构造超越方程的解析解

2023-01-11 20:20 作者:艾琳娜的糖果屋 0人读过 | 我要投稿

最近看到一个很无聊的问题求解方程 $%0Ax%5E2%3D%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20%0A$ ，很显然该方程有两个零点一个是0，另一个零点无法初等表达出来，只需要数值计算即可，当然要是强行凑出一个解析形式也不是不可以。我们可以考虑函数

$f%5Cleft(%20z%20%5Cright)%20%3D%5Cfrac%7Bz%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7Bz%5E2-%5Cln%20%5Cleft(%201%2Bz%20%5Cright)%7D%0A%0A$

它的一阶极点就是超越方程的一阶零点，考虑锁孔围道并规定 $%0A-%5Cpi%20%3Carg%5Cleft(%201%2Bz%20%5Cright)%20%3C%5Cpi%20%0A%0A$

容易估计在大圆上的积分是 $%0A2%5Cpi%20i%0A%0A$ ，小圆上的积分是0，于是由留数定理有

$%0A2%5Cpi%20i%2B%5Cint_%7B%5Cinfty%7D%5E1%7B%5Cfrac%7B-x%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7Bx%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20-i%5Cpi%7D%7D-dx%2B%5Cint_1%5E%7B%5Cinfty%7D%7B%5Cfrac%7B-x%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7Bx%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20%2Bi%5Cpi%7D-dx%7D%3D2%5Cpi%20ires%5Cleft(%20f%5Cleft(%20z%20%5Cright)%20%2C0%2Cx_0%20%5Cright)%20%0A$

$%0Ares%5Cleft(%20f%5Cleft(%20z%20%5Cright)%20%2Cz_k%20%5Cright)%20%3D%5Cfrac%7Bz%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7B%5Cleft(%20z%5E2-%5Cln%20%5Cleft(%201%2Bz%20%5Cright)%20%5Cright)%20%5Cprime%7D%5Cmid_%7Bz%3Dz_k%7D%5E%7B%7D%3D%5Cfrac%7B1%2Bz_k%7D%7B2z_k%2B1-%5Csqrt%7B3%7D%7D%0A%0A%0A$

$%0A%5Cint_%7B%5Cinfty%7D%5E1%7B%5Cfrac%7B-x%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7Bx%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20-i%5Cpi%7D%7D-dx%2B%5Cint_1%5E%7B%5Cinfty%7D%7B%5Cfrac%7B-x%2B%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7Bx%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20%2Bi%5Cpi%7D-dx%7D%3D-2%5Cpi%20i%5Cint_1%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20x-%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright)%7D%7B%5Cleft(%20x%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20%5Cright)%20%5E2%2B%5Cpi%20%5E2%7Ddx%7D%0A%0A$

于是我们得到一个简单方程

$%0A1-%5Cint_1%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20x-%5Cfrac%7B1%2B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright)%7D%7B%5Cleft(%20x%5E2-%5Cln%20%5Cleft(%20x-1%20%5Cright)%20%5Cright)%20%5E2%2B%5Cpi%20%5E2%7Ddx%7D%3D%5Cfrac%7B1%7D%7B1-%5Csqrt%7B3%7D%7D%2B%5Cfrac%7B1%2Bx_0%7D%7B2x_0%2B1-%5Csqrt%7B3%7D%7D%0A%0A$

再对积分换元整理就得到一个形式上的解析解

$%0Ax_0%3D%5Cfrac%7B%5Csqrt%7B3%7D%2B1-%5Cleft(%20%5Csqrt%7B3%7D-1%20%5Cright)%20%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20t%2B%5Cfrac%7B1-%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright)%7D%7B%5Cleft(%20%5Cleft(%20t%2B1%20%5Cright)%20%5E2-%5Cln%20t%20%5Cright)%20%5E2%2B%5Cpi%20%5E2%7Ddt%7D%7D%7B2%2B%5Csqrt%7B3%7D-2%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20t%2B%5Cfrac%7B1-%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright)%7D%7B%5Cleft(%20%5Cleft(%20t%2B1%20%5Cright)%20%5E2-%5Cln%20t%20%5Cright)%20%5E2%2B%5Cpi%20%5E2%7Ddt%7D%7D%0A%0A$

数值验证

选取不同的函数不同的围道还能得到更多的表达。

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