現在の数式のサイズは 3 です。 サイズを選択 2 3 4 5 6 7
 Tweet このページ

# オイラーの公式

$3\displaystyle{ {e}^{ \mathbb{{i}}{\theta} }=\cos {\theta}+\mathbb{{i}}\sin {\theta} }$

## ■導出計算

$3\displaystyle{{e}^{x}}$  をマクローリン展開すると，

$3\displaystyle{ {e}^{x}=1+ x+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{x}^{2}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3! \hspace{2}}{x}^{3}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 4! \hspace{2}}{x}^{4}+ }$$3\displaystyle{ \cdots \cdots + \frac{\hspace{2}1\hspace{2}}{\hspace{2} n! \hspace{2}}{x}^{n}+ \cdots \cdots }$ ･･････(1)

(1)の$3x$ に形式的に$3\displaystyle{\mathbb{{i}}{\theta}}$ を代入して計算してみると，

$3\displaystyle{\displaystyle{{e}^{ {i}{\theta} }=1+ \( {i}{\theta} \) + \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{ \( {i}{\theta} \) }^{2}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3! \hspace{2}}{ \( {i}{\theta} \) }^{3}}}$$3\displaystyle{ + \frac{\hspace{2}1\hspace{2}}{\hspace{2} 4! \hspace{2}}{ \( {i}{\theta} \) }^{4}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}{ \( {i}{\theta} \) }^{5}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}{ \( {i}{\theta} \) }^{6}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2}7\hspace{2}}{ \( {i}{\theta} \) }^{7} }$$3\displaystyle{ + \cdots \cdots }$

$3\displaystyle{ =1+ {i}{\theta}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{{\theta}}^{2}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3! \hspace{2}}{i}{{\theta}}^{3}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 4! \hspace{2}}{{\theta}}^{4} }$$3\displaystyle{ + \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}{i}{{\theta}}^{5}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}{{\theta}}^{6}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}7\hspace{2}}{i}{{\theta}}^{7}+ \cdots \cdots }$

$3\displaystyle{ = \( 1- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{{\theta}}^{2}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 4! \hspace{2}}{{\theta}}^{4}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}{{\theta}}^{6}+ \cdots \cdots \) }$$3\displaystyle{ + \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}{ \( {i}{\theta} \) }^{5}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}{ \( {i}{\theta} \) }^{6}+ \frac{\hspace{2}1\hspace{2}}{\hspace{2}7\hspace{2}}{ \( {i}{\theta} \) }^{7}+ \cdots \cdots }$ ･･････(2)

$3\displaystyle{\cos {\theta}}$  のマクローリン展開は，

$3\displaystyle{\cos {\theta}=1- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2! \hspace{2}}{{\theta}}^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 4! \hspace{2}}{{\theta}}^{4}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}{{\theta}}^{6}+\cdots \cdots }$  ･･････(3)

$3\displaystyle{\sin {\theta}}$  のマクローリン展開は，

$3\displaystyle{\sin {\theta}={\theta}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 3! \hspace{2}}{{\theta}}^{3}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}{{\theta}}^{5}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}7\hspace{2}}{{\theta}}^{7}+\cdots \cdots }$  ･･････(4)

(2)に(3)，(4)を代入すると，

$3\displaystyle{{e}^{ \mathbb{{i}}{\theta} }=\cos {\theta}+\mathbb{{i}}\sin {\theta}}$

となり，オイラーの公式（Euler's Formula）が得られた．

ホーム>>カテゴリー分類>>複素数 >>複素解析>>オイラーの公式