Newtons Method, help calling anon function inside of function file

Ok so I have been tasked with writing a file myNewton which finds the root of any function given the listed inputs. I am maybe misunderstanding how to implement the method in general, but my matlab related question pertains more to the nature of calling anony functions inside of function files. I do not do this here but I anticipate that the best way to solve this problem is to do that. I am basically confused! Any help is appreciated, here is what I have so far.

function p = myNewton(f,fprime,x0,tol)
%input f, anonymous function for root finding problem
%input fprime, anonlymous function, the derivative of f
%input x0, an initial guess
%input tol, a tolerance (method will stop when successive iterates are within tol of each other)
%output p, a root f(p) = 0
    x(1) = x0 - (f(x0)/fprime(x0));
    k = 2;
    ex(1) = abs(x(1)-x0);%error
    while (ex(k-1) >= tol) %while my error is greater than tolerance
        x(k) = x(k-1) - (f(x(k-1))/fprime(x(k-1)));
        k = k+1;
    end
end

code to call function;

%As a test, the root of f(x) = x+2^x is -0.64118574
% define f, fprime, an initial guess (you can use 1), and use a tolderance of 1e-6
% send these into the function and store the result as the variable p
format long % to display more digits
x0=1;
tol=1e-6;
myNewton('x+2^x','diff(x+2^x)',x0,tol)

Answer 1

[Walter Roberson]

You are calling the anonymous function properly in your code.

Your problem is that you are not passing in an anonymous function. You are passing in character vectors.

myNewton(@(x) x+2^x, @(x) 2^x * log(2) + 1, x0, tol)

You cannot ask to diff(x+2^x) unless you have the symbolic toolbox.

[Peter M Hogan]

thank you. I wasnt sure if i could do what you did. I initially tried to define it like this;

f= @(x) x+2^x;
fprime= @(x) 2^x*log(2)+1;
x0=1
tol=1e-6
myNewton(f,fprime,x0,tol);

but i dont think that worked as well.

I am getting this error;

Index exceeds the number of array elements (1). Error in myNewton (line 10) while (ex(k-1) >= tol)

I understand what that means sort of but I am not sure how to fix that. I then get the error that

[Walter Roberson]

Your while loop assigns new values to x(k) but it does not calculate and assign new values to ex(k) so as soon as k increments then you run into difficulties.

[Peter M Hogan]

function p = myNewton(f,fprime,x0,tol)

%input f, anonymous function for root finding problem
%input fprime, anonlymous function, the derivative of f
%input x0, an initial guess
%input tol, a tolerance (method will stop when successive iterates are within tol of each other)
%output p, a root f(p) = 0
    x(1) = x0 - (f(x0)/fprime(x0));
    k = 2;
    ex(1) = abs(x(1)-x0);%error
    while (ex(k-1) >= tol) %while my error is greater than tolerance
        x(k) = x(k-1) - (f(x(k-1))/fprime(x(k-1)));
        k = k+1;
        ex(k)=abs(x(k)-x0);
    end
    p=x(k);
end

this is my new code. if you have any insights, i want p to express the root of the function. i am not sure where i am going wrong.

[Walter Roberson]

Where do you assign ex(2) ?

When k = 2 you use ex(k-1) which is ex(1), which you have assigned to specifically, so that part is good. You then increment k from 2 to 3, and assign to ex(3). Your while loop then tests ex(3-1) which is ex(2) but you did not assign to ex(2)