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hartnn Group TitleBest ResponseYou've already chosen the best response.3
8(1/1^2 + 1/3^2+1/5^2 +....)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
pi^2/8, right ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
the eights cancel
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
or f(x) = x for 0<x<= pi/2 = pix 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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
or simply f(x) = [(pix)/2 ]^2 in [0,2pi] find F.S expansion
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
2nd one is simpler i guess.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
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 \[ T (x) = {8 \over\pi^2 \ }\sum_{n=1,3,5 ..}{(1)^{n1 \over 2} \over n^2} \sin (n x) \\ T(\pi/2) = 1 \\ {8 \over \pi^2 } {(1)^{n1} \over (2n1)^2} \sin((2n1)\pi/2) = 1 \\ {8 \over \pi^2} ({1^2 + 1/3^2 + 1/5^2 + .. .})=1\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
was the the Q realting Fourier Transform ?? i thot it was....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i can do the fourier cosines series method i was wondering if there was another way ,
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
yeah, as i said the 2nd way i shown is simpler...u want the formulas ? or know it ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i have worked it out
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\large{\color{grey}{π^2}}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
Plot[8/Pi^2 Sum[(1)^(n  1)/(2 n  1)^2 Sin[(2 n  1) x], {n, 1, 20}], {x, 0, 2 Pi}]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
\[ T(x) = {2 \pi} \arcsin(\sin(x))\] http://www.wolframalpha.com/input/?i=Plot+2%2Fpi+sin^1%28sin%28x%29%29
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
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\[8A=\pi^2\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
yep!! i realized that :)
 one year ago
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