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sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1356270940251:dw

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge \frac{\pi^2}{6}\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1which method do u want? i know the one using Fourier Transform...

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Any easy method.... I havent learn Fourier Transform

KorcanKanoglu
 2 years ago
Best ResponseYou've already chosen the best response.0btw, you can write your question with the equation buttom, then you can copy the script and paste it to the question box, it will emerge as an equation :)

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0U can do it using Fourier Transform .... I will try to catch it

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1Fourier Series expansion for f(x) in [0,2pi] \(f(x)=a_0/2+\sum \limits_{n=1}^\infty a_n\cos nx+\sum \limits_{n=1}^\infty b_n\sin nx\) \[a_0=(1/\pi)\int \limits_0^{2\pi}f(x)dx\] \[a_n=(1/\pi)\int \limits_0^{2\pi}f(x)\cos nxdx\] \[a_0=(1/\pi)\int \limits_0^{2\pi}f(x)\sin nxdx\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1here, use \(f(x)=(\frac{\pix}{2})^2\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1and find Fourier series Expansion of f(x) u should get, bn=0 an=1/n^2

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1sorry, the last one is bn \[b_n=(1/\pi)\int \limits_0^{2\pi}f(x)\sin nxdx\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1this really is the longer way to do it...

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Thanx for the reply will see it later
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