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

$3$\displaystyle{\log $ 1+x $ =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{+\cdots }$

■導出

$3$\displaystyle{f $x$ =\log $ 1+x $ }$ とおく．　　 $3$\displaystyle{f $0$ =\log 1=0}$

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

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

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

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

$3$\displaystyle{\displaystyle{\begin{array}{lll} {f}^{ \left({5}\right) }\left({x}\right)=- \left({ - 4 }\right)\left({ 3! }\right){ \left({ 1+x }\right) }^{ - 5 }{ \left({ 1+x }\right) }^{\prime } & \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}=4!{ \left({ 1+x }\right) }^{ - 5 } & \vspace{6}\\ \end{array}}}$ 　　 $3$\displaystyle{{f}^{ $5$ } $0$ =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

$3$\displaystyle{\log \left({ 1+x }\right)=0+1\cdot x+\frac{\hspace{2} \left({ - 1 }\right) \hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2}}$ $3$\displaystyle{+\frac{\hspace{2} 2! \hspace{2}}{\hspace{2} 3! \hspace{2}}{x}^{3}}$ $3$\displaystyle{+\frac{\hspace{2} - 3! \hspace{2}}{\hspace{2} 4! \hspace{2}}{x}^{4}}$ $3$\displaystyle{+\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}+\cdots }$

■収束半径

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

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

$3$\displaystyle{x}$ の値が収束半径と等しいとき，

$3$\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}+\cdots \cdots }$

の収束性について検討する．

$3$\displaystyle{x=1}$ のとき

$3$\displaystyle{\begin{array}{lll}\frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}+\cdots \cdots & \vspace{6}\\ \text{\hspace{5}}= $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ + $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} $ + $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}} $ +\cdots \cdots & \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2} 2- 1 \hspace{2}}{\hspace{2} 1\cdot 2 \hspace{2}}+\frac{\hspace{2} 4- 3 \hspace{2}}{\hspace{2} 3\cdot 4 \hspace{2}}+\frac{\hspace{2} 6- 5 \hspace{2}}{\hspace{2} 5\cdot 6 \hspace{2}}+\cdots \cdots & \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1\cdot 2 \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 3\cdot 4 \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 5\cdot 6 \hspace{2}}+\cdots \cdots & \vspace{6}\\ \text{\hspace{5}}={ \lim }\limits_{ n\rightarrow \infty }{\sum }\limits_{ k=1 }^{n} \frac{\hspace{2}1\hspace{2}}{\hspace{2} $ 2k- 1 $ 2k \hspace{2}} & \vspace{6}\\ \text{\hspace{5}}< { \lim }\limits_{ n\rightarrow \infty }{\sum }\limits_{ k=1 }^{n} \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 2k- 1 $ }^{2} \hspace{2}} \begin{array} & & & & \vspace{6}\\ \end{array} $ {}^\bullet\limits{ }_\bullet {}^\bullet \frac{\hspace{2}1\hspace{2}}{\hspace{2} \( 2k- 1 $ 2k \hspace{2}}< \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 2k- 1 $ }^{2} \hspace{2}} \) & \vspace{6}\\ \text{\hspace{5}}< 1+{ \lim }\limits_{ n\rightarrow \infty }\int _{1}^{n} \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ 2x- 1 $ }^{2} \hspace{2}}dx & \vspace{6}\\ \text{\hspace{5}}=1+{ \lim }\limits_{ n\rightarrow \infty } \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2x- 1 \hspace{2}} \] _{1}^{n}& \vspace{6}\\ \text{\hspace{5}}=1+{ \lim }\limits_{ n\rightarrow \infty } $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2n- 1 \hspace{2}}+1 $ & \vspace{6}\\ \text{\hspace{5}}=2& \vspace{6}\\ \end{array}}$

となり，収束する．

$3$\displaystyle{ x=- 1 }$ のとき

$3$\displaystyle{\begin{array}{lll}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}1\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}- - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}- \cdots \cdots & \vspace{6}\\ \text{\hspace{5}}={ \lim }\limits_{ n\rightarrow \infty }{\sum }\limits_{ k=1 }^{n} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}k\hspace{2}} $ & \vspace{6}\\ \text{\hspace{5}}> { \lim }\limits_{ n\rightarrow \infty }\int _{0}^{n} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} x+1 \hspace{2}} $ dx & \vspace{6}\\ \text{\hspace{5}}={ \lim }\limits_{ n\rightarrow \infty } \[ - \log $ x+1 $ \] _{0}^{n}& \vspace{6}\\ \text{\hspace{5}}={ \lim }\limits_{ n\rightarrow \infty } \[ - \log $ n+1 $ - \log 1 \] & \vspace{6}\\ \text{\hspace{5}}={ \lim }\limits_{ n\rightarrow \infty } \{ - \log $ n+1 $ \} & \vspace{6}\\ \text{\hspace{5}}=- \infty& \vspace{6}\\ \end{array}}$

となり，発散する．以上より，

が収束する $3$\displaystyle{x}$ の範囲は

$3$\displaystyle{- 1< x\leq 1}$

となる．

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最終更新日： 2022年12月7日