欧拉公式

${\displaystyle e^{ix}=\cos x+i\sin x}$

${\displaystyle x=\pi }$ 时，尤拉公式变为${\displaystyle {{{e}^{{i}\,{\pi }}}+{1}}=0}$，即尤拉恒等式

历史

${\displaystyle {\frac {1}{1+x^{2}}}={\frac {1}{2}}\left({\frac {1}{1-ix}}+{\frac {1}{1+ix}}\right).}$

${\displaystyle \int {\frac {dx}{1+ax}}={\frac {1}{a}}\ln(1+ax)+C,}$

${\displaystyle ix=\ln(\cos x+i\sin x).}$

1740 年左右，欧拉把注意力从对数转向指数函数，得到了以他命名的欧拉公式。欧拉公式通过比较指数的级数展开和三角函数得到（其实此证法存在问题，原因见验证方法，但结论正确。），于1748 年发表[6][5]

形式

${\displaystyle e^{ix}=\cos x+i\sin x}$

${\displaystyle \sin x={\frac {e^{ix}-e^{-ix}}{2i}}}$${\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}}$

${\displaystyle x=\pi \,}$时，欧拉公式的特殊形式为

${\displaystyle e^{i\pi }+1=0\,}$

证明

${\displaystyle r={\sqrt {u^{2}+v^{2}}}\in \mathbb {R} ,\theta =\arctan({\frac {v}{u}})\in \mathbb {R} }$

${\displaystyle (1+{\frac {a+bi}{n}})^{n}=[(1+{\frac {a}{n}})+i{\frac {b}{n}}]^{n}=r_{n}(\cos \theta _{n}+i\sin \theta _{n})}$

${\displaystyle r_{n}=[(1+{\frac {a}{n}})^{2}+({\frac {b}{n}})^{2}]^{\frac {n}{2}},\theta _{n}=n\arctan {\frac {\frac {b}{n}}{1+{\frac {a}{n}}}}}$

{\displaystyle {\begin{aligned}\lim _{n\rightarrow \infty }\ln r_{n}&=\lim _{n\rightarrow \infty }[{\frac {n}{2}}\ln(1+{\frac {2a}{n}}+{\frac {a^{2}+b^{2}}{n^{2}}})]\\&=\lim _{n\rightarrow \infty }[{\frac {n}{2}}({\frac {2a}{n}}+{\frac {a^{2}+b^{2}}{n^{2}}})]\\&=a\\\end{aligned}}}

${\displaystyle \lim _{n\rightarrow \infty }r_{n}=\lim _{n\rightarrow \infty }e^{\ln r_{n}}=e^{a}}$

{\displaystyle {\begin{aligned}\lim _{n\rightarrow \infty }\theta _{n}&=\lim _{n\rightarrow \infty }(n\arctan {\frac {\frac {b}{n}}{1+{\frac {a}{n}}}})\\&=\lim _{n\rightarrow \infty }(n{\frac {\frac {b}{n}}{1+{\frac {a}{n}}}})\\&=b\\\end{aligned}}}

${\displaystyle \lim _{n\rightarrow \infty }(1+{\frac {a+bi}{n}})^{n}=e^{a}(\cos b+i\sin b)}$

${\displaystyle e^{a+ib}=e^{a}(\cos b+i\sin b)}$

${\displaystyle a=0}$，可得欧拉公式。

验证方法

${\displaystyle e^{x}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots }$
${\displaystyle \cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots }$
${\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots }$
${\displaystyle x=iz\,}$代入${\displaystyle e^{x}\,}$可得：
{\displaystyle {\begin{aligned}e^{iz}&=1+iz+{\frac {(iz)^{2}}{2!}}+{\frac {(iz)^{3}}{3!}}+{\frac {(iz)^{4}}{4!}}+{\frac {(iz)^{5}}{5!}}+{\frac {(iz)^{6}}{6!}}+{\frac {(iz)^{7}}{7!}}+{\frac {(iz)^{8}}{8!}}+\cdots \\&=1+iz-{\frac {z^{2}}{2!}}-{\frac {iz^{3}}{3!}}+{\frac {z^{4}}{4!}}+{\frac {iz^{5}}{5!}}-{\frac {z^{6}}{6!}}-{\frac {iz^{7}}{7!}}+{\frac {z^{8}}{8!}}+\cdots \\&=\left(1-{\frac {z^{2}}{2!}}+{\frac {z^{4}}{4!}}-{\frac {z^{6}}{6!}}+{\frac {z^{8}}{8!}}-\cdots \right)+i\left(z-{\frac {z^{3}}{3!}}+{\frac {z^{5}}{5!}}-{\frac {z^{7}}{7!}}+\cdots \right)\\&=\cos z+i\sin z\end{aligned}}}

${\displaystyle f(x)\,}$之导数为：
{\displaystyle {\begin{aligned}f'(x)&={\frac {(-\sin x+i\cos x)\cdot e^{ix}-(\cos x+i\sin x)\cdot i\cdot e^{ix}}{(e^{ix})^{2}}}\\&={\frac {-\sin x\cdot e^{ix}-i^{2}\sin x\cdot e^{ix}}{(e^{ix})^{2}}}\\&={\frac {-\sin x\cdot e^{ix}+\sin x\cdot e^{ix}}{(e^{ix})^{2}}}\\&=0\end{aligned}}}
${\displaystyle [a,b]\in I}$${\displaystyle c\in (a,b)}$
${\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}.}$拉格朗日中值定理
${\displaystyle \because f'(x)=0}$
${\displaystyle \therefore f'(c)=0}$
${\displaystyle f(a)=f(b)}$

${\displaystyle f(x)=f(0)}$
${\displaystyle {\frac {\cos x+i\sin x}{e^{ix}}}={\frac {\cos 0+i\sin 0}{e^{0}}}=1}$

${\displaystyle e^{ix}=\cos x+i\sin x}$

${\displaystyle {\frac {d}{dx}}e^{ix}=ie^{ix}=iy}$
{\displaystyle {\begin{aligned}{\frac {d}{dx}}(\cos x+i\sin x)&=-\sin x+i\cos x\\&=i(i\sin x+\cos x)\\&=iy\end{aligned}}}
${\displaystyle e^{i0}=e^{0}=1}$
${\displaystyle \cos 0+i\sin 0=1+i(0)=1}$

${\displaystyle x=0}$时，原函数的值相等，所以以上两个函数相等。
${\displaystyle e^{ix}=\cos x+i\sin x}$

cis函数

${\displaystyle \operatorname {cis} \theta =\cos \theta +i\sin \theta }$
${\displaystyle \operatorname {cis} \theta =e^{i\theta }}$

${\displaystyle \theta }$值为复数时，cis函数仍然是有效的，所以有些人可利用cis函数将欧拉公式推广到更复杂的版本。[2]

证明和角公式

{\displaystyle {\begin{aligned}e^{i(\alpha +\beta )}&=\cos(\alpha +\beta )+i\sin(\alpha +\beta )=e^{i\alpha +i\beta }\\&=e^{i\alpha }\times e^{i\beta }\\&=(\cos \alpha +i\sin \alpha )\times (\cos \beta +i\sin \beta )\\&=(\cos \alpha \times \cos \beta +i\sin \alpha \times i\sin \beta )+(i\sin \alpha \times \cos \beta +\cos \alpha \times i\sin \beta )\\&=(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+i(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\\\end{aligned}}}

${\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$
${\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }$

在复分析的应用

${\displaystyle z=x+iy=|z|(\cos \phi +i\sin \phi )=|z|e^{i\phi }\,}$
${\displaystyle {\bar {z}}=x-iy=|z|(\cos \phi -i\sin \phi )=|z|e^{-i\phi }\,}$

${\displaystyle x=\mathrm {Re} \{z\}\,}$为实部
${\displaystyle y=\mathrm {Im} \{z\}\,}$为虚部
${\displaystyle |z|={\sqrt {x^{2}+y^{2}}}}$z
${\displaystyle \phi =\mathrm {atan2} {(y,x)}}$，其中${\displaystyle \mathrm {atan2} {(y,x)}={\begin{cases}\arctan \left({\frac {y}{x}}\right)&\qquad x>0\\\pi +\arctan \left({\frac {y}{x}}\right)&\qquad y\geq 0,x<0\\-\pi +\arctan \left({\frac {y}{x}}\right)&\qquad y<0,x<0\\{\frac {\pi }{2}}&\qquad y>0,x=0\\-{\frac {\pi }{2}}&\qquad y<0,x=0\\{\text{undefined}}&\qquad y=0,x=0\end{cases}}}$

参考资料

1. ^ Eulers Formula. 密苏里科技大学. [2021-06-13].
2. Moskowitz, Martin A. A Course in Complex Analysis in One Variable. World Scientific Publishing Co. 2002: 7. ISBN 981-02-4780-X.
3. ^ Feynman, Richard P. The Feynman Lectures on Physics, vol. I. Addison-Wesley. 1977: 22-10. ISBN 0-201-02010-6.
4. ^ Bernoulli, Johann. Solution d'un problème concernant le calcul intégral, avec quelques abrégés par rapport à ce calcul [Solution of a problem in integral calculus with some notes relating to this calculation]. Mémoires de l'Académie Royale des Sciences de Paris. 1702, 1702: 197–289.
5. John Stillwell. Mathematics and Its History. Springer. 2002 [2018-07-17]. （原始内容存档于2019-06-04）.
6. ^ Leonard Euler (1748) Chapter 8: On transcending quantities arising from the circle页面存档备份，存于互联网档案馆） of Introduction to the Analysis of the Infinite, page 214, section 138 (translation by Ian Bruce, pdf link from 17 century maths).
7. ^ 张, 筑生. 数学分析新讲（第一册）. 北京大学出版社. 1990.