Prove that the iteration in the Babylonian method above converges quadratically to the square root of S

Asked Dec 30 '22 at 01:54

Active Dec 30 '22 at 02:02

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Prove that the iteration in the Babylonian method above converges quadratically to the square root of $S$. In particular, show that the error $\epsilon_n = \frac{x_n}{\sqrt{S}} - 1$ satifies $\epsilon_{n+1} = \frac{\epsilon_n^2}{2(\epsilon_n +1)}$

Note we were given $x_{n+1} = \frac{1}{2}(x_n + \frac{S}{x_n}) $

My atempt: $\epsilon_{n+1} = \frac{x_{n+1}}{\sqrt{S}} -1 =\frac{x_n + \frac{S}{x_n}}{2\sqrt{S}}-1$ .

I need to show that this equals $\frac{\epsilon_n^2}{2(\epsilon_n +1)}$ but I dont know where to go from here, this i tried to expand the fraction but couldn't get anywhere, any help would be great.

edited Dec 30 '22 at 02:02

asked Dec 30 '22 at 01:54

Cole Hendrickson

1

Is $x$ supposed to be $S$? – angryavian Dec 30 '22 at 02:00
@angryavian yes sorry I corrected it. – Cole Hendrickson Dec 30 '22 at 02:03
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Plug in $x_n = \sqrt{S}(1+\epsilon_n)$ into your last expression and do some simplification. – angryavian Dec 30 '22 at 02:14
For the babylonian/Heron/Newton method for the square root, see also https://math.stackexchange.com/questions/2455260/prove-fixed-point-iteration-of-gx-frac12x-fraca2x-converges-to, https://math.stackexchange.com/questions/2289703/finding-the-square-root-newton-raphson-convergence-proof – Lutz Lehmann Dec 30 '22 at 09:42

Prove that the iteration in the Babylonian method above converges quadratically to the square root of S

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