Error while writing code for bounding phase algorithm

x0=0;

step_size=input('Enter step size ');

a=x0-abs(step_size);

b=x0+abs(step_size);

F_x0=feval(objF,x0);

F_a=feval(objF,a);

F_b=feval(objF,b);

figure; hold on;

if (F_a>=F_x0 && F_x0>=F_b)

      step_size = 1*step_size;

else

          step_size = -step_size;

end

k=0;

F_p = feval(objF,k);

F_q = feval(objF,k+1);

   for k = 0:10
       if F_q<F_p
          x(0)=zeros(1, 1000000);
          x(k+1)=x(k)+2^k*step_size;
       else 
          xleft=x(k-1);
          xright=x(k+1);
       end
  end

ObjF is

function y=objF(x)
y=x^5-5*x^3-20*x+5;

Now matlab states that there are errors at F_x0=feval(objF,x0); and at y=x^5-5*x^3-20*x+5;

I absolutely cannot fathom the reason for these error messages. Can someone help me?

Answer 1

[Prem Kumar Tiwari]

It would've been great if you could've stated the error that Matlab shows. However, the issue is most probably because in a call to feval, the first argument has to be either a string that is name of the function like 'objF' or handle of the function like '@objF' without quotes of course.

To learn more about feval, check out the documentation here at : feval - Documentation

[Walter Roberson]

Right.

These days if you are given a function handle, you should typically use function syntax instead of feval. Like if

F = @objF;

then instead of

F_x0=feval(objF,x0);

you should code

F_x0 = F(x0);

feval() should be reserved for one of two cases:

1) You are doing some wizardry involving the symbolic engine, in which case your command would start feval(symengine, ....)

2) You are specifically trying to design a program that has to be able to accept function names as character vectors instead of the better function handles. Using function names as character vectors has a bunch of limitations, and is not recommended, but some people are given mandates to program for compatibility with MATLAB 4 before function handles existed.

Answer 2

[ADITYA SHARMA]

%Try this code.

clc;

clear;

global nf;

nf = 0;

k = 0;

syms x;

fx = input('Enter function in x: ');

b = 2; %Enter base for the exponent;

disp('f(x) = '); disp(fx);

while (1)

x0 = input('Enter initial guess value: ');

d = input('Enter a positive value for increment: ');

xi = [x0-d; x0; x0+d];

fx1 = eval(subs(fx,x,xi(1)));

fx2 = eval(subs(fx,x,xi(2)));

fx3 = eval(subs(fx,x,xi(3)));

fx_eval = [fx1; fx2; fx3];

disp('The column vector[x0-d; x0; x0+d] is xi = ');disp(xi);

disp('The column vector[f(x0-d); f(x0); f(x0+d)] is f(x) = '); disp(fx_eval);

if (fx1>=fx2 && fx2>=fx3)

break;

elseif (fx1<=fx2 && fx2<=fx3)

d = -1*d;

break;

else

disp('Enter new set of intial values: ');

end

end

xi(1) = x0-d;

fprintf('Intial value x0 = %.3f\n',x0);

fprintf('Increment d = %.3f\n',d);

fprintf('=====================================================================================================================================================|\n');

fprintf('\t|Iteration number| \t\t\t\t |x_values|\t\t\t\t\t\t\t\t |Function Evaluations at x_values|\t\t\t\t\t\t |Length|\n');

fprintf('=====================================================================================================================================================|\n');

while(1)

xi(3) = xi(2)+d*b^k;

fx1 = eval(subs(fx,x,xi(1)));

fx2 = eval(subs(fx,x,xi(2)));

fx3 = eval(subs(fx,x,xi(3)));

fx_eval = [fx1; fx2; fx3];

%fx3 = eval(subs(fx,x,xi(3)));

fprintf('\tk = %3d\t\t | \t',k);

fprintf('\t%f\t ',xi);

fprintf('|');

fprintf('\t\t%f',fx_eval); fprintf('\t|'); fprintf('\t\t%f',(xi(3)-xi(2)));

fprintf('\n');fprintf('-----------------------------------------------------------------------------------------------------------------------------------------------------|\n');

% disp('xi = ');disp(xi); disp('f(x) = '); disp(fx_eval);

if (fx3<fx2)

k = k+1;

xi(1) = xi(2); xi(2) = xi(3);

fx1 = fx2;fx2 = fx3;

nf = nf+1;

else

if d>0

fprintf('Minimum point lies in the interval (%.3f,%.3f)\n',xi(1),xi(3));

else

fprintf('Minimum point lies in the interval (%.3f,%.3f)\n',xi(3),xi(1));

end

fprintf('No. of iterations = %d \n',k);

fprintf('No. of function evaluations = %d \n',nf);

break;

end

end