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