べき級数に展開する問題

■問題

次の関数をべき級数展開（マクローリン展開）をせよ．

$3$\displaystyle{ \sqrt{ 1+x } }$

■答

$3$\displaystyle{f $x$ =\sqrt{ 1+x }={ $ 1+x $ }^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }}$ とおく．
$3$\displaystyle{f $0$ ={1}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }=1}$

$3$\displaystyle{\begin{array}{lll}{f}^{\prime } $x$ = $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }{ $ 1+x $ }^{\prime }& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}= $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }\cdot 1& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{ $ 1+x $ }^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }& \vspace{6}\\ \end{array}}$
$3$\displaystyle{{f}^{\prime } $0$ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}}$

$3$\displaystyle{\begin{array}{lll}{f}^{\prime\prime } $x$ = $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} }{ $ 1+x $ }^{\prime }& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}= $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} }& \vspace{6}\\ \end{array}}$
$3$\displaystyle{{f}^{\prime\prime } $0$ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}}$

$3$\displaystyle{\begin{array}{lll}{f}^{\prime\prime\prime } $x$ = $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} $ $ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}5\hspace{2}}{\hspace{2}2\hspace{2}} }{ $ 1+x $ }^{\prime }& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}=\frac{\hspace{2}3\hspace{2}}{\hspace{2}8\hspace{2}}{ $ 1+x $ }^{ - \frac{\hspace{2}5\hspace{2}}{\hspace{2}2\hspace{2}} }& \vspace{6}\\ \end{array}}$
$3$\displaystyle{{f}^{\prime\prime\prime } $0$ =\frac{\hspace{2}3\hspace{2}}{\hspace{2}8\hspace{2}}}$

$3$\displaystyle{\begin{array}{lll}{f}^{ $4$ } $x$ = $ \frac{\hspace{2}3\hspace{2}}{\hspace{2}8\hspace{2}} $ $ - \frac{\hspace{2}5\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}2\hspace{2}} }{ $ 1+x $ }^{\prime }& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}=- \frac{\hspace{2} 15 \hspace{2}}{\hspace{2} 16 \hspace{2}}{ $ 1+x $ }^{ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}2\hspace{2}} }& \vspace{6}\\ \end{array}}$
$3$\displaystyle{{f}^{ $4$ } $0$ =- \frac{\hspace{2} 15 \hspace{2}}{\hspace{2} 16 \hspace{2}}}$

$3$\displaystyle{\begin{array}{lll}{f}^{ $5$ } $x$ = $ - \frac{\hspace{2} 15 \hspace{2}}{\hspace{2} 16 \hspace{2}} $ $ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}2\hspace{2}} $ { $ 1+x $ }^{ - \frac{\hspace{2}9\hspace{2}}{\hspace{2}2\hspace{2}} }{ $ 1+x $ }^{\prime }& \vspace{6}\\ \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}=\frac{\hspace{2} 105 \hspace{2}}{\hspace{2} 32 \hspace{2}}{ $ 1+x $ }^{ - \frac{\hspace{2}9\hspace{2}}{\hspace{2}2\hspace{2}} }& \vspace{6}\\ \end{array}}$
$3$\displaystyle{{f}^{ $5$ } $0$ =\frac{\hspace{2} 105 \hspace{2}}{\hspace{2} 32 \hspace{2}}}$

したがって，マクローリン展開の公式

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

に代入して

$3$\displaystyle{ \sqrt{ 1+x }=1+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}x+\frac{\hspace{2} \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }\right) \hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2} }$ $3$\displaystyle{ +\frac{\hspace{2} \frac{\hspace{2}3\hspace{2}}{\hspace{2}8\hspace{2}} \hspace{2}}{\hspace{2} 3! \hspace{2}}{x}^{3}+\frac{\hspace{2} \left({ - \frac{\hspace{2} 15 \hspace{2}}{\hspace{2} 16 \hspace{2}} }\right) \hspace{2}}{\hspace{2} 4! \hspace{2}}{x}^{4}+\frac{\hspace{2} \frac{\hspace{2} 105 \hspace{2}}{\hspace{2} 32 \hspace{2}} \hspace{2}}{\hspace{2} 5! \hspace{2}}{x}^{5}+\cdots }$

$3$\displaystyle{ =1+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}8\hspace{2}}{x}^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 16 \hspace{2}}{x}^{3}- \frac{\hspace{2}5\hspace{2}}{\hspace{2} 128 \hspace{2}}{x}^{4} }$ $3$\displaystyle{ +\frac{\hspace{2}7\hspace{2}}{\hspace{2} 256 \hspace{2}}{x}^{5}+\cdots }$

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最終更新日： 2025年4月25日