若干常见的傅里叶级数

Eufisky posted @ 2014年4月10日 17:45 in 数学分析 , 1400 阅读

关于Fourier级数，我们有以下几种常用表达式

\begin{align*}&x = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\cos \left[ {\left( {2n - 1} \right)x} \right]}}{{{{\left( {2n - 1} \right)}^2}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {nx} \right)}}{n}} \qquad 0 \le x < \pi \\&{x^2}=\frac{{{\pi ^2}}}{3}+4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{\cos \left( {nx} \right)}}{{{n^2}}}} \qquad - \pi \le x \le \pi \\&\cos \left( {ax} \right) = \frac{{\sin \left( {a\pi } \right)}}{\pi }\left( {\frac{1}{a} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2a}}{{{a^2} - {n^2}}}\cos \left( {nx} \right)} } \right)\qquad - \pi \le x \le \pi \\&\cot x = \frac{1}{x} + \sum\limits_{n = 1}^\infty {\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}}\qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\frac{1}{{\sin x}} = \frac{1}{x} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}} \qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\left| {\cos x} \right| = \frac{2}{\pi } + \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\left| {\sin x} \right| = \frac{2}{\pi } - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\cos \left( {nx} \right)}}{n}} = \ln \left( {2\cos \frac{x}{2}} \right)\qquad - \pi < x < \pi \\&\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {nx} \right)}}{n}} = - \ln \left( {2\sin \frac{x}{2}} \right)\qquad 0 < x < 2\pi \\&{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{k = 1}^\infty {\frac{{a\cos kx - k\sin kx}}{{{k^2} + {a^2}}}} } \right)\qquad x \in \left( {0,2\pi } \right),a \ne 0\end{align*}

根据这些高端级数我们有以下结论：

结论一：\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.\]

解：由Fourier级数

\[{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty {\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\quad \;\;\;{\mkern 1mu} x \in \left( {0,2\pi } \right),a \ne 0\]

将$\displaystyle f(x)=e^{ax}(0<x<2\pi)$延拓为$\displaystyle(-\infty,+\infty)$上以$\displaystyle 2\pi$为周期的函数$\displaystyle \widetilde{f}(x)$，则必须有

\[ \widetilde{f}(0)=\widetilde{f}(2\pi)=\frac{f(2\pi-0)+f(0+0)}{2}=\frac{e^{2\pi a}+e^{0\cdot a}}{2}=\frac{e^{2\pi a}+1}{2}.\]

由此得到

\[\widetilde{f}(x)= \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty{\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\qquad {\mkern 1mu} - \infty < x < + \infty \]

令$\displaystyle a=1,x=0$，我们立得

\[\frac{{{e^{2\pi }} + 1}}{2} = \frac{{{e^{2\pi }} - 1}}{\pi }\left( {\frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} } \right).\]

即

\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.\]

一般地，我们有

\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {a^2}}}} = \frac{\pi }{{2a}}\frac{{{e^{2\pi }} + 1}}{{{e^{2a\pi }} - 1}} - \frac{1}{{2{a^2}}}.\]