Half Range Fourier Series: Sine and Cosine Series

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A half range Fourier series of a function f(x) is a Fourier series of f(x) in the interval [0, L] although the function is defined for [-L, L]. Here, we learn about half range Fourier series with examples.

Previously, we have learnt about the Fourier series representation of a periodic function of period 2L or 2π. Now, the question is the following: If the function is given for only the half of the period, that is, f(x) is defined for [0, L] then what will be the Fourier series representation? If if exists then it is called the half range Fourier series expansion.

Types of Half Range Expansion

There are two types of half-range expansions:

  1. Fourier cosine series
  2. Fourier sine series.

Fourier Cosine Series

Let f(x) be a piecewise continuous function over [0, L]. Then the Fourier cosine series of f(x) is given by

f(x) = a0 + $\displaystyle \sum_{n=1}^\infty$ an cos$\big( \dfrac{n\pi x}{L}\big)$.

where the Fourier coefficients are as follows:

a0 = $\dfrac{1}{L} \int_0^L$ f(x) dx

an = $\dfrac{2}{L} \displaystyle \int_0^L$ f(x) cos$\big( \dfrac{n\pi x}{L}\big)$ dx, for n = 1, 2, 3, …

Fourier Sine Series

Let f(x) be a piecewise continuous function over [0, L]. Then the Fourier sine series of f(x) is given by

f(x) = $\displaystyle \sum_{n=1}^\infty$ bn sin$\big( \dfrac{n\pi x}{L}\big)$.

where

bn = $\dfrac{2}{L} \displaystyle \int_0^L$ f(x) sin$\big( \dfrac{n\pi x}{L}\big)$ dx, for n = 1, 2, 3, …

Solved Examples

Question 1: Find the Fourier cosine and sine series of the following function:

f(x) = k, 0≤x≤5.

Answer:

The Fourier cosine series of f(x) is given by

f(x) = a0 + $\displaystyle \sum_{n=1}^\infty$ an cos$\big( \dfrac{n\pi x}{L}\big)$

where L = 5 and the Fourier coefficients are as follows:

a0 = $\dfrac{1}{L} \displaystyle \int_0^L$ f(x) dx = $\dfrac{1}{5} \displaystyle \int_0^5$ k dx = $\dfrac{1}{5} \Big[ kx\Big]_0^5$ = k.

and

an = $\dfrac{2}{L} \displaystyle \int_0^L$ f(x) cos$\big( \dfrac{n\pi x}{L}\big)$ dx

= $\dfrac{2}{5} \displaystyle \int_0^5$ k cos$\big( \dfrac{n\pi x}{5}\big)$ dx

= $\dfrac{2k}{5} \Big[ \dfrac{\sin(\frac{n\pi x}{5})}{\frac{n\pi}{5}} \Big]_0^5$

= $\dfrac{2k}{n\pi} \Big( \sin n\pi – \sin 0 \Big)$

= 0.

Therefore, the Fourier cosine series of f(x) is f(x) = k.

On the other hand, the Fourier sine series of f(x) is given by

f(x) = $\displaystyle \sum_{n=1}^\infty$ bn sin$\big( \dfrac{n\pi x}{5}\big)$

where

bn = $\dfrac{2}{5} \displaystyle \int_0^5$ f(x) sin$\big( \dfrac{n\pi x}{5}\big)$ dx

= $\dfrac{2}{5} \displaystyle \int_0^5$ k sin$\big( \dfrac{n\pi x}{5}\big)$ dx

= $\dfrac{2k}{5} \Big[ – \dfrac{\cos(\frac{n\pi x}{5})}{\frac{n\pi}{5}} \Big]_0^5$

= $\dfrac{2k}{n\pi} \Big( \cos 0 -\cos n\pi \Big)$

= $\dfrac{2k}{n\pi} \Big[ 1 -(-1)^n \Big]$.

Therefore, the Fourier sine series of f(x) is f(x) = $\displaystyle \sum_{n=1}^\infty$ $\dfrac{2k}{n\pi} \Big[ 1 -(-1)^n \Big]$ sin$\big( \dfrac{n\pi x}{5}\big)$.

Also Read:

Practice Problems:

Find the Fourier cosine and sine series of the following functions.

1. f(x) = $\begin{cases}x, & \text{ if } 0< x \leq \frac{\pi}{2} \\ \pi-x, & \text{ if } \frac{\pi}{2} < x < \pi. \end{cases}$

2. f(x) = $\begin{cases}1, & \text{ if } 0\leq x \leq 1 \\ -1, & \text{ if } 1 < x < 2 \\ 0, & \text{ if } 2 \leq x < 3. \end{cases}$

FAQs

Q1: What is the half-range Fourier series?

Answer: If a function f(x) is defined over half of the range, say [0, L], instead of the full range [-L, L], then its Fourier series expansion on [0, L] is called the half-range Fourier series. This is a series of sine terms only or of cosine terms only, called the half range Fourier sine series or the half range Fourier cosine series respectively.

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