Loop with Bisection Method Error (for may not be aligned etc.)

I have a bisection method function that is running inside a loop to generate solutions for multiple values p. However, I'm getting an error that says "FOR may not be aligned with its matching end" yet there are the correct number of end statements for the loops in my code. What can I do to solve this?

k = 0.05;
p = 1:.2:4;

%preallocate solution vectors
x1 = zeros(length(p),1);
approx_err1 = zeros(length(p),1);
iter = zeros(length(p),1);

for p = 1:.2:4

      fun = @x x/(1-x) * sqrt((2p)/(2+x)) - k;
      es = 1*10^-4; %stopping tolerance
      xl = 0.01; xu = 0.1; xroot = xl; %initial guesses

      %bisection function
      for iter 1:50

          xr_old = xroot;
          xroot = (xl + xu)/2;
          ea = (xroot - xr_old)/xroot;
          test = fun(xl) * fun(xroot);

          if abs(ea) <= es, break, end

          if test < 0
              xu = xroot;

          elseif test > 0
             xl = xroot;

          else 
              ea = 0; %test = 0, means found exact root

          end    
      end
      %end bisection function

      x1 (p,:) = xroot;
      approx_err1 (p,:) = ea;
      iter (p,:) = iter;
end

Answer 1

[Geoff Hayes]

Josh - FOR may not be aligned with its matching end is a warning only and can be ignored for the moment. In fact, once you resolve the errors, this warning message will disappear. With line

 fun = @x x/(1-x) * sqrt((2p)/(2+x)) - k;

there is a Parse error at x: usage might be invalid MATLAB syntax. For anonymous functions, you need to wrap the input variable(s) in parentheses like

 fun = @(x) x/(1-x) * sqrt((2p)/(2+x)) - k;

This will lead to another error message invalid syntax at p. What is the intent here? Do you wish to multiply p by 2 or divide or ..? You must supply the correct operator.

Once that has been corrected, there will be another parse error at the line

 for iter 1:50

You are missing an equals sign to indicate which values the indexing variable iter is meant to take on

 for iter=1:50

And that is it. The warning message disappears and you should be good to execute the code....except for the

 Subscript indices must either be real positive integers or logicals.
 Error in xxxx (line 27)
      x1 (p,:) = root;

which is telling you that your p is not a real positive integer. In this case it is a double since p is the indexing variable of the outer for loop

 for p = 1:.2:4

What are you trying to assign to x1 and why are you trying to use p? What is intended here?