Re: Smarter way to calculate middle-right terms of continued fraction partial sums
- To: mathgroup at smc.vnet.net
- Subject: [mg70169] Re: Smarter way to calculate middle-right terms of continued fraction partial sums
- From: "Diana" <diana.mecum at gmail.com>
- Date: Fri, 6 Oct 2006 01:58:39 -0400 (EDT)
- References: <eflefm$dg8$1@smc.vnet.net><eg2erm$7mn$1@smc.vnet.net>
Maxim, I have found a way to multiply e(z) as defined by any polynomial in T. I learned that I can do a simple partial sum by partial sum multiplication. I don't have to perform a solve at each partial sum iteration. So, my problem with computer power and efficiency has been solved. This allows me to evaluate (x/y) e(theta/f) where x, y, f are polynomials in T, f != 0, and theta = 1. When the characteristic q = 2, theta might also equal T, T + 1, or T^2 - T. > Very neat way to do it, by the way, how did you come up with it? I didn't. I was using the algorithm suggested by Dr. Dinesh Thakur. My thesis is an exposition of three of his articles on the exponential function in the function field case. I am now trying to figure out how to add continued fractions. By this, I mean that I am trying to evaluate (x/y) e(theta/f) + (z/w), where x, y, z, w, and f are polynomials in T. I see that you can place z/w in the a_0 partial quotient location. I would like to be able to convert this form of the continued fraction to a continued fraction with 0 in the a_0 term, if this is possible. Thank you very much for your time. Diana