1、用割线法解非线性方程组自动化学院 1011203050 陈晓祺拟牛顿法解下列方程组 0sin1.0co5. )(si21xx先将拟牛顿法的程序代码如下Functionr,m=mulVlineF,x0,A, eps)Format long;If nargin=3Eps=1e-4;EndX0=transpose(x0);X1=tanspose(x1);N=length(x0);Fx=subs(F,findsym(F),x0);Fx1=subs(F,findsym(F),x1);H=x0-x1;J=zeros(n,n);Xt=x1;Xt(1)=x0(1);J(:,1)=(subs(F,findsym
2、(F),xt)-sbus(F,findsym(F),x1)/h(1);For i=2:nXt=x1;Xt(1:i)=x0(1:i);Xt_m=xl;Xt_m(1:i-1)=x0(1:i-1);J(:,i)=(subs(F,findsym(F),xt)-sbus(F,findsym(F),xt_m)/h(1);EndR=x1-inv(J)*fx1;M=1;Tol=1;While tolepsX0=x1;X1=r;Fx=subs(F,findsym(F),x0);Fx1=subs(F,findsym(F),x1);H=x0-x1;J=zeros(n,n);Xt=x1;Xt(1)=x0(1);J(:
3、,1)=(subs(F,findsym(F),xt)-subs(F,findsym(F),x1)/h(1);For i=2:nXt=x1;Xt(1:i)=x0(1:i);Xt_m=x1;Xt_m(1:i-1)x0(1:i-1);J(:,1)=(subs(F,findsym(F),xt)-subs(F,findsym(F),xt_m)/h(1);EndR=x1-inv(J)*fx1;Tol=norm(r-x1);M=m+1;If(m100000)Disp(fail);Return;EndEnd然后直接调用该方法Syms x yZ=x2+y2-5;2*x-y-3;r,m=mulline(z,0 1,4,3)r,m=mulline(z,-3 1,4,3)得到解为（2,1） （0.4，-2.2）
