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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.38(1/1^2 + 1/3^2+1/5^2 +....)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3or 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

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.32nd one is simpler i guess.

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3was the the Q realting Fourier Transform ?? i thot it was....

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i can do the fourier cosines series method i was wondering if there was another way ,

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i have worked it out

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large{\color{grey}{π^2}}\]

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

experimentX
 2 years ago
Best 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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0just 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\[8A=\pi^2\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3yep!! i realized that :)
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