Newton method for nonlinear equations

조회 수: 10 (최근 30일)
Ana Garvanlieva
Ana Garvanlieva 2015년 3월 13일
댓글: Torsten 2015년 3월 24일
I have the following system of non-linear equations:
f(x)={(x(1)^5 + x(2)^3*x(3)^4 + 1)
(x(1)^2*x(2)*x(3))
(X(3)^4 - 1)}
Do you know, and can you help me with the code for the Newton method. As help I have instructions to note some difficulties with convergence and "As a remedy implement a damped Newton modification using the Armijo-Goldstein criterion."
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이전 댓글 5개 표시
Torsten
Torsten 2015년 3월 17일
I wonder why you don't use the Jacobian you calculated in an earlier thread:
n=20;
f=@(x)[x(1)^5+x(2)^3*x(3)^4+1 ; x(1)^2*x(2)*x(3) ; x(3)^4-1 ];
Df=@(x)[5*x(1)^4 3*x(2)^2 4*x(3)^3
2*x(1)*x(2)*x(3) x(1)^2*x(3) x(1)^2*x(2)
0 0 4*x(3)^3];
x = [.1;.1;.1]; % starting guess
for i = 1: n
Dx = -Df ( x )\ f( x ); % solve for increment
x = x + Dx; % add on to get new guess
f (x ); % see if f(x ) is really zero
end
Best wishes
Torsten.
Ana Garvanlieva
Ana Garvanlieva 2015년 3월 17일
Thank you for replaying Torsten... So i did this: in Command window i wrote:
syms x1 x2 x3 J = jacobian([x1^5 + x2^3*x3^4 + 1; x1^2*x2*x3; x3^4-1],[x1;x2;x3]);
And i run the code from your post. It gives me:
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 6.940734e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 7.145906e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 7.359285e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 7.579619e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 7.806710e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 8.040648e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 8.281610e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 8.529801e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 8.785447e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 9.048804e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 9.320218e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 9.600300e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 9.890515e-020.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.019509e-019.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.052724e-019.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.092946e-019.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.154018e-019.
> In proba at 8
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 1.281592e-019.
> In proba at 8

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Torsten
Torsten 2015년 3월 17일
You don't need to write the above lines in the command window - just execute the code above.
Can you output x and f(x) in the for-loop ? What values do you get ?
The x-values should either converge towards (-1,0,1) or towards (0,-1,1).
The f(x)-values should converge towards (0,0,0).
If this is not the case, use the hint to implement the damped Newton method.
In principle, this means that you replace the line
x = x + Dx; % add on to get new guess
by
x = x + lambda*Dx; % add on to get new guess
where 0 < lambda < 1 is calculated according to some rule.
Best wishes
Torsten.
  댓글 수: 10
이전 댓글 8개 표시
Torsten
Torsten 2015년 3월 24일
No, it's not ok. As I said before, the values of x must tend towards (-1,0,1) or towards (0,-1,1).
The f(x)-values should converge towards (0,0,0).
Maybe this code works better:
n=20;
f=@(x)[x(1)^5+x(2)^3*x(3)^4+1 ; x(1)^2*x(2)*x(3) ; x(3)^4-1 ];
Df=@(x)[5*x(1)^4 3*x(2)^2 4*x(3)^3
2*x(1)*x(2)*x(3) x(1)^2*x(3) x(1)^2*x(2)
0 0 4*x(3)^3];
T=@(x)0.5*(f(x))'*f(x);
tau=0.5;
x = [-.01;-.01;-.01]; % starting guess
for i = 1: n
Dx = -Df (x)\f(x); % solve for increment
lambda=1.0;
while T(x+lambda*Dx)-T(x) > -0.5*lambda*T(x)
lambda=tau*lambda;
end
x=x+lambda*Dx;
f(x); % see if f(x) is really zero
fprintf('value of x: %d\n', x);
fprintf('value of function: %d\n', f(x));
end
Best wishes
Torsten.
Torsten
Torsten 2015년 3월 24일
Jacobian is wrong ; use this version:
n=20;
f=@(x)[x(1)^5+x(2)^3*x(3)^4+1 ; x(1)^2*x(2)*x(3) ; x(3)^4-1 ];
Df=@(x)[5*x(1)^4 3*x(2)^2*x(3)^4 x(2)^3*4*x(3)^3
2*x(1)*x(2)*x(3) x(1)^2*x(3) x(1)^2*x(2)
0 0 4*x(3)^3];
T=@(x)0.5*(f(x))'*f(x);
tau=0.5;
x = [-.01;-.01;-.01]; % starting guess
for i = 1: n
Dx = -Df (x)\f(x); % solve for increment
lambda=1.0;
while T(x+lambda*Dx)-T(x) > -0.5*lambda*T(x)
lambda=tau*lambda;
end
x=x+lambda*Dx;
f(x); % see if f(x) is really zero
fprintf('value of x: %d\n', x);
fprintf('value of function: %d\n', f(x));
end
x should converge towards (-1,0,-1), f(x) should converge towards (0,0,0).
Best wishes
Torsten.

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