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May 18, 2014 at 10:59 comment added user21820 @Hurkyl: I don't know what method you're referring to, but there is a simple way to accelerate the convergence of this sequence directly, since the numerator and denominator of its convergents are described by recurrence relations and hence the $k$-th convergent is expressible as the corresponding matrix raised to the $k$-th power and then multiplied by some matrix corresponding to the initial values of the recurrences. The matrix exponentiation can be done by repeated squaring.
Dec 22, 2013 at 7:38 comment added user14972 If this turns into an odd method I remember seeing once, there turns out to be a simple way to step quadratically through the sequence. i.e. after $n$ steps, you are looking at $\epsilon_{2^n}$, not $\epsilon_n$.
Oct 25, 2013 at 14:26 history edited Eric Jablow CC BY-SA 3.0
Fix a stray factor of 2.
Oct 25, 2013 at 1:00 history edited Eric Jablow CC BY-SA 3.0
Fix typo.
Oct 24, 2013 at 22:44 comment added Eric Jablow This should be slower than Newton's method for square roots. With Newton, $\epsilon_{n+1}\approx k\epsilon_n^2$.
Oct 24, 2013 at 21:50 comment added Max I wonder how quickly this converges compared to Newton's method.
Oct 24, 2013 at 21:01 history answered Eric Jablow CC BY-SA 3.0