フーリエ級数

■問題

周期関数 $3$\displaystyle{f $x$ =x}$ ( $3$\displaystyle{- l\leq x< l}$ ) ， $3$\displaystyle{f $x$ =f $ x+2l $ }$ のフーリエ級数を求めよ．

■ヒント

フーリエ係数

$3$\displaystyle{ \hspace{1}{a}_{0}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2{\pi} \hspace{2}}\displaystyle{ \int _{ - {\pi} }^{{\pi}} f $x$ dx } }$

$3$\displaystyle{ \hspace{1}{a}_{n}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\int _{ - {\pi} }^{{\pi}} f $x$ \cos nx dx }$ 　　　　( $3$\displaystyle{ n=1,2,3\text{\hspace{1}}\cdot \text{\hspace{1}}\cdot \text{\hspace{1}}\cdot }$ )

$3$\displaystyle{ \hspace{1}{b}_{n}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\int _{ - {\pi} }^{{\pi}} f $x$ \sin nxdx }$ 　　　　( $3$\displaystyle{ n=1,2,3\text{\hspace{1}}\cdot \text{\hspace{1}}\cdot \text{\hspace{1}}\cdot }$ )

を求める

■解き方

$3$\displaystyle{\hspace{1}{a}_{0}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2l \hspace{2}}\displaystyle{ \int _{ - l }^{l} f $x$ dx }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2l \hspace{2}}\displaystyle{ \int _{ - l }^{l} xdx }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2l \hspace{2}} \[ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2} \] _{ - l }^{l}}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2l \hspace{2}} \{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{l}^{2}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{ $ - l $ }^{2} \} }$

$3$\displaystyle{=0}$

$3$\displaystyle{\hspace{1}{a}_{n}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}l\hspace{2}}\displaystyle{ \int _{ - l }^{l} f $x$ \cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}xdx }}$

$3$\displaystyle{f $x$ =x}$ は奇関数．よって $3$\displaystyle{ f $x$ \cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x }$ は奇関数となる．

したがって

$3$\displaystyle{\hspace{1}{a}_{n}=0}$

となる．

$3$\displaystyle{\hspace{1}{b}_{n}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}l\hspace{2}}\displaystyle{ \int _{ - l }^{l} f $x$ \sin \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}xdx }}$

$3$\displaystyle{f $x$ =x}$ 奇関数．よって $3$\displaystyle{ f $x$ \sin \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x }$ は偶関数となる．

したがって

$3$\displaystyle{\hspace{1}{b}_{n}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}l\hspace{2}}\cdot 2\displaystyle{ \int _{0}^{l} x\sin \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}xdx }}$

$3$\displaystyle{=\frac{\hspace{2}2\hspace{2}}{\hspace{2}l\hspace{2}}\displaystyle{ \int _{0}^{l} x{ $ - \frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}}\cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x $ }^{\prime }dx }}$

$3$\displaystyle{=\frac{\hspace{2}2\hspace{2}}{\hspace{2}l\hspace{2}} \{ \[ - \frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}}x\cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x \] _{0}^{l}- \displaystyle{ \int _{0}^{l} 1\cdot $ - \frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}}\cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x $ dx } \} }$

$3$\displaystyle{=\frac{\hspace{2}2\hspace{2}}{\hspace{2}l\hspace{2}} \[ - \frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}} \{ l{ $ - 1 $ }^{n}- 0 \} +\frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}}\displaystyle{ \int _{0}^{l} \cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}xdx } \] }$

$3$\displaystyle{=- \frac{\hspace{2} 2l \hspace{2}}{\hspace{2} n{\pi} \hspace{2}}{ $ - 1 $ }^{n}+\frac{\hspace{2}2\hspace{2}}{\hspace{2} n{\pi} \hspace{2}} \[ \frac{\hspace{2}l\hspace{2}}{\hspace{2} n{\pi} \hspace{2}}\sin \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}} \] _{ 0 }^{l}}$

$3$\displaystyle{=- \frac{\hspace{2} 2l \hspace{2}}{\hspace{2} n{\pi} \hspace{2}}{ $ - 1 $ }^{n}}$

$3$\displaystyle{f $x$ \sim \hspace{1}{a}_{0}+\displaystyle{{\sum }\limits_{ n=1 }^{\infty} $ \hspace{1}{a}_{n}\cos \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x+\hspace{1}{b}_{n}\sin \frac{\hspace{2} n{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x $ }}$

$3$\displaystyle{\sim \frac{\hspace{2} 2l \hspace{2}}{\hspace{2}{\pi}\hspace{2}} $ \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}l\hspace{2}}x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\sin \frac{\hspace{2} 2{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\sin \frac{\hspace{2} 3{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\sin \frac{\hspace{2} 4{\pi} \hspace{2}}{\hspace{2}l\hspace{2}}x+\cdots $ }$

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最終更新日： 2025年10月31日