Preliminary

Euler's formula

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

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

• $f^{\prime }(\theta )=0$
• $f(0)=1$

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 $$⟨x,y⟩=\sum _{j=-\infty }^{\infty }{x}_j{\bar{y}}_j.$$ and for $x,y\in {\mathbb{C}}^N$, the inner product defined as $$⟨x,y⟩=\sum _{j=1}^N{x}_j{\bar{y}}_j.$$

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

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 $$⟨v_k,v_l⟩=||v_k||\cdot ||v_l||\cdot {\delta }_{k,l},$$ where $$||v_k||=\sqrt{⟨v_k,v_k⟩}={\left(\sum _{j=0}^{n-1}{v}_{k,j}\right)}^{\frac{1}{2}}$$

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 $\forall f\in {\mathbb{C}}^n$, $$f=\sum _{k=0}^{n-1}\frac{⟨f,v_k⟩}{⟨v_k,v_k⟩}{v}_k=\sum _{k=0}^{n-1}\frac{⟨f,v_k⟩}{||v_k||}\cdot \frac{v_k}{||v_k||}.$$

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, $$f=\frac{1}{||v|{|}^2}\left[\begin{array}{c}{v}_0{v}_1\cdots v_{n-1}\end{array}\right]\left[\begin{array}{c}⟨f,v_0⟩\\ ⟨f,v_1⟩\\ ⋮\\ ⟨f,v_{n-1}⟩\end{array}\right]=\frac{1}{||v|{|}^2}\left[\begin{array}{c}{v}_0{v}_1\cdots v_{n-1}\end{array}\right]{\left[\begin{array}{c}{\bar{v}}_0{\bar{v}}_1\cdots {\bar{v}}_{n-1}\end{array}\right]}^{T}f$$ that is $$\left[\begin{array}{c}{f}_0\\ f_1\\ \\ f_{n-1}\end{array}\right]=\frac{1}{||v|{|}^2}\left[\begin{array}{cccc}{v}_{0,0}& v_{1,0}& \cdots & v_{n-1,0}\\ v_{0,1}& v_{1,1}& \cdots & v_{n-1,1}\\ ⋮& ⋮& ⋮& ⋮\\ v_{0,n-1}& v_{1,n-1}& \cdots & v_{n-1,n-1}\end{array}\right]\left[\begin{array}{cccc}{\bar{v}}_{0,0}& {\bar{v}}_{0,1}& \cdots & {\bar{v}}_{0,n-1}\\ {\bar{v}}_{1,0}& {\bar{v}}_{1,1}& \cdots & {\bar{v}}_{1,n-1}\\ ⋮& ⋮& ⋮& ⋮\\ {\bar{v}}_{n-1,0}& {\bar{v}}_{n-1,1}& \cdots & {\bar{v}}_{n-1,n-1}\end{array}\right]\left[\begin{array}{c}{f}_0\\ f_1\\ \\ f_{n-1}\end{array}\right].$$

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 $$⟨v_k(x),v_l(x)⟩={\int }_a^b{v}_k(x){\bar{v}}_l(x)dx=||v_k||\cdot ||v_l||\cdot {\delta }_{k,l},$$ where $\forall k\in \mathbb{Z}$, $$||v_k||={\left({\int }_a^b{v}_k(x){\bar{v}}_k(x)dx\right)}^{\frac{1}{2}}.$$ Then $$f(x)=\sum _{k=-\infty }^{\infty }\frac{⟨f(x),v_k(x)⟩}{||v_k||}\cdot \frac{v_k(x)}{||v_k||}=\sum _{k=-\infty }^{\infty }\frac{1}{||v_k||}\left({\int }_a^{b}f(x){\bar{v}}_k(x)dx\right)\cdot \frac{v_k(x)}{||v_k||}.$$

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

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

$$f(x)={\int }_{-\infty }^{+\infty }\frac{1}{||v(t)||}⟨f(x),v(x,t)⟩\cdot \frac{v(x,t)}{||v(t)||}dt={\int }_{-\infty }^{+\infty }\frac{1}{||v(t)||}\left({\int }_{-\infty }^{+\infty }f(x)\bar{v}(x,t)dx\right)\cdot \frac{v(x,t)}{||v(t)||}dt.$$ where $\forall t\in \mathbb{R}$, $$||v(t)||={\left({\int }_{-\infty }^{+\infty }v(x,t)\bar{v}(x,t)dx\right)}^{\frac{1}{2}}.$$