8+8/9+8/25+...=

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8(1/1^2 + 1/3^2+1/5^2 +....)
pi^2/8, right ?
the eights cancel

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oh yeah...
use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html
1 Attachment
or f(x) = x for 0
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
was the the Q realting Fourier Transform ?? i thot it was....
*this
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
ok...
\[\large{\color{grey}{π^2}}\]
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 :)

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