Re: A simple 11c/15c minichallenge (phi) Message #20 Posted by Gerson W. Barbosa on 21 Aug 2009, 4:56 p.m., in response to message #19 by Jeff O.
Until yesterday I hadn't looked up any reference either. I was trying to simplify some constants so that they took up less steps on the 11C. For instance, I had found 4 LN HYP SIN was equivalent to 1.875, which was a gain (3 steps instead of 5). For other constants I would do the backward sequence: k HYP^{1} SIN e^{x}. I don't remember for what reason, since there's no way to get .5 with less than two steps, I did .5 HYP^{1} SIN e^{x}. I recognized the number on the display as phi...
The following is somewhat interesting:
It appears the continued fraction form for 1/n+sqrt(1+(1/n)^2), which is equivalent to e^sinh^{1}(1/n), where n is an integer greater than 1, is always [1; n1, 1, 1, n1, 1, 1, n1, 1, 1, ...]. When n=2, the continued fraction form for the expression will be [1; 1, 1, 1, 1, 1, ...], that is, phi.
