用向量分析推导欧拉恒等式与复变三角函数

2021-04-26 00:40 作者:中国大黄鸭鸭 0人读过 | 我要投稿

第一节　基础知识

1 微基分

具体内容略，2333~~~

【up主】还有，上面正确的写法应该是“积微成著”的“积”，不是“搞基搞比利”的“基”，2333~

觉得看着枯燥？不如买一包薯条吧

矢函数？？屎函数？？

第二节　定义复变三角函数

　　在推导复变三角函数之前，我们先讨论一下三角函数到底是什么。我想下面这张图，应该已经很好地定义了 $%5Csin%20%CE%B8$ 和 $%5Ccos%20%CE%B8$ 。下图主要依赖于几何思维，而非代数思维，而三角函数的出处就是几何。因此它对定义复变三角函数，掌握了绝对发言权，自然也比解析延拓更有发言权。

　　根据基本的微分几何常识，可知 $%7C%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%7C%3D1$ 恒成立。积分式中的“ $%C2%B1$ ”，是因为计算向量模长时，会求出 $2$ 个根，而此处 $%CE%B8$ 可能等于正根，也可能等于负根。

接下来的内容，就是为了检验红字部分是否恒成立。

听说鸭子的阴茎是螺旋形的，貌似可以用三角函数描述

第三节　证明图中 $%5Cvec%20r%20%5Cleft(%201%20%5Cright)%3D%5Cleft(%20%5Ccos%20%CE%B8%20%2C%20%5Csin%20%CE%B8%20%5Cright)$

　　注：本节暂时只考虑 $%CE%B8%E2%88%88%5Cmathbb%20R$ 的情况。复变三角函数的推导过程在第四节。

$%E2%88%B5%5Cfrac%7Bd%20%5Cvec%20r%7D%7Bdt%7D%3D%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5E2%20%5Cvec%20r%7D%7Bdt%5E2%7D%3D%CE%B8%5E2%20%5Cvec%20k%E2%88%A7%20%5Cleft(%20k%E2%88%A7%5Cvec%20r%20%5Cright)%3D%CE%B8%5E2%20%5Cleft(%20%5Cleft%3C%5Cvec%20r%2C%5Cvec%20k%20%5Cright%3E%20%5Cvec%20k%20-%5Cleft%3C%5Cvec%20k%2C%5Cvec%20k%20%5Cright%3E%20%5Cvec%20r%20%5Cright)%3D-%CE%B8%5E2%20%5Cvec%20r$

　　记 $y%3D%5Cvec%20r_n%2Cn%E2%88%88%5Cmathbb%20Z_%2B$ ，则：

$%5Cfrac%7Bd%5E2%20y%7D%7Bdt%5E2%7D%3D-%CE%B8%5E2y%5C%20%E2%91%B4%5C%20%5C%20$

　　此时定义 $p%3Dp(t)%3D%5Cfrac%7Bdy%7D%7Bdt%7D%5C%20%E2%91%B5%5C%20%5C%20$ ，则

$%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7Bdp%7D%7Bdx%7D%3D%5Cfrac%7Bdp%7D%7Bdy%7D%5Cfrac%7Bdy%7D%7Bdx%7D%3Dp%5Cfrac%7Bdp%7D%7Bdy%7D%5C%20%E2%91%B6%5C%20%5C%20$

　　将 $%5C%20%E2%91%B6%5C%20%5C%20$ 代入 $%5C%20%20%E2%91%B4%5C%20%5C%20$ ，得

$p%5Cfrac%7Bdp%7D%7Bdy%7D%3D-%CE%B8%5E2y$

　　等式两边乘 $dy$ 并求积分，得

$%5Cint%20pdp%3D-%CE%B8%5E2%5Cint%20ydy$

　　化简，得

$p%5E2%3DC_1%5E2-%CE%B8%5E2y%5E2$

　　上式开平方并求倒数，得

$%5Cfrac%201p%3D%5Cfrac%201%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D$

　　将 $%5C%20%E2%91%B5%5C%20%5C%20$ 代入，得

$%5Cfrac%20%7Bdt%7D%7Bdy%7D%3D%5Cfrac%201%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D$

　　等式两边乘 $dy$ 并求积分，得

$%5Cint%20dt%20%3D%20%5Cint%20%5Cfrac%20%7Bdy%7D%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D%3D%5Cfrac%201%7BC_1%7D%5Cint%20%5Cfrac%20%7Bdy%7D%7B%5Csqrt%7B1-%20%5Cleft(%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%5Cright)%5E2%20%7D%7D$

　　记 $%5Csin%20%CF%89%3D%5Cfrac%7B%CE%B8y%7D%7BC_1%7D$ ，于是 $%5Ccos%20%CF%89%3D%5Csqrt%7B1-%20%5Cleft(%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%5Cright)%5E2%20%7D$ 。则：

$C_1t%20%3D%5Cint%20%5Cfrac%20%7Bd%20%5Csin%20%CF%89%7D%7B%5Ccos%20%CF%89%7D%20%3D%20%5Cint%20%5Cfrac%20%7B%5Ccos%20%CF%89%7D%7B%5Ccos%20%CF%89%7Dd%CF%89%3D%CF%89%2BC_2%3D%20%5Csin%5E%7B-1%7D%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%2BC_2$

$%E2%88%B4y%3D%20%5Cfrac%20%7BC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%7D%CE%B8$

小心，别把西瓜汁蹭到衣服上~~

　　此时可以基本确定：

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%7D%CE%B8%2C%5Cfrac%20%7BC_3%20%5Csin%20%5Cleft(%20C_3%20t-C_4%20%5Cright)%7D%CE%B8%20%5Cright%5D%5C%20%E2%91%B7%5C%20%5C%20$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%5E2%20%5Ccos%20%5Cleft(%20C_1%20t-C_2%20%5Cright)%20%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Ccos%20%5Cleft(%20C_3%20t-C_4%20%5Cright)%20%7D%CE%B8%20%5Cright%5D$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5C%7B%20%5Cfrac%20%7BC_1%5E2%20%5Csin%20%5Cleft%5B%2090%C2%B0-%20%5Cleft(%20C_1t-C_2%20%5Cright)%20%5Cright%5D%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Csin%20%5Cleft%5B%2090%C2%B0-%20%5Cleft(%20C_3t-C_4%20%5Cright)%20%5Cright%5D%20%7D%CE%B8%20%5Cright%5C%7D$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%5E2%20%5Csin%20%5Cleft(%2090%C2%B0%2BC_2-C_1t%20%20%5Cright)%20%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Csin%20%5Cleft(%2090%C2%B0%2BC_4-C_3t%20%5Cright)%20%7D%CE%B8%20%5Cright%5D%5C%20%E2%91%B8%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r%20%5Cleft(%20t%20%5Cright)$

$%E2%88%B4%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20-C_3%20%5Csin%20%5Cleft(%20C_3t-C_4%20%5Cright)%20%EF%BC%8CC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%20%5Cright%5D%5C%20%E2%91%B9%5C%20%5C%20$

　　由 $%5C%20%E2%91%B8%5C%20%5C%20$ 与 $%5C%20%E2%91%B9%5C%20%5C%20$ ，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%3D-C_3%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%26%20%E2%91%BA%5C%5C%0AC_1%3D%5C%20%5C%20%5C%20%5Cfrac%7BC_3%5E2%7D%CE%B8%20%26%20%E2%91%BB%5C%5C%0AC_3%3D-%5Cfrac%7BC_1%5E2%7D%CE%B8%20%20%20%20%20%20%26%20%E2%91%BC%5C%5C%0AC_4%3D-C_2-90%C2%B0%20%20%20%20%20%20%20%20%20%20%20%20%20%20%26%20%E2%91%BD%5C%20%5C%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　将 $%5C%20%E2%91%BA%5C%20%5C%20$ 代入 $%5C%20%E2%91%BB%5C%20%5C%20$ 与 $%5C%20%E2%91%BC%5C%20%5C%20$ ，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_3%3D-%5Cfrac%7BC_3%5E2%7D%CE%B8%20%5C%5C%0AC_1%3D%5C%20%5C%20%5C%20%5Cfrac%7BC_1%5E2%7D%CE%B8%20%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　约去公因式，得

$%5C%20%E2%91%BE%5C%20%5C%20$ $%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%3D%5C%20%5C%20%5C%20%CE%B8%20%5C%5C%0AC_3%3D-%CE%B8%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　将 $%5C%20%E2%91%BD%5C%20%5C%20$ 和 $%5C%20%E2%91%BE%5C%20%5C%20$ 代入 $%5C%20%E2%91%B7%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Csin%20%5Cleft(%20%CE%B8t-C_2%20%5Cright)%2C%5Csin%20%5Cleft(%20%CE%B8t%2BC_2%2B90%C2%B0%20%5Cright)%20%5Cright%5D%5C%20%E2%91%BF%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%20%5Cleft(%200%20%5Cright)%20%3D%20%5Cvec%20i$

　　记 $%E2%88%80k%2C%E2%88%80l%20%E2%88%88%20%5Cmathbb%20Z$ ，则

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0A0%CE%B8-C_2%20%20%20%20%20%20%20%20%26%20%3D90%C2%B0(4k%2B1)%20%20%5C%5C%0A0%CE%B8%2BC_2%2B90%C2%B0%20%26%20%3D%2090%C2%B0(4l%2B0)%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　化简，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_2%20%3D%2090%C2%B0%20%5Cleft(%20-4k-1%20%5Cright)%20%5C%5C%0AC_2%20%3D%2090%C2%B0%20%5Cleft(%20%5C%20%5C%20%5C%204l%5C%20-1%20%5Cright)%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　代入 $%5C%20%E2%91%BF%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Csin%20%5Cleft(%20%CE%B8t%2B90%C2%B0%20%5Cright)%2C-%5Csin%20%5Cleft(%20-%CE%B8t%20%5Cright)%20%5Cright%5D$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft(%20%5Ccos%20%CE%B8t%2C%5Csin%20%CE%B8t%20%5Cright)%5C%20%E2%92%80%5C%20%5C%20$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%201%20%5Cright)%20%3D%20%5Cleft(%20%5Ccos%20%CE%B8%2C%5Csin%20%CE%B8%20%5Cright)$

　　另外，这个结论也可以验证——将 $%5C%20%E2%92%80%5C%20%5C%20$ 代入图中的向量微分方程即可。

第四节复变三角函数如何计算？

　　将图中的 $%CE%B8$ 换成 $i%CE%B8%E2%88%88%5Cmathbb%20I$ ，可得

$%5Cfrac%7Bd%20%5Cvec%20r%7D%7Bdt%7D%3Di%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5E2%20%5Cvec%20r%7D%7Bdt%5E2%7D%3D%CE%B8%5E2%20%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5Cvec%20r%7D%7Bdt%7D%3D%C2%B1%CE%B8%20%5Cvec%20r$

　　记 $y%3D%5Cvec%20r_n%2Cn%E2%88%88%5Cmathbb%20Z_%2B$ ，则：

$%5Cfrac%7Bdy%7D%7Bdt%7D%3D%C2%B1%CE%B8y$

$%E2%88%B4%5Cfrac%7Bdt%7Dy%5Cfrac%7Bdy%7D%7Bdt%7D%3D%C2%B1%5Cfrac%7Bdt%7Dy%CE%B8y$

$%E2%88%B4%5Cfrac%7Bdy%7Dy%3D%C2%B1%CE%B8dt$

$%E2%88%B4%5Cint%20%5Cfrac%7Bdy%7Dy%3D%C2%B1%CE%B8%5Cint%20dt$

$%E2%88%B4%5Cln%20y%3D%C2%B1%CE%B8t%2BC_0$

$%E2%88%B4y_1%3DC_1e%5E%7B%CE%B8t%7D%2Cy_2%3DC_2e%5E%7B-%CE%B8t%7D$

　　根据二阶常系数线性齐次微分方程的通解规则，由于 $y_1%2Cy_2$ 线性无关，因此 $y%3DAy_1%2BBy_2$ ，即：

$y%3DC_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft(%20C_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%2C%20C_3e%5E%7B%CE%B8t%7D%2BC_4e%5E%7B-%CE%B8t%7D%20%5Cright)$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%3D%CE%B8%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%2C%20C_3e%5E%7B%CE%B8t%7D-C_4e%5E%7B-%CE%B8t%7D%20%5Cright)%5C%20%E2%92%81%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20i%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r%20%5Cleft(%20t%20%5Cright)$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cfrac%20i%CE%B8%20%5Cvec%20k%E2%88%A7%5Cvec%20r%E2%80%99$

　　由 $%5C%20%E2%92%81%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3Di%5Cleft(%20-C_3e%5E%7B%CE%B8t%7D%2BC_4e%5E%7B-%CE%B8t%7D%20%2C%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%2C%20%5Cright)$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft%5B%20C_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%2C%20i%20%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%20%5Cright)%20%5Cright%5D%5C%20%E2%92%82%5C%20%5C%20$

你正在吃的是奥利奥还是丹麦曲奇？

　　再根据 $%5C%20%E2%92%82%5C%20%5C%20$ 可得：

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%2BC_2%3D1%5C%5C%0Ai%5Cleft(C_1-C_2%20%5Cright)%3D0%0A%5Cend%7Barray%7D%0A%5Cright.$

　　解得：

$C_1%3DC_2%3D%5Cfrac%2012$

　　代入 $%5C%20%E2%92%82%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft%5B%20%5Cfrac%2012%20%5Cleft(%20e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%20%5Cright)%2C%20%5Cfrac%20i2%20%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%20%5Cright)%20%5Cright%5D$