Question

The Fourier series of the function,

f(x) = 0, -π < x ≤ 0

= π - x, 0 < x < π

in the interval [= π, π] is

\(f\left( x \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos x}}{1^2} + \frac{{\cos 3x}}{{{3^2}}} \ldots \ldots \ldots } \right] + \left[ {\frac{{\sin x}}{1} + \frac{{\sin 2x}}{2} + \frac{{\sin 3x}}{3} + \ldots \ldots \ldots } \right]\)

The convergence of the above Fourier series at x = 0 gives

1. \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^2}}} = \frac{{{\pi ^2}}}{6}\)
2. \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^2}}} = \frac{{{\pi ^2}}}{{12}}\)
3. \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\)
4. \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}} = \frac{\pi }{4}\)

Answer 1

Correct Answer - Option 3 : \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\)

The given fourier series is,

\(f\left( x \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos x}}{1^2} + \frac{{\cos 3x}}{{{3^2}}} \ldots \ldots \ldots } \right] + \left[ {\frac{{\sin x}}{1} + \frac{{\sin 2x}}{2} + \frac{{\sin 3x}}{3} + \ldots \ldots \ldots } \right]\)

\(f\left( 0 \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{1}}{1^2} + \frac{{1}}{{{3^2}}} + \frac{{1}}{{{5^2}}} \ldots \ldots \ldots } \right]\)

The convergence of f(x) at x = 0 is valid if

\(f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}\)

\(f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}\)

where \(f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to 0\;} \left( {\pi - x} \right) = \pi\)

\(f\left( 0 \right) = \frac{\pi}{2}\)

\(\frac{\pi }{2} = \frac{\pi }{4} + \frac{2 }{\pi} \left[ {\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \ldots } \right]\)

\( \frac{1}{1} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \ldots =\frac{{{\pi^2}}}{8}\)

\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{{\left( {2n - 1} \right)}^2}}} = \frac{\pi^2 }{8}\)