Representing the square root of a number n as an infinite fraction. Alright, I want to understand how I can use solving quadratics as continued fractions to approximate the square root of a number.

Question

inkyvoyd · Answer

if x^2-n=0,
x=n/x
using recursion,
x=n/(n/x)
x=n/(n/(n/x))
How exactly would I make an approximation?

inkyvoyd · Answer

More info at wikipedia's page on solving quadratics with continued fractions.

anonymous · Answer

maybe like this:

$$x^2 - n = 0$$$$x^2 = n$$

$$x = \frac{n}{x}$$

$$2x = \frac{n}{x} + x $$

$$x= \frac{(n/x) + x}{2}$$
$$x_{n+1}= \frac{(n/x_{n}) + x_{n}}{2} $$

anonymous · Answer

pretty proud of that, just came up with it on the spot :D

inkyvoyd · Answer

OMG THATS THE BABYLONIAN METHOD

anonymous · Answer

so eg n = 2
start with $$x_0 = 1$$
x is approximately:
(2/(((2/1)+1)/2))) +  ((2/1)+1)/2)/2 (Bracket overload)

anonymous · Answer

babylonian? cool!

anonymous · Answer

i just tried making a quadratic with x as a root

anonymous · Answer

is it important?