Module III: Fourier Series
1. Dirichlet's Conditions
Conditions for Fourier Series to Exist
A function \( f(x) \) can be represented by a Fourier series if:
- \( f(x) \) is single-valued and periodic
- \( f(x) \) has a finite number of discontinuities in one period
- \( f(x) \) has a finite number of maxima and minima in one period
- \( \int |f(x)| \, dx \) over one period is finite (absolutely integrable)
At a point of discontinuity, Fourier series converges to \( \frac{f(x^+) + f(x^-)}{2} \)
2. Definition of Fourier Series
Period 2π
\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \]
Euler's Formulae (Period 2π):
\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \]
\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]
\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]
Period 2l
\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos\frac{n\pi x}{l} + b_n \sin\frac{n\pi x}{l} \right] \]
Euler's Formulae (Period 2l):
\[ a_0 = \frac{1}{l} \int_{-l}^{l} f(x) \, dx \]
\[ a_n = \frac{1}{l} \int_{-l}^{l} f(x) \cos\frac{n\pi x}{l} \, dx \]
\[ b_n = \frac{1}{l} \int_{-l}^{l} f(x) \sin\frac{n\pi x}{l} \, dx \]
3. Parseval's Identity
Period 2π:
\[ \frac{1}{\pi} \int_{-\pi}^{\pi} [f(x)]^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) \]
Period 2l:
\[ \frac{1}{l} \int_{-l}^{l} [f(x)]^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) \]
Parseval's Identity is used to find the sum of certain infinite series
4. Fourier Series of Period 2π Functions
Find Fourier series of \( f(x) = x \) in \( (-\pi, \pi) \)
- Check symmetry: \( f(-x) = -x = -f(x) \) → Odd function
So \( a_0 = 0 \) and \( a_n = 0 \) (only \( b_n \) exists)
- Find \( b_n \):
\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx = \frac{2}{\pi} \int_0^{\pi} x \sin(nx) \, dx \]
(since integrand is even)
- Using integration by parts: \( \int x \sin(nx) \, dx = -\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \)
\[ b_n = \frac{2}{\pi} \left[ -\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \right]_0^{\pi} \]
\[ = \frac{2}{\pi} \left[ -\frac{\pi\cos(n\pi)}{n} + 0 - 0 + 0 \right] = \frac{-2\cos(n\pi)}{n} = \frac{-2(-1)^n}{n} \]
- \( b_n = \frac{2(-1)^{n+1}}{n} \)
\[ f(x) = x = 2\left[ \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots \right] \]
Find Fourier series of \( f(x) = x^2 \) in \( (-\pi, \pi) \)
- Check symmetry: \( f(-x) = (-x)^2 = x^2 = f(x) \) → Even function
So \( b_n = 0 \) (only \( a_0 \) and \( a_n \) exist)
- Find \( a_0 \):
\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{2}{\pi} \int_0^{\pi} x^2 \, dx = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3} \]
- Find \( a_n \):
\[ a_n = \frac{2}{\pi} \int_0^{\pi} x^2 \cos(nx) \, dx \]
Using integration by parts twice:
\[ a_n = \frac{2}{\pi} \left[ \frac{x^2 \sin(nx)}{n} + \frac{2x \cos(nx)}{n^2} - \frac{2\sin(nx)}{n^3} \right]_0^{\pi} \]
\[ = \frac{2}{\pi} \cdot \frac{2\pi \cos(n\pi)}{n^2} = \frac{4(-1)^n}{n^2} \]
\[ f(x) = x^2 = \frac{\pi^2}{3} + 4\left[ -\frac{\cos x}{1^2} + \frac{\cos 2x}{2^2} - \frac{\cos 3x}{3^2} + \cdots \right] \]
5. Fourier Series of Period 2l Functions
Find Fourier series of \( f(x) = x \) in \( (-2, 2) \)
- Here \( 2l = 4 \), so \( l = 2 \)
- Check symmetry: \( f(-x) = -x = -f(x) \) → Odd function
So \( a_0 = 0 \) and \( a_n = 0 \)
- Find \( b_n \):
\[ b_n = \frac{1}{l} \int_{-l}^{l} f(x) \sin\frac{n\pi x}{l} \, dx = \frac{1}{2} \int_{-2}^{2} x \sin\frac{n\pi x}{2} \, dx \]
\[ = \frac{2}{2} \int_0^{2} x \sin\frac{n\pi x}{2} \, dx \] (integrand is even)
- Let \( u = \frac{n\pi x}{2} \), then by integration by parts:
\[ b_n = \left[ -\frac{2x\cos\frac{n\pi x}{2}}{n\pi} + \frac{4\sin\frac{n\pi x}{2}}{n^2\pi^2} \right]_0^{2} \]
\[ = -\frac{4\cos(n\pi)}{n\pi} = \frac{-4(-1)^n}{n\pi} = \frac{4(-1)^{n+1}}{n\pi} \]
\[ f(x) = \frac{4}{\pi}\left[ \sin\frac{\pi x}{2} - \frac{1}{2}\sin\frac{2\pi x}{2} + \frac{1}{3}\sin\frac{3\pi x}{2} - \cdots \right] \]
6. Fourier Series of Even and Odd Functions
| Property |
Even Function: \( f(-x) = f(x) \) |
Odd Function: \( f(-x) = -f(x) \) |
| \( a_0 \) |
\( \frac{2}{\pi} \int_0^{\pi} f(x) \, dx \) |
\( 0 \) |
| \( a_n \) |
\( \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx) \, dx \) |
\( 0 \) |
| \( b_n \) |
\( 0 \) |
\( \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \, dx \) |
| Series has |
Only cosine terms |
Only sine terms |
Find Fourier series of \( f(x) = |x| \) in \( (-\pi, \pi) \)
- Check symmetry: \( f(-x) = |-x| = |x| = f(x) \) → Even function
So \( b_n = 0 \), only cosine terms
- Find \( a_0 \):
\[ a_0 = \frac{2}{\pi} \int_0^{\pi} x \, dx = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi \]
- Find \( a_n \):
\[ a_n = \frac{2}{\pi} \int_0^{\pi} x \cos(nx) \, dx = \frac{2}{\pi} \left[ \frac{x\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \right]_0^{\pi} \]
\[ = \frac{2}{\pi} \left[ 0 + \frac{\cos(n\pi)}{n^2} - \frac{1}{n^2} \right] = \frac{2}{\pi n^2}[(-1)^n - 1] \]
For even \( n \): \( a_n = 0 \). For odd \( n \): \( a_n = \frac{-4}{\pi n^2} \)
\[ |x| = \frac{\pi}{2} - \frac{4}{\pi}\left[ \frac{\cos x}{1^2} + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + \cdots \right] \]
7. Half Range Fourier Series
When to Use
When \( f(x) \) is defined only in \( (0, l) \) or \( (0, \pi) \), we extend it as even or odd to get full period
Half Range Cosine Series (Even Extension)
In \( (0, \pi) \):
\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) \]
\[ a_0 = \frac{2}{\pi} \int_0^{\pi} f(x) \, dx, \quad a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx) \, dx \]
In \( (0, l) \):
\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\frac{n\pi x}{l} \]
\[ a_0 = \frac{2}{l} \int_0^{l} f(x) \, dx, \quad a_n = \frac{2}{l} \int_0^{l} f(x) \cos\frac{n\pi x}{l} \, dx \]
Find half range cosine series of \( f(x) = x \) in \( (0, \pi) \)
- Find \( a_0 \):
\[ a_0 = \frac{2}{\pi} \int_0^{\pi} x \, dx = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi \]
- Find \( a_n \):
\[ a_n = \frac{2}{\pi} \int_0^{\pi} x \cos(nx) \, dx = \frac{2}{\pi} \left[ \frac{x\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \right]_0^{\pi} \]
\[ = \frac{2}{\pi} \left[ \frac{\cos(n\pi) - 1}{n^2} \right] = \frac{2[(-1)^n - 1]}{\pi n^2} \]
For even \( n \): \( a_n = 0 \). For odd \( n \): \( a_n = \frac{-4}{\pi n^2} \)
\[ x = \frac{\pi}{2} - \frac{4}{\pi}\left[ \cos x + \frac{\cos 3x}{9} + \frac{\cos 5x}{25} + \cdots \right] \quad (0 < x < \pi) \]
Half Range Sine Series (Odd Extension)
In \( (0, \pi) \):
\[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \]
\[ b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \, dx \]
In \( (0, l) \):
\[ f(x) = \sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{l} \]
\[ b_n = \frac{2}{l} \int_0^{l} f(x) \sin\frac{n\pi x}{l} \, dx \]
Find half range sine series of \( f(x) = x \) in \( (0, \pi) \)
- Find \( b_n \):
\[ b_n = \frac{2}{\pi} \int_0^{\pi} x \sin(nx) \, dx \]
Using integration by parts:
\[ b_n = \frac{2}{\pi} \left[ -\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \right]_0^{\pi} \]
\[ = \frac{2}{\pi} \left[ -\frac{\pi\cos(n\pi)}{n} \right] = \frac{-2(-1)^n}{n} = \frac{2(-1)^{n+1}}{n} \]
\[ x = 2\left[ \sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \frac{\sin 4x}{4} + \cdots \right] \quad (0 < x < \pi) \]
Find half range sine series of \( f(x) = 1 \) in \( (0, 2) \)
- Here \( l = 2 \)
- Find \( b_n \):
\[ b_n = \frac{2}{2} \int_0^{2} 1 \cdot \sin\frac{n\pi x}{2} \, dx = \int_0^{2} \sin\frac{n\pi x}{2} \, dx \]
\[ = \left[ -\frac{2\cos\frac{n\pi x}{2}}{n\pi} \right]_0^{2} = -\frac{2}{n\pi}[\cos(n\pi) - 1] \]
\[ = \frac{2}{n\pi}[1 - (-1)^n] \]
For even \( n \): \( b_n = 0 \). For odd \( n \): \( b_n = \frac{4}{n\pi} \)
\[ 1 = \frac{4}{\pi}\left[ \sin\frac{\pi x}{2} + \frac{1}{3}\sin\frac{3\pi x}{2} + \frac{1}{5}\sin\frac{5\pi x}{2} + \cdots \right] \quad (0 < x < 2) \]
Quick Reference: All Formulas
| Type |
Formula |
| Period 2π Series |
\( f(x) = \frac{a_0}{2} + \sum (a_n\cos nx + b_n\sin nx) \) |
| Period 2l Series |
\( f(x) = \frac{a_0}{2} + \sum \left(a_n\cos\frac{n\pi x}{l} + b_n\sin\frac{n\pi x}{l}\right) \) |
| Parseval (2π) |
\( \frac{1}{\pi}\int_{-\pi}^{\pi}[f(x)]^2dx = \frac{a_0^2}{2} + \sum(a_n^2 + b_n^2) \) |
| Even function |
\( b_n = 0 \), use \( \frac{2}{\pi}\int_0^{\pi} \) |
| Odd function |
\( a_0 = a_n = 0 \), use \( \frac{2}{\pi}\int_0^{\pi} \) |
| Half Range Cosine |
\( f(x) = \frac{a_0}{2} + \sum a_n\cos\frac{n\pi x}{l} \) |
| Half Range Sine |
\( f(x) = \sum b_n\sin\frac{n\pi x}{l} \) |