傅里叶级数（Fourier Series）_determine the coefficients (an and bn) of the four-优快云博客

博客介绍了傅里叶级数，指出任何周期函数可用正弦和余弦级数表示。阐述了基本定义，包括函数周期、狄利克雷积分等条件。还介绍了偶函数和奇函数的傅里叶级数形式，并通过两个实例展示了如何计算傅里叶系数。

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Baron Jean Baptiste Joseph Fourier (1768−1830) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.

Basic Definition

A function f(x) is said to have period P if f(x+P)=f(x) for all x. Let the function f(x) has period $2\pi$ . In this case, it is enough to consider behavior of the function on the interval $[-\pi, \pi]$ .

Suppose that the function f(x) with period $2\pi$ is absolutely integrable on $[-\pi, \pi]$ so that the following so-called Dirichlet integral is finite: $\int _{-\pi}^{\pi}|f(x)|dx< \infty$ ;
Suppose also that the function f(x) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).

If the conditions 1 and 2 are satisfied, the Fourier series for the function f(x) exists and converges to the given function. At a discontinuity $x_0$ , the Fourier Series converges to $lim_{\varepsilon \rightarrow 0}\frac{1}{2}[f(x_0-\varepsilon )-f(x_0+\varepsilon )]$ . The Fourier series of the function f(x) is given by

$f(x)=\frac{a_{0}}{2}+\sum ^\infty _{n=1}[a_ncos(nx)+b_nsin(nx)]$ ,

where the Fourier coefficients $a_0$ , $a_n$ , and $b_n$ are defined by the integrals

$a_0=\frac{1}{\pi}\int^{\pi}_{-\pi} f(x)dx, \quad a_n=\frac{1}{\pi}\int^{\pi}_{-\pi} f(x)cos(nx)dx, \quad b_n=\frac{1}{\pi}\int^{\pi}_{-\pi} f(x)sin(nx)dx$ .

Sometimes alternative forms of the Fourier series are used. Replacing $a_n$ and $b_n$ by the new variables $d_{n}$ and $\varphi _n$ or $d_{n}$ and $\theta _n$ , where

$d_n=\sqrt {a^2_n+b^2_n}, \quad tan\varphi _n=\frac{a_n}{b_n}, \quad tan\theta _n=\frac{b_n}{a_n}$ ,

we can write:

$f(x)=\frac{a_{0}}{2}+\sum ^\infty _{n=1}[d_nsin(nx+\varphi _n)] \quad or \quad f(x)=\frac{a_{0}}{2}+\sum ^\infty _{n=1}[d_ncos(nx+\theta _n)]$ .

Fourier Series of Even Functions

The Fourier series expansion of an even function f(x) with the period of $2\pi$ does not involve the terms with sines and has the form:

$f(x)=\frac{a_{0}}{2}+\sum ^\infty _{n=1}[a_ncos(nx)]$ ,

where the Fourier coefficients are given by the formulas

$a_0=\frac{1}{\pi}\int^{\pi}_{-\pi} f(x)dx, \quad a_n=\frac{2}{\pi}\int^{\pi}_{0} f(x)cos(nx)dx$ .

Fourier Series of Odd Functions

Accordingly, the Fourier series expansion of an odd $2\pi$ -periodic function f(x) consists of sine terms only and has the form:

$f(x)=\sum ^\infty _{n=1}[b_nsin(nx)]$ ,

where the coefficients $b_n$ are

$b_n=\frac{2}{\pi}\int^{\pi}_{0} f(x)sin(nx)dx$ .

Examples

Example 1

Let the function f(x) be $2\pi$ -periodic and suppose that it is presented by the Fourier series:

$f(x)=\frac{a_0}{2}+\sum^\infty _{n=1}(a_ncosnx+b_nsinnx)$

Calculate the coefficients a0, an, and bn.

Solution 1

To define a0, we integrate the Fourier series on the interval [−π,π]:

$\int ^\pi_{-\pi}f(x)dx=\pi a_0+\sum^\infty _{n=1}[a_n\int ^\pi_{-\pi}cosnxdx+b_n\int ^\pi_{-\pi}sinnxdx]$

For all n>0,

$\int ^\pi_{-\pi}cosnxdx=(\frac{sinnx}{n})|^\pi_{-\pi}=0 \quad and \int ^\pi_{-\pi}sinnxdx=(-\frac{cosnx}{n})|^\pi_{-\pi}=0$ .

Therefore, all the terms on the right of the summation sign are zero, so we obtain

$\int ^\pi_{-\pi}f(x)dx=\pi a_0 \quad or \quad a_0=\frac{1}{\pi}\int ^\pi_{-\pi}f(x)dx$ .

In order to find the coefficients an, we multiply both sides of the Fourier series by cosmx and integrate term by term:

$\int ^\pi_{-\pi}f(x)cosmxdx=\frac{a_0}{2}\int ^\pi_{-\pi}cosmxdx+\sum^\infty_{n=1}[a_n\int ^\pi_{-\pi}cosnxcosmxdx+b_n\int ^\pi_{-\pi}sinnxcosmxdx]$ .

The first term on the right side is zero. Then, using the well-known trigonometric identities, we have

$\int ^\pi_{-\pi}f(x)sinnxcosmxdx=\frac{1}{2}\int ^\pi_{-\pi}[sin(n+m)x+sin(n-m)x]dx=0$ ,

$\int ^\pi_{-\pi}f(x)cosnxcosmxdx=\frac{1}{2}\int ^\pi_{-\pi}[cos(n+m)x+cos(n-m)x]dx=0$ ,

if $m\neq n$ . In case when m=n, we can write:

$\int ^\pi_{-\pi}f(x)sinnxcosmxdx=\frac{1}{2}\int ^\pi_{-\pi}[sin(2m)x+sin0]dx=0,\Rightarrow \int ^\pi_{-\pi}sin^2mxdx=\frac{1}{2}[(-\frac{cos2mx}{2m})|^\pi_{-\pi}]=0$ ;

$\int ^\pi_{-\pi}f(x)cosnxcosmxdx=\frac{1}{2}\int ^\pi_{-\pi}[cos(2m)x+cos0]dx=0,\Rightarrow \int ^\pi_{-\pi}cos^2mxdx=\frac{1}{2}[(\frac{sin2mx}{2m})|^\pi_{-\pi}+2\pi]=\frac{1}{4m}[sin(2m\pi)-sin(2m(-2\pi))]+\pi=\pi$ .

Thus,

$\int ^\pi_{-\pi}f(x)cosmxdx=a_m\pi, \Rightarrow a_m=\frac{1}{\pi}\int ^\pi_{-\pi}f(x)cosmxdx, m=1,2,3,...$

Similarly, multiplying the Fourier series by sinmx and integrating term by term, we obtain the expression for bm:

$b_m=\frac{1}{\pi}\int ^\pi_{-\pi}f(x)sinmxdx, m=1,2,3,...$ .

Rewriting the formulas for an, bn, we can write the final expressions for the Fourier coefficients:

$a_n=\frac{1}{\pi}\int ^\pi_{-\pi}f(x)cosnxdx, \quad b_n=\frac{1}{\pi}\int ^\pi_{-\pi}f(x)sinnxdx$ .

Example 2

Find the Fourier series for the square 2π-periodic wave defined on the interval [−π,π]:

$f(x)=\left\{\begin{matrix} 0, \quad if -\pi\leq x\leq 0\\ 1, \quad if 0<x\leq\pi \end{matrix}\right.$

Solution 2

First we calculate the constant a0:

$a_0=\frac{1}{\pi}\int^\pi_{-\pi}f(x)dx=\frac{1}{\pi}\int^\pi_{0}1dx=\frac{1}{\pi}\pi=1$ .

Find now the Fourier coefficients for n≠0:

$a_n=\frac{1}{\pi}\int^\pi_{-\pi}f(x)cosnxdx=\frac{1}{\pi}\int^\pi_{-\pi}1\cdot cosnxdx=\frac{1}{\pi}[(\frac{sinnx}{n})|^\pi_{-\pi}]=\frac{1}{n\pi}\cdot0=0$ ,

$b_n=\frac{1}{\pi}\int^\pi_{-\pi}f(x)sinnxdx=\frac{1}{\pi}\int^\pi_{0}1\cdot sinnxdx=\frac{1}{\pi}[(-\frac{cosnx}{n})|^\pi_{0}]=-\frac{1}{n\pi}\cdot(cosn\pi-cos0)=\frac{1-cosn\pi}{n\pi}$ .