1. Fourier Series - Prelude and Trigonometry

Fourier Series - Prelude

$\bullet$ From Sinusoid ( $sin(kx)$ ) to Square Wave

1. Assume a function $y = f(x)$

defined on $(-\pi, \pi]$
discontinuous at $x = 0$

y = f(x) = \begin{cases} &-\frac{\pi}{4} &\quad (-\pi < x \le 0) \\ &\frac{\pi}{4} &\quad (0 < x \le \pi) \end{cases}

2. The Fourier Series of $sin(kx)$

\begin{align*} & y = f(x) = \sum_{k = 1}^{\infty} b_k sin(kx) \\ &\cdot where\ b_k = \begin{cases} &b_{2k} = 0 \\ &b_{2k-1} = \frac{1}{2k - 1} \end{cases} \end{align*}

3. Expand $y$

\begin{align*} y = f(x) &= \sum_{k = 1}^{\infty} \frac{1}{2k - 1} sin(2k - 1)x \\ &= sinx + \frac{1}{3} sin3x + \frac{1}{5}sin5x + \ldots \end{align*}

The shapes of three sinusoids

$sinx + \frac{1}{3}sin3x + \frac{1}{5}sin5x$

$\sum_{k = 1}^{10000} \frac{1}{2k - 1} sin(2k - 1)x$

Tip

As we accumulate more sinusoids, $f(x)$ approaches a square wave
$n = 300$ is the boundary where our eyes can barely tell the difference between sinusoids and a square wave

Gibb’s Phenomenon

Overshoot occurs near the jump of discontinuity in the singal
It’s an error with about $9\%$ of the jump size
It’s fixed when $n \rightarrow \infty$

Trignometry Recap

1. Properties

$\cdot$ Even function

cos(-x) = cos(x)

$\cdot$ Odd function

\begin{align*} &sin(-x) = -sin(x) \\ &tan(-x) = -tan(x) \end{align*}

2. Basic Formulas

\begin{align*} &(i)&\quad cos^2x + sin^2x = 1 \\ &(ii)&\quad tanx = \frac{sinx}{cosx} \\ &(iii)&\quad 1 + tan^2x = \frac{1}{cos^2x} \\ \end{align*}

3. Addition Theorem

\begin{align*} &(i)&\quad cos(\alpha \pm \beta) = cos\alpha \cdot cos\beta \mp sin\alpha \cdot sin\beta \\ &(ii)&\quad sin(\alpha \pm \beta) = sin\alpha \cdot cos\beta \pm cos\alpha \cdot sin\beta \\ \end{align*}

4. Double Angle Formulas

\begin{align*} cos2\alpha &= cos^2\alpha - sin^2\alpha \\ &=2cos^2\alpha - 1 \\ &=1-2sin^2\alpha \\ \end{align*}

sin2\alpha = 2sin\alpha \cdot cos\alpha

5. Half Angle Formulas

\begin{align*} &(i)&\quad cos^2\alpha = \frac{(1 + cos2\alpha)}{2} \\ &(ii)&\quad sin^2\alpha = \frac{(1 - cos2\alpha)}{2} \\ \end{align*}

6. Product $\rightarrow$ Sum Formulas

\begin{align*} &(i)&\quad cos\alpha \cdot cos\beta = \frac{1}{2} \{cos(\alpha + \beta) + cos(\alpha - \beta)\} \\ &(ii)&\quad sin\alpha \cdot sin\beta = \frac{1}{2} \{cos(\alpha + \beta) - cos(\alpha - \beta)\} \\ &(iii)&\quad sin\alpha \cdot cos\beta = \frac{1}{2} \{sin(\alpha + \beta) - sin(\alpha - \beta)\} \\ \end{align*}

Periodic Functions

Given $x \in \R$ and $\exists L \in K$ , $f(x)$ is a periodic function of $2L$

\begin{align*} &f(x + 2L) = f(x) \\ \end{align*}

1. $sin(kx)$

The period of $sin(x)$ is $2\pi$

\begin{align*} &y = sin(x) \Rightarrow L = 2\pi \\ &y = sin(2x) \Rightarrow L = \frac{2\pi}{2} \\ &y = sin(3x) \Rightarrow L = \frac{2\pi}{3} \\ &\ldots \\ &y = sin(kx) \Rightarrow L = \frac{2\pi}{k} \\ \end{align*}

By adding every multiple the period, we derive a function with $L = 2\pi$

\begin{align*} y = \sum^{\infty}_{k = 1} b_k sin(kx) \end{align*}

2. Approximate the Function of Square Wave

Square Wave

\begin{align*} y &= f(x) \quad (-\pi < x \le \pi) \\ &= \begin{cases} &\frac{-\pi}{4} &\quad (-\pi < x \le 0)\\ &\frac{\pi}{4} &\quad (0 < x \le \pi) \end{cases} \end{align*}

Fourier Series

y = \sum_{k = 1}^{\infty} \frac{sin(2k-1)x}{2k-1}

Tip

Both functions have the period of $2\pi$
The discontinuities are shown with Gibb’s Overshoot.

Even and Odd Functions

1. Even function

\begin{align*} &\cdot f(-x) = f(x) \\ &\cdot \int_{-a}^{a} f(x)dx = 2\int_{0}^{a}f(x)dx \end{align*}

$cos(x)$

\int_{-\pi}^{\pi} cos(x)dx = 2\int_{0}^{\pi}cos(x)dx

2. Odd function

\begin{align*} &\cdot f(-x) = -f(x) \\ &\cdot \int_{-a}^{a} f(x)dx = 0 \end{align*}

$sin(x)$

\int_{-\pi}^{\pi} sin(x)dx = 0

Important

Properties

\begin{align*} &\cdot \quad even \times even = even \\ &\cdot \quad odd \times odd = even \\ &\cdot \quad even \times odd = odd \end{align*}

Trignometry and Integration

\begin{align} &\int_{-\pi}^{\pi} cos(mx)dx = 0 \quad \quad \int_{-\pi}^{\pi} sin(mx)dx = 0 \quad \quad \\ &\int_{-\pi}^{\pi} sin(mx) \cdot cos(nx)dx = 0 \quad \quad \\ &\int^{\pi}_{-\pi} cos(mx) \cdot cos(nx)dx = \begin{cases} \pi \quad(m = n) \\ 0 \quad(m \ne n) \end{cases} \\ &\int^{\pi}_{-\pi} sin(mx) \cdot sin(nx)dx = \begin{cases} \pi \quad(m = n) \\ 0 \quad(m \ne n) \end{cases} \end{align} \\

where

$n, m \in \mathbb{N}$

(1) Integration of the $sin$ and $cos$ functions

$\cdot \int_{-\pi}^{\pi} cos(mx)dx = 0$

\begin{align*} \int_{-\pi}^{\pi} cos(mx)dx &= \frac{1}{m} [sin(mx)]^{\pi}_{-\pi} \\ &= \frac{1}{m} [sin(m\pi) - sin(-m\pi)] \\ &= \frac{1}{m} [sin(m\pi) + sin(m\pi)] \\ &\because sin(m\pi) = 0 \\ &\therefore \frac{1}{m} [sin(m\pi) + sin(m\pi)] = 0 \\ &\Rightarrow 0 \end{align*}

$\cdot \int_{-\pi}^{\pi} sin(mx)dx = 0$

\begin{align*} \int_{-\pi}^{\pi} sin(mx)dx &= -\frac{1}{m} [cos(mx)]^{\pi}_{-\pi} \\ &= -\frac{1}{m} [cos(m\pi) - cos(-m\pi)] \\ &= -\frac{1}{m} [cos(m\pi) - cos(m\pi)] \\ &\because cos(m\pi) = -1 \\ &\therefore \frac{1}{m} [cos(m\pi) + cos(m\pi)] = 0 \\ &\Rightarrow 0 \end{align*}

(2) $\int_{-\pi}^{\pi} sin(mx) \cdot cos(nx)dx = 0$

\begin{align*} &\int_{-\pi}^{\pi} \underbrace{sin(mx)}_{odd\ func} \cdot \underbrace{cos(nx)}_{even\ func} dx \\ &\because odd\times even = odd \\ &\therefore \int_{-\pi}^{\pi} sin(mx) \cdot cos(nx)dx = 0 \end{align*}

(3) $\int^{\pi}\_{-\pi} cos(mx) \cdot cos(nx)dx = \begin{cases} \pi \quad(m = n) \\ 0 \quad(m \ne n) \end{cases} \\$

$\cdot (m = n)$

\begin{align*} &\int^{\pi}_{-\pi} cos(mx) \cdot cos(nx)dx \\ &= \int^{\pi}_{-\pi} cos^2(mx) dx\\ &= \int^{\pi}_{-\pi} cos^2(mx) dx\quad\quad (\text{Half angle formula})\\ &= \frac{1}{2}[\int^{\pi}_{-\pi} dx + \underbrace{\int^{\pi}_{-\pi} cos(2mx)dx}_{0}] \\ &= \pi + 0 \\ &= \pi \end{align*}

$\cdot (m \ne n)$

\begin{align*} &\int^{\pi}_{-\pi} cos(mx) \cdot cos(nx)dx \\ &= \frac{1}{2}\int^{\pi}_{-\pi} cos(mx + nx) + cos(mx-nx)dx\\ &= \frac{1}{2}[\int^{\pi}_{-\pi} cos(mx + nx) + \int^{\pi}_{-\pi} cos(mx-nx)dx]\\ &= \frac{1}{2}[\underbrace{\int^{\pi}_{-\pi} cos(m + n)xdx}_{0}+ \underbrace{\int^{\pi}_{-\pi} cos(m-n)xdx}_{0}]\\ &= 0 \end{align*}

(4) $\int^{\pi}_{-\pi} sin(mx) \cdot sin(nx)dx = \begin{cases} \pi \quad(m = n) \\ 0 \quad(m \ne n) \end{cases} \\$

Tip

The proof is the same as the one given in (3)

Inner Product and Norm

Given two functions $f$ and $g$ that are continuous on the closed interval $[-\pi, \pi]$

\begin{align*} &\cdot \text{The inner product of $f$ and $g$: } \\ &\begin{align} (f, g) = \int_{-\pi}^{\pi} f(x)g(x) dx \quad\quad \end{align} \\ \end{align*}

\begin{align*} &\cdot \text{The norm (magnitude) of $f$ is: } \\ &\begin{align} ||f|| = \sqrt{(f, f)} = \sqrt{\int_{-\pi}^{\pi} \{f(x)\}^2 dx} \quad\quad \end{align} \end{align*}

Orthogonality relations of trigonometric functions

Constant vs. cosine and sine

\int_{-\pi}^{\pi} \cos(mx)\, dx = 0 \quad \Rightarrow \quad (1, \cos(mx)) = 0

\int_{-\pi}^{\pi} \sin(mx)\, dx = 0 \quad \Rightarrow \quad (1, \sin(mx)) = 0

So, the constant function $1$ is orthogonal to both $\cos(mx)$ and $\sin(mx)$ .

Sine vs. cosine

\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\, dx = 0 \quad \Rightarrow \quad (\sin(mx), \cos(nx)) = 0

Thus, sine and cosine are orthogonal to each other.

Cosine vs. cosine

\int_{-\pi}^{\pi} \cos(mx)\cos(nx)\, dx = \begin{cases} \pi & m = n \\[6pt] 0 & m \ne n \end{cases} \quad \Rightarrow \quad (\cos(mx), \cos(nx)) = \begin{cases} \pi & m = n \\[6pt] 0 & m \ne n \end{cases}

Sine vs. sine

\int_{-\pi}^{\pi} \sin(mx)\sin(nx)\, dx = \begin{cases} \pi & m = n \\[6pt] 0 & m \ne n \end{cases} \quad \Rightarrow \quad (\sin(mx), \sin(nx)) = \begin{cases} \pi & m = n \\[6pt] 0 & m \ne n \end{cases}

1. Fourier Series - Prelude and Trigonometry

Fourier Series - Prelude

∙\bullet∙ From Sinusoid (sin(kx)sin(kx)sin(kx)) to Square Wave

1. Assume a function y=f(x)y = f(x)y=f(x)

2. The Fourier Series of sin(kx)sin(kx)sin(kx)

3. Expand yyy