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Jun 19, 2013 at 18:39 comment added Ben Wieland Yes, you're right, it should be $a_{n+1}=q_n^{\mu-2}$. Then $q_{n+1}\sim q_n^{\mu-1}$ and the reciprocal error is $q_{n+1}q_n\sim q_n^\mu$.
Jun 19, 2013 at 17:39 comment added jsm Thanks, I'm happy to specify numbers by their continued fraction expansion. But don't you mean $\mu-2$ instead of $\mu-1$? (sorry if I'm being stupid here)
Jun 19, 2013 at 1:55 history answered Ben Wieland CC BY-SA 3.0