Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following problem:


Find the constants $c_n$ so that $$ \frac{1}{\sin^2 z } = \sum_{n= -\infty} ^ {+\infty} \frac{c_n}{(z-\pi n)^2} $$ and the series converges uniformly on every bounded set after dropping finitely many terms. Justify all your claims. Hint: Use Liouville's theorem to prove the equality.


Let $c_n = 1 $ for all $n$.

Let $\displaystyle f(z) := \frac{1}{\sin^2 z}$ and $\displaystyle g(z) := \sum_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $.

Begin by showing that $h$ is an analytic function which converges uniformly on compact subsets of $\mathbb{C} \backslash \mathbb{Z}$.

Suppose $K$ is a compact set which contains no integers. Define $$ \delta_n = \inf_{z \in K} |\pi-z/n| =\frac{1}{n} \inf_{z \in K} |\pi n-z|, $$ where the infimum exists because $K$ is compact. Also, compactness implies boundedness. Thus as $n \to \pm \infty$, we have $\delta_n \to \pi$.

Therefore, for sufficiently large $n$, we have $\delta_n > 2$, and $$ \frac{1}{|z \pm n | ^2} \leq \frac{1}{\delta_n ^2 n^2} < \frac{1}{4n^2}. $$ By the Weierstrass M-test, $g$ converges absolutely uniformly on $K$. Since each term is analytic on $K$, we conclude that the series converges to an analytic function on $\mathbb{C} \backslash \mathbb{Z}$.

Clearly the only poles of $g(z)$ are at $\pi n$ for each integer $n$, with corresponding principal part $\frac{1}{(z-\pi n)^2}$.

For each integer $n$, we have $\sin^2 (\pi n) = \frac{d}{dz} \sin^2(\pi n) = 0$ and $\frac{d^2}{dz^z} \sin^2 (\pi n) \neq 0$, so $f(z) = 1/ \sin^2 (z)$ also has a pole of order two at each integer multiple of $\pi$, and no other poles.
Furthermore, the principal part of $f(z)$ is, using the Laurent formulas and contour integration, equal to $\displaystyle \frac{1}{(z-n\pi)^2}$.

Thus $h(z) := f(z)-g(z)$ has removable singularities at the points $n\pi$.

Note that both $f$ and $g$ are periodic with period $\pi$. That is, $$ f(z) = f(z +\pi) \quad \text{ and } \quad g(z) = g(z+\pi) \quad \text{ for all } z \in \mathbb{C}\backslash\mathbb{Z}. $$ Thus, since $h$ is bounded on the square $\{ z: |\operatorname{Re} z| < \pi, |\operatorname{Im} z |< \pi \}$ and periodic, we may conclude that $h$ is bounded on the set $\{ z: |\operatorname{Im} z |< \pi \}$. To show boundedness on the entire plane, we show that it holds on the vertical strip (with center removed) $$ S = \{ z: 0 \leq \operatorname{Re} z \leq \pi, | \operatorname{Im} z | \geq \pi \}. $$ For $z$ in $S$, \begin{align*} \sum_{n= -\infty} ^ {+ \infty} |z-\pi n | ^{-2} &= \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(x-\pi n)^2 + y^2} &\leq \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi (n-1))^2 +y^2} \end{align*} \begin{align*} & = \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2} &< 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2} &\leq 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +\pi ^2}. \end{align*} Thus $g$ is bounded on the set. It is easy to show that $f$ is also bounded on $S$. Thus the difference $f-g$ is also bounded. By the periodicity of $h$, the function is bounded on the entire plane, and is constant by Liouville's theorem.

I'm wondering if my reasoning so far is valid; and if so, how to show that the relevant constant is in fact zero.

Thanks.

share|improve this question

2 Answers 2

up vote 0 down vote accepted

A discussion on the relevant constant:

Consider $z=it$.It is easy to figure out that $f(z)$ tends to zero when $t$ tends to infinity.So we just need to derive $\lim_{t\rightarrow\infty}g(it)=0$.

$\eqalign{ |g(it)|\leq\frac{1}{t^2}+\sum_{n=1}^{\infty}\frac{1}{t^2+\pi^2n^2} } $

The first term tends to zero when $t$ tends to infinity.

If $t\geq\pi^2N^2$ for some integer $N$ then $t^2+\pi^2n^2\geq\pi^2n^2t$ for all $n\leq N$.

Hence $\eqalign{ |g(it)|\leq\frac{1}{t^2}+\frac{1}{t}\sum_{n=1}^{N}\frac{1}{\pi^2n^2}+\sum_{n=N+1}^{\infty}\frac{1}{\pi^2n^2} } $

Let $N$ tends to infinity and the result follows.

share|improve this answer

Equation $(8)$ from this answer says $$ \begin{align} \pi\cot(\pi z) &=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\\ &=\frac1z+\sum_{k=1}^\infty\left(\frac{1}{z-k}+\frac{1}{z+k}\right)\tag{1} \end{align} $$ The convergence is uniform on compact subsets of $\mathbb{C}$, so we can differentiate to get $$ -\pi^2\csc^2(\pi z)=-\frac{1}{z^2}-\sum_{k=1}^\infty\left(\frac{1}{(z-k)^2}+\frac{1}{(z+k)^2}\right)\tag{2} $$ A wee bit o' manipulation, and a change of variables, gives us $$ \csc^2(z)=\sum_{k=-\infty}^\infty\frac{1}{(z-k\pi)^2}\tag{3} $$

share|improve this answer
    
@ec92: You might also be able to adapt that argument to your problem directly. –  robjohn Jul 2 '12 at 23:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.