${\displaystyle sinx}$ のマクローリン展開

${\displaystyle sinx=x-\frac{1}{3!}{x}^3+\frac{1}{5!}{x}^5-\frac{1}{7!}{x}^7+\cdots }$  

■導出 

${\displaystyle f\left(x\right)=sinx}$ とおく． ${\displaystyle f\left(0\right)=sin0=0}$

${\displaystyle f^{\text{'}}\left(x\right)=cosx}$　　${\displaystyle f^{\text{'}}\left(0\right)=cos0=1}$

${\displaystyle f^{\text{'}\text{'}}\left(x\right)=-sinx}$　　${\displaystyle f^{\text{'}\text{'}}\left(0\right)=-sin0=0}$

${\displaystyle f^{\text{'}\text{'}\text{'}}\left(x\right)=-cosx}$　　${\displaystyle f^{\text{'}\text{'}\text{'}}\left(0\right)=-cos0=-1}$

${\displaystyle f^{\left(4\right)}\left(x\right)=sinx}$　　${\displaystyle f^{\left(4\right)}\left(0\right)=sin0=0}$

${\displaystyle f^{\left(5\right)}\left(x\right)=cosx}$　　${\displaystyle f^{\left(5\right)}\left(0\right)=cos0=1}$

である（以下，これの繰り返し）．すなわち，${\displaystyle m=0,1,2\cdots }$に対して  

${\displaystyle f^{\left(n\right)}\left(0\right)=\{\begin{array}{l}0\begin{array}{ccc}& \left(n=2m\right)& \end{array}\\ 1\begin{array}{cccc}& \left(n=4m+1\right)& & \end{array}\\ -1\begin{array}{cccc}& \left(n=4m+3\right)& & \end{array}\end{array}}$  

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 \sin x=0+1\cdot x+\frac{0}{2!}{x}^2+\frac{\left(-1\right)}{3!}{x}^3}$${\displaystyle +\frac{0}{4!}{x}^4}$${\displaystyle +\frac{1}{5!}{x}^5}$${\displaystyle +\frac{0}{6!}{x}^6+\frac{\left(-1\right)}{7!}{x}^7+\cdots }$

${\displaystyle =x-\frac{1}{3!}{x}^3+\frac{1}{5!}{x}^5-\frac{1}{7!}{x}^7+\cdots }$

■収束半径 

${\displaystyle a_n=\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}}$ , ${\displaystyle a_{n+1}=\frac{{\left(-1\right)}^{n+1}}{\left(2n+3\right)!}}$

${\displaystyle \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|}$ ${\displaystyle =\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(-1\right)}^{n+1}}{\left(2n+3\right)!}}{\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}}\right|}$ ${\displaystyle =\underset{n\to \infty }{\lim }\left|\frac{{\left(-1\right)}^{n+1}}{{\left(-1\right)}^n}\cdot \frac{\left(2n+1\right)!}{\left(2n+3\right)!}\right|}$ ${\displaystyle =\underset{n\to \infty }{\lim }\left|-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right|}$ ${\displaystyle =0}$

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．

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最終更新日： 2022年12月7日