# ${\left(1+x\right)}^{\alpha }$のマクローリン展開

${\left(1+x\right)}^{\alpha }=1+\alpha x+\frac{\alpha \left(\alpha -1\right)}{2!}{x}^{2}$$+\frac{\alpha \left(\alpha -1\right)\left(\alpha -2\right)}{3!}{x}^{3}$$+\cdots \cdots$

## ■導出

$f\left(x\right)={\left(1+x\right)}^{\alpha }$

$f\left(0\right)={1}^{\alpha }=1$

$\begin{array}{l}{f}^{\prime }\left(x\right)=\alpha \cdot {\left(1+x\right)}^{\alpha -1}{\left(1+x\right)}^{\prime }\\ \text{ }\text{ }\text{}\text{}=\alpha {\left(1+x\right)}^{\alpha -1}\\ \text{ }\text{ }\text{}\text{}=\alpha {\left(1+x\right)}^{\alpha -1}\end{array}$

${f}^{\prime }\left(0\right)=\alpha$

$\begin{array}{l}{f}^{″}\left(x\right)=\alpha \left(\alpha -1\right){\left(1+x\right)}^{a-2}{\left(1+x\right)}^{\prime }\\ \text{ }\text{ }=\alpha \left(\alpha -1\right){\left(1+x\right)}^{a-2}\end{array}$

${f}^{″}\left(0\right)=\alpha \left(\alpha -1\right)$

$\begin{array}{l}{f}^{‴}\left(x\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right){\left(1+x\right)}^{a-3}{\left(1+x\right)}^{\prime }\\ \text{ }\text{ }=\alpha \left(\alpha -1\right)\left(\alpha -2\right){\left(1+x\right)}^{a-3}\end{array}$

${f}^{‴}\left(0\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)$

$\begin{array}{l}{f}^{\left(4\right)}\left(x\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right){\left(1+x\right)}^{a-4}{\left(1+x\right)}^{\prime }\\ \text{ }\text{ }=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right){\left(1+x\right)}^{a-4}\end{array}$ 　　${f}^{\left(4\right)}\left(0\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right)$

$\begin{array}{l}{f}^{\left(5\right)}\left(x\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right)\left(\alpha -4\right){\left(1+x\right)}^{a-5}{\left(1+x\right)}^{\prime }\\ \text{ }\text{ }=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right)\left(\alpha -4\right){\left(1+x\right)}^{a-5}\end{array}$ 　　${f}^{\left(5\right)}\left(0\right)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(\alpha -3\right)\left(\alpha -4\right)$

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

${\left(1+x\right)}^{\alpha }=1+\alpha x+\frac{\alpha \left(\alpha -1\right)}{2!}{x}^{2}$$+\frac{\alpha \left(\alpha -1\right)\left(\alpha -2\right)}{3!}{x}^{3}$$+\cdots \cdots$

## ■収束半径

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

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

よって，収束半径 $R$

$R=\frac{1}{1}=1$

となる．

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