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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 7.4 years ago by | modified 23 months ago by |
$$a_0 =\dfrac 1{2\pi}\int\limits_0^{2\pi} f(2) dx =\dfrac 1{2\pi} \int\limits_0^{2\pi} \dfrac {(3x^2-6\pi x + 2\pi^2)}{12} dx = \dfrac 1{24\pi}[x^3-3\pi x^2 + 2\pi x^2 ]_0^{2\pi}$$
$=\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 …