${\displaystyle log\left(1+x\right)}$ のマクローリン展開

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

■導出 

${\displaystyle f\left(x\right)=log\left(1+x\right)}$ とおく．　　 ${\displaystyle f\left(0\right)=log1=0}$

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

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

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

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

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

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 

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

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

■収束半径 

${\displaystyle a_n=\frac{{\left(-1\right)}^{n-1}\left(n-1\right)!}{n!}}$ ， ${\displaystyle a_{n+1}=\frac{{\left(-1\right)}^{n}n!}{\left(n+1\right)!}}$

よって，収束半径 ${\displaystyle R}$ は

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

となる．

${\displaystyle x}$ の値が収束半径と等しいとき， 

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

の収束性について検討する． 

${\displaystyle x=1}$ のとき

となり，収束する． 

${\displaystyle x=-1}$ のとき

${\displaystyle -\frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\cdots \cdots }$

${\displaystyle =\underset{n\to \infty }{\lim }{\displaystyle \sum }_{k=1}^n\left(-\frac{1}{k}\right)}$

${\displaystyle >\underset{n\to \infty }{\lim }{{\displaystyle \int }}_0^n\left(-\frac{1}{x+1}\right)dx}$

${\displaystyle =\underset{n\to \infty }{\lim }{\left[-\log \left(x+1\right)\right]}_0^n}$

${\displaystyle =\underset{n\to \infty }{\lim }\left[-\log \left(n+1\right)-\log 1\right]}$

${\displaystyle =\underset{n\to \infty }{\lim }\left\{-\log \left(n+1\right)\right\}}$

${\displaystyle =-\infty }$

となり，発散する．以上より

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

が収束する ${\displaystyle x}$ の範囲は 

${\displaystyle -1<x\le 1}$  

となる． 

 

ホーム>>カテゴリー分類>>数列>>級数展開>>マクローリン展開>> ${\displaystyle log\left(1+x\right)}$ のマクローリン展開

最終更新日： 2026年8月18日