${\displaystyle e^x}$ のマクローリン展開

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

■導出 

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

${\displaystyle f^{\text{'}}\left(x\right)=e^x}$ ， ${\displaystyle f^{\text{'}}\left(0\right)=e^0=1}$

${\displaystyle f^{″}\left(x\right)=e^x}$ ， ${\displaystyle f^{″}\left(0\right)=e^0=1}$

${\displaystyle f^{‴}\left(x\right)=e^x}$ ， ${\displaystyle f^{‴}\left(0\right)=e^0=1}$

${\displaystyle f^{\left(4\right)}\left(x\right)=e^x}$ ， ${\displaystyle f^{\left(4\right)}\left(0\right)=e^0=1}$

${\displaystyle f^{\left(5\right)}\left(x\right)=e^x}$ ， ${\displaystyle f^{\left(5\right)}\left(0\right)=e^0=1}$

したがって，マクローリン展開の公式 

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

に代入して

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

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

■収束半径 

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

${\displaystyle \underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{lim}\left|\frac{\frac{1}{\left(n+1\right)!}}{\frac{1}{n!}}\right|}$ ${\displaystyle =\underset{n\to \infty }{lim}\left|\frac{n!}{\left(n+1\right)!}\right|}$ ${\displaystyle =\underset{n\to \infty }{lim}\left|\frac{1}{n+1}\right|=0}$

よって，収束半径 ${\displaystyle R}$ ${\displaystyle \left(R=\frac{1}{\alpha }\right)}$ は ${\displaystyle \alpha \to 0}$ より ${\displaystyle R\to \infty }$

${\displaystyle R=\infty }$

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>> ${\displaystyle e^x}$ のマクローリン展開

最終更新日： 2026年8月18日