# $log\left(1+x\right)$ のマクローリン展開

$log\left(1+x\right)=x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}$$+\cdots$

## ■導出

$f\left(x\right)=log\left(1+x\right)$ とおく．　　$f\left(0\right)=log1=0$

$\begin{array}{l}{f}^{\text{'}}\left(x\right)=\frac{1}{1+x}{\left(1+x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\frac{1}{1+x}\cdot 1\\ \text{ }\text{ }\text{ }={\left(1+x\right)}^{-1}\end{array}$　　${f}^{\text{'}}\left(0\right)=1$

$\begin{array}{l}{f}^{\text{'}\text{'}}\left(x\right)=\left(-1\right){\left(1+x\right)}^{-2}{\left(1+x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=\left(-1\right){\left(1+x\right)}^{-2}\cdot 1\\ \text{ }\text{ }\text{ }=-{\left(1+x\right)}^{-2}\end{array}$　　${f}^{\text{'}\text{'}}\left(0\right)=-1$

$\begin{array}{l}{f}^{\text{'}\text{'}\text{'}}\left(x\right)=\left(-2\right)\left(-1\right){\left(1+x\right)}^{-3}{\left(1+x\right)}^{\text{'}}\\ \text{ }\text{ }\text{ }=2!{\left(1+x\right)}^{-3}\end{array}$　　${f}^{\text{'}\text{'}\text{'}}\left(0\right)=2!$

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

$\begin{array}{l}{f}^{\left(5\right)}\left(x\right)=-\left(-4\right)\left(3!\right){\left(1+x\right)}^{-5}{\left(1+x\right)}^{\text{'}}\hfill \\ \text{ }\text{ }\text{ }=4!{\left(1+x\right)}^{-5}\hfill \end{array}$　　${f}^{\left(5\right)}\left(0\right)=4!$

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

$=x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}+\cdots$

## ■収束半径

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

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

よって，収束半径 $R$

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

となる．

$x$ の値が収束半径と等しいとき，

$x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}+\cdots \cdots$

の収束性について検討する．

$x=1$のとき

$\begin{array}{l}\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \cdots \\ \text{ }=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots \cdots \\ \text{ }=\frac{2-1}{1\cdot 2}+\frac{4-3}{3\cdot 4}+\frac{6-5}{5\cdot 6}+\cdots \cdots \\ \text{ }=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots \cdots \\ \text{ }=\underset{n\to \infty }{lim}\sum _{k=1}^{n}\frac{1}{\left(2k-1\right)2k}\\ \text{ }<\underset{n\to \infty }{lim}\sum _{k=1}^{n}\frac{1}{{\left(2k-1\right)}^{2}}\begin{array}{cccc}& & & \end{array}\left(\because \frac{1}{\left(2k-1\right)2k}<\frac{1}{{\left(2k-1\right)}^{2}}\right)\\ \text{ }<1+\underset{n\to \infty }{lim}{\int }_{1}^{n}\frac{1}{{\left(2x-1\right)}^{2}}dx\\ \text{ }=1+\underset{n\to \infty }{lim}{\left[-\frac{1}{2x-1}\right]}_{1}^{n}\\ \text{ }=1+\underset{n\to \infty }{lim}\left(-\frac{1}{2n-1}+1\right)\\ \text{ }=2\end{array}$

となり，収束する．

$x=-1$ のとき

$\begin{array}{l}-\frac{1}{1}-\frac{1}{2}-\frac{1}{3}--\frac{1}{4}-\cdots \cdots \\ \text{ }=\underset{n\to \infty }{lim}\sum _{k=1}^{n}\left(-\frac{1}{k}\right)\\ \text{ }>\underset{n\to \infty }{lim}{\int }_{0}^{n}\left(-\frac{1}{x+1}\right)dx\\ \text{ }=\underset{n\to \infty }{lim}{\left[-log\left(x+1\right)\right]}_{0}^{n}\\ \text{ }=\underset{n\to \infty }{lim}\left[-log\left(n+1\right)-log1\right]\\ \text{ }=\underset{n\to \infty }{lim}\left\{-log\left(n+1\right)\right\}\\ \text{ }=-\infty \end{array}$

となり，発散する．以上より，

$x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}+\cdots \cdots$

が収束する$x$の範囲は

$-1

となる．

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