Matlab Bisection Algorithm code

% I'm trying to create a proper bisection code, and would like some advice on whether this code will...
%run properly. I am not sure how to test it, since every time I write a function on the 
%command window, it tells me "x" is an unrecognized variable

function[r,resarray] = bisect(f,a,b,tol,N)
    %f is my anonymous function, a and b are my guesses for where the root
    %might be, tol if the tolerance(in this case 10^-6) and N is the
    %maximum number of iterations
    f = f(a);
    
    k = 1;
    
    while (k<= N && abs(f)>tol)
       
        c = 0.5*(a+b);
       
        f = f(c);
       
        if f *f(a) <0
            b = c;
       
        else 
            a = c;
        end
        k = k+1;
    end
    r = c;
end

[Steven Lord]

Can you show us how you're trying to call it and the full and exact text of any warning and/or error messages you receive?

I'm guessing you're doing this as part of a homework assignment. Does your textbook have any worked examples that you can try to run using your code to check that you receive the same results?

Nowhere do you assign a value to resarray so if you ever call your function with two outputs it will error.

[Aaron Millan]

Hello! This is how I am attempting to call the function

bisect(sin(x),pi/2,1.5*pi,10^-6,100)

And this is the error message:

Unrecognized function or variable 'x'.

Answer 1

[Steven Lord]

This

bisect(sin(x),pi/2,1.5*pi,10^-6,100)

attempts to call the sin function with the contents of the variable x as input and use whatever that function call returns as the first input to your bisect function. From that error message x doesn't exist and so MATLAB can't evaluate that function call, but even if it did exist the output likely would be a numeric array (if x was a numeric array.)

You want to give bisect a function it can call with inputs of its choice, and for that the easy thing to do is to pass an anonymous function or a regular function handle.

bisect(@(x) sin(x), pi/2,1.5*pi,10^-6,100)
bisect(@sin, pi/2,1.5*pi,10^-6,100)

What's an example where you might want to call a function and pass whatever it returns to bisect? One possible scenario is the "bind pattern", a function that returns a function handle.

bindK = @(k) @(x) x.^k; % This function handle creates and returns a function handle

f = bindK(2) % a function handle that computes x.^2
g = bindK(pi) % computes x.^pi

bisect(bindK(3), ...) % Use bisect on essentially @(x) x.^3

[Aaron Millan]

What a helpful answer! Thank you!