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I'm looking this recursive formula to approximate $\sqrt{2}$

\begin{align*} x_1 &= 2 \\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \end{align*}

I noticed an easier way to approximate by hand would be to think of each $x_k$ as a rational number $x_k = \frac{a_k}{b_k}$ which when substituted into the formula above produces

\begin{align*} x_1 &= \frac{a_1}{b_1} =\frac{2}{1} \\ x_{n+1} &= \frac{a_{n+1}}{b_{n+1}} = \frac{a_n^2 + 2b_n^2}{2a_nb_n} \end{align*}

But what's interesting is that

\begin{align*} x_1 = &\frac{2}{1} \\ &x_2 = \frac{3}{2} \\ x_3 = \frac{9+8}{12} \\ x_4 &= \frac{289 + 288}{408} \\ x_5 = \frac{332929 + 332928}{470832} \end{align*}

So it appears that after $n = 2$, that $a_{n+1} = 2{a_n}^2 -1$. If this were true for all $n > 2$ then I could save a lot of time computationally, and that's what I want to prove. From there on out the numerator would only be dependent on the previous numerator. But I'm having a hard time proving that. I've tried induction but it seems like everything comes back to depending on a $b,$-term.

Zduff
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  • Can't you rearrange your second equation to get $x_{n+1} = \frac{1}{2} (\frac{x_n^2 + 2}{x_n}) = \frac{x_n^2 + 2}{2x_n}$? Then we can see that this equals $2x_n = \frac{x_n^2 + 2}{x_{n+1}}$. – Toby Mak Nov 12 '17 at 22:35
  • Nice observation! But I don't think you'd save any computational effort. – lhf Nov 13 '17 at 00:29
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    Possible duplicate of https://math.stackexchange.com/questions/1733530/newtons-method-for-square-roots-jumps-through-the-continued-fraction-converge. See also https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Using_Pell.27s_equation. – lhf Nov 13 '17 at 00:38

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