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

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

■導出

$3$\displaystyle{f $x$ ={ $ 1+x $ }^{{\alpha}}}$

$3$\displaystyle{f $0$ ={1}^{{\alpha}}=1}$

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

$3$\displaystyle{{f}^{\prime } $0$ ={\alpha}}$

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

$3$\displaystyle{{f}^{\prime\prime } $0$ ={\alpha} $ {\alpha}- 1 $ }$

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

$3$\displaystyle{{f}^{\prime\prime\prime } $0$ ={\alpha} $ {\alpha}- 1 $ $ {\alpha}- 2 $ }$

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

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

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

に代入して

■収束半径

$3$\displaystyle{\hspace{1}{a}_{n}=\frac{\hspace{2} {\alpha} $ {\alpha}- 1 $ $ {\alpha}- 2 $ \cdots $ {\alpha}- n+1 $ \hspace{2}}{\hspace{2} n! \hspace{2}}}$ , $3$\displaystyle{\hspace{1}{a}_{ n+1 }=\frac{\hspace{2} {\alpha} $ {\alpha}- 1 $ $ {\alpha}- 2 $ \cdots $ {\alpha}- n $ \hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}}}$

$3$\displaystyle{ {\lim}\limits_{ n\rightarrow \infty } \mid \frac{\hspace{2}\hspace{1}{a}_{ n+ 1 }\hspace{2}}{\hspace{2}\hspace{1}{a}_{n}\hspace{2}}\mid }$ $3$\displaystyle{ ={ \lim }\limits_{ n\rightarrow \infty } \| \frac{\hspace{2} \frac{\hspace{2} \alpha $ \alpha - 1 $ $ \alpha - 2 $ \cdots $ \alpha - n $ \hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}} \hspace{2}}{\hspace{2} \frac{\hspace{2} \alpha $ \alpha - 1 $ $ \alpha - 2 $ \cdots $ \alpha - n+1 $ \hspace{2}}{\hspace{2} n! \hspace{2}} \hspace{2}} \| }$ $3$\displaystyle{ ={ \lim }\limits_{ n\rightarrow \infty } \| \frac{\hspace{2} n! \hspace{2}}{\hspace{2} $ n+1 $ ! \hspace{2}}\cdot \frac{\hspace{2} \alpha $ \alpha - 1 $ $ \alpha - 2 $ \cdots $ \alpha - n $ \hspace{2}}{\hspace{2} \alpha $ \alpha - 1 $ $ \alpha - 2 $ \cdots $ \alpha - n+1 $ \hspace{2}} \| }$ $3$\displaystyle{ ={ \lim }\limits_{ n\rightarrow \infty } \| \frac{\hspace{2} \alpha - n \hspace{2}}{\hspace{2} n+1 \hspace{2}} \| }$ $3$\displaystyle{ ={ \lim }\limits_{ n\rightarrow \infty } \| \frac{\hspace{2} \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}n\hspace{2}}- 1 \hspace{2}}{\hspace{2} 1+\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}} \hspace{2}} \| }$ $3$\displaystyle{ =1 }$

よって，収束半径 $3$\displaystyle{R}$ は

$3$\displaystyle{R=\frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}=1}$

となる．

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最終更新日： 2023年7月27日