# Fourier-Serials

Today study something about Fourier serials

## Taylor Serial

As we all know,Taylor formula give us the way for expressing complicated function as simple polynomial function.

We should clearly remember the Taylor’s formula as follows:

$f(x)=f(x_0)+f’(x_0)(x-x_0)\cdots f^{(n)}(x_0)(x-x_0)^n\cdots$ ①

We use polynomial function to matching $f(x)$.

Especially when $x_0=0$ ,we call it as Maclaurin expansion.

For instance,we all know $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}$ ②

## Fourier Serial

### backgroud

But some periodic function may not be matched beautifully by using polynomial function,so we usually match that function by using the basic periodic function namely $\sin x$ and $\cos x$.

### basic formula

$f(x)=a_0+\sum _{i=1}^{n}(a_i\cos (nx)+b_i\sin(nx))$

### Trigonometric functions

If we integrating the function in $[-\pi,\pi]$,we will acquire 0 in most cases,including

• $\int_{-\pi}^{\pi}\sin(kx)=0$
• $\int_{-\pi}^{\pi}\cos(kx)=0$
• $\int_{-\pi}^{\pi}\cos(kx)\cos(nx)=0(n\neq k)$
• $\int_{-\pi}^{\pi}\sin(kx)\cos(nx)=0(allows\ n=k)$
• $\int_{-\pi}^{\pi}\sin(kx)\sin(nx)=0(n\neq k)$

This is the orthogonality of trigonometric functions.

### expansion

We expand the function by using the formula,but we need to certain ${a_i}$ and ${b_i}$.

$f(x)=a_0+\sum _{i=1}^{n}(a_i\cos (nx)+b_i\sin(nx))$

Firstly,we integrating the function on both sides of equation in $[-\pi,\pi ]$.

$\int_{-\pi}^{\pi}f(x)dx=\int_{-\pi}^{\pi}a_0dx+\sum {i=1}^{n}\int{-\pi}^{\pi}(a_i\cos (nx)+b_i\sin(nx))dx$.

According to the attribution of trigonometric functions,we know the equation could be simplify to

$\int_{-\pi}^{\pi}f(x)dx=\int_{-\pi}^{\pi}a_0dx$.

Lastly we got the value of $a_0$ and $a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx$.

Another equation writes the formula as:

$f(x)=\frac{a_0}{2}+\sum _{i=1}^{n}(a_i\cos (nx)+b_i\sin(nx))$,

so $a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx$.