Question

3. Use pi/sin(pi.z) to find the partial fraction development of 1 / (cos pi* z) and show that it leads to pi/4=1- 1/3 + 1/5 - 1/7 +... .

Answer 1

$\frac{1}{sinz}=cotz+tan(z/2)$

$\pi tan(\pi z)=8z\displaystyle{\sum^{\infin}_{n=0}}\frac{1}{(2n+1)^2-4z^2}$

$\pi cot(\pi z)=\frac{1}{z}+2z\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2-n^2}$

$\frac{\pi}{sin(\pi z)}=\frac{1}{z}+2z[\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2-n^2}-\displaystyle{\sum^{\infin}_{n=0}}\frac{1}{z^2-(2n+1)^2}]=$

$=\frac{1}{z}+2z[\frac{1}{z^2-1^2}+\frac{1}{z^2-2^2}+\frac{1}{z^2-3^2}+...-\frac{2}{z^2-1^2}-\frac{2}{z^2-1^2}-\frac{2}{z^2-1^2}-...]=$

$=\frac{1}{z}+\displaystyle{\sum^{\infin}_{n=1}}(-1)^{n-1}\frac{2z}{n^2-z^2}$

$cos(\pi z)=sin(\pi(1/2- z))$

$\frac{1}{cos(\pi z)}=\pi [\frac{2}{1-2z}+(\frac{2}{1+2z}-\frac{2}{3-2z})-(\frac{2}{3+2z}-\frac{2}{5-2z})+...]=$

$=\pi [\frac{4\cdot1}{1^2-4z^2}-\frac{4\cdot3}{3^2-4z^2}+\frac{4\cdot5}{5^2-4z^2}-...]$

$=4\pi \displaystyle{\sum^{\infin}_{n=0}}(-1)^{n}\frac{2n+1}{(2n+1)^2-4z^2}$

if z = 0, tnen:

$\pi/4=1-1/3+1/5-1/7+...$