# $sinx$ のマクローリン展開

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

## ■導出

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

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

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

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

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

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

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

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

したがって，マクローリン展開の公式

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

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

## ■収束半径

${a}_{n}=\frac{{\left(-1\right)}^{n}}{\left(2n+1\right)!}$ , ${a}_{n+1}=\frac{{\left(-1\right)}^{n+1}}{\left(2n+3\right)!}$

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

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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