Can you show that $$ \sum_{n=-\infty}^\infty (x+n\pi)^{-2} = \sin^{-2}(x) $$ It is noted that if you make the substitution $x=x+2\pi$ the relation remains unchanged, but how can you show that relation?
Asked
Active
Viewed 194 times
1
-
1possible duplicate of Compact form of the series $\sum\limits_{n=-\infty}^{\infty} {1\over (x-n)^2}$ – Hans Lundmark Feb 28 '15 at 18:50
-
i think expression is wrong as for $x=0$ r.h.s tends to infinity but l.h.s is equal to $\frac{1}{3}$ – avz2611 Feb 28 '15 at 19:03
-
oh i forgot $n=0$ also comes – avz2611 Feb 28 '15 at 19:06
1 Answers
6
If we start from the Weierstrass product for the sine function: $$\frac{\sin(\pi z)}{\pi z} = \prod_{n=1}^{+\infty}\left(1-\frac{z^2}{n^2}\right)\tag{1}$$ we just have to consider $\frac{d^2}{dz^2}\log(\cdot)$ of both sides.
Jack D'Aurizio
- 353,855