Understand the concepts of Fourier series and determine Fourier expansions of periodic functions over full and half-range intervals.
A function \( f(x) \) is periodic with period \( T > 0 \) if:
The smallest positive value of \( T \) is called the fundamental period.
Joseph Fourier (1807) discovered that any reasonably well-behaved periodic function can be expressed as an infinite sum of sines and cosines. This is the foundation of Fourier analysis and has enormous applications in signal processing, heat conduction, vibration analysis, and electrical engineering.
The trigonometric functions satisfy these key orthogonality relations over \( (0, 2\pi) \) or \( (-\pi, \pi) \):
A function \( f(x) \) defined on \( (a, b) \) can be expanded as a Fourier series if it satisfies Dirichlet's conditions:
At a point of discontinuity \( x = c \), the Fourier series converges to the average of the left-hand and right-hand limits:
where \( f(c^-) \) = limit from left, \( f(c^+) \) = limit from right.
If \( f(x) = x \) on \( (-\pi, \pi) \), then at \( x = \pi \) (a discontinuity in the periodic extension), the Fourier series converges to:
\[ \frac{f(\pi^-) + f(\pi^+)}{2} = \frac{\pi + (-\pi)}{2} = 0 \]If \( f(x) \) is periodic with period \( 2\pi \) and defined on \( (0, 2\pi) \):
where \( n = 1, 2, 3, \ldots \)
If \( f(x) \) is periodic with period \( 2\pi \) and defined on \( (-\pi, \pi) \):
| Integral | Result |
|---|---|
| \( \displaystyle\int_0^{2\pi} \cos nx\,dx \) | \( 0 \) |
| \( \displaystyle\int_0^{2\pi} \sin nx\,dx \) | \( 0 \) |
| \( \displaystyle\int_0^{2\pi} x\cos nx\,dx \) | \( 0 \) |
| \( \displaystyle\int_0^{2\pi} x\sin nx\,dx \) | \( -\dfrac{2\pi}{n} \) |
| \( \displaystyle\int_0^{2\pi} x^2\cos nx\,dx \) | \( \dfrac{4\pi}{n^2} \) |
| \( \displaystyle\int_0^{2\pi} x^2\sin nx\,dx \) | \( -\dfrac{4\pi^2}{n} \) |
Problem: Find the Fourier series of \( f(x) = x \) on \( (0, 2\pi) \).
Solution:
Step 1: \( a_0 \)
\[ a_0 = \frac{1}{2\pi}\int_0^{2\pi} x\,dx = \frac{1}{2\pi}\cdot\frac{(2\pi)^2}{2} = \pi \]Step 2: \( a_n \)
\[ a_n = \frac{1}{\pi}\int_0^{2\pi} x\cos nx\,dx \]Using integration by parts: \( u = x \), \( dv = \cos nx\,dx \):
\[ = \frac{1}{\pi}\left[\frac{x\sin nx}{n} + \frac{\cos nx}{n^2}\right]_0^{2\pi} = \frac{1}{\pi}\left[0 + \frac{1 - 1}{n^2}\right] = 0 \]Step 3: \( b_n \)
\[ b_n = \frac{1}{\pi}\int_0^{2\pi} x\sin nx\,dx \] \[ = \frac{1}{\pi}\left[-\frac{x\cos nx}{n} + \frac{\sin nx}{n^2}\right]_0^{2\pi} = \frac{1}{\pi}\left[-\frac{2\pi}{n}\right] = -\frac{2}{n} \]Step 4: Fourier Series
\[ \boxed{x = \pi - 2\sum_{n=1}^{\infty}\frac{\sin nx}{n} = \pi - 2\left(\sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + \cdots\right)} \]Problem: Find the Fourier series of \( f(x) = x^2 \) on \( (-\pi, \pi) \).
Solution:
Step 1: \( a_0 \) (\( f \) is even, so integrals simplify)
\[ a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{1}{\pi}\int_0^{\pi} x^2\,dx = \frac{1}{\pi}\cdot\frac{\pi^3}{3} = \frac{\pi^2}{3} \]Step 2: \( a_n \) (even function)
\[ a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\cos nx\,dx = \frac{2}{\pi}\int_0^{\pi} x^2\cos nx\,dx \]Using integration by parts twice:
\[ = \frac{2}{\pi}\left[\frac{2x\cos nx}{n^2} + \frac{(n^2 x^2 - 2)\sin nx}{n^3}\right]_0^{\pi} = \frac{4(-1)^n}{n^2} \]Step 3: \( b_n = 0 \) (odd integrand × even function = odd, integrates to 0)
Step 4: Fourier Series
\[ \boxed{x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx} \] \[ = \frac{\pi^2}{3} - 4\left(\cos x - \frac{\cos 2x}{2^2} + \frac{\cos 3x}{3^2} - \cdots\right) \]Problem: Find the Fourier series of \( f(x) = \begin{cases} 0 & -\pi < x < 0 \\ x & 0 < x < \pi \end{cases} \)
Solution:
Step 1: \( a_0 \)
\[ a_0 = \frac{1}{2\pi}\int_{-\pi}^{0} 0\,dx + \frac{1}{2\pi}\int_0^{\pi} x\,dx = \frac{1}{2\pi}\cdot\frac{\pi^2}{2} = \frac{\pi}{4} \]Step 2: \( a_n \)
\[ a_n = \frac{1}{\pi}\int_0^{\pi} x\cos nx\,dx = \frac{1}{\pi}\left[\frac{x\sin nx}{n} + \frac{\cos nx}{n^2}\right]_0^{\pi} = \frac{(-1)^n - 1}{\pi n^2} \]So \( a_n = \begin{cases} 0 & n \text{ even} \\ -\dfrac{2}{\pi n^2} & n \text{ odd} \end{cases} \)
Step 3: \( b_n \)
\[ b_n = \frac{1}{\pi}\int_0^{\pi} x\sin nx\,dx = \frac{1}{\pi}\left[-\frac{x\cos nx}{n} + \frac{\sin nx}{n^2}\right]_0^{\pi} = -\frac{(-1)^n}{n} = \frac{(-1)^{n+1}}{n} \]Step 4: Fourier Series
\[ f(x) = \frac{\pi}{4} - \frac{2}{\pi}\sum_{\substack{n=1\\n \text{ odd}}}^{\infty}\frac{\cos nx}{n^2} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx \]For a function with period \( 2l \) defined on \( (0, 2l) \):
Problem: Find the Fourier series of \( f(x) = x \) on \( (0, 2l) \).
Solution:
Step 1: \( a_0 \)
\[ a_0 = \frac{1}{2l}\int_0^{2l} x\,dx = \frac{1}{2l}\cdot\frac{(2l)^2}{2} = l \]Step 2: \( a_n \)
\[ a_n = \frac{1}{l}\int_0^{2l} x\cos\frac{n\pi x}{l}\,dx \]Using integration by parts with \( u = x \), \( dv = \cos\frac{n\pi x}{l}\,dx \):
\[ = \frac{1}{l}\left[\frac{l x\sin(n\pi x/l)}{n\pi} + \frac{l^2\cos(n\pi x/l)}{n^2\pi^2}\right]_0^{2l} = \frac{1}{l}\cdot\frac{l^2(1 - 1)}{n^2\pi^2} = 0 \]Step 3: \( b_n \)
\[ b_n = \frac{1}{l}\int_0^{2l} x\sin\frac{n\pi x}{l}\,dx \] \[ = \frac{1}{l}\left[-\frac{l x\cos(n\pi x/l)}{n\pi} + \frac{l^2\sin(n\pi x/l)}{n^2\pi^2}\right]_0^{2l} = \frac{1}{l}\left[-\frac{2l^2}{n\pi}\right] = -\frac{2l}{n\pi} \]Step 4: Fourier Series
\[ \boxed{x = l - \frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin\frac{n\pi x}{l}} \]Problem: Find the Fourier series of \( f(x) = x^2 \) on \( (-l, l) \).
Solution:
Since \( f \) is even, \( b_n = 0 \).
Step 1: \( a_0 \)
\[ a_0 = \frac{1}{2l}\int_{-l}^{l} x^2\,dx = \frac{1}{l}\int_0^{l} x^2\,dx = \frac{l^2}{3} \]Step 2: \( a_n \)
\[ a_n = \frac{2}{l}\int_0^{l} x^2\cos\frac{n\pi x}{l}\,dx \]Using integration by parts twice:
\[ a_n = \frac{4l^2(-1)^n}{n^2\pi^2} \]Step 3: Fourier Series
\[ \boxed{x^2 = \frac{l^2}{3} + \frac{4l^2}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos\frac{n\pi x}{l}} \]Problem: Find the Fourier series of \( f(x) = x \) on \( (-\pi, \pi) \).
Solution:
Since \( f \) is odd, \( a_0 = 0 \), \( a_n = 0 \).
\( b_n \):
\[ b_n = \frac{2}{\pi}\int_0^{\pi} x\sin nx\,dx = \frac{2}{\pi}\left[-\frac{x\cos nx}{n} + \frac{\sin nx}{n^2}\right]_0^{\pi} = \frac{2(-1)^{n+1}}{n} \]Fourier Series:
\[ \boxed{x = 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx = 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right)} \]| Type | Condition | Examples |
|---|---|---|
| Even | \( f(-x) = f(x) \) | \( x^2 \), \( \cos x \), \( |x| \) |
| Odd | \( f(-x) = -f(x) \) | \( x \), \( x^3 \), \( \sin x \) |
If \( f(x) \) is even on \( (-l, l) \):
(Only cosine terms — cosine series)
If \( f(x) \) is odd on \( (-l, l) \):
(Only sine terms — sine series)
| Function Type | \( a_0 \) | \( a_n \) | \( b_n \) | Series Form |
|---|---|---|---|---|
| Even | \( \frac{1}{l}\int_0^l f\,dx \) | \( \frac{2}{l}\int_0^l f\cos\frac{n\pi x}{l}\,dx \) | 0 | Cosine only |
| Odd | 0 | 0 | \( \frac{2}{l}\int_0^l f\sin\frac{n\pi x}{l}\,dx \) | Sine only |
| General | \( \frac{1}{2l}\int_{-l}^l f\,dx \) | \( \frac{1}{l}\int_{-l}^l f\cos\frac{n\pi x}{l}\,dx \) | \( \frac{1}{l}\int_{-l}^l f\sin\frac{n\pi x}{l}\,dx \) | Both |
Problem: Find the Fourier series of \( f(x) = x^2 \) on \( (-\pi, \pi) \) using even symmetry.
Solution:
\( f \) is even → \( b_n = 0 \)
\[ a_0 = \frac{1}{\pi}\int_0^{\pi} x^2\,dx = \frac{\pi^2}{3} \] \[ a_n = \frac{2}{\pi}\int_0^{\pi} x^2\cos nx\,dx = \frac{4(-1)^n}{n^2} \] \[ \boxed{x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx} \]Problem: Find the Fourier series of \( f(x) = x^3 \) on \( (-\pi, \pi) \).
Solution:
\( f \) is odd → \( a_0 = 0 \), \( a_n = 0 \)
\[ b_n = \frac{2}{\pi}\int_0^{\pi} x^3\sin nx\,dx \]Using integration by parts three times:
\[ = \frac{2}{\pi}\left[-\frac{\pi^3\cos n\pi}{n} + \frac{3\pi^2\sin n\pi}{n^2} + \frac{6\pi\cos n\pi}{n^3} - \frac{6\sin n\pi}{n^4}\right]_0^{\pi} \] \[ = \frac{2}{\pi}\left[\frac{(-1)^{n+1}\pi^3}{n} + \frac{6\pi(-1)^n}{n^3}\right] = 2(-1)^{n+1}\pi^2\left(\frac{1}{n} - \frac{6}{n^3\pi^2}\right) \]So the Fourier series is \( x^3 = \sum_{n=1}^{\infty} b_n \sin nx \).
Problem: Find the Fourier series of \( f(x) = e^x \) on \( (-\pi, \pi) \).
Solution:
Since \( f \) is neither even nor odd, both \( a_n \) and \( b_n \) are non-zero.
\[ a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^x\,dx = \frac{e^{\pi} - e^{-\pi}}{2\pi} = \frac{\sinh\pi}{\pi} \] \[ a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} e^x\cos nx\,dx = \frac{(-1)^n(e^{\pi} - e^{-\pi})}{\pi(1 + n^2)} = \frac{2(-1)^n\sinh\pi}{\pi(1+n^2)} \] \[ b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} e^x\sin nx\,dx = \frac{(-1)^{n+1}n(e^{\pi} - e^{-\pi})}{\pi(1+n^2)} = \frac{2(-1)^{n+1}n\sinh\pi}{\pi(1+n^2)} \]Therefore:
\[ e^x = \frac{\sinh\pi}{\pi} + \frac{2\sinh\pi}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n(\cos nx - n\sin nx)}{1+n^2} \]When a function \( f(x) \) is defined only on \( (0, l) \), we can extend it to \( (-l, l) \) in two ways:
If \( f(x) \) is defined on \( (0, l) \) and extended as an even function:
If \( f(x) \) is defined on \( (0, l) \) and extended as an odd function:
Problem: Find the half-range cosine series of \( f(x) = x \) on \( (0, \pi) \).
Solution:
Step 1: \( a_0 \)
\[ a_0 = \frac{1}{\pi}\int_0^{\pi} x\,dx = \frac{\pi}{2} \]Step 2: \( a_n \)
\[ a_n = \frac{2}{\pi}\int_0^{\pi} x\cos nx\,dx = \frac{2}{\pi}\left[\frac{x\sin nx}{n} + \frac{\cos nx}{n^2}\right]_0^{\pi} = \frac{2[(-1)^n - 1]}{\pi n^2} \]So \( a_n = \begin{cases} 0 & n \text{ even} \\ -\dfrac{4}{\pi n^2} & n \text{ odd} \end{cases} \)
Step 3: Series
\[ \boxed{x = \frac{\pi}{2} - \frac{4}{\pi}\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{\cos nx}{n^2}} \] \[ = \frac{\pi}{2} - \frac{4}{\pi}\left(\cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + \cdots\right) \]Problem: Find the half-range sine series of \( f(x) = x \) on \( (0, \pi) \).
Solution:
\[ b_n = \frac{2}{\pi}\int_0^{\pi} x\sin nx\,dx = \frac{2}{\pi}\left[-\frac{x\cos nx}{n} + \frac{\sin nx}{n^2}\right]_0^{\pi} = -\frac{2\cos n\pi}{n} = \frac{2(-1)^{n+1}}{n} \]Series:
\[ \boxed{x = 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx = 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right)} \]Problem: Find the half-range sine series of \( f(x) = 1 \) on \( (0, \pi) \).
Solution:
\[ b_n = \frac{2}{\pi}\int_0^{\pi} \sin nx\,dx = \frac{2}{\pi}\left[-\frac{\cos nx}{n}\right]_0^{\pi} = \frac{2(1 - (-1)^n)}{\pi n} \]So \( b_n = \begin{cases} 0 & n \text{ even} \\ \dfrac{4}{\pi n} & n \text{ odd} \end{cases} \)
Series:
\[ \boxed{1 = \frac{4}{\pi}\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{\sin nx}{n} = \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)} \]This is a classic result: \( 1 = \dfrac{4}{\pi}\left(\sin x + \dfrac{\sin 3x}{3} + \dfrac{\sin 5x}{5} + \cdots\right) \) for \( 0 < x < \pi \).
Problem: Find the half-range cosine series of \( f(x) = x \) on \( (0, l) \).
Solution:
Even extension → cosine series.
\[ a_0 = \frac{1}{l}\int_0^l x\,dx = \frac{l}{2} \] \[ a_n = \frac{2}{l}\int_0^l x\cos\frac{n\pi x}{l}\,dx = \frac{2}{l}\left[\frac{lx\sin(n\pi x/l)}{n\pi} + \frac{l^2\cos(n\pi x/l)}{n^2\pi^2}\right]_0^l \] \[ = \frac{2l[(-1)^n - 1]}{n^2\pi^2} = \begin{cases} 0 & n \text{ even} \\ -\dfrac{4l}{n^2\pi^2} & n \text{ odd} \end{cases} \]Series:
\[ \boxed{x = \frac{l}{2} - \frac{4l}{\pi^2}\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{1}{n^2}\cos\frac{n\pi x}{l}} \]Parseval's identity relates the integral of the square of a function to the sum of the squares of its Fourier coefficients. It's essentially the conservation of energy in the frequency domain.
Problem: Using the Fourier series \( x^2 = \dfrac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^2}\cos nx \) on \( (-\pi, \pi) \), evaluate \( \sum_{n=1}^{\infty}\dfrac{1}{n^2} \).
Solution:
Here \( a_0 = \dfrac{\pi^2}{3} \), \( a_n = \dfrac{4(-1)^n}{n^2} \), \( b_n = 0 \).
By Parseval's identity:
\[ \frac{1}{\pi}\int_{-\pi}^{\pi} x^4\,dx = 2\left(\frac{\pi^2}{3}\right)^2 + \sum_{n=1}^{\infty}\frac{16}{n^4} \]LHS: \( \dfrac{1}{\pi}\cdot\dfrac{2\pi^5}{5} = \dfrac{2\pi^4}{5} \)
So: \( \dfrac{2\pi^4}{5} = \dfrac{2\pi^4}{9} + 16\sum \dfrac{1}{n^4} \)
\[ 16\sum\frac{1}{n^4} = \frac{2\pi^4}{5} - \frac{2\pi^4}{9} = 2\pi^4\cdot\frac{4}{45} = \frac{8\pi^4}{45} \] \[ \boxed{\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}} \]Problem: Using \( x = 2\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}\sin nx \) on \( (-\pi, \pi) \), evaluate \( \sum_{n=1}^{\infty}\dfrac{1}{n^2} \).
Solution:
Here \( a_0 = 0 \), \( a_n = 0 \), \( b_n = \dfrac{2(-1)^{n+1}}{n} \).
\[ \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \sum_{n=1}^{\infty}\frac{4}{n^2} \]LHS: \( \dfrac{1}{\pi}\cdot\dfrac{2\pi^3}{3} = \dfrac{2\pi^2}{3} \)
\[ \frac{2\pi^2}{3} = 4\sum\frac{1}{n^2} \implies \boxed{\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}} \]Problem: If \( f(x) \) represents a periodic voltage signal, interpret Parseval's identity physically.
Solution:
For an electrical signal \( V(t) \), the quantity \( \frac{1}{T}\int_0^T V^2\,dt \) is the mean square voltage. Parseval's identity states that this equals the sum of the mean squares of the harmonic components:
\[ \frac{1}{T}\int_0^T V^2\,dt = 2a_0^2 + \sum_{n=1}^{\infty}(a_n^2 + b_n^2) \]This is the power conservation principle: the total power in a signal equals the sum of powers in its harmonic components.
| Field | Application |
|---|---|
| Signal Processing | Decomposing signals into frequency components; filtering, modulation, compression (MP3, JPEG) |
| Heat Conduction | Solving the heat equation with periodic boundary conditions |
| Vibration Analysis | Modal analysis of mechanical structures; resonance |
| Electrical Engineering | Analyzing AC circuits; harmonic distortion |
| Communication Systems | Bandwidth analysis, Fourier transform, spectrum |
| Image Processing | 2D Fourier transform for image filtering |
| Quantum Mechanics | Wave function decomposition |
| Acoustics | Sound spectrum analysis |
| Topic | Key Formula / Concept |
|---|---|
| Periodic Function | \( f(x + T) = f(x) \) |
| Fourier Series (Period 2π) | \( f(x) = a_0 + \sum(a_n\cos nx + b_n\sin nx) \) |
| \( a_0 \) on \( (0, 2\pi) \) | \( \dfrac{1}{2\pi}\int_0^{2\pi} f(x)\,dx \) |
| \( a_n \) on \( (0, 2\pi) \) | \( \dfrac{1}{\pi}\int_0^{2\pi} f(x)\cos nx\,dx \) |
| \( b_n \) on \( (0, 2\pi) \) | \( \dfrac{1}{\pi}\int_0^{2\pi} f(x)\sin nx\,dx \) |
| Fourier Series (Period 2l) | \( f(x) = a_0 + \sum\!\left(a_n\cos\frac{n\pi x}{l} + b_n\sin\frac{n\pi x}{l}\right) \) |
| \( a_0 \) on \( (-l, l) \) | \( \dfrac{1}{2l}\int_{-l}^{l} f(x)\,dx \) |
| \( a_n \) on \( (-l, l) \) | \( \dfrac{1}{l}\int_{-l}^{l} f(x)\cos\frac{n\pi x}{l}\,dx \) |
| \( b_n \) on \( (-l, l) \) | \( \dfrac{1}{l}\int_{-l}^{l} f(x)\sin\frac{n\pi x}{l}\,dx \) |
| Even Function | \( b_n = 0 \), only cosine terms |
| Odd Function | \( a_0 = a_n = 0 \), only sine terms |
| Half-Range Cosine | \( a_0 = \dfrac{1}{l}\int_0^l f\,dx \), \( a_n = \dfrac{2}{l}\int_0^l f\cos\frac{n\pi x}{l}\,dx \) |
| Half-Range Sine | \( b_n = \dfrac{2}{l}\int_0^l f\sin\frac{n\pi x}{l}\,dx \) |
| Discontinuity | Series converges to \( \dfrac{f(c^-) + f(c^+)}{2} \) |
| Parseval's Identity | \( \dfrac{1}{l}\int_{-l}^l f^2\,dx = 2a_0^2 + \sum(a_n^2 + b_n^2) \) |
| Function | Fourier Series | Interval |
|---|---|---|
| \( x \) | \( 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx \) | \( (-\pi, \pi) \) |
| \( x \) | \( \pi - 2\sum_{n=1}^{\infty}\frac{\sin nx}{n} \) | \( (0, 2\pi) \) |
| \( x^2 \) | \( \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx \) | \( (-\pi, \pi) \) |
| \( x^2 \) | \( \frac{4\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{\cos nx}{n^2} \) | \( (0, 2\pi) \) |
| \( |x| \) | \( \frac{\pi}{2} - \frac{4}{\pi}\sum_{\substack{n\text{ odd}}}\frac{\cos nx}{n^2} \) | \( (-\pi, \pi) \) |
| \( 1 \) | \( \frac{4}{\pi}\sum_{\substack{n\text{ odd}}}\frac{\sin nx}{n} \) | \( (0, \pi) \) |
Find the Fourier series of \( f(x) = x \) on \( (0, 2\pi) \).
Find the Fourier series of \( f(x) = x^2 \) on \( (-\pi, \pi) \).
Find the Fourier series of \( f(x) = x \) on \( (0, 2l) \).
Find the Fourier series of \( f(x) = \begin{cases} -k & -\pi < x < 0 \\ k & 0 < x < \pi \end{cases} \)
Find the half-range cosine series of \( f(x) = x \) on \( (0, \pi) \).
Find the half-range sine series of \( f(x) = 1 \) on \( (0, \pi) \).
Using Parseval's identity on the Fourier series of \( x^2 \) on \( (-\pi, \pi) \), find \( \sum_{n=1}^{\infty}\dfrac{1}{n^4} \).
State the value to which the Fourier series of \( f(x) = x \) on \( (-\pi, \pi) \) converges at \( x = \pi \).
Already worked out in Section III, Example 1:
\[ x = \pi - 2\sum_{n=1}^{\infty}\frac{\sin nx}{n} \]Already worked out in Section III, Example 2:
\[ x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx \]Already worked out in Section IV, Example 1:
\[ x = l - \frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin\frac{n\pi x}{l} \]\( f \) is odd → \( a_0 = a_n = 0 \)
\[ b_n = \frac{2}{\pi}\int_0^{\pi} k\sin nx\,dx = \frac{2k}{\pi}\left[-\frac{\cos nx}{n}\right]_0^{\pi} = \frac{2k(1 - (-1)^n)}{\pi n} \]So \( b_n = \begin{cases} 0 & n \text{ even} \\ \dfrac{4k}{\pi n} & n \text{ odd} \end{cases} \)
\[ \boxed{f(x) = \frac{4k}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)} \]Already worked out in Section VI, Example 1:
\[ x = \frac{\pi}{2} - \frac{4}{\pi}\left(\cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + \cdots\right) \]Already worked out in Section VI, Example 3:
\[ 1 = \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right) \]For \( x^2 = \dfrac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^2}\cos nx \):
\( a_0 = \dfrac{\pi^2}{3} \), \( a_n = \dfrac{4(-1)^n}{n^2} \), \( b_n = 0 \)
By Parseval's identity:
\[ \frac{1}{\pi}\int_{-\pi}^{\pi} x^4\,dx = 2\left(\frac{\pi^2}{3}\right)^2 + \sum_{n=1}^{\infty}\frac{16}{n^4} \] \[ \frac{2\pi^4}{5} = \frac{2\pi^4}{9} + 16\sum\frac{1}{n^4} \] \[ 16\sum\frac{1}{n^4} = \frac{2\pi^4}{5} - \frac{2\pi^4}{9} = \frac{8\pi^4}{45} \] \[ \boxed{\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}} \]At \( x = \pi \), the periodic extension of \( f(x) = x \) has a jump from \( \pi^- \) to \( -\pi^+ \).
\[ \text{Convergence value} = \frac{f(\pi^-) + f(\pi^+)}{2} = \frac{\pi + (-\pi)}{2} = \boxed{0} \]| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO6 Components | Covered in Section |
|---|---|
| Understand periodic functions | I |
| Apply Dirichlet's conditions | II |
| Find Fourier series over \( (0, 2\pi) \) | III |
| Find Fourier series over \( (-l, l) \) | IV |
| Use even/odd symmetry | V |
| Find half-range sine/cosine series | VI |
| Apply Parseval's identity | VII |
⭐ End of Unit VI Notes ⭐
Fourier Series and Periodic Functions
MTH165 – Mathematics for Engineers
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