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5th root with Newton-Raphson code

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GEORGIOS BEKAS
GEORGIOS BEKAS 2018-3-11
编辑： John D'Errico 2018-3-11
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

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采纳的回答

GEORGIOS BEKAS
GEORGIOS BEKAS 2018-3-11
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
  1 个评论
John D'Errico
John D'Errico 2018-3-11
编辑：John D'Errico 2018-3-11
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.

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更多回答（1 个）

John D'Errico
John D'Errico 2018-3-11
编辑：John D'Errico 2018-3-11
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.
  1 个评论
GEORGIOS BEKAS
GEORGIOS BEKAS 2018-3-11
编辑：GEORGIOS BEKAS 2018-3-11
while abs(xold-xnew)<10^-3
xnew = xold - (xold*xold*xold*xold*xold)/(5*xold*xold*xold*xold)
xold=xnew
end

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