IGNOU PHE 5 2026 - Obtain the Fourier series for the following periodic function which ha

Question

Obtain the Fourier series for the following periodic function which has a period of $2\pi$ :

$$f(x) = \pi^2 - x^2 \text{ for } -\pi < x < \pi$

29 Jan 2026

Answer :

Word Count : 440

The given function is ( f(x) = \pi^2 - x^2 ) with period ( 2\pi ). Since ( f(x) ) is even (( f(-x) = f(x) )), its Fourier series contains only cosine terms: [ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx), ] where [ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) , dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) \cos(nx) , dx. ] Step 1: Compute ( a_0 ) [ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) , dx = \frac{1}{\pi} \left[ \int_{-\pi}^{\pi} \pi^2 dx - \int_{-\pi}^{\pi} x^2 dx \right]. ] [ \int_{-\pi}^{\pi} \pi^2 dx = \pi^2 \cdot (2\pi) = 2\pi^3, ] [ \int_{-\pi}^{\pi} x^2 dx = 2 \int_{0}^{\pi} x^2 dx = 2 \cdot \frac{\pi^3}{3} = \frac{2\pi^3}{3}. ] [ a_0 = \frac{1}{\pi} \left( 2\pi^3 - \frac{2\pi^3}{3} \right) = \frac{1}{\pi} \cdot \frac{4\pi^3}{3} = \frac{4\pi^2}{3}. ] Step 2: Compute ( a_n ) for (

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