5th root with Newton-Raphson code

This is a code I have created to calculate numerically the 5th root. Something is wrong.

x = 0.05;
x_old = 500;
iter = 0;
while abs(x_old-x) > 10^-3 && x ~= 0
    x_old = x;
    x = x - (5*x*x*x*x);
    iter = iter + 1;
end

Answer 1

[GEORGIOS BEKAS]

If n is the number whose 5th root has to be calculated:

function f = fifth_root(n)
  xold=10;
  for i = 1:100
  xnew = xold - (xold*xold*xold*xold*xold-n)/(5*xold*xold*xold*xold);
  xold=xnew;
  end
  f=xnew
  end

[John D'Errico]

Yes. That will work. And very good of you to accept your own answer when I spent the time to find the bug in your code and explain how to fixit.

Answer 2

[John D'Errico]

What is wrong? I'd start by pointing out that this is NOT Newton's method as you wrote it.

https://en.wikipedia.org/wiki/Newton%27s_method

Newton (or Newton/Raphson if you wish) has

xnew = xold - f(xold)/f'(xold)

So, if f(x) = x^5 - y, then what should you have written there? So, y is the value you want to compute a root of. Clearly, the 5th root of y will produce zero in f, as I wrote it.

So now you need to write the Newton iteration to use fas I wrote it, as well as the derivative of f.

[GEORGIOS BEKAS]

while abs(xold-xnew)<10^-3
xnew = xold - (xold*xold*xold*xold*xold)/(5*xold*xold*xold*xold)
xold=xnew
end