The Fourier series of the function, \[ array*20c f ( x ) = 0,& - < x 0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - x,&0 < x < array\] in the interval [ - ,\, ] is f ( x ) = /4 + 2/ [ x1^2 + 3x3^3 + \,... ] + [ x1…

The Fourier series of the function, \[ array*20c f ( x ) = 0,& - < x…

Engineering Maths · Q20

Transform Theory

Engineering and GATE · Engineering Maths · question 20

Q20

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

[- π, π]

f (x) = \frac{π}{4} + \frac{2}{π} [\frac{cos x}{1 ^{2}} + \frac{cos 3x}{3 ^{3}} + ...] + [\frac{sin x}{1} + \frac{sin 2x}{2} + \frac{sin 3x}{3} + ...]

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

n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6}

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n ^{2}} = \frac{π ^{2}}{12}

n = 1 \sum \infty \frac{1}{( 2n - 1 ) ^{2}} = \frac{π ^{2}}{8}

Answer

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{2n - 1} = \frac{π}{4}

Answer: Option C

Answer: Option C

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