Alternative Newton-Method using for-loop

I have a task that requires me to create an alternative Newton-Raphson Method - not using a derivative function - described by:

where the delta value is given and its very close to 0. The input data should be a function f, the initial approach x0, the required minimum accuracy TolX for the stop criterion | (xk+1 —xk) / xk+1 | and the maximum number of iterations MaxIter. The results should be the approximation of the root z, the value of the function in the root fz and a vector containing iter iterations.

This is the code I have so far:

function [z, fz, iter] = quasinewton(f, x0, TolX, MaxIter)

% f = x^3-x^2-x+1 TolX = 1e-2

delta = 1*10^(-5);

iter = zeros(MaxIter+1,1); %Pre-allocates memory for iter vector

iter(1) = x0; %x0 = -0.2

for k = 1:MaxIter

fz = feval(f, iter(k));

dfdx = fz / (feval(f, iter(k) + delta) - fz);

iter(k+1) = iter(k) - (dfdx * delta); %Calculates new iteration

err = (iter(k+1)-iter(k))/iter(k+1); %calculates the error of sucessive iterations

fz = feval(f, iter(k+1));

if abs(err) < TolX, break; end

end

z = iter(k+1);

if k == MaxIter

fprintf('The best in %d iterations\n', MaxIter)

end

It keeps giving me the wrong iterations and instead of converging to 0 it increases iteration after iteration. I really don't understand what could be wrong. I'm new in Matlab. Any suggestions?

Answer 1

[Alan Stevens]

It works for me ok using the function f = x^3-x^2-x+1 and initial guess x0 = -2. However, replace

if k == MaxIter
fprintf('The best in %d iterations\n', MaxIter)
end

by

fprintf('The best in %d iterations is f = %f at x = %f \n', k, f(z), z)

[Dasneves]

Thank you very much for dedicating a little bit of your time to respond me. I have now detected what I was doing wrong when running the function in the command window. I've also done what you suggested. :)

Answer 2

[Alan Stevens]

This

f = @(x) x^3-x^2-x+1;
MaxIter = 100;
TolX = 1e-2;
x0 = -0.2;
delta = 1*10^(-5);
iter = zeros(MaxIter+1,1); %Pre-allocates memory for iter vector
iter(1) = x0; %x0 = -0.2
for k = 1:MaxIter
    fz = feval(f, iter(k));
    dfdx = fz / (feval(f, iter(k) + delta) - fz);
    iter(k+1) = iter(k) - (dfdx * delta); %Calculates new iteration
    err = (iter(k+1)-iter(k))/iter(k+1); %calculates the error of sucessive iterations
    fz = feval(f, iter(k+1));
    if abs(err) < TolX, break; end
end
z = iter(k+1);
fprintf('The best in %d iterations is f = %f at x = %f \n', k, f(z), z)

results in

The best in 9 iterations is f = 0.000102 at x = 1.007135