MATLAB: Using ‘fsolve’

fsolveturning points

I have the function f(x,y) = x.^(2-x.^(0.5))+y.^(2-y.^(0.5)) and I found the Jacobian and Hessian Matrices. Now I need to find the turning point using the function "fsolve" and stating its nature.

Can anyone help me?

Thanks in advance.

Best Answer

Hi,

why not solve it symbolic:

syms f(x,y)
f(x,y) =  x.^(2-x.^(0.5))+y.^(2-y.^(0.5));
[xsol,ysol] = vpasolve(diff(f,x) + diff(f,y) == 0, [x,y], [1 3; 1 3]);
zsol = subs(f,[x,y],[xsol,ysol]);
% plot results
fsurf(f)
hold on
scatter3(double(xsol),double(ysol),double(zsol),'or','LineWidth',2,'MarkerFaceColor','r')

Related Solutions

MATLAB: How to find total derivative

Hey Kanav,

Check out "diff" in the Symbolic Math Toolbox. One thing to note is that for the form you want, you need to explicitly define y as a function of x.

syms x y(x)
diff(cos(x) - sin(y)*x,x)

returns

- sin(y(x)) - sin(x) - x*cos(y(x))*diff(y(x), x)

-Adam

MATLAB: How to calculate the minimum of a function of two variables

One approach:

syms x y 
f(x,y) = exp(x-y) + x^2 + y^2;
Dfx = diff(f, x)
D2fx = diff(Dfx,x)
Dfy = diff(f, y)
D2fy = diff(Dfy, y)
Inflpt = solve(Dfx, Dfy, [x,y])
xs = Inflpt.x
ys = Inflpt.y
Check = [D2fx(xs,ys); D2fy(xs,ys)]
figure
fsurf(f, double([xs-2 xs+2    ys-2 ys+2]))

Since ‘Check’ is positive for both functions, ‘Inflpt’ is a minimum.

EDIT —

Added plot figure —

Best Answer

Related Solutions

MATLAB: How to find total derivative

MATLAB: How to calculate the minimum of a function of two variables

Related Question