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十甫寸木蟲 (正式寫手)
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[求助]
擬牛頓法求解非線性方程組——老是出錯,跪求大神指點~(或提供一種新的算法)
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function [r,m]=mulVNewton(F,x0,A,eps) %方程組:F %方程組的初始解:x0 % 初始A矩陣:A %解的精度:eps %求得的一組解:r %迭代步數(shù):m clear clc syms x1 x2 x3 f1 f2 f3 D=input('請輸入頻帶的實測數(shù)據(jù)[q,r,v]:'); q=D(:,1);%第一列 r=D(:,2); v=D(:,3); %將實測數(shù)據(jù)代入方程組中 nn=length(D); F1=0;F2=0;F3=0; for i=1:nn f1=(v(i)-(x1)*q(i)^(x2)*r(i)^(x3))*q(i)^(x2)*r(i)^(x3); f2=(v(i)-(x1)*q(i)^(x2)*r(i)^(x3))*q(i)^(x2)*r(i)^(x3)*log(q(i)); f3=(v(i)-(x1)*q(i)^(x2)*r(i)^(x3))*q(i)^(x2)*r(i)^(x3)*log(r(i)); F1=F1+f1;F2=F2+f2;F3=F3+f3; end disp('F1='); disp(F1); disp('F2='); disp(F2); disp('F3='); disp(F3); F(1)=F1;F(2)=F2;F(3)=F3; x=[x1,x2,x3]; x0=[50,0.6,-1.4]; A=[1,0,0; 0,1,0;0,0,1]; if nargin==1 A=eye(length(x0)); %A取為單位陣 eps=1.0e-4; else if nargin==2 eps=1.0e-4; end end x0 = transpose(x0); Fx = subs(F, findsym(F),x0); r=x0-Fx/A; m=1; tol=1; while tol>eps x0=r; Fx = subs(F, findsym(F),x0); r=x0-Fx/A; y=r-x0; Fr = subs(F, findsym(F),r); z= Fr-Fx; A1=A+(z-A*y)*transpose(y)/norm(y); %調(diào)整A A=A1; m=m+1; if(m>100000) %迭代步數(shù)控制 disp('迭代步數(shù)太多,可能不收斂!'); return; end tol=norm(r-x0); end 輸入數(shù)據(jù)為[q,r,v]: [5.6,15.62,1.431 5.6,22.36,1.067 5.6,29.73,0.582 5.6,36.4,0.516 5.6,43.17,0.348 9.4,20.59,1.433 9.4,27.86,0.964 9.4,35.44,0.632 9.4,49.03,0.376 12.8,27.31,1.204 5.6,18.03,1.586 5.6,32.57,0.633 5.6,39.29,0.601 5.6,46.1,0.418 11.6,22.36,1.332 11.6,29.73,0.911 11.6,37.36,0.549 11.6,44.15,0.479 11.6,50.99,0.386] |

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