... and I did omit a right paren, but I think you will see where it goes. /Kip
Kip Murray wrote:
Henry,
I do not have code but can point you to formulas in the Handbook of
Mathematical Functions by Abramowitz and Stegun, Chapter 3 Elementary
Mathematical Methods, Section 3.10 Theorems on Continued Fractions (page
19 in my edition). Be sure to read both columns of formulas.
There is no unique continued fraction product for two continued
fractions, but you could try the following to find _a_ product. First
given two infinite continued fractions f and g use the recursive matrix
multiplication given in formula (3) of Section 3.10 to find the
numerators and denominators of the n^th convergents
f_n = A_n/B_n
n = 1,2,3,...
g_n = C_n/D_n
(conventional notation, _n means subscript n). The convergents converge
to the value of the continued fraction.
Let
E_n = A_n * C_n
n = 1,2,3...
F_n = B_n * D_n
Then, to find a continued fraction h (this is your product) with
convergents
h_n = E_n/F_n n = 1,2,3...
solve the matrix recursion (3) for for the entries e_n and f_n in
h = f_0 + e_1
---------
f_1 + e_2
--------
f_2 + ...
(I hope that comes out right on your screen.)
The recursion (3) written for the continued fraction h is
(E_n , F_n) = ((E_(n-1) , F_(n-1)) ,. (E_(n-2) , F_(n-2)) o (f_n , e_n)
using a mix of standard notation and J, where o is +/ . *
Starting values are E_(-1) = 1, E_0 = f_0, F_(-1) = 0, F_0 = 1. You can
use f_0 = 0.
Please get the correct formulas from Abramowitz and Stegun -- in case I
mistyped something. And of course you have %. for solving for (f_n , e_n).
Kip Murray
Henry Rich wrote:
Does anyone have J code for multiplying two numbers expressed as
simple continued fractions, producing a continued-fraction result?
Henry Rich
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