TITLE: Is this (classical?) exercice missing a hypothesis?
QUESTION [9 upvotes]: A friend just told me about an exercice he was given quite a few years ago, but he wasn't sure wether he remembered all the hypothesis correctly. Does anybody recognize this?

Let $f$ be a smooth real-valued function on $\Bbb R$ such that for all $n\in\Bbb N,~\sup_{\Bbb R}|f^{(n)}|=1$ and $f(0)=1$, then $f=\cos$.

He wasn't sure wether theose were the exact hypothesis needed for this to work. He noticed that $f$ is automatically a power series with infinite radius of convergence, but wondered wether one ought to impose $f$ to be even, and wether the condition on the suprermums of the derivatives is correct.
NOTE. If you know the hypothesis, please post them as an answer, and if you want to post a solution, please hide the text so that one needs to scroll over it to reveal it, thanks!

REPLY [3 votes]: There is a 1980 paper by John Roe titled A characterization of the sine function, with the following abstract. I suspect the specific result here follows from Roe's theorem.

If a function and all its derivatives and integrals are absolutely
  uniformly bounded, then the function is a sine function with period
  2π.

http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2081636


TITLE: How to evaluate the integral involving two Bessel functions and following elementary function?
QUESTION [2 upvotes]: Let's have the integral
$$
\int \limits_{0}^{\infty} J_{0}(kr) J_{0}(kr')\frac{kdk}{k^2 + a^2}.
$$
How to evaluate it? I failed when was trying using integral representations for the Bessel functions.

REPLY [5 votes]: Let me first state the answer: for $r<r'$
$$\mathcal{I}(r,r',a)=\int_0^{\infty}J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=I_0(ar)K_0(ar').\tag{1}$$
More general formulas can be found in Prudnikov-Brychkov-Marychev, Vol. 2.
To show (1) (for generalizations the procedure is similar), note that
\begin{align}
&\left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-a^2\right)\mathcal{I}(r,r',a)=\\&=
\int_0^{\infty}\left(\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-a^2\right)J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=\\
&=-\int_0^{\infty}(k^2+a^2)J_0(kr)J_0(kr')\frac{kdk}{k^2+a^2}=\\
&=-\int_0^{\infty}kJ_0(kr)J_0(kr')dk=\\
&=0.\tag{2}
\end{align}
At the last step we have used the known orthogonality relations for Bessel functions.
Thus we see that $\mathcal{I}(r,r',a)$ solves the modified Bessel equation w.r.t. $ar$ and should therefore be a linear combination of its two independent solutions. Moreover since it is regular at $r=0$, it can only be proportional to $I_0(ar)$.
Similarly, $\mathcal{I}(r,r',a)$ solves the modified Bessel equation w.r.t. $ar'$ and vanishes at infinity, hence it can only be proportional to $K_0(ar')$. Therefore,
$$\mathcal{I}(r,r',a)=\operatorname{const}\cdot I_0(ar) K_0(ar').$$
The remaining constant can be fixed in several ways - e.g. using the known asymptotic behavior of Bessel functions.


TITLE: How to find the values of $k$ if $2k \in \mathbb{Z}$ and $\frac{2}{k}\in \mathbb{Z}$
QUESTION [0 upvotes]: Let $k \in \mathbb{R}$. What are the possible values of $k$ Under the condition that $2k \in \mathbb{Z}$ and $\frac{2}{k} \in \mathbb{Z}$. I know that the answer is $k \in \lbrace \frac{1}{2}, 1, 2 , -\frac{1}{2}, -1, -2\rbrace$, But I don't know how to prove it ?
Thanks for any help!

REPLY [2 votes]: Notice that $2k \cdot \frac{2}{k} = 4$, and since by assumption both $2k$ and $k/2$ are integers, you can see that $2k \in \{\pm 1, \pm 2, \pm 4 \}$ and hence $k$ is in the set you described.


TITLE: How can i evaluate this integral with or without using CAS $\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$
QUESTION [2 upvotes]: $$\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$$
Firstly i used substitution $\pi x=t; \mathrm dx=\frac{\mathrm dt}{\pi}$
$$\pi \int_0^{\infty}\frac{\mathrm dt}{(t^2+\pi^2)\cosh t}$$
writing $\cosh t =\frac{e^{t}+e^{-t}}{2}=\frac{1+e^{-2t}}{2e^{-t}}$
$$2\pi\int_0^{\infty}\frac{e^{-t}}{(t^2+\pi^2)(1+e^{-2t})}$$
substituting $e^{-t}=v;-e^{-t}\mathrm dt=\mathrm dv$
$$2\pi\int_0^{1}\frac{\mathrm dv}{(\ln^2 v+\pi^2)(1+v^2)}$$

REPLY [1 votes]: Recall the following two results:
$$\int_0^\infty  e^{-bt}\cos\left(xt \right)\,dt=\frac{b}{b^2+x^2}\tag{1}$$
$$\int_0^\infty  \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx=\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \tag{2}$$

$$
\begin{aligned}
\int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,dx&= \frac 1b\int_0^\infty \frac{1}{\cosh\left(ax \right)}\,\int_0^\infty  e^{-bt}\cos\left(xt \right)\,dt\,dx\\
&= \frac 1b\int_0^\infty  e^{-bt}\int_0^\infty  \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx\,dt\\
&= \frac 1b\int_0^\infty  e^{-bt}\left(\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \right)\,dt\\
&= \frac{\pi}{2 a b}\int_0^\infty  \frac{ e^{-bt}} {\cosh\left(\frac{\pi t}{2a} \right)} \,dt\\
&= \frac{1}{b}\int_0^\infty  \frac{ e^{-\frac{2ab}{\pi}t}} {\cosh\left(t \right)} \,dt \qquad \left(\frac{2ab}{\pi}w=t\right)\\
&= \frac{2}{b}\int_0^\infty  \frac{ e^{-ct}} {e^{t}+e^{-t}} \,dt \qquad \left(c=\frac{2ab}{\pi}\right)\\
&= \frac{2}{b}\int_0^\infty  \frac{ e^{-ct}e^{-t}} {1+e^{-2t}} \,dt\\
&= \frac{2}{b}\int_0^\infty   e^{-ct}e^{-t}\left(\sum_{k=0}^\infty (-1)^{k}e^{-2kt} \right) \,dt\\
&= \frac{2}{b}\int_0^\infty   e^{-ct}\left(\sum_{k=0}^\infty (-1)^{k}e^{-(2k+1)t} \right) \,dt\\
&= \frac{2}{b}\int_0^\infty   e^{-ct}\left(\sum_{k=1}^\infty (-1)^{k-1}e^{-(2k-1)t} \right) \,dt\\
&= \frac{2}{b}\sum_{k=1}^\infty (-1)^{k-1}\,\int_0^\infty  e^{-\left(c+(2k-1)\right)t}  \,dt\\
&= \frac{2}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{\frac{2ab}{\pi}+(2k-1)}\\
&=\frac{2 \pi}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2ab+(2k-1)\pi} \qquad \blacksquare
\end{aligned}
$$

$$
\begin{aligned}
\int_0^\infty \frac{1}{(1+x^2)\cosh\left( \pi x \right)}\,dx&=2 \pi\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 \pi+(2k-1)\pi}\\
&=2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k+1}\\
&=-2 \sum_{k=1}^\infty \frac{(-1)^{k}}{2k+1}\\
&=2-2 \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1}\\
&=2-2 \frac \pi 4 \\
&=2-\frac \pi 2 \qquad \blacksquare
\end{aligned}
$$

$$
\begin{aligned}
\int_0^\infty \frac{1}{(\pi^2+x^2)\cosh\left(  x \right)}\,dx&=\frac{2 \pi}{\pi}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 \pi+(2k-1)\pi}\\
&=\frac 2 \pi \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k+1}\\
&=-\frac 2 \pi \sum_{k=1}^\infty \frac{(-1)^{k}}{2k+1}\\
&=\frac 2 \pi-\frac 2 \pi \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1}\\
&=\frac 2 \pi-\frac 2 \pi \cdot \frac \pi 4 \\
&=\frac 2 \pi-\frac 1 2 \qquad \blacksquare
\end{aligned}
$$

$$
\begin{aligned}
&\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+\frac{1}{2}}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k\int_0^1t^{k+1/2-1}dt\\
&=\frac{1}{2}\int_0^1t^{-1/2}\left(\sum_{k=0}^{\infty}(-1)^kt^k\right)dt\\
&=\frac{1}{2}\int_0^1\frac{t^{-1/2}}{1+t}dt\\
&=\frac{1}{4}\left(\psi\left( \frac{3}{4}\right)-\psi\left( \frac{1}{4}\right) \right)\\
&=\frac{1}{4}\left(\pi \cot\left( \frac{\pi}{4}\right) \right)\\
&=\frac{\pi}{4}\\
\end{aligned}
$$
Where We used the results
$$\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left( \frac{x}{2}\right) \right)$$
and
$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$


TITLE: Uniserial modules are determined up to isomorphism by their composition series?
QUESTION [4 upvotes]: I have a question on uniserial modules. Here, every modules are finitely generated modules over a finite dimensional algebra $A$ over a field $K$.
In the book Elements of the Representation Theory of Associative Algebras: Volume 1 , there is a remark as below. (p.164)

We also notice that uniserial modules are determined up to isomorphism
by their composition series, that is, if M and N are uniserial modules and
have the same composition factors in the same order, then they are isomorphic.
An isomorphism is constructed by an obvious induction on the
common composition length of M and N.

But I cannot figure out how to make an induction step. Obviously, we have an isomorphism $M/radM \cong N/radN$ and by induction hypothesis, have an isomorphism $radM\cong radN$, but I don't know how to make an isomorphism between $M$ and $N$ from these informations. How can I show that those two are isomorphic?

REPLY [3 votes]: This is not true in general. For example, the Klein four group has infinitely many $2$-dimensional uniserial modules over a field of characteristic $2$. I'm not quite sure what the authors mean in this section. It doesn't look like they are considering specific classes of algebras, where this might well be true.
An explicit example for $G=\langle x,y\rangle$ Klein four is to map
$$ x\mapsto \begin{pmatrix}1&1\\0&1\end{pmatrix}$$
and
$$ y\mapsto \begin{pmatrix}1&\lambda\\0&1\end{pmatrix}$$
for any $\lambda$ in the field. It is an easy exercise to see that there is no invertible matrix that centralizes $x$ and maps $y$ from one such matrix to any other, hence these representations are all inequivalent.


TITLE: Geometric intuition behind locality of morphisms of locally ringed spaces
QUESTION [3 upvotes]: Let $(X,\mathcal O_X)$ be a locally ringed space. The examples I have in mind are sheaves of continuous/$C^k$/smooth/etc. functions over a suitable topological space.
The direction $f^\sharp:\mathcal O_Y\rightarrow f_\ast\mathcal O_X$ of a morphism $(X,\mathcal O_X)\rightarrow (Y,\mathcal O_Y)$ is clear to me, but I am struggling to understand the geometric significance of respecting the maximal ideals. I don't know commutative algebra so excuse the question if it's a well discussed triviality.
First, let's switch from $f^\sharp:\mathcal O_Y\rightarrow f_\ast\mathcal O_X$ to $f^\flat:f^\ast\mathcal O_Y\rightarrow \mathcal O_X$. Now we can look at a map between the stalks at $x\in X$:
$$\mathcal O_{Y,f(x)}\rightarrow \mathcal O_{X,x}$$
The maximal ideal $\mathfrak m_{f(x)}\vartriangleleft \mathcal O_{Y,f(x)}$ is comprised of the nonzero germs at $f(x)$. These are just the locally non-invertible functions at $f(x)$. To map $\mathfrak m_{f(x)}$ into $\mathfrak m_x$ is to say we do not accidentally invert any nonunits in $\mathcal O_{Y,f(x)}$.
This sounds like a good thing at the formal level, but what are some interesting geometric examples in which not being a local homomorphism has unpleasant consequences?

REPLY [4 votes]: Perhaps it's more helpful to think about curves if you're not comfortable with commutative algebra.  If you consider a curve $f(x,y)=0$ over $\mathbb{C}$, then the sheaf of regular functions on the curve has stalks $(\mathbb{C}[x,y]/(f(x,y)))_{(x-a,y-b)}$, which is the coordinate ring localized at the point $P = (a,b)$. So each element of the stalk is an equivalence class of polynomials on the curve $f(x,y)$ with possible poles at points not equal to $P$.  In other words, we look polynomial functions in $x$ and $y$, group them together by which ones evaluate the same on the points of the curve, and allow them to have finitely many poles away from $P$, since we only care about the local information in stalks.
When we want to map another curve $g(x,y)$ to the curve $f(x,y)$, then an important property that we want to have is that every time we have a function $p(x,y)$ on the curve $f(x,y)$, taking the curve to $\mathbb{C}$, then we get a function from $g(x,y)$ to $\mathbb{C}$.  If you think about it, this is a very natural property to want: after all, we have a map from $g$ to $f$ and from $f$ to $\mathbb{C}$.  This, you said, is pretty clear to you, and gives the direction of the map $f^\sharp$ from $\mathcal{O}_{f} \to f_* \mathcal{O}_{g}$.
Then, suppose that we broke our rule, and the induced map on stalks gave some function in the maximal ideal of the stalk of $\mathcal{O}_f$ at $P$ which maps to some function which is not in the maximal ideal of the stalk of $\mathcal{O}_g$ at $Q \in f^{-1}(P)$  (and let's suppose there is some point $Q$, so that something interesting happens).  Then we took a very nice function on the curve $f(x,y)$ to $\mathbb{C}$ which evaluated to zero at $P$ and we got a function from $g$ to $\mathbb{C}$ which does not evaluate to zero at $Q$.  This should make not very much sense, because if we literally had functions, not just sections in a sheaf, and could compose the function $f$ to $\mathbb{C}$ with the function taking the curve $g$ to $f$ then, $Q$ would map to $P$ under the curve map and $P$ would map to $0$ under the polynomial in the maximal ideal of the stalk.  So, in order for the morphism of locally ringed spaces to emulate the behavior we see on varieties, it makes sense to require that the induced map on the stalks map relate the maximal ideals in this way.