べき級数に展開する問題

■問題

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt[3]{ 1+2x } \hspace{2}}}$

■ヒント

$3$\displaystyle{{ $ 1+x $ }^{{\alpha}}}$ のマクローリン展開の公式

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

を用いる．

■答

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt[3]{ 1+2x } \hspace{2}}={ \left({ 1+2x }\right) }^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} } }$

したがって， $3$\displaystyle{{ $ 1+x $ }^{{\alpha}}}$ のマクローリン展開の公式の $3$\displaystyle{x}$ を $3$\displaystyle{ 2x }$ ， $3$\displaystyle{{\alpha}}$ を $3$\displaystyle{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }$ に置き換える．

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt[3]{ 1+2x } \hspace{2}} }$ $3$\displaystyle{=1- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ 2x }\right)+\frac{\hspace{2} - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- 1 }\right) \hspace{2}}{\hspace{2} 2! \hspace{2}}{\left({ 2x }\right)}^{2}+\frac{\hspace{2} - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- 1 }\right)\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- 2 }\right) \hspace{2}}{\hspace{2} 3! \hspace{2}}{\left({ 2x }\right)}^{3}+\cdots }$

$3$\displaystyle{=1- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ - \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}} }\right)\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot {2}^{2}{x}^{2}+- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ - \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}} }\right)\left({ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} }\right)\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}\cdot {2}^{3}{x}^{3}+\cdots }$

$3$\displaystyle{=1- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}x+\frac{\hspace{2}8\hspace{2}}{\hspace{2}9\hspace{2}}{x}^{2}- \frac{\hspace{2} 112 \hspace{2}}{\hspace{2} 81 \hspace{2}}{x}^{3}+\cdots }$

■別解

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

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

$3$\displaystyle{\begin{array}{lll}{f}^{\prime\prime } $x$ = $ - \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} $ $ - \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}} $ { $ 1+2x $ }^{ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} }{ $ 1+2x $ }^{\prime }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}= $ \frac{\hspace{2} 16 \hspace{2}}{\hspace{2}9\hspace{2}} $ { $ 1+2x $ }^{ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} }& \vspace{6}\\ \end{array}}$ $3$\displaystyle{{f}^{\prime\prime } $0$ =\frac{\hspace{2} 16 \hspace{2}}{\hspace{2}9\hspace{2}}}$

$3$\displaystyle{\begin{array}{lll}{f}^{\prime\prime\prime } $x$ = $ \frac{\hspace{2} 16 \hspace{2}}{\hspace{2}9\hspace{2}} $ $ - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} $ { $ 1+2x $ }^{ - \frac{\hspace{2} 10 \hspace{2}}{\hspace{2}3\hspace{2}} }{ $ 1+2x $ }^{\prime }& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}= $ - \frac{\hspace{2} 224 \hspace{2}}{\hspace{2} 27 \hspace{2}} $ { $ 1+2x $ }^{ - \frac{\hspace{2} 10 \hspace{2}}{\hspace{2}3\hspace{2}} }& \vspace{6}\\ \end{array}}$ $3$\displaystyle{{f}^{\prime\prime\prime } $0$ =- \frac{\hspace{2} 224 \hspace{2}}{\hspace{2} 27 \hspace{2}}}$

したがって，マクローリン展開の公式

$3$\displaystyle{f $x$ =f $0$ +{f}^{\prime } $0$ x+\frac{\hspace{2} {f}^{ \prime \prime } $0$ \hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2}+\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{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt[3]{ 1+2x } \hspace{2}} }$ $3$\displaystyle{ =1+ $ - \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} $ x+\frac{\hspace{2} $ \frac{\hspace{2} 16 \hspace{2}}{\hspace{2}9\hspace{2}} $ \hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2}+\frac{\hspace{2} $ - \frac{\hspace{2} 224 \hspace{2}}{\hspace{2} 27 \hspace{2}} $ \hspace{2}}{\hspace{2} 3! \hspace{2}}{x}^{3}+\cdots }$

$3$\displaystyle{ =1- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}x+\frac{\hspace{2}8\hspace{2}}{\hspace{2}9\hspace{2}}{x}^{2}- \frac{\hspace{2} 112 \hspace{2}}{\hspace{2} 81 \hspace{2}}{x}^{3}+\cdots }$

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最終更新日： 2026年6月16日