${\displaystyle \frac{1}{1-x}}$ のマクローリン展開

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

■導出 

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

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

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

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

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

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

したがって，マクローリン展開の公式 

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

に代入して

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

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

■収束半径 

${\displaystyle a_n=1}$ , ${\displaystyle a_{n+1}=1}$

${\displaystyle \underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{lim}\left|\frac{1}{1}\right|=1}$

よって，収束半径 ${\displaystyle R}$ は

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

となる．

　

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最終更新日： 2026年8月18日