# $\frac{1}{1-x}$ のマクローリン展開

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

## ■導出

$f\left(x\right)=\frac{1}{1-x}={\left(1-x\right)}^{-1}$とおく．　　$f\left(0\right)={1}^{-1}=1$

$\begin{array}{l}{f}^{\text{'}}\left(x\right)=\left(-1\right){\left(1-x\right)}^{-2}{\left(1-x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\left(-1\right){\left(1-x\right)}^{-2}\left(-1\right)\\ \text{ }\text{ }\text{ }=1\cdot {\left(1-x\right)}^{-2}\end{array}$　　　　 ${f}^{\text{'}}\left(0\right)=1\text{ }\text{ }$

$\begin{array}{l}{f}^{\text{'}\text{'}}\left(x\right)=\left(-2\right)\cdot 1\cdot {\left(1-x\right)}^{-3}{\left(1-x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=2\cdot 1\cdot {\left(1-x\right)}^{-3}\\ \text{ }\text{ }\text{ }=\left(2!\right){\left(1-x\right)}^{-3}\end{array}$　　${f}^{\text{'}\text{'}}\left(0\right)=2!$

$\begin{array}{l}{f}^{\text{'}\text{'}\text{'}}\left(x\right)=\left(-3\right)\left(2!\right){\left(1-x\right)}^{-4}{\left(1-x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\left(3!\right){\left(1+x\right)}^{-4}\end{array}$　　${f}^{\text{'}\text{'}\text{'}}\left(0\right)=3!$

$\begin{array}{l}{f}^{\left(4\right)}\left(x\right)=\left(-4\right)\left(3!\right){\left(1-x\right)}^{-5}{\left(1-x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\left(4!\right){\left(1+x\right)}^{-5}\end{array}$　　${f}^{\left(4\right)}\left(0\right)=4!$

$\begin{array}{l}{f}^{\left(5\right)}\left(x\right)=\left(-5\right)\left(4!\right){\left(1-x\right)}^{-6}{\left(1-x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\left(5!\right){\left(1-x\right)}^{-6}\end{array}$　　${f}^{\left(5\right)}\left(0\right)=5!$

したがって，マクローリン展開の公式

$f\left(x\right)=f\left(0\right)+{f}^{\text{'}}\left(0\right)x+\frac{{f}^{\text{'}\text{'}}\left(0\right)}{2!}{x}^{2}$$+\frac{{f}^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^{3}$$+\cdots \cdots$$+\frac{{f}^{\left(n\right)}\left(0\right)}{n!}{x}^{n}$$+\cdots \cdots$

に代入して

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

$=1+x+{x}^{2}+{x}^{3}+{x}^{4}+\cdots$

## ■収束半径

${a}_{n}=1$ , ${a}_{n+1}=1$

$\underset{n\to \infty }{lim}|\frac{{a}_{n+1}}{{a}_{n}}|=\underset{n\to \infty }{lim}|\frac{1}{1}|=1$

よって，収束半径 $R$

$R=\frac{1}{1}=1$

となる．

## ■インターラクティブなグラフ

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