非線性方程組解法(一)Newton's method 阿新 • • 發佈：2022-03-05 Untitled 非線性方程組解法(一)Newton's methodformat long;x0=[1,1]';X=x0;d=1;num=0;F=[1,1]';while d > 10^(-6) DF=[2*x0(1)-1,2*x0(2);2*x0(1),-2*x0(2)-1]; F=-[x0(1)^2+x0(2)^2-x0(1);x0(1)^2-x0(2)^2-x0(2)]; s=DF\F; x1=x0+s; X=[X,x1]; d=max(abs(x1-x0)); num=num+1; x0=x1;endfigureplot(0:num,X,'.-')xlabel('迭代次數'),ylabel( '結果');legend('x1','x2')title('初始值為(1,0)時')syms x1 x2 x3a=[x1 x2 x3];F=[-1/81*cos(x1)+1/9*x2^2+1/3*sin(x3)-x1 1/3*sin(x1)+1/3*cos(x3)-x2 -1/9*cos(x1)+1/3*x2+1/6*sin(x3)-x3];x0=[1,1,1];[X,num,x0,ERR] = newton(F,a,x0);figure(1)plot(0:num,X','.-');xlabel('迭代次數'),ylabel('結果');legend('x1','x2','x3')title('初始值為(1,1,1)時' )figure(2)plot(1:num,ERR','.-');xlabel('迭代次數'),ylabel('殘差');title('初始值為(1,1,1)時')function [X,num,x0,ERR] = newton(fun,v,x0)%NEWTON 法求解非線性方程組DF=jacobian(fun,v);%求雅可比矩陣X=x0;num=0;err=1000;ERR=[];while err > 10^(-6) df=double(subs(DF,v,x0)); f=-double(subs(fun,v,x0)); s=df\f; x1=x0+s'; X=[X;x1]; Tf=double(subs(fun,v,x0)); err=max(abs(Tf-0)); num=num+1; x0=x1; ERR=[ERR,err];endend
