Why is this variable undefined?

I have a function that uses gradient ascent to calculate a maximum, but I keep getting an error. fhandle is a function handle, x0 is a length 2 vector with initial guesses, tolx is the tolerance, and maxiter is the maximum iterations. This is the part of the function that gives me an error:

function [x,fval,exitFlag]=gradient_ascent_2D(fhandle,x0,TolX,MaxIter)
x1=x0(1);
x2=x0(2);
iter=0;
a=length(x0);
for i=1:MaxIter
    iter=iter+1

     *** dx1=(feval(fhandle,x1+.001*x1)-feval(fhandle,x1))/.001

      dx2=(feval(fhandle,x2+.001*x2)-feval(fhandle,x2))/.001
      dx=[dx1 dx2]
      if norm(dx)<Tolx
          x(1)=x1+dx1*.001
          x(2)=x2+dx2*.001
          break
      end

The stars mark the line the error occurs on. This is what the error reads:

Attempted to access x0(2); index out of bounds because numel(x0)=1.

Error in funcl (line 2)
test=x0(1)^3-7*x0(1)^2+4*x0(1)*x0(2)-2*x0(2)^2;

Error in gradient_ascent_2D (line 85)
    dx1=(feval(fhandle,b)-feval(fhandle,x1))/.001

For some reason x0(2) isn't defined because the length of x0 is 1, but I am inputting it as a length 2 vector. Any ideas what the problem is, or an easier way to do this?

[Guillaume]

Can you show the code for function funcl, as that's where the problem is.

As a side note there's no need for feval in your code, you could just do:

dx1 = (fhandle(x1+.001*x1) - fhandle(x1)) / 0.001;

[Andrew]

function test=funcl(x0)
    test=x0(1)^3-7*x0(1)^2+4*x0(1)*x0(2)-2*x0(2)^2;
end

It's just a simple function I've been using to test my gradient ascent function. Thanks, I didn't realize that either.

[Andrew]

I took out the feval part, now it can calculate dx1 and dx2, but it runs into the same problem later in the function. It happens on these lines:

if dx(1)<0
        xnew(1)=x0(1)-dx(1)*.001
        ** x0(1)=gs_max(fhandle,x0(1),xnew(1))

with the same error, x0 is not length 2. gs_max is another function I call to find the maximum using golden section between 2 bounds.

[Image Analyst]

Save me some trouble, and read http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup.

Answer 1

[Image Analyst]

Put these as the first lines in your gradient_ascent_2D function

whos x0
x0  % Don't use a semi colon

What do you see in the command window?

[Andrew]

>> gradient_ascent_2D(@funcl,[.29 .23],.001,1000) Name Size Bytes Class Attributes

x0        1x2                16  double

x0 =

    0.2900    0.2300

and then the same error as before.

Answer 2

[Guillaume]

You define xo as a vector of length 2 in gradient_ascend_2D, but the problem is in funcl, where you must have defined a different x0, this one of length 1.

If you want to use the xo of gradient_ascend_2D in funcl, you'll have to pass it as an argument.

[Guillaume]

Having now seen the code for funcl, I'll say use more descriptive variable names than x0, x1, x2, etc,

The problem is that the x0 local to funcl is not the same x0 local to gradient_ascend_2D. When you calculate dx1, the x0 local to funcl is just x1 from gradient_... while for dx2 it's just x2.

There seems to be some confusion on the purpose of all these xk variables.