フーリエ級数の問題

■問題

周期関数

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

のフーリエ級数を求めよ．

■答

${\displaystyle f\left(x\right)\sim \frac{4}{\pi }\left(\sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\frac{1}{7}\sin 7x+\cdot \cdot \cdot \right)}$

■ヒント

フーリエ係数

${\displaystyle a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)cosnxdx}$  　　　　( ${\displaystyle n=0,1,2···}$ )

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

を求める．

■解説

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

${\displaystyle =\frac{1}{\pi }\left({\displaystyle {\int }_{-\pi }^0\left(-1\right)\cos nxdx}+{\displaystyle {\int }_0^{\pi }\cos nxdx}\right)}$

◇ $n\ne 0$ のとき

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

${\displaystyle =\frac{1}{\pi }\left(0+0\right)}$

${\displaystyle =0}$

◇ $n=0$ のとき

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

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

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

${\displaystyle =0}$

　

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

${\displaystyle =\frac{1}{\pi }\left({\displaystyle {\int }_{-\pi }^0\left(-1\right)\sin nxdx}+{\displaystyle {\int }_0^{\pi }\sin nxdx}\right)}$

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

${\displaystyle =\frac{1}{\pi }\left[\frac{1}{n}\left\{1-\cos \left(-n\pi \right)\right\}+\frac{1}{n}\left\{1-\cos n\pi \right\}\right]}$

余弦関数は偶関数なので ${\displaystyle \cos \left(-n\pi \right)=\cos n\pi }$

${\displaystyle =\frac{1}{\pi }\left[\frac{1}{n}\left\{1-\cos n\pi \right\}+\frac{1}{n}\left\{1-\cos n\pi \right\}\right]}$

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

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

${\displaystyle =\left\{\begin{array}{cc}0& \left(n\text{が偶数}\right)\\ {\displaystyle \frac{4}{n\pi }}& \left(n\text{が奇数}\right)\end{array}\right.}$

以上より

${\displaystyle f\left(x\right)\sim \frac{4}{\pi }\left(\sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\frac{1}{7}\sin 7x+\cdot \cdot \cdot \right)}$

となる．

■フーリエ級数の可視化

　

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最終更新日： 2026年8月15日