# $cosx$ のマクローリン展開

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

## ■導出

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

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

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

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

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

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

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

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

したがって，マクローリン展開の公式

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

に代入して

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

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

## ■収束半径

${a}_{n}=\frac{{\left(-1\right)}^{n}}{\left(2n\right)!}$${a}_{n+1}=\frac{{\left(-1\right)}^{n+1}}{\left(2n+1\right)!}$

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

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

$R=\infty$

となる．

## ■インターラクティブなグラフ

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