How can I plot a graph according to the results

This is my code for find the minimum of a function using Newton's Method. Here I need to plot a graph according to the results. But I cannot figure out which parameters to use. Hope for a little help)

This is the example of the graph which I need.

Thank you.

clc
clear

%Newton method
func = @(x) 10.*x(1).^2 - 4.*x(1).*x(2) + 7.*x(2).^2 - 4.* sqrt5 (5.*x(1) - x(2)) - 16 ;

e = 0.01;
i=0;
n=0;

b = [-20*sqrt(5); 4*sqrt(5)];
x =[0 10]';
[H,antiGrad,F]=QF(x(1),x(2));
n = n + 7;

norm_aG = norm(antiGrad);
Xmin=-inv(H)*b;
Points(:,i+1)=x;
funcvalues(1)=F;

while(norm_aG > e)
i=i+1;
del=linsolve(H,antiGrad);
x=x+del;
[H,antiGrad,F]=QF(x(1),x(2));
n=n+7;
norm_aG=norm(antiGrad);
Points(:,i+1)=x;
funcvalues(i+1)=F;

end

fprintf('Precision parameter %.4f\n',e);

Precision parameter 0.0100

fprintf('Point of minimum [%.4f ,%.4f]\n', x(1),x(2));

Point of minimum [2.2361 ,0.0000]

fprintf('Min: %10.4f \n', F);

Min:   -66.0000 

fprintf('Iterations: %d \n',i );

Iterations: 1 

fprintf('Number of calculated objective function values: %d\n\n',n );

Number of calculated objective function values: 14

% drawGraph(Xmin, Points, func);

function [H,antiGrad,I]=QF(x,y)
derivative_x = 20*x - 4*y - 20*sqrt(5);
derivative_y = - 4*x + 14*y + 4*sqrt(5);
antiGrad = -[derivative_x,derivative_y]';
H = [20 -4 ; -4 14];
I = x.^2.*10 + 4.*x.*y + 7.*y.^2 - 4.*sqrt(5).*(5*x-y) - 16;

%contour(x, y, I, func,'r', 'showText', 'on');
%plot(Points(1,:), Points(2, :), 'g-*');

end

[Janidu]

I want to plot like above mentioned graph. Yes iteration is one. So there should be one red circle like that.

[Torsten]

I think it will be a rotated ellipse instead of a circle.

Answer 1

[Torsten]

clc
clear

%Newton method
func = @(x,y) 10*x.^2 - 4*x.*y + 7*y.^2 - 4*sqrt(5)*(5*x-y) - 16 ;

e = 0.01;
i=0;
n=0;

b = [-20*sqrt(5); 4*sqrt(5)];
x =[0 10]';
[H,antiGrad,F]=QF(x(1),x(2));
n = n + 7;

norm_aG = norm(antiGrad);
Xmin=-inv(H)*b;
Points(:,i+1)=x;
funcvalues(i+1)=F;

while(norm_aG > e)
i=i+1;
del=linsolve(H,antiGrad);
x=x+del;
[H,antiGrad,F]=QF(x(1),x(2));
n=n+7;
norm_aG=norm(antiGrad);
Points(:,i+1)=x;
funcvalues(i+1)=F;
end

hold on
plot(Points(1,:),Points(2,:))
fimplicit(@(x,y)func(x,y)-funcvalues(1),[-15 15])
hold off

fprintf('Precision parameter %.4f\n',e);

Precision parameter 0.0100

fprintf('Point of minimum [%.4f ,%.4f]\n', x(1),x(2));

Point of minimum [2.2361 ,0.0000]

fprintf('Min: %10.4f \n', F);

Min:   -66.0000 

fprintf('Iterations: %d \n',i );

Iterations: 1 

fprintf('Number of calculated objective function values: %d\n\n',n );

Number of calculated objective function values: 14

% drawGraph(Xmin, Points, func);

function [H,antiGrad,I]=QF(x,y)
derivative_x = 20*x - 4*y - 20*sqrt(5);
derivative_y = - 4*x + 14*y + 4*sqrt(5);
antiGrad = -[derivative_x,derivative_y]';
H = [20 -4 ; -4 14];
I = x.^2.*10 - 4.*x.*y + 7.*y.^2 - 4.*sqrt(5).*(5*x-y) - 16;

%contour(x, y, I, func,'r', 'showText', 'on');
%plot(Points(1,:), Points(2, :), 'g-*');

end

[Janidu]

Thank you very much!

[Torsten]

I had to correct I in your code:

I = x.^2.*10 - 4.*x.*y + 7.*y.^2 - 4.*sqrt(5).*(5*x-y) - 16;

instead of

I = x.^2.*10 + 4.*x.*y + 7.*y.^2 - 4.*sqrt(5).*(5*x-y) - 16;

[Janidu]

ahh got it! Thank you very much again!!