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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
the eights cancel
 one year ago

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

hartnnBest 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

hartnnBest 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

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

experimentXBest 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

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

UnkleRhaukusBest 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

hartnnBest 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

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

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

experimentXBest 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

experimentXBest 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

mukushlaBest 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

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