Steepest Ascent Method to find maximum

I am looking for a problem in my implementation for steepest ascent method. My answer is in a long equation form with h in it. But I am calculating h in the loop still ther is no value of it for the answer. The solution is at the end of the function

function STPAscent(x,theta,TC)

x0 =1500;

theta0=pi/6;

i=0;

i_max=100;

ea=5;

es=1;

while (es<double(subs(ea)) && i<i_max)

fx = tan(theta0)/(2*100^2*(cos(theta0))^2) + (2*9.81*x0)/(2*100^2*(cos(theta0))^2); % derivative wrt "x"

ftheta= x0*(sin(theta0)*(100*sin(theta0)+981*x0*cos(theta0))+50)/(1000000*(cos(theta0))^4); % derivative wrt "theta"

syms h

x=x0+fx*h;

theta=theta0+ftheta*h;

f= 1000 + (x*tan(theta))/(2*100^2*(cos(theta))^2) + (9.81*x^2)/(2*100^2*(cos(theta))^2);

c=diff(f,h);

assume(h,'clear')

h=solve(c==0,h,'PrincipalValue',true);

x_new=x+fx*h;

theta_new=theta0+ftheta*h;

ea=abs((x_new-x0)/x_new)*100;

x0=x_new;

theta0=theta_new;

i=i+1;

end

fxtheta = 1000 + (x0*tan(theta0))/(2*100^2*(cos(theta0))^2) + (9.81*x0^2)/(2*100^2*(cos(theta0))^2)

simplify(fxtheta)

end

Command Window

Warning: Unable to solve symbolically. Returning a numeric

solution using vpasolve.

> In sym/solve (line 304)

In STPAscent (line 18)

fxtheta =

(981*((8836235812531391*h)/4503599627370496 + 1499.9993954193889930311396801317)^2)/(2000000*cos(pi/6 - 0.52363352427033321420523588398365)^2) + (tan(pi/6 - 0.52363352427033321420523588398365)*((8836235812531391*h)/4503599627370496 + 1499.9993954193889930311396801317))/(20000*cos(pi/6 - 0.52363352427033321420523588398365)^2) + 1000

ans =

0.0018882263675909642813601409431637*h^2 + 2.8871384744661982638760370739896*h + 2103.6241090862540686603690416302

[Torsten]

There is a weird mix of symbolic and numeric variables in your code.

Updating x as

x_new=x+fx*h;

has two different h's in it: The x contains a symbolic h, the h in f*h is still symbolic, but has a value that comes from

h=solve(c==0,h,'PrincipalValue',true);

In my opinion, it's nonsense to use symbolic maths in numerical optimization because it slows down the computational speed and gives problems from the mix as in your code above.

[Syed Abdul Rafay]

But h is unknown and if I don't use symbolic function the code doeasn't proceed with error unrecognized variable 'h'. I have to use the h value calculate in x new to get a updated value of x .

[Torsten]

You can define a function dependent on an unknown h for which you want to find the root and use "fzero" or "fsolve" to solve for h.

That's exactly what "solve" will do in your case because it doesn't find an analytical solution for h.