupid) mistakes in it, but as it turns out l'Hopital's rule is not helpful, and actually there is a much easier answer. Write the expression as
$$\frac{e^{-(\ln x)(\sin x)}-e^{-x\ln \sin x}}{x}$$
$$=(\sin x)^{-x}\frac{e^{-(\ln x)(\sin x)+x\ln \sin x}-1}{x}$$
Now $(\sin x)^{x}\to 1$ is easily verified. The remainder can be written as, $$\frac{e^{f(x)}-1}{x}=\frac{e^{f(x)}-1}{f(x)}\frac{f(x)}{x}$$ where the first factor again limits to $1$. Thus we are reduced to evaluating
$$\ln(\sin x)-\frac{\sin x}{x}\ln x=\ln(\frac{\sin x}{x})-(x\ln x)(\frac{\sin x-x}{x^2})$$ and its easy to see that all elements limit to zero.


TITLE: Exercise 32 from chapter 4 (“Hilbert Spaces: An Introduction”) of Stein & Shakarchi's “Real Analysis” 2
QUESTION [2 upvotes]: I have a question about the following problem in Stein and Shakarchi's book.

Consider the operator $T:L^2([0,1]) \to L^2([0,1])$ defined by $$T(f)(t) = tf(t).$$
  (a) Prove that $T$ is a bounded linear operator with $T = T^*$, but that $T$ is not compact.
  (b) However, show that $T$ has no eigenvectors.

First of all, I don't really get how this operator works. Could you guys give me insight into this operator and guide me through these problems? Thanks!

REPLY [3 votes]: That operator $T$ that is defined above is taking any function $f\in L^2[0,1]$ to the function $tf(t)\in L^2[0,1]$.
$\textbf {Boundedness of T}:$ Note that
$$ \|Tf\|_2^2=\int_{0}^{1}t^2|f(t)|^2dt\leq\int_{0}^{1}|f(t)|^2dt=\|f\|_2^2.$$
Thus, $T$ is a bounded linear operator. Moreover, $\|T\|\leq 1$. (In fact, you can show that $\|T\|=1$.)
$\textbf{T=T*}$: $$\langle Tf,g\rangle =\int_{0}^{1}tf(t)\overline{g(t)}dt= \int_{0}^{1}f(t)\overline{tg(t)}dt = \langle f,Tg\rangle.$$
So, by the uniqueness of the adjoint operator, it follows that $T=T^*$.
$\textbf{T has no eigenvectors}$: First of all recall that an eigenvector is nonzero by definition. Let us assume that $f$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$. Then
$$ tf(t) = Tf(t) = \lambda f(t),$$
for all $t\in[0,1]$, whence it follows that $f=0$ almost everywhere. This contradiction proves that $T$ has no eigenvector.
$\textbf{T is not compact}$: For this part have a look at the following
Compactness of  Multiplication Operator on $L^2$


TITLE: Proving how to apply the adjoint to a dyad
QUESTION [1 upvotes]: I have to prove that  $(|\Psi\rangle \langle \Phi|)^\dagger= |\Phi\rangle \langle \Psi|$
Hint: Prove that $\langle f|(|\Psi\rangle \langle \Phi|)^\dagger|g\rangle = ... =  \langle f||\Phi\rangle \langle \Psi||g\rangle$

REPLY [1 votes]: The notation obsecures the meaning but I'll try to cope with it by writing the definition of the adjoint of $A$ as
$$
\langle f|A|g\rangle = \overline{\langle g|A^\dagger|f\rangle}\\
$$
The given operator is in a tensor (dyad) form. It is to be understood as scaling the lefthand ket by the inner product of the right bra with the argument, (on which the operator is applied)
$$
\langle f||\Phi\rangle \langle \Psi||g\rangle = \overline{\langle \Phi||f\rangle \langle g||\Psi\rangle}
= \overline{ \langle g||\Psi\rangle \langle \Phi||f\rangle}
=  \overline{ \langle g|(|\Phi\rangle \langle\Psi |)^\dagger|f\rangle}
$$
In a more clear notation and for better explanation we do it without the Dirac notation but the meaning of the operator is as above
$$
A = |\Phi\rangle \langle \Psi| = \Psi\langle\Phi,\cdot\rangle
$$
this acts on $x$ like
$$
Ax = \Psi\langle\Phi,x\rangle
$$
The Definition of the adjoint of $A$ is
$$
\langle f, Ag \rangle = \langle A^\dagger f, g \rangle
$$
Proving the statement
$$
\langle f, Ag \rangle
= \langle f, \Psi\langle\Phi,g\rangle \rangle
= \langle  f, \Psi \rangle \langle\Phi,g\rangle
=  \langle \overline{\langle  f, \Psi \rangle} \Phi,g\rangle
=  \langle \langle  \Psi, f \rangle \Phi,g\rangle
=  \langle\Phi \langle  \Psi, f \rangle ,g\rangle
=  \langle A^\dagger f  ,g\rangle
$$
Where
$$
A^\dagger = \Phi \langle  \Psi, \cdot \rangle = |\Psi\rangle \langle \Phi|
$$

REPLY [1 votes]: Let $A = |\Psi\rangle \langle \Phi|$, then
$$
\langle f|A^\dagger|g\rangle = \langle g|A|f\rangle^* = \big(\langle g|\Psi\rangle \langle \Phi|f\rangle\big)^* =  \langle g|\Psi\rangle^* \langle \Phi|f\rangle^* = \langle\Psi|g\rangle\langle f|\Phi\rangle = \langle f|\Phi\rangle \langle\Psi|g\rangle,
$$
so $A^\dagger = |\Phi\rangle \langle\Psi|$.


TITLE: A base in which all primes end with $5$ different symbols?
QUESTION [1 upvotes]: In base $10$, all prime numbers (a part $2$ and $5$) end with $1,3,7$ or $9$, i.e. with four different symbols.
Is there a base in which all prime numbers end with $5$ different symbols (or also with $5$ distinct groups of symbols)? If yes, which base?
Thanks for your help! I apologize for such a trivial question!
NOTE: This question is related to this one.

REPLY [1 votes]: Hint $ $ The number of residues coprime to $n> 2$ is even: $ $ negation reflection  $\,x\mapsto -x\pmod {\!n}\,$  partitions them into pairs (since it has no fixed points: $\,-a\equiv a\,\Rightarrow\, n\mid 2a,\,$ contra $(n,a)=1)$.
Remark $\ $ Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.


TITLE: Prove columns of Jacobian Matrix of $T(r,\varphi,\theta) := (r\ \cos \varphi \cos \theta ,\ r\ \sin \varphi \cos \theta ,r\ \sin \theta)$ orthogonal
QUESTION [0 upvotes]: Let $f:\mathbb{R^3} \to \mathbb{R}$ be a differentiable function.
For $r > 0, \varphi \in [0,2\pi]$ and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ we look at the functions
$$T(r,\varphi,\theta) := (r\ \cos \varphi \cos \theta ,\ r\ \sin \varphi \cos \theta ,r\ \sin \theta)$$ and
$$g(r,\varphi, \theta) =  f(T(r,\varphi, \theta))$$
How can one prove that the columns of the Jacobian Matrix of $T$ are orthogonal perpendicular to one another?
For which values of $r,\varphi,\theta$ is $T'$ regular and how do $\nabla f$ and $\nabla g$ calculate into each other?
I know that Laplace's equation is given by
$$\nabla^2u = u_{xx}+u_{yy}+u_{zz} = 0$$
and in another thread Laplace's equation in spherical coordinates is proven.
I found this on the internet:

but I still don't know what exactly needs to be done to prove that the columns of the Jacobian Matrix of T are orthogonal (perpendicular to one another) and how to go on with the other question..

REPLY [0 votes]: Arrange the derivatives $\frac{\partial T}{\partial r}$,$\frac{\partial T}{\partial \varphi}$ and $\frac{\partial T}{\partial \theta}$ as columns and do the inner product among them.


TITLE: If f is continuous and 0 < f(x) < g(x) on the interval [0, ∞) and $\int_0^\infty g(x) dx = M < \infty$ then $\int_0^\infty f(x) dx$ exists?
QUESTION [1 upvotes]: If f is continuous and $0 < f(x) < g(x)$ on the interval $[0, ∞)$ and $\int_0^\infty g(x) dx = M < \infty$ then $\int_0^\infty f(x) dx$ exists.
True or False, and why? I'm not sure what the M means or how to use it.

REPLY [1 votes]: This is true. $M$ is a real number for the value of integral. Let $h(t)=\int_0^tf(x)\: dx$. Since
$$
0<h(t)=\int_0^t f(x)\: dx<\int_0^t g(x)\: dx<\int_0^\infty g(x)\: dx < \infty
$$
$h(t)$ is monotone increasing with upper bound. So
$$
\lim_{t\to\infty}h(t)=\int_0^\infty f(x)\: dx
$$
exists.