Question

Q.1: Find the fourier series of f(x)=│x│ on [-π, π]
Q.2: Find the fourier series of f(x)=2-x2 on (-2<x<2)

Answer 1

Answer on Question #57660 – Math – Functional Analysis

Q.1: Find the Fourier series of $f(x) = |x|$ on $[- \pi, \pi]$

Q.2: Find the Fourier series of $f(x) = 2 - x^2$ on $(-2 < x < 2)$

Solution

1.

$$a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |x| \, dx = \frac{1}{\pi} \int_{0}^{\pi} x \, dx = \frac{\pi}{2}$$ $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \cos nx \, dx = \frac{2}{\pi} \int_{0}^{\pi} x \cos nx \, dx = \frac{2}{\pi} \left[ \frac{x \sin nx}{n} + \frac{\cos nx}{n^2} \right]_{0}^{\pi} = -2 \frac{[1 - (-1)^n]}{\pi n^2}.$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \sin nx \, dx$$

Before we dive into the integration observe that $|x|$ defines an even function while $\sin nx$ defines an odd function. Since the product of an even and an odd function is odd, it follows that $b_n = 0$ for all $n$. This simple observation saves us from having to perform a rather laborious integration.

The Fourier series is given by

$$f(x) \sim \frac{\pi}{2} - \sum_{n=1}^{\infty} 2 \frac{[1 - (-1)^n]}{\pi n^2} \cos nx.$$

2.

$$a_0 = \frac{1}{4} \int_{-2}^{2} 2 \, dx - \frac{1}{4} \int_{-2}^{2} x^2 \, dx = 2 \frac{1}{4} (x)_{-2}^{2} - \frac{1}{4} \left( \frac{1}{3} x^3 \right)_{-2}^{2} = 2 - \frac{1}{12} (8 - (-8)) = 2 - \frac{4}{3} = \frac{2}{3}.$$ $$a_n = \frac{1}{2} \int_{-2}^{2} 2 \cos \left( \frac{\pi nx}{2} \right) \, dx - \frac{1}{2} \int_{-2}^{2} x^2 \cos \left( \frac{\pi nx}{2} \right) \, dx = \frac{16}{\pi^2 n^2} (-1)^{n+1}.$$ $$\frac{1}{2} \int_{-2}^{2} 2 \cos \left( \frac{\pi nx}{2} \right) \, dx = \int_{-2}^{2} \cos \left( \frac{\pi nx}{2} \right) \, dx = \frac{2}{n\pi} \left( \sin \left( \frac{\pi nx}{2} \right) \right)_{-2}^{2} = \frac{2}{n\pi} (\sin(\pi n) - \sin(-\pi n)) = \frac{2}{n\pi} (0 - 0) = 0$$ $$\begin{array}{l} \frac{1}{2} \int_{-2}^{2} x^{2} \cos \left(\frac{\pi n x}{2}\right) dx = \frac{1}{2} \left(\frac{2}{n \pi} \left(x^{2} \sin \left(\frac{\pi n x}{2}\right)\right)_{-2}^{2} - \frac{4}{n \pi} \int_{-2}^{2} x \sin \left(\frac{\pi n x}{2}\right) dx\right) \\ = \frac{1}{2} \left(\frac{8}{n \pi} (\sin (\pi n) - \sin (-\pi n)) - \frac{4}{n \pi} \left(- \frac{2}{n \pi} \left(x \cos \left(\frac{\pi n x}{2}\right)\right)_{-2}^{2} + \frac{2}{n \pi} \int_{-2}^{2} \cos \left(\frac{\pi n x}{2}\right) dx\right)\right) \\ = \frac{1}{2} \left(\frac{8}{n \pi} (0 - 0) - \frac{4}{n \pi} \left(- \frac{2}{n \pi} \left(x \cos \left(\frac{\pi n x}{2}\right)\right)_{-2}^{2} + \frac{2}{n \pi} \int_{-2}^{2} \cos \left(\frac{\pi n x}{2}\right) dx\right)\right). \end{array}$$

We already know that

$$\int_{-2}^{2} \cos \left(\frac{\pi n x}{2}\right) dx = 0.$$ $$\frac{1}{2} \int_{-2}^{2} x^{2} \cos \left(\frac{\pi n x}{2}\right) dx = \frac{4}{\pi^{2} n^{2}} (2 \cos \pi n - (-2) \cos (-\pi n)) = \frac{16}{\pi^{2} n^{2}} (-1)^{n}.$$

Thus

$$a_{n} = 0 - \frac{16}{\pi^{2} n^{2}} (-1)^{n} = \frac{16}{\pi^{2} n^{2}} (-1)^{n+1}.$$ $$b_{n} = -\frac{1}{2} \int_{-2}^{2} (2 - x^{2}) \sin \left(\frac{\pi n x}{2}\right) dx$$

Before we dive into the integration observe that $2 - x^{2}$ defines an even function while $\sin \left(\frac{\pi n x}{2}\right)$ defines an odd function. Since the product of an even and an odd function is odd, it follows that $b_{n} = 0$ for all $n$. This simple observation saves us from having to perform a rather laborious integration.

The Fourier series is given by

$$f(x) \sim \frac{2}{3} + \sum_{n=1}^{\infty} \frac{16}{\pi^{2} n^{2}} (-1)^{n+1} \cos \left(\frac{\pi n x}{2}\right).$$

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