関数のべき級数展開

■問題

$3$\displaystyle{\log $ 1- x $ }$

■ヒント

$3$\displaystyle{f $x$ =\log $ 1- x $ \text{\hspace{1} \hspace{1} }}$ とおいて微分を5回行い $3$\displaystyle{{f}^{\prime } $x$ }$ ～ $3$\displaystyle{{f}^{ $5$ } $x$ }$ を求める．

それぞれに $3$\displaystyle{\text{\hspace{1} \hspace{1} }x=0\text{\hspace{1} \hspace{1} }}$ を代入して $3$\displaystyle{f $0$ }$ ～ $3$\displaystyle{{f}^{ $5$ } $0$ }$ を求める．

最後に $3$\displaystyle{f $0$ }$ ～ $3$\displaystyle{{f}^{ $5$ } $0$ }$ 値をマクローリン展開の式に代入する．

■答

$3$\displaystyle{f $x$ =\log $ 1- x $ \text{\hspace{1} \hspace{1} }}$ とおいて， $3$\displaystyle{{f}^{\prime } $x$ }$ から $3$\displaystyle{{f}^{ $5$ } $x$ }$ を求める．

$3$\displaystyle{{f}^{\prime } $x$ }$ $3$\displaystyle{=\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} 1- x \hspace{2}}}$

$3$\displaystyle{{f}^{\prime\prime } $x$ }$ $3$\displaystyle{={ $ \frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} 1- x \hspace{2}} $ }^{\prime }}$ $3$\displaystyle{={ \{ $ - 1 $ { $ 1- x $ }^{ - 1 } \} }^{\prime }\text{\hspace{1} \hspace{1} }}$ $3$\displaystyle{= $ - 1 $ \{ $ - 1 $ { $ 1- x $ }^{ - 2 }\times $ - 1 $ \} \text{\hspace{1} \hspace{1} }}$ $3$\displaystyle{= $ - 1 $ { $ 1- x $ }^{ - 2 }}$ $3$\displaystyle{=\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} { $ 1- x $ }^{2} \hspace{2}}}$

$3$\displaystyle{{f}^{\prime\prime\prime } $x$ }$ $3$\displaystyle{={ \{ \frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} { $ 1- x $ }^{2} \hspace{2}} \} }^{\prime }}$ $3$\displaystyle{={ \{ $ - 1 $ { $ 1- x $ }^{ - 2 } \} }^{\prime }}$ $3$\displaystyle{= $ - 1 $ \{ $ - 2 $ { $ 1- x $ }^{ - 3 }\times $ - 1 $ \} }$ $3$\displaystyle{= $ - 2 $ { $ 1- x $ }^{ - 3 }}$ $3$\displaystyle{=\frac{\hspace{2} - 2 \hspace{2}}{\hspace{2} { $ 1- x $ }^{3} \hspace{2}}}$

$3$\displaystyle{{f}^{ $4$ } $x$ }$ $3$\displaystyle{={ \{ \frac{\hspace{2} - 2 \hspace{2}}{\hspace{2} { $ 1- x $ }^{3} \hspace{2}} \} }^{\prime }}$ $3$\displaystyle{={ \{ $ - 2 $ { $ 1- x $ }^{ - 3 } \} }^{\prime }}$ $3$\displaystyle{= $ - 2 $ \{ $ - 3 $ { $ 1- x $ }^{ - 4 }\times $ - 1 $ \} }$ $3$\displaystyle{= $ - 6 $ { $ 1- x $ }^{ - 4 }}$ $3$\displaystyle{=\frac{\hspace{2} - 6 \hspace{2}}{\hspace{2} { $ 1- x $ }^{4} \hspace{2}}}$

$3$\displaystyle{{f}^{ $5$ } $x$ }$ $3$\displaystyle{={ \{ \frac{\hspace{2} - 6 \hspace{2}}{\hspace{2} { $ 1- x $ }^{4} \hspace{2}} \} }^{\prime }\text{\hspace{1} \hspace{1} }}$ $3$\displaystyle{={ \{ $ - 6 $ { $ 1- x $ }^{ - 4 } \} }^{\prime }}$ $3$\displaystyle{= $ - 6 $ \{ $ - 4 $ { $ 1- x $ }^{ - 5 }\times $ - 1 $ \} }$ $3$\displaystyle{= $ - 24 $ { $ 1- x $ }^{ - 5 }\text{\hspace{1} \hspace{1} }}$ $3$\displaystyle{=\frac{\hspace{2} - 24 \hspace{2}}{\hspace{2} { $ 1- x $ }^{5} \hspace{2}}}$

$3$\displaystyle{\text{\hspace{1} \hspace{1} }x=0\text{\hspace{1} \hspace{1} }}$ のとき $3$\displaystyle{f $x$ }$ から $3$\displaystyle{{f}^{ $5$ } $x$ }$ の値は

$3$\displaystyle{f $x$ =\log $ 1- x $ }$ 　　 $3$\displaystyle{f $0$ =0}$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }\text{\hspace{1} }f \(0$ =\log $ 1+0 $ =\log 1\text{\hspace{1} }\text{\hspace{1} } \) }$

$3$\displaystyle{{f}^{\prime } $x$ =\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} 1- x \hspace{2}}}$ 　　 $3$\displaystyle{{f}^{\prime } $0$ =- 1}$ 　　 $3$\displaystyle{ ${f}^{\prime } \(0$ =\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} 1- 0 \hspace{2}}=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}\text{\hspace{1} }\text{\hspace{1} }\) }$

$3$\displaystyle{{f}^{\prime\prime } $x$ =\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} { $ 1- x $ }^{2} \hspace{2}}}$ 　　 $3$\displaystyle{{f}^{\prime\prime } $0$ =- 1}$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }\text{\hspace{1} }{f}^{\prime\prime } \(0$ =\frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} { $ 1- 0 $ }^{2} \hspace{2}}=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {1}^{2} \hspace{2}}=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}\text{\hspace{1} }\text{\hspace{1} } \) }$

$3$\displaystyle{{f}^{\prime\prime\prime } $x$ =\frac{\hspace{2} - 2 \hspace{2}}{\hspace{2} { $ 1- x $ }^{3} \hspace{2}}}$ 　　 $3$\displaystyle{{f}^{\prime\prime\prime } $0$ =- 2}$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }\text{\hspace{1} }{f}^{\prime\prime\prime } \(0$ =\frac{\hspace{2} - 2 \hspace{2}}{\hspace{2} { $ 1- 0 $ }^{3} \hspace{2}}=- \frac{\hspace{2}2\hspace{2}}{\hspace{2} {1}^{3} \hspace{2}}=- \frac{\hspace{2}2\hspace{2}}{\hspace{2}1\hspace{2}}\text{\hspace{1} }\text{\hspace{1} } \) }$

$3$\displaystyle{{f}^{ $4$ } $x$ =\frac{\hspace{2} - 6 \hspace{2}}{\hspace{2} { $ 1- x $ }^{4} \hspace{2}}}$ 　　 $3$\displaystyle{{f}^{ $4$ } $0$ =- 6}$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }\text{\hspace{1} }{f}^{ \(4$ } $0$ =\frac{\hspace{2} - 6 \hspace{2}}{\hspace{2} { $ 1- 0 $ }^{4} \hspace{2}}=- \frac{\hspace{2}6\hspace{2}}{\hspace{2} {1}^{4} \hspace{2}}=- \frac{\hspace{2}6\hspace{2}}{\hspace{2}1\hspace{2}}\text{\hspace{1} }\text{\hspace{1} } \) }$

$3$\displaystyle{{f}^{ $5$ } $x$ =\frac{\hspace{2} - 24 \hspace{2}}{\hspace{2} { $ 1- x $ }^{5} \hspace{2}}}$ 　　 $3$\displaystyle{{f}^{ $5$ } $0$ =- 24}$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }\text{\hspace{1} }{f}^{ \(5$ } $0$ =\frac{\hspace{2} - 24 \hspace{2}}{\hspace{2} { $ 1- 0 $ }^{5} \hspace{2}}=- \frac{\hspace{2} 24 \hspace{2}}{\hspace{2} {1}^{5} \hspace{2}}=- \frac{\hspace{2} 24 \hspace{2}}{\hspace{2}1\hspace{2}}\text{\hspace{1} }\text{\hspace{1} } \) }$

　　 $3$\displaystyle{\vdots }$

となる．マクローリン展開の式に $3$\displaystyle{f $0$ }$ から $3$\displaystyle{{f}^{ $5$ } $0$ }$ の値を代入する．

$3$\displaystyle{\log $ 1- x $ }$ $3$\displaystyle{=f $0$ +{f}^{\prime } $0$ x+\frac{\hspace{2} {f}^{\prime\prime } $0$ \hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2}}$ $3$\displaystyle{+\frac{\hspace{2} {f}^{\prime\prime\prime } $0$ \hspace{2}}{\hspace{2} 3! \hspace{2}}{x}^{3}}$ $3$\displaystyle{+\cdots }$ $3$\displaystyle{+\frac{\hspace{2} {f}^{ $n$ } $0$ \hspace{2}}{\hspace{2} n! \hspace{2}}{x}^{n}}$ $3$\displaystyle{+\cdots }$

$3$\displaystyle{=0+ $ - 1 $ x+ \frac{\hspace{2} - 1 \hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2}+ \frac{\hspace{2} - 2 \hspace{2}}{\hspace{2}6\hspace{2}}{x}^{3}}$ $3$\displaystyle{+ \frac{\hspace{2} - 6 \hspace{2}}{\hspace{2}24\hspace{2}}{x}^{4}}$ $3$\displaystyle{+ \frac{\hspace{2} - 24 \hspace{2}}{\hspace{2}120\hspace{2}}{x}^{5}+ \cdots }$

$3$\displaystyle{=- x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}{x}^{4}}$ $3$\displaystyle{- \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}+ \cdots }$

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最終更新日： 2024年5月28日