How Newton's method is reduced to Babylonian?

Question

I can't figure out how Newton's formula gets transformed into Babylonian in this article: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots

How is this possible?

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^2-S}{2x_n}=\frac{1}{2}(x_n+\frac{S}{x_n})$

I mean, I feel like I'm missing some basic algebraic knowledge to simplify $x_n-\frac{x_n^2-S}{2x_n}$ into $\frac{1}{2}(x_n+\frac{S}{x_n})$.

P.S. I need this to make my own square root function implementation, but I don't want to use code that I don't understand.

Answer 1

You're probably overthinking it; you can manipulate

$$ x_n - \frac{x_n^2 - S}{2x_n}$$

by finding a common denominator and then subtracting two fractions:

$$ x_n - \frac{x_n^2 - S}{2x_n} = \frac{2 x_n^2}{2x^n} - \frac{x_n^2 - S}{2x_n} = \frac{2x_n^2 - (x_n^2 - S)}{2x_n} = \frac{x_n^2 + S}{2x_n} $$

Answer 2

Thanks to Hurkly (and I have to admit, simbolab site too) now I see the full picture:

$$ x_n - \frac{x_n^2 - S}{2x_n} = \frac{2 x_n^2}{2x^n} - \frac{x_n^2 - S}{2x_n} = \frac{2x_n^2 - (x_n^2 - S)}{2x_n} = \frac{x_n^2 + S}{2x_n}= $$ Basic fraction rule -----> $$\frac{x_n^2}{2x_n}+\frac{S}{2x_n}=$$ Cancel x-s -----> $$\frac{x_n}{2}+\frac{S}{2x_n}=$$ Take out the common factor -----> $$\frac{1}{2}(x_n+\frac{S}{x_n}).$$

Seems like it was 6-7th grade problem... Sorry if I wasted somebody's time. This site seems to be full of very serious questions.

Sometimes I feel sad that wikipedia and most of other math sites/books (exept mathisfun, ofc) don't show the full transformation for people like me... Then again, they are made for people that know math very good, to know it even better.