preliminary.md

Preliminary

Euler's formula

The Euler's formula refers to $$ e^{i\theta} = \operatorname{cos}\theta + i\operatorname{sin}\theta . $$

The proof of this formula can be introduced by investigating the function $$ f(\theta) = e^{-i\theta} \left( \operatorname{cos}\theta + i\operatorname{sin}\theta\right) $$ with the facts that

• $f'(\theta) = 0$
• $f(0) = 1$

which conclude that $\forall \theta\in\mathbb{R}, f(\theta) = 1$.

Beware of the following facts,

• $e^{i\theta}$ is $2\pi$ periode function,
• $|e^{i\theta}| = 1$.

they will be the key to step into the Fourier kingdom.

Inner product

For $x,y\in l^2(\mathbb{Z})$, the inner product defined as $$ \langle x, y \rangle = \sum_{j=-\infty}^{\infty} x_j \bar y_j. $$ and for $x,y\in \mathbb{C}^N$, the inner product defined as $$ \langle x, y \rangle = \sum_{j=1}^N x_j \bar y_j. $$

For $f,g \in L^2(\mathbb{R})$, the inner product defined as $$ \langle f, g \rangle = \int f \bar g. $$ and for $f,g \in L^2(a,b)$, the inner product defined as $$ \langle f, g \rangle = \int_a^b f \bar g. $$

Orthogonal Basis

Discrete case, the space $\mathbb{C}^n$

If $ {v_k}0^{n-1} = (v_0, v_1, \dots, v{n-1})$ is an orthogonal basis of space $\mathbb{C}^n$, which means $$ \langle v_k, v_l \rangle = ||v_k||\cdot||v_l||\cdot\delta_{k,l}, $$ where $$ ||v_k|| = \sqrt{ \langle v_k, v_k \rangle }

\left( \sum_{j=0}^{n-1} v_{k,j} \right)^{\frac{1}{2}} $$

Note that $ v_k = (v_{k,0}, v_{k,1},\dots v_{k,n-1}) \in\mathbb{C}^n $ and $ v_{k,j}\in\mathbb{C}$.

then $\forall f \in \mathbb{C}^n$, $$ f = \sum_{k = 0}^{n-1} \frac{\langle f,v_k\rangle}{\langle v_k, v_k\rangle} v_k = \sum_{k = 0}^{n-1} \frac{\langle f,v_k\rangle}{||v_k||} \cdot \frac{v_k}{||v_k||} . $$

It is an orthonormal bases if $\forall k, ||v_k|| = 1$.

Suppose $\forall k\in\mathbb{Z}, ||v_k||=||v||$, the above formula can be rewritten in matrix form, $$ f =

\frac{1}{||v||^2} \begin{bmatrix} v_0 , v_1 , \cdots , v_{n-1} \end{bmatrix} \begin{bmatrix} \langle f, v_0 \rangle \ \langle f, v_1 \rangle \ \vdots \ \langle f, v_{n-1} \rangle \ \end{bmatrix}

\frac{1}{||v||^2} \begin{bmatrix} v_0 , v_1 , \cdots , v_{n-1} \end{bmatrix} \begin{bmatrix} \bar v_0 , \bar v_1 , \cdots , \bar v_{n-1} \end{bmatrix}^T f $$ that is $$ \begin{bmatrix} f_0 \ f_1 \ \ f_{n-1} \ \end{bmatrix}

\frac{1}{||v||^2} \begin{bmatrix} v_{0,0} & v_{1,0} & \cdots & v_{n-1,0} \ v_{0,1} & v_{1,1} & \cdots & v_{n-1,1} \ \vdots & \vdots & \vdots & \vdots \ v_{0,n-1} & v_{1,n-1} & \cdots & v_{n-1,n-1} \ \end{bmatrix} \begin{bmatrix} \bar v_{0,0} & \bar v_{0,1} & \cdots & \bar v_{0,n-1} \ \bar v_{1,0} & \bar v_{1,1} & \cdots & \bar v_{1,n-1} \ \vdots & \vdots & \vdots & \vdots \ \bar v_{n-1,0} & \bar v_{n-1,1} & \cdots & \bar v_{n-1,n-1} \ \end{bmatrix} \begin{bmatrix} f_0 \ f_1 \ \ f_{n-1} \ \end{bmatrix} . $$

The construction of orthogonal bases in space $\mathbb{C}^n$ equals to find square matrix $V$ such that $V\bar V^T=||v||^2$.

The following shows the intuition to extend $ k \in \mathbb{N}$, i.e. orthogonal bases $ {v_k}_{0}^{\infty}$,

the space $L^1(a,b)$

For $f \in L^1(a,b)$, extending to $k\in\mathbb{Z}$, $v_k(x)\in L^1(a,b)$, respect to $$ \langle v_k(x), v_l(x) \rangle = \int_{a}^{b} v_k(x) \bar v_l(x) \operatorname{d}! x = ||v_k|| \cdot ||v_l|| \cdot \delta_{k,l}, $$ where $\forall k\in \mathbb{Z}$, $$ ||v_k|| = \left( \int_a^b v_k(x)\bar v_k(x)\operatorname{d}! x \right)^{\frac{1}{2}}. $$ Then $$ f(x) = \sum_{k = -\infty}^{\infty} \frac{\langle f(x), v_k(x)\rangle }{||v_k||}\cdot \frac{v_k(x)}{||v_k||} = \sum_{k = -\infty}^{\infty} \frac{1}{||v_k||}\left( \int_{a}^{b} f(x) \bar v_k(x)\operatorname{d}! x \right) \cdot \frac{v_k(x)}{||v_k||}. $$

the space $L^1(\mathbb{R})$

For $f \in L^1(\mathbb{R})$, extending to $t\in\mathbb{R}$, which is $v(x,t)\in L^1(\mathbb{R})$,

$$ f(x) = \int_{-\infty}^{+\infty} \frac{1}{||v(t)||} \langle f(x), v(x,t)\rangle \cdot \frac{v(x,t)}{||v(t)||} \operatorname{d}! t = \int_{-\infty}^{+\infty} \frac{1}{||v(t)||} \left( \int_{-\infty}^{+\infty} f(x)\bar v(x,t)\operatorname{d}! x \right) \cdot \frac{v(x,t)}{||v(t)||} \operatorname{d}! t. $$ where $\forall t\in\mathbb{R}$, $$ ||v(t)|| = \left(\int_{-\infty}^{+\infty} v(x,t)\bar v(x,t)\operatorname{d}! x\right)^{\frac{1}{2}}. $$