The Fourier series of the function, [*20c f( x ) = 0, - < x 0 = - x,0 < x < ] in the interval [ - , ] is f( x ) = 4 + 2 [ cos x1^2 + cos 3x3^3 + ... ] + [ sin x1 + sin 2x2 + sin 3x3 + ... ] The convergence of the above Fourier series at x = 0 gives

limits_n = 1^fty 1( 2n - 1 )^2 = ^28

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Transform Theory

The Fourier series of the function,
\[\begin{array}{*{20}{c}} {{\text{f}}\left( {\text{x}} \right) = 0,}&{ - \pi < {\text{x}} \leqslant 0} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \pi - {\text{x,}}}&{0 < {\text{x}} < \pi } \end{array}\] in the interval $$\left[ { - \pi ,\,\pi } \right]$$ is $${\text{f}}\left( {\text{x}} \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos {\text{x}}}}{{{1^2}}} + \frac{{\cos {\text{3x}}}}{{{3^3}}} + \,...} \right] + \left[ {\frac{{\sin {\text{x}}}}{1} + \frac{{\sin {\text{2x}}}}{2} + \frac{{\sin {\text{3x}}}}{3} + \,...} \right]$$
The convergence of the above Fourier series at x = 0 gives

A) $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{6}} $$ B) $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{{12}}} $$ C) $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\left( {{\text{2n}} - 1} \right)}^2}}} = \frac{{{\pi ^2}}}{8}} $$ D) $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{\text{2n}} - 1}} = \frac{\pi }{4}} $$

Answer & Solution

Correct Answer: Option C

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