Minimize a function using Newton's Method

I am trying to minimise the function stated below using Newton's method, however I am not able to display a plot which illustrates points as they iterate down towards the minimum:

% Minimise z =(3-X).^2 + 30*((Y-(X.^2)).^2) with a starting point of x=y=0

% My implementation:

X = -4:0.01:4; % Declare range of x values 

[X,Y] = meshgrid(X);  % return 2-D grid coordinates
f =(3-X).^2 + 30*((Y-(X.^2)).^2); % Initiallize function for plotting
surf(X,Y,f,'FaceColor','y','FaceAlpha',0.3,'EdgeColor','none'); % Plot function 

syms X Y;
f =(3-X).^2 + 30*((Y-(X.^2)).^2);

x(1) = 0; % Initial value of x 
y(1) = 0;%  Initial value of y

df_dx = diff(f, X);
df_dy = diff(f, Y);
J = [subs(df_dx,[X,Y], [x(1),y(1)]) subs(df_dy, [X,Y], [x(1),y(1)])]; % Gradient
ddf_ddx = diff(df_dx,X);
ddf_ddy = diff(df_dy,Y);
ddf_dxdy = diff(df_dx,Y);
ddf_ddx_1 = subs(ddf_ddx, [X,Y], [x(1),y(1)]);
ddf_ddy_1 = subs(ddf_ddy, [X,Y], [x(1),y(1)]);
ddf_dxdy_1 = subs(ddf_dxdy, [X,Y], [x(1),y(1)]);
H = [ddf_ddx_1, ddf_dxdy_1; ddf_dxdy_1, ddf_ddy_1]; % Hessian
B = inv(H);

for i = 1:60
    A = [x(i),y(i)]';
    x(i+1) = A(1)-B(1,:)*J';
    y(i+1) = A(2)-B(2,:)*J';
    i = i+1;
    J = [subs(df_dx,[X,Y], [x(i),y(i)]) subs(df_dy, [X,Y], [x(i),y(i)])]; % Update Jacobian
    ddf_ddx_1 = subs(ddf_ddx, [X,Y], [x(i),y(i)]);
    ddf_ddy_1 = subs(ddf_ddy, [X,Y], [x(i),y(i)]);
    ddf_dxdy_1 = subs(ddf_dxdy, [X,Y], [x(i),y(i)]);
    H = [ddf_ddx_1, ddf_dxdy_1; ddf_dxdy_1, ddf_ddy_1]; % Update Hessian
    B = inv(H); % New Search Direction
end

% Plot:
hold on;

plot3(x,y,f,'o','Markersize',3,'Color','red'); % Illustrate descent 

Error using plot3
Data must be numeric, datetime, duration or an array convertible to double.

xlabel x; ylabel y; zlabel z;
title('Newton Method');
hold off
grid on;

Answer 1

[Matt J]

Perhaps you should be using the method that you accepted yesterday,

https://www.mathworks.com/matlabcentral/answers/1694790-gradient-descent-for-function-with-multiple-variables?s_tid=srchtitle

[Shikhar Singh]

Indeed I could use the Gradient Descent method, however I am trying to implement Newton's method as well so that I am able to compare and comment on which approach is more effective.

[Matt J]

Yes, but the method for displaying the trajectory taken by the algorithm should work just the same, regardless of whether it's Newton's method or a different method.

[Shikhar Singh]

I have also tried implementing the same method for displaying the trajectory taken by the algorithm:

hold on;
plot(x,y,f,'o','Markersize',3,'Color','red'); % Illustrate descent 
xlabel x; ylabel y; zlabel z;
title('Gradient Descent');
hold off
grid on;

However, it outputs the following error:

"Error using plot3 "

"Data must be numeric, datetime, duration or an array convertible to double."

[Matt J]

The code you have posted does not call plot3(), so it's hard to see how you would have gotten that error message.

[Shikhar Singh]

I have updated the code above with plot3 implemented.

[Matt J]

You cannot pass symbolic variables (like f) to plot3. In your previous post, you computed a sequence of z values that you fed to plot3,

z(i+1) = ((3-x(i+1)).^2) + (30.*(y(i+1)-(x(i+1).^2)).^2);

You have not done a similar thing here.