フーリエ級数の問題

■問題

周期関数

${\displaystyle f\left(x\right)=\left\{\begin{array}{cc}-1& \left(-1\leqq x<0\right)\\ 2& \left(0\leqq x<1\right)\end{array}\right.}$，${\displaystyle f\left(x+2\right)=f\left(x\right)}$

のフーリエ級数を求めよ．

■解答

${\displaystyle f\left(x\right)\sim \frac{1}{2}+\frac{6}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x+\cdots \cdots \right)}$

■ヒント

フーリエ係数

${\displaystyle a_0=\frac{1}{2\pi }{\displaystyle {\int }_{-\pi }^{\pi }f\left(x\right)dx}}$

${\displaystyle a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)cosnxdx}$ 　　　　(${\displaystyle n=1,2,3···}$ )

${\displaystyle b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)sinnxdx}$ 　　　　(${\displaystyle n=1,2,3···}$ )

を求める

■解き方

${\displaystyle a_0=\frac{1}{2}{\displaystyle {\int }_{-1}^{1}f\left(x\right)dx}}$

${\displaystyle =\frac{1}{2}\left\{{\displaystyle {\int }_{-1}^0\left(-1\right)dx+{\displaystyle {\int }_0^{1}2dx}}\right\}}$

${\displaystyle =\frac{1}{2}\left\{-{\left[x\right]}_{-1}^0+2{\left[x\right]}_0^1\right\}}$

${\displaystyle =\frac{1}{2}\left\{-1+2·1\right\}}$

${\displaystyle =\frac{1}{2}}$

　

${\displaystyle a_n=\frac{1}{1}{\displaystyle {\int }_{-1}^{1}f\left(x\right)\cos n\pi x}dx}$

${\displaystyle ={\displaystyle {\int }_{-1}^0\left(-\cos n\pi x\right)dx}+{\displaystyle {\int }_0^{1}2\cos n\pi xdx}}$

${\displaystyle ={\left[-\frac{1}{n\pi }\sin n\pi x\right]}_{-1}^0+{\left[\frac{2}{n\pi }\sin n\pi x\right]}_0^1}$

${\displaystyle =0+0}$

${\displaystyle =0}$

${\displaystyle b_n=\frac{1}{1}{\displaystyle {\int }_{-1}^{1}f\left(x\right)\sin n\pi xdx}}$

${\displaystyle ={\displaystyle {\int }_{-1}^0\left(-\sin n\pi x\right)dx}+{\displaystyle {\int }_0^{1}2\sin n\pi xdx}}$

${\displaystyle ={\left[\frac{1}{n\pi }\cos n\pi x\right]}_{-1}^0+2{\left[-\frac{1}{n\pi }\cos n\pi x\right]}_0^1}$

${\displaystyle =\frac{1}{n\pi }\left(1-{\left(-1\right)}^n\right)-\frac{2}{n\pi }\left({\left(-1\right)}^n-1\right)}$

${\displaystyle =\frac{3}{n\pi }\left(1-{\left(-1\right)}^n\right)}$

${\displaystyle =\left\{\begin{array}{cc}0& n\text{が偶数のとき}\\ {\displaystyle \frac{6}{n\pi }}& n\text{が奇数のとき}\end{array}\right.}$

以上より　

${\displaystyle f\left(x\right)\sim \frac{1}{2}+\frac{6}{\pi }\left(\sin \pi x+\frac{1}{3}\sin 3\pi x+\frac{1}{5}\sin 5\pi x+\cdots \cdots \right)}$

 

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最終更新日： 2023年7月7日