Nonlinear system using newton

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Aldo
Aldo 2016-11-10
编辑： Torsten 2016-11-10
I am supposed to find the solution to these two functions, and according to wolfram alpha there are two intersections between these two functions.
x=[0 2]'; iter=0; dxnorm=1;
disp(' x f J dx')
while dxnorm>0.5e-9 & iter<10
f=[((x(1)-4)/5)^2 + ((x(2)-6)/7)^2-1
10*(x(1)^10-5*x(1).^2+6*x(1)-1)-x(2)
];
J=[2/5*((x(1)-4)/5) 2/7*((x(2)-6)/7)
10*(3*x(1).^2-10*x(1)+6) -1];
dx=-J\f;
disp([x f J dx]), disp(' ')
x=x+dx;
iter= iter+1;
end
x, iter
my code give these solutions though, what am I doing wrong. And how do you solve it with 6 decimals?
Best regards

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采纳的回答

John D'Errico
John D'Errico 2016-11-10
Your code gave the correct solution!
Why do you think that Newton's method will yield both solutions from one starting value? That is clearly impossible. You should know, since you apparently wrote the code. You did, right?
Optimization tools will find ONE solution at best from ONE set of starting values. Sometimes they will diverge, or fail to converge for variety of reasons.
So just pick a different point to start it from.

更多回答（1 个）

Torsten
Torsten 2016-11-10
编辑：Torsten 2016-11-10
You'll get all solutions if you insert y from the second equation in the first equation and use MATLAB's "roots" command to determine the zeros of the resulting polynomial of order 6.
Best wishes
Torsten.

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