Here's the question you clicked on:
UnkleRhaukus
8+8/9+8/25+...=
8(1/1^2 + 1/3^2+1/5^2 +....)
use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html
or f(x) = x for 0<x<= pi/2 = pi-x for pi/2 <= x < pi find its half range cosine series, it comes out to be f(x) = pi/4 - (2/pi)[cos 2x/1^2 + cos6x/3^2 + cos 10x/5^2 +...] thn put x=pi/2
or simply f(x) = [(pi-x)/2 ]^2 in [0,2pi] find F.S expansion
2nd one is simpler i guess.
Define a function T(x) this way T(x) = arctan(2/pi)x for -pi/2<=x<pi/2 T(x) = -arctan(2/pi)x for pi/2<=x<3pi/2 \[ T (x) = {8 \over\pi^2 \ }\sum_{n=1,3,5 ..}{(-1)^{n-1 \over 2} \over n^2} \sin (n x) \\ T(\pi/2) = 1 \\ {8 \over \pi^2 } {(-1)^{n-1} \over (2n-1)^2} \sin((2n-1)\pi/2) = 1 \\ {8 \over \pi^2} ({1^2 + 1/3^2 + 1/5^2 + .. .})=1\]
was the the Q realting Fourier Transform ?? i thot it was....
i can do the fourier cosines series method i was wondering if there was another way ,
yeah, as i said the 2nd way i shown is simpler...u want the formulas ? or know it ?
i have worked it out
\[\large{\color{grey}{π^2}}\]
Plot[8/Pi^2 Sum[(-1)^(n - 1)/(2 n - 1)^2 Sin[(2 n - 1) x], {n, 1, 20}], {x, 0, 2 Pi}]
\[ T(x) = {2 \pi} \arcsin(\sin(x))\] http://www.wolframalpha.com/input/?i=Plot+2%2Fpi+sin^-1%28sin%28x%29%29
just for fun\[\zeta(2)=\frac{\pi^2}{6}\]\[\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\frac{1}{4}\zeta(2)+A\]where\[A=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}\]and \[A=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}\]so ur expression becomes\[8A=\pi^2\]
yep!! i realized that :)