Answer to Question #258283 in Complex Analysis for yuvasri

Question #258283

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 +... .


1
Expert's answer
2021-11-01T16:56:14-0400

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


πtan(πz)=8zn=01(2n+1)24z2\pi tan(\pi z)=8z\displaystyle{\sum^{\infin}_{n=0}}\frac{1}{(2n+1)^2-4z^2}


πcot(πz)=1z+2zn=11z2n2\pi cot(\pi z)=\frac{1}{z}+2z\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2-n^2}


πsin(πz)=1z+2z[n=11z2n2n=01z2(2n+1)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}]=


=1z+2z[1z212+1z222+1z232+...2z2122z2122z212...]==\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}-...]=


=1z+n=1(1)n12zn2z2=\frac{1}{z}+\displaystyle{\sum^{\infin}_{n=1}}(-1)^{n-1}\frac{2z}{n^2-z^2}


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


1cos(πz)=π[212z+(21+2z232z)(23+2z252z)+...]=\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})+...]=


=π[41124z243324z2+45524z2...]=\pi [\frac{4\cdot1}{1^2-4z^2}-\frac{4\cdot3}{3^2-4z^2}+\frac{4\cdot5}{5^2-4z^2}-...]


=4πn=0(1)n2n+1(2n+1)24z2=4\pi \displaystyle{\sum^{\infin}_{n=0}}(-1)^{n}\frac{2n+1}{(2n+1)^2-4z^2}


if z = 0, tnen:


π/4=11/3+1/51/7+...\pi/4=1-1/3+1/5-1/7+...


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