${\displaystyle cosx}$ のマクローリン展開

${\displaystyle cosx=1-\frac{1}{2!}{x}^2+\frac{1}{4!}{x}^4-\frac{1}{6!}{x}^6+\cdots }$

■導出 

${\displaystyle f\left(x\right)=cosx}$ とおく．　　${\displaystyle f\left(0\right)=cos0=1}$

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

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

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

${\displaystyle f^{\left(4\right)}\left(x\right)=cosx}$　　${\displaystyle f^{\left(4\right)}\left(0\right)=cos0=1}$

${\displaystyle f^{\left(5\right)}\left(x\right)=-sinx}$　　${\displaystyle f^{\left(5\right)}\left(0\right)=-sin0=0}$

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  

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

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

${\displaystyle =1-\frac{1}{2!}{x}^2+\frac{1}{4!}{x}^4-\frac{1}{6!}{x}^6+\cdots }$

■収束半径 

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

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

 

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最終更新日： 2026年8月18日