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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
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