matlab解方程组解析解

您所在的位置:网站首页 二元组形式 matlab解方程组解析解

matlab解方程组解析解

2022-12-20 16:41| 来源: 网络整理| 查看: 265

问题描述

对于一个如下形式的方程组 { y 1 = f 1 ( x 1 , x 2 , . . . , x n ) y 2 = f 2 ( x 1 , x 2 , . . . , x n ) . . . y m = f m ( x 1 , x 2 , . . . , x n ) \begin{cases} y_1 = f_1(x_1, x_2, ..., x_n)\\ y_2 = f_2(x_1, x_2, ..., x_n) \\ ...\\ y_m = f_m(x_1, x_2, ..., x_n) \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧​y1​=f1​(x1​,x2​,...,xn​)y2​=f2​(x1​,x2​,...,xn​)...ym​=fm​(x1​,x2​,...,xn​)​ 需要求解其反函数,即 { x 1 = g 1 ( y 1 , y 2 , . . . , y m ) x 2 = g 2 ( y 1 , y 2 , . . . , y m ) . . . x n = g m ( y 1 , y 2 , . . . , y m ) \begin{cases} x_1 = g_1(y_1, y_2, ..., y_m)\\ x_2 = g_2(y_1, y_2, ..., y_m) \\ ...\\ x_n = g_m(y_1, y_2, ..., y_m) \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧​x1​=g1​(y1​,y2​,...,ym​)x2​=g2​(y1​,y2​,...,ym​)...xn​=gm​(y1​,y2​,...,ym​)​

matlab求解

可以通过matlab的slove函数可以实现上述功能

syms x1 x2 .... xn, y1 y2 ... ym eq1 = y1 == f_1(x_1, x_2, ..., x_n) eq2 = y2 == f_2(x_1, x_2, ..., x_n) ... eqm = ym == f_m(x_1, x_2, ..., x_n) [x1_, x2_, ..., xn_] = solve(eq1, eq2, eq3, ..., eqm, x_1, x_2, ..., x_n) 具体实例 syms kappa d phi varepsilon_0 varepsilon_a varepsilon_b varepsilon_c equ1 = varepsilon_a == kappa*d*cos(-phi) + varepsilon_0; equ2 = varepsilon_b == kappa*d*cos(-phi+2/3*pi) + varepsilon_0; equ3 = varepsilon_c == kappa*d*cos(-phi+4/3*pi) + varepsilon_0; result = solve(equ1,equ2,equ3, kappa, phi, varepsilon_0)


【本文地址】


今日新闻


推荐新闻


CopyRight 2018-2019 办公设备维修网 版权所有 豫ICP备15022753号-3