求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos

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求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos

求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos
求非线性方程组的“解析解”
-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0
-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0
-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)
其中x0,y0,z0已知,但不是某个具体的常数.所以就解出这个方程的“解析解”,a,b,c用x0,y0,z0的反三角函数表示.用matlab或mathmatics具体如何编程,希望给出可以运行的源代码.

求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos
程序很简单
[a,b,c]=solve('-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0','-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0','-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)','a','b','c');
但是matlab没有解出解析解.

求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos function [r,n]=mulSimNewton(F,x0,eps) % 用简化牛顿法求非线性方程组的一组解% 非线性方程组:F% 初始解:x0% 解的精度:eps% 求得的一组解:r% 迭代步数:n% 初始迭代一组解:x0=[x1:xn]if nargin==2eps=1.0e- 非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co 求Lingo高手帮帮忙,解非线性方程组.经常使用lingo的哥哥些帮帮忙,用“集”和for语言对解非线性方程组编程,如: 求用matlab解非线性方程组,可以复制的.求大神,高分悬赏 关于matlab中fsolve的使用当用fsolve解非线性方程组时,x=fsolve(fun,x0),x0是初始矩阵,麻烦最好举个例子说明下, 非线性方程组何时无解 matlab如何解非线性超标定方程我有一个三个未知数,九个方程的非线性方程组,cos(x3)*sin(x2)*sin(x1)-sin(x3)*cos(x1)=-0.9944 ;sin(x3)*sin(x2)*sin(x1)+cos(x3)*cos(x1)=-0.0870;cos(x2)*sin(x1)=-0.0606;cos(x3)*sin(x2)*cos(x1)+sin( MATLAB用高斯消去法解非线性方程组的代码 Newton法求非线性方程组的根 要求:(1)求满足条件的有根区间 (2)对分法二,三次缩小有根区间 (3)选取初值x0,使f(xo)*f(xo')>0 (4)Newton法 用matlab求解非线性耦合微分方程组其中m=1、k=0.1、a=1,而初始条件x0、y0、x'、y'是后输入的那种,求完整程序 求问,用matlab 解符号非线性方程组的时候出现 matlab Explicit solution could not be foundsyms a b c d e f g h i j k l m n o;p=f*cos(k)+g*sin(a)+o-i*sin(c)-h*sin(b);q=f*sin(k)+g*cos(a)+h*cos(b)+i*cos(c)-n;r=f*cos(k)+g*sin(a)+j*sin(d)+o-e 如何使用MATLAB解非线性方程组 怎么用matlab解非线性方程组 怎样利用解析式求方程组的解 matlab解非线性方程组,fun1.m 算出的是一个含有xyz的方程组,ex1.m来解方程组,如何做到?用X=fsolve('fun',X0,option) 的话,初值是个矩阵形式,如何解决啊?xyz用x(1)x(2)x(3)替换? 怎么做这个非线性回归方程,求K值如图的非线性回归方程,x,x0和t都已知,怎么用excel做出能得到k值的非线性回归方程 求cos x在x0处的极限