Fourier级数
函数的Fourier级数的展开
Euler--Fourier公式
我们探讨这样一个问题：
假设\(f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{k}coskt+b_{k}sinkt\)
Euler--Fourier公式:
\(a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx\)
\(a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x \mathrm{d} x, \quad n=0,1,2, \cdots\)
\(b_{n}=\frac{1}{\pi} \int_{-x}^{\pi} f(x) \sin n x \mathrm{d} x, \quad n=1,2, \cdots\)
\[\int_{-\pi}^{\pi}cosmx=0\]
\[\int_{-\pi}^{\pi}sinmx=0\]
\[\int_{-\pi}^{\pi}sinnxcosmx=0\]
\[\int_{-\pi}^{\pi}cosnxcosmx=0(n \neq m)\]
\[\int_{-\pi}^{\pi}cosnxcosmx=\pi(n = m)\](n=m时，cos0x=1，\(\rightarrow \frac{1}{2} \int_{-\pi}^{\pi}1dx\))
\[\int_{-\pi}^{\pi}sinnxcosmx=\pi(n = m)\]
利用三角公式：
\[cosmtcosnt=\frac{1}{2}[cos(m-n)t+cos(m+n)t]\]
正弦级数和余弦级数
注意奇函数如果在零点有定义的话，\(f(0)=0\)
\(f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} (a_{n}cosnx+b_{n}sinnx)\)
正弦级数表达式：\[f(x)=\sum_{n=1}^{\infty}b_{n}sinnx\]
同样余弦级数：\[f(x)=\frac{a_{0}}{2}+\sum_{n=0}^{\infty}a_{n}cosnx\]
Fourier级数习题：
1.1f(t)=\(\frac{A}{2}(sint+|sint|)\)展开的Fourier级数
(\(\int_{-\pi}^{\pi}|sint|dt=4\))
\(a_{0}=4 \times \frac{A}{2}\)
\(a_{n}\)=
\(\int_{-\pi}^{\pi}|sint|cosntdt\)
这里我们操作一下：\[a_{n}=\int_{0}^{\pi}sintcosntdt\]
学会利用三角变换：
\[sintcosnt=\frac{1}{2}[sin(t-nt)+sin(t+nt)]\]
也可以用分部积分法来计算
\[
\frac{1}{n} \sin t \sin n t \bigg|_{0}^{\pi}+\frac{1}{n} \int_{0}^{\pi} \cos t \sin n t
\]
\[
\frac{n^{2}-1}{n^{2}} \int_{0}^{\pi} \sin t \cos n t d t=\left.\frac{1}{n} \sin t \sin n t\right|_{0} ^{\pi}+\left.\frac{1}{n} \cos t \cos n t\right|_{0} ^{\pi}
\]
\(a_{n}=\int_{0}^{\pi}sintcosntdt=-\frac{2(cos(n\pi+1))}{n^2-1}\)}(详细写的话，分为奇数和偶数)
\[\int_{0}^{\pi}sintcosntdt=-\frac{2(cos(n\pi+1))}{n^2-1}\]
\(b_{n}\)=
{\(\int_{-\pi}^{\pi}sintsinntdt+|sint|sinntdt\)}
注意\(\int_{-\pi}^{\pi}|sint|sinntdt\)=0(偶函数)
\[\int_{-\pi}^{\pi}sintsinntdt=-\frac{2sin(\pi n)}{n^2 -1}\](n分奇偶数来考虑)
f(t)=\(A|sin t|\)
与1的类似：注意\(a_{n}=0\)
\[ f(x)=\left\{
\begin{aligned}
1 \quad x\in[-\pi,0),\\
0 \quad x\in[0,\pi)
\end{aligned}
\right.
\]的Fourier级数
\(f(x)=sgn(x),x \in(-\pi,\pi)\)展开成Fourier级数
sgn(x)为奇函数 \(a_{0}=0,a_{n}=0\)
利用正弦公式\(b_{n}=\int_{-\pi}^{0}-sin(nt)dt=\frac{1-cos(\pi n)}{n}\)
\(f(x)=\frac{x^{2}}{2}-\pi^{2}\)
\(f(x)\)为偶函数，考虑\(a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)cosnxdx\)\
\(\frac{1}{\pi} \int_{-\pi}^{\pi}(\frac{x^{2}}{2}-\pi^{2})cosnxdx\)
先考虑\(\int_{-\pi}^{\pi} \pi^{2}cosnxdx\),由于对应的原函数sinnx里面\(sin n\pi =0\)\
接下来我们考虑\(\int_{-\pi}^{\pi} \frac{x^{2}}{2}cosnxdx\)
我们利用分部积分:\(sinnx \frac{x^2}{2} \bigg|_{-\pi}^{\pi}-\frac{1}{n}\frac{1}{\pi}\int_{-\pi}^{\pi}xsinnxdx\)
\(\int_{-\pi}^{\pi}xsinnxdx\)我们采用分部积分\(-\frac{1}{n}xcosnx \bigg|_{\pi}^{\pi}=\frac{2cosnx}{n^{2}}\)
\[ f(x)=\left\{
\begin{aligned}
ax \quad x\in[-\pi,0),\\
bx \quad x\in[0,\pi)
\end{aligned}
\right.
\]
正弦级数与余弦级数的习题：
\(f(x)=x(x \in[0,\pi])\)分别展开成正弦级数和余弦级数
正弦级数：
\(f(x)=e^{-2x},x\in[0,\pi]\)
\(b_{n}=\) {\(\int_{0}^{\pi}e^{-2x}sinnx dx\)}
=\(\frac{n-e^{-2 \pi}ncos(\pi n)}{n^2+4}\)
\[ f(x)=\left\{
\begin{aligned}
cos\frac{\pi x}{2} \quad x\in[0,1),\\
0 \quad x\in[1,2]
\end{aligned}
\right.
\]
余弦级数：
\(f(x)=e^{x},x \in [0,\pi]\)
\(f(x)=x-\frac{\pi}{2}+|x-\frac{\pi}{2}|,x \in [0,\pi]\)