関数のべき級数展開

■問題

$3$\displaystyle{\log \frac{\hspace{2} 1+2x \hspace{2}}{\hspace{2} 1- 2x \hspace{2}}}$

■ヒント

対数の性質の $3$\displaystyle{\hspace{1}{\log }_{a}\frac{\hspace{2}R\hspace{2}}{\hspace{2}S\hspace{2}}=\hspace{1}{\log }_{a}R- \hspace{1}{\log }_{a}S}$ を利用して式を変形する．

各項を $3$\displaystyle{ f $x$ =\log $ 1+x $ \text{\hspace{1}}\text{\hspace{1}} }$ の形に変形し，べき級数展開を行う．

$3$\displaystyle{ f $x$ =\log $ 1+x $ \text{\hspace{1}}\text{\hspace{1}} }$ のべき級数展開は，

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

となる．

これの $3$\displaystyle{\text{\hspace{1}\hspace{1}}x\text{\hspace{1}}\text{\hspace{1}}}$ に $3$\displaystyle{ \text{\hspace{1}}\text{\hspace{1}}2x\text{\hspace{1}}\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}- 2x\text{\hspace{1}}\text{\hspace{1}} }$ をそれぞれ代入して計算する．

■答

$3$\displaystyle{\log \frac{\hspace{2} 1+2x \hspace{2}}{\hspace{2} 1- 2x \hspace{2}}=\log $ 1+2x $ - \log $ 1- 2x $ }$

（対数の性質を利用する．）

各項を別々に計算すると，

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

$3$\displaystyle{=2x- \frac{\hspace{2}4\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2}+\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- \frac{\hspace{2} 16 \hspace{2}}{\hspace{2}4\hspace{2}}{x}^{4}+\frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}}$ $3$\displaystyle{- \frac{\hspace{2} 64 \hspace{2}}{\hspace{2}6\hspace{2}}{x}^{6}+\cdot \cdot \cdot }$

$3$\displaystyle{=2x- 2{x}^{2}+\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- 4{x}^{4}+\frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}- }$ $3$\displaystyle{\frac{\hspace{2} 32 \hspace{2}}{\hspace{2}3\hspace{2}}{x}^{6}+\cdot \cdot \cdot }$

$3$\displaystyle{ \log $1- 2x$=\log \{ 1+ $ - 2x $ \} }$

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

$3$\displaystyle{=- 2x- \frac{\hspace{2}4\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2}- \frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- \frac{\hspace{2} 16 \hspace{2}}{\hspace{2}4\hspace{2}}{x}^{4}- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}}$ $3$\displaystyle{- \frac{\hspace{2} 64 \hspace{2}}{\hspace{2}6\hspace{2}}{x}^{6}- \cdot \cdot \cdot }$

$3$\displaystyle{=- 2x- 2{x}^{2}- \frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- 4{x}^{4}- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}}$ $3$\displaystyle{- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}3\hspace{2}}{x}^{6}- \cdot \cdot \cdot }$

となるので

$3$\displaystyle{\begin{array}{lll}\text{\hspace{5}}\text{\hspace{5}}\log \left({ 1+2x }\right)\text{\hspace{5}}=2x- 2{x}^{2}+\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- 4{x}^{4}+\frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}3\hspace{2}}{x}^{6}+\cdot \cdot \cdot & \vspace{6}\\ - \underline{ \)\text{\hspace{5}}\log \left({ 1- 2x }\right)\text{\hspace{5}}=- 2x- 2{x}^{2}- \frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}- 4{x}^{4}- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}- \frac{\hspace{2} 32 \hspace{2}}{\hspace{2}3\hspace{2}}{x}^{6}- \cdot \cdot \cdot }& \vspace{6}\\ \text{\hspace{5}}\log \left({ 1+2x }\right)- \log \left({ 1- 2x }\right)=2\left({ 2x+\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}+\frac{\hspace{2} 32 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}+\cdot \cdot \cdot }\right)& \vspace{6}\\ \end{array}}$

よって，

$3$\displaystyle{ \log \frac{\hspace{2} 1+2x \hspace{2}}{\hspace{2} 1- 2x \hspace{2}}=4x+\frac{\hspace{2} 16 \hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}+\frac{\hspace{2} 64 \hspace{2}}{\hspace{2}5\hspace{2}}{x}^{5}+\cdot \cdot \cdot }$

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最終更新日： 2024年5月28日