2

The question

Consider the representation: $$ \frac{π^2}{\sin^2(πz)} = \sum_{n∈ℤ}\frac{1}{(z+n)^2} \tag{0}$$ valid for all $z ∈ ℂ \setminus ℤ$. Deduce that the Taylor series of $z^2/\sin^2(z)$ for $|z|<π$ is: $$ \frac{z^2}{\sin^2{z}} = 1 + 2\sum_{m=1}^∞(2m-1) \frac{ζ(2m)}{π^{2m}}z^{2m} \tag{0.1}$$

Deduce also that: $$ ζ(2) = \sum_{n=1}\frac{1}{n^2} = \frac{π^2}{6}, \qquad ζ(4) = \sum_{n=1}^∞\frac{1}{n^4} = \frac{π^4}{90} $$

Finally explain why for all $m ≥ 1$, we have: $$ \frac{ζ(2m)}{π^{2m}} = \frac{1}{π^{2m}}\sum_{n=1}^∞\frac{1}{n^{2m}} $$

What I've tried

For the first and third question I have no idea of how to start. And for the second question I've been able to see that because of the sine infinite products expansion one can have:

$$ \sin(πz) = πz\prod_{n=1}^∞\left(1 - \frac{z^2}{n^2}\right) $$ $$ πz - \frac{π^3}{3!} + … = πz \left( 1 - \sum_{n=1}^∞\frac{z^2}{n^2} + … \right) $$ And then: $$ \frac{π^3}{3!} = π\sum_{n=1}^∞\frac{1}{n^2} $$ so necessarily we have $\frac{π^2}{6} = \sum_{n=1}^∞\frac{1}{n^2}$. But that calculus does not come from the expression given at the beginning of the question.

Any ideas for the 3 questions?

[Edit] - based on what Daniel Fischer♦ suggested

If we want to expand $\frac{z^2}{(z+n)^2}$ into a series, calculating its successive derivatives we get: $$\left( \frac{z^2}{(z+n)^2} \right)^{(j)}\Big|_{z=0} = (-1)^j\frac{(j-1)n^2}{n^j}\tag{1},\quad \text{with}\quad j>1$$ and $0$ when $j=0,1$.

Then one can express the fraction as follows: $$ \frac{z^2}{(z+n)^2} = \sum_{j=2}^∞(-1)^j\frac{(j-1)}{n^j}z^j \tag{2}$$ And summing over $n$ gives: $$ \sum_{j=2}^∞(-1)^j(j-1)z^j \sum_{n∈ℤ} \frac{1}{n^j} \tag{3}$$ If $j$ is odd the sum vanishes, and when it's even we get twice $ζ(2j)$. So $(3)$ equals to: $$ 2·\sum_{j=1}^∞(2j-1)z^{2j}ζ(2j) $$

Recalling $(0)$ we have: $$ \frac{(πz)^2}{\sin^2(πz)} = 2·\sum_{j=1}^∞(2j-1)z^{2j}ζ(2j) $$ so: $$ \frac{z}{\sin^2(z)} = 2·\sum_{j=1}^∞(2j-1)\frac{ζ(2j)}{π^{2j}}z^{2j} \tag{4}$$

But $(4)$ differs from $(0.1)$ in a unity. What am I doing wrong?

Jorge Martín Pérez
  • 373
  • 1
  • 10
  • 1
    See here for more details. – Dietrich Burde Apr 24 '16 at 18:46
  • 1
    You know how to expand $\frac{z^2}{(z+n)^2}$ into a series for fixed $n\in \mathbb{Z}\setminus {0}$. Do that, and sum over $n$. – Daniel Fischer Apr 24 '16 at 18:49
  • 1
    Are you sure that the representation of $\frac{\pi^2}{\sin^2(\pi z)}$ is correct? For real $z \in (-\epsilon, \epsilon)$ (with $\epsilon \in (0,1)$) the right-hand side converges (in particular it converges in $0$), but the left-hand side explodes in $0$. If your formula is correct, then it exhibits a very strange behaviour. What could cause it? – Alex M. Apr 24 '16 at 18:52
  • @AlexM. see that the statement is valid for $z ∈ ℂ \setminus ℤ$ – Jorge Martín Pérez Apr 24 '16 at 18:59
  • Ah, totally missed that, thank you. – Alex M. Apr 24 '16 at 19:01
  • @DanielFischer I've tried what you suggested (as shown in the edit). But I'm missing the +1 term, and don't know what I'm doing wrong. – Jorge Martín Pérez Apr 24 '16 at 22:14
  • The expansion $(2)$ works only for $n \neq 0$. For $m > 0$, adding $(2)$ for $n = m$ and for $n = -m$ gives $$\frac{z^2}{(z+m)^2} + \frac{z^2}{(z-m)^2} = 2\sum_{j = 1}^{\infty} (2j-1)\frac{z^{2j}}{m^{2j}}.$$ Summing over $m > 0$ gives $$\sum_{n\in \mathbb{Z}\setminus {0}} \frac{z^2}{(z+n)^2} = 2\sum_{j = 1}^{\infty} (2j-1)\zeta(2j) z^{2j}.$$ So on the right hand side of $(4)$, the term for $n = 0$, $\frac{z^2}{(z+0)^2} = 1$, is still missing. – Daniel Fischer Apr 24 '16 at 22:47

1 Answers1

1

Given: $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ we have: $$ \log\sin(\pi z)-\log(\pi z) = \sum_{n\geq 1}\log\left(1-\frac{z^2}{n^2}\right)\tag{2}$$ and by differentiating both sides twice we get: $$ -\frac{\pi^2}{\sin^2(\pi z)}+\frac{1}{z^2}=-\sum_{n\geq 1}\left[-\frac{1}{(z-n)^2}-\frac{1}{(z+n)^2}\right]\tag{3}$$ or: $$ \frac{\pi^2}{\sin^2(\pi z)}=\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}\tag{4} $$ where both sides are meromorphic functions with double poles at the integers, with residue zero and singular part behaving like $\frac{1}{z^2}$. So: $$ \frac{\pi^2 z^2}{\sin^2(\pi z)} = 1+\sum_{n\neq 0}\sum_{m\geq 1}(2m-1)n^{-2m}x^{2m}=1+2\sum_{m\geq 1}(2m-1)\zeta(2m) x^{2m}\tag{5} $$ where both sides are regular functions in a neighbourhood of zero. The values of $\zeta(2m)$ can so be computed through the Taylor series of the LHS.

Jack D'Aurizio
  • 353,855