Question

Obtain the Fourier series for the following periodic function which has a period of 2π:

f(x) = x^2 for -π ≤x ≤π

Answer 1

Answer on Question #66843 - Math - Differential Equations

Question

Obtain the Fourier series for the following periodic function which has a period of $2\pi$: $f(x) = x^2$ for $-\pi \leq x \leq \pi$.

Solution

For the periodic function $f(x) = x^2$ which has a period of $2\pi$, $-\pi \leq x \leq \pi$, the Fourier series is given by

$$f(x) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty} \left(a_n \cos nx + b_n \sin nx\right)$$

where

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx$$

are the Fourier coefficients.

Find $a_0$:

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx = \left. \frac{1}{\pi} \frac{x^3}{3} \right|_{-\pi}^{\pi} = \frac{1}{3\pi} \left(\pi^3 - (-\pi)^3\right) = \frac{2\pi^2}{3}$$

Find $a_n$:

$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos(nx) \, dx$$

here we used that the integrand, $x^2 \cos(nx)$, is even. To find $a_n$ we use integration by parts

$$\begin{aligned} a_n &= \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos(nx) \, dx = \frac{2}{\pi n} \int_{0}^{\pi} x^2 \, d(\sin(nx)) = \frac{2}{\pi n} x^2 \sin(nx) \Big|_{0}^{\pi} - \frac{2}{\pi n} \int_{0}^{\pi} 2x \sin(nx) \, dx = \\ &= \frac{2}{\pi n} \left(\pi^2 \sin(n\pi) - 0\right) - \frac{4}{\pi n^2} \int_{0}^{\pi} x \, d(-\cos(nx)) = 0 + \frac{4}{\pi n^2} \int_{0}^{\pi} x \, d(\cos(nx)) = \\ &= \frac{4}{\pi n^2} x \cos(nx) \Big|_{0}^{\pi} - \frac{4}{\pi n^2} \int_{0}^{\pi} \cos(nx) \, dx = \frac{4}{\pi n^2} (\pi \cos(n\pi) - 0) - \frac{4}{\pi n^3} \sin(nx) \Big|_{0}^{\pi} = \\ &= \frac{4}{n^2} \cos(n\pi) - \frac{4}{\pi n^3} (\sin(n\pi) - \sin 0) = \frac{4}{n^2} (-1)^n - 0 = (-1)^n \frac{4}{n^2} \end{aligned}$$

Find $b_n$:

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) \, dx = 0$$

since the integrand, $x^2 \sin(nx)$, is odd.

So

$$f(x) = x^2 = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx$$

Answer: Fourier series is

$$f(x) = x^2 = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx$$

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