Find fourier series for $f(x) =\dfrac {3x^2 6\pi x + 2\pi^2}{12}, (0,2\pi) $ Hence deduce that $\dfrac 1{1^2}+\dfrac 1{2^2}+\dfrac 1{3^2} + --- =\dfrac {\pi^2}6$

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written 8.2 years ago by

teamques10 ★ 69k

modified 2.8 years ago by

rahulsahu8516 • 30

$a_0 =\dfrac 1{2\pi}\int\limits_0^{2\pi} f(2) dx =\dfrac 1{2\pi} \int\limits_0^{2\pi} \dfrac {(3x^2-6\pi x + 2\pi^2)}{12} dx = \dfrac 1{24\pi}[x^3-3\pi x^2 + 2\pi x^2 ]_0^{2\pi}$

$=\dfrac 1{24\pi} [8\pi^3-12\pi^3+4\pi^3]=0 \\ an =\dfrac 1{\pi} \int\limits_0^{2\pi}f(x) \cos(nx) dx =\dfrac 1\pi \int\limits_0^{2\pi}\dfrac {(3x^2-\pi x +2\pi^2)}{12}\cos nx \space dx \\ \dfrac 1{12\pi}[ (3x^2-6\pi x +2\pi^2 ) \dfrac {8mnx}n -(6x-6\pi)(\dfrac {-\cos n x}{n^2})+ (6) (\dfrac {-\sin nx}{n^3})]_0^{2\pi} \\ =\dfrac 1{12\pi} [\dfrac {6\pi}{n^2}+6\pi/n^2] =\dfrac 1{n^2} \\ bn =\dfrac 1\pi \int\limits_0^{2\pi} f(x) \sin nx \space dx =\dfrac 1{12\pi} \int\limits_0^{2\pi} (3x^2-6\pi x +2\pi^2 ) \sin (nx) dx \\ =\dfrac 1{12\pi} [(3x^2-6\pi x+2\pi^2)(\dfrac {-\cos nx}n)-(6x-6\pi)(\dfrac {-\sin nx}{n^2})+(6) (\dfrac {\cos nx}{n^3})]_0^{2\pi} \\ =\dfrac 1{12\pi}[ (12\pi^2-12\pi^2+2\pi^2) (-\dfrac 1n)+ \dfrac 6{n^3}-(2\pi^2)(-\dfrac 1n)-\dfrac 6{n^3}]=0 \\ f(x) =\dfrac {(3x^2-6\pi x + 2\pi^2)}{12} =\sum\limits_{n=0}^\infty \dfrac 1{n^2}\cos (nx) \\ put \space x=0 \\ \dfrac {\pi^2}6 =\dfrac 1{1^2}+\dfrac 1{2^2}+\dfrac 1{3^2}----$

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