べき級数に展開する問題

■問題

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

■答

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

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

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

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

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

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

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} 2+x \hspace{2}} }$

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

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

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\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}1\hspace{2}}{\hspace{2} 32 \hspace{2}}{x}^{4}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 64 \hspace{2}}{x}^{5}+\cdots }$

■別解

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

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

ここで

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

の式の $3$\displaystyle{x}$ を $3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}x }$ に置き換える

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

ホーム>>カテゴリー分類>>数列>>級数展開>>問題演習>>べき級数に展開する問題

学生スタッフ作成
最終更新日： 2022年6月5日