Fourier Series of Mod x

The Fourier series of mod x in [-π, π] is given by |x| = π/2 -4/π [cosx/1² + cos3x/3² + cos5x/5² + …]. In this article, we learn how to find the Fourier series of mod x.

Fourier Series of |x|

Question: Find the Fourier series representation of f(x) = |x| in -π<x<π.

Answer:

The Fourier series representation of f(x) = |x| is as follows:

f(x) = |x| = a₀ + $\displaystyle \sum_{n=1}^\infty \Big[ a_n \cos(nx) + b_n \sin(nx) \Big]$

where the Fourier coefficients a₀, a_n, b_n are given by

a0 = $\dfrac{1}{2\pi} \displaystyle \int_{-\pi}^\pi f(x) ~ dx$ = $\dfrac{1}{2\pi} \displaystyle \int_{-\pi}^\pi |x| ~ dx$ = $\dfrac{2}{2\pi} \displaystyle \int_{0}^\pi |x| ~ dx$ as |x| is an even function. = $\dfrac{1}{\pi} \Big[ \dfrac{x^2}{2}\Big]_{0}^\pi$ = $\dfrac{\pi}{2}$.

Also,

an = $\dfrac{1}{\pi} \displaystyle \int_{-\pi}^\pi f(x) \cos(nx) ~ dx$ = $\dfrac{1}{\pi} \displaystyle \int_{-\pi}^\pi |x| \cos(nx) ~ dx$ = $\dfrac{2}{\pi} \displaystyle \int_{0}^\pi |x| \cos(nx) ~ dx$ as |x|cos(nx) is an even function. = $\dfrac{2}{\pi} \displaystyle \int_{0}^\pi x \cos(nx) ~ dx$ = $\dfrac{1}{\pi}$ $\Big[ \dfrac{x \sin(nx)}{n} – \displaystyle \int \dfrac{\sin(nx)}{n}\Big]_{0}^\pi$, obtained using integration by parts. = $\dfrac{1}{\pi}$ $\Big[ \dfrac{x \sin(nx)}{n} + \dfrac{\cos(nx)}{n^2}\Big]_{0}^\pi$ = $\dfrac{1}{\pi}$ $\Big[ \left( 0 + \dfrac{\cos(n\pi)}{n^2} \right)$ $- \left( 0 + \dfrac{1}{n^2} \right) \Big]$ as we know cos(nπ) = (-1)n and sin(nπ) = 0. = $\dfrac{2}{n^2 \pi} [(-1)^n -1]$.

And,

b_n = $\dfrac{1}{\pi} \displaystyle \int_{-\pi}^\pi f(x) \sin(nx) ~ dx$ = 0, since f(x) = |x| is an even function and sin(nx) is an odd function which makes the product f(x) sin(nx) odd.

Therefore, the Fourier series expansion of f(x) = |x| is given as follows:

f(x) = $\dfrac{\pi}{2}+$ $\displaystyle \sum_{n=1}^\infty \dfrac{2}{n^2 \pi} \big( (-1)^n – 1\big)$ cos(nx)

⇒ |x| = $\dfrac{\pi}{2}- \dfrac{2}{\pi} \Big[ \dfrac{2\cos x}{1^2} + 0$ $+\dfrac{2\cos 3x}{3^2} + 0 +\dfrac{2\cos 5x}{5^2}+\cdots \Big]$

⇒ |x| = $\dfrac{\pi}{2}-$ $ \dfrac{4}{\pi} \Big[ \dfrac{\cos x}{1^2} +\dfrac{\cos 3x}{3^2}$ $+\dfrac{\cos 5x}{5^2}+\cdots \Big]$

Also Study:

• Fourier Series: Definition, Formula, Solved Examples
• Fourier Series of Even and Odd Functions
• Periodic Functions
• Fourier Series of a Periodic Function
• Half range Fourier Sine & Cosine series

FAQs