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Fourier series of the periodic function (period 2π) defined by

$$f(x) = \begin{cases} 0, -p < x < 0\\x, 0 < x < p \end{cases} \text { is }\\ \frac{\pi}{4} + \sum \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]$$

But putting $x = \pi$, we get the sum of the series

$$ 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \text { is }$$ 

  1. $\frac{{\pi }^2 }{4}$
  2. $\frac{{\pi }^2 }{6}$
  3. $\frac{{\pi }^2 }{8}$
  4. $\frac{{\pi }^2 }{12}$
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out of syllabus now.
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1 Answer

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First of all, Fourier series of given function is calculated wrong here.

Refer This :


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Why can't we answer by seeing series instead of solving
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Answer:

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