How is this method for computing square roots a manifestation of Newton's method?

Question

I discovered that we can use Newton's method to conveniently compute the square root of a positive integer.

Newton's method stipulates that we keep iterating as such:

$$ x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

However, apparently, we can just compute the square root by keep iterating on

$$ x = \frac{1}{2} \left( x + \frac{a}{x} \right) $$

where $a$ is the number we are finding the square root of and $x$ is our eventual answer.

I don't understand how this method for computing square roots is a manifestation of Newton's method though.

Did you try to put $f(x)=x^2-a$ in Newton's method and simplifying the resulting expression? — Harald Hanche-Olsen, Jun 05 '14 at 19:36

score 4 · Accepted Answer · answered Jun 05 '14 at 19:36

4

Let $f(x) = x^2 - a$. Then

$$x - \frac{f(x)}{f'(x)} = x - \frac{x^2-a}{2x} = \frac{2x^2 - (x^2-a)}{2x} = \frac{x^2+a}{2x} = \frac{1}{2}\left(x + \frac{a}{x}\right).$$

answered Jun 05 '14 at 19:36

Daniel Fischer

1 Answers1