Unit-3: Sequences & Series + Fourier Series - Complete Notes
Unit-3: Sequence and Series + Fourier Series (Complete Theory + Formulas + Solved Examples)
1. Sequence
A sequence is a function whose domain is the set of natural numbers.Example: Arithmetic Sequence: \( a, a+d, a+2d, \dots \)
Geometric Sequence: \( a, ar, ar^2, \dots \)
2. Series
Sum of terms of a sequence. Infinite series: \( S = \sum_{n=1}^{\infty} u_n \)3. Convergence, Divergence and Oscillation of a Series
- Convergent: \( s_n \to s \) (finite limit)
- Divergent: \( s_n \to \pm \infty \)
- Oscillatory: no limit
(Note: Not sufficient)
4. Tests for Convergence of Positive Term Series
| Test Name | Statement | Conclusion |
|---|---|---|
| Comparison Test | If \( 0 \leq u_n \leq v_n \) and \( \sum v_n \) converges → \( \sum u_n \) converges If \( \sum v_n \) diverges → \( \sum u_n \) diverges |
Use standard series: p-series, geometric |
| Limit Comparison Test | If \( \lim_{n \to \infty} \frac{u_n}{v_n} = L > 0 \) (finite, non-zero) then both series behave same way |
Very useful |
| D’Alembert’s Ratio Test | \( \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right| = L \) L < 1 → convergent L > 1 → divergent L = 1 → inconclusive |
Best for exponential/factorial terms |
| Raabe’s Test | If ratio test fails (L=1), compute \( \lim_{n \to \infty} n \left( \frac{u_n}{u_{n+1}} - 1 \right) = R \) R > 1 → convergent R < 1 → divergent R = 1 → inconclusive |
Higher order test |
| Cauchy’s Root Test | \( \lim_{n \to \infty} \sqrt[n]{|u_n|} = L \) Same conclusions as ratio test |
Useful for \( u_n = a_n^{b_n} \) |
Example 1 (Ratio Test): Test convergence of \( \sum \frac{n!}{n^n} \)
\( \left| \frac{u_{n+1}}{u_n} \right| = \frac{(n+1)! /(n+1)^{n+1}}{n!/n^n} = \frac{n^n}{(n+1)^n} \cdot (n+1) = \left( \frac{n}{n+1} \right)^n (n+1) \to e^{-1} \cdot \infty ? \)
Wait: \( \left( \frac{n}{n+1} \right)^n = \left( \frac{1}{1 + 1/n} \right)^n \to e^{-1} \), then × (n+1) → ∞ → L = ∞ > 1
Divergent
\( \left| \frac{u_{n+1}}{u_n} \right| = \frac{(n+1)! /(n+1)^{n+1}}{n!/n^n} = \frac{n^n}{(n+1)^n} \cdot (n+1) = \left( \frac{n}{n+1} \right)^n (n+1) \to e^{-1} \cdot \infty ? \)
Wait: \( \left( \frac{n}{n+1} \right)^n = \left( \frac{1}{1 + 1/n} \right)^n \to e^{-1} \), then × (n+1) → ∞ → L = ∞ > 1
Divergent
Example 2 (Raabe’s Test): \( \sum \frac{1}{n (\ln n)^p} \) (p>1)
Ratio test gives L=1 → apply Raabe
\( \frac{u_n}{u_{n+1}} = \frac{(n+1)(\ln(n+1))^p}{n (\ln n)^p} \approx 1 + \frac{1}{n} \)
\( n \left( \frac{u_n}{u_{n+1}} - 1 \right) \approx n \cdot \frac{1}{n} = 1 \)
R = 1 → inconclusive (actually converges for p>1 by integral test)
But for p=1: diverges (harmonic)
Ratio test gives L=1 → apply Raabe
\( \frac{u_n}{u_{n+1}} = \frac{(n+1)(\ln(n+1))^p}{n (\ln n)^p} \approx 1 + \frac{1}{n} \)
\( n \left( \frac{u_n}{u_{n+1}} - 1 \right) \approx n \cdot \frac{1}{n} = 1 \)
R = 1 → inconclusive (actually converges for p>1 by integral test)
But for p=1: diverges (harmonic)
5. Fourier Series
Any periodic function f(x) with period 2L (or 2π) can be expressed as: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right) \]6. Fourier Coefficients (Period 2L)
\[ a_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx \] \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n\pi x}{L} \, dx \quad (n \geq 1) \] \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n\pi x}{L} \, dx \]7. Even and Odd Functions
- Even function: f(-x) = f(x) → only cosine terms (b_n = 0)
- Odd function: f(-x) = -f(x) → only sine terms (a_n = 0)
8. Half-Range Fourier Series (Interval 0 to L)
| Type | Expansion | Coefficients |
|---|---|---|
| Half-Range Cosine Series (even extension) |
\( f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos \frac{n\pi x}{L} \) | \( a_n = \frac{2}{L} \int_0^L f(x) \cos \frac{n\pi x}{L} \, dx \) |
| Half-Range Sine Series (odd extension) |
\( f(x) = \sum_{n=1}^{\infty} b_n \sin \frac{n\pi x}{L} \) | \( b_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \, dx \) |
Example 3: Find Fourier series of f(x) = x, -π < x < π (period 2π)
Odd function → only sine terms
\( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin nx \, dx = \frac{2}{\pi} \int_0^{\pi} x \sin nx \, dx \)
Integration by parts: \( \int x \sin nx \, dx = -\frac{x \cos nx}{n} + \frac{\sin nx}{n^2} \)
From 0 to π: \( b_n = \frac{2}{\pi} \left[ -\frac{\pi \cos n\pi}{n} \right] = \frac{2}{\pi} \left( -\frac{\pi (-1)^n}{n} \right) = \frac{2 (-1)^{n+1}}{n} \)
\( f(x) = \sum_{n=1}^{\infty} \frac{2 (-1)^{n+1}}{n} \sin nx \)
Odd function → only sine terms
\( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin nx \, dx = \frac{2}{\pi} \int_0^{\pi} x \sin nx \, dx \)
Integration by parts: \( \int x \sin nx \, dx = -\frac{x \cos nx}{n} + \frac{\sin nx}{n^2} \)
From 0 to π: \( b_n = \frac{2}{\pi} \left[ -\frac{\pi \cos n\pi}{n} \right] = \frac{2}{\pi} \left( -\frac{\pi (-1)^n}{n} \right) = \frac{2 (-1)^{n+1}}{n} \)
\( f(x) = \sum_{n=1}^{\infty} \frac{2 (-1)^{n+1}}{n} \sin nx \)
Example 4 (Half-Range Cosine): f(x) = x, 0 < x < π
Cosine series:
\( a_n = \frac{2}{\pi} \int_0^{\pi} x \cos nx \, dx \)
\( a_0 = \frac{2}{\pi} \int_0^{\pi} x \, dx = \pi \)
\( a_n = \frac{2}{\pi} \left[ \frac{x \sin nx}{n} \bigg|_0^{\pi} - \frac{1}{n} \int_0^{\pi} \sin nx \, dx \right] = \frac{2(-1)^n - 2}{n^2 \pi} \)
\( x = \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^n - 1)}{n^2 \pi} \cos nx \quad (0 < x < \pi) \)
Cosine series:
\( a_n = \frac{2}{\pi} \int_0^{\pi} x \cos nx \, dx \)
\( a_0 = \frac{2}{\pi} \int_0^{\pi} x \, dx = \pi \)
\( a_n = \frac{2}{\pi} \left[ \frac{x \sin nx}{n} \bigg|_0^{\pi} - \frac{1}{n} \int_0^{\pi} \sin nx \, dx \right] = \frac{2(-1)^n - 2}{n^2 \pi} \)
\( x = \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^n - 1)}{n^2 \pi} \cos nx \quad (0 < x < \pi) \)
Example 5 (Square Wave): f(x) =
\begin{cases}
-1 & -\pi < x < 0 \\
1 & 0 < x < \pi
\end{cases}
Odd function → sine series
\( b_n = \frac{2}{\pi} \int_0^{\pi} 1 \cdot \sin nx \, dx = \frac{2}{\pi} \left[ -\frac{\cos nx}{n} \right]_0^{\pi} = \frac{2}{\pi n} (1 - (-1)^n) \)
So b_n = 4/(πn) for n odd, 0 for n even
\( f(x) = \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{\sin (2k-1)x}{2k-1} \)
Odd function → sine series
\( b_n = \frac{2}{\pi} \int_0^{\pi} 1 \cdot \sin nx \, dx = \frac{2}{\pi} \left[ -\frac{\cos nx}{n} \right]_0^{\pi} = \frac{2}{\pi n} (1 - (-1)^n) \)
So b_n = 4/(πn) for n odd, 0 for n even
\( f(x) = \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{\sin (2k-1)x}{2k-1} \)
Dirichlet Conditions for Fourier Series Convergence
f(x) must be:- Periodic
- Single-valued
- Piecewise continuous with finite discontinuities
- At discontinuity: converges to average of left & right limits