Here is how I would generate the terms of the finite continued fraction
1 + 2
--
3 + 4
--
5 + 6
--
7
cf =: {.@] , {.@] + [: ([: %`+/ 0 ,~ ,/)\ (,. }.)
2 4 6 cf 1 3 5 7x
1 5r3 29r19 233r151
The underlying idea is that if tk is the linear fractional transformation ak
% bk + ]
then the nth term is b0 + t1 t2 ... tn 0 . ("term" is defined below)
--Kip Murray
Sent from my iPad
> On Feb 23, 2014, at 4:17 PM, km <[email protected]> wrote:
>
> The generalized continued fraction
>
> b0 + a1
> ---------
> b1 + a2
> ---------
> b2 + a3
> --------
> b3 + ...
>
> is the infinite sequence
>
> b0 , b0 + a1%b1 , b0 + a1%b1 + a2%b2 , b0 + a1%b1 + a2%b2 + a3%b3 , ...
>
> where between the commas I am assuming J's right-to-left evaluation. The nth
> term would be
>
> b0 + a1%b1 + a2%b2 + a3%b3 + ... + an%bn (still using right to left
> evaluation).
>
> What is a good way to calculate this nth term in J?
>
>
> For a half-page introduction to generalized continued fractions see
>
> http://people.math.sfu.ca/~cbm/aands/page_19.htm
>
> Continued fractions for ln(1+z) and ln((1+z)%(1-z)) are given here
>
> http://people.math.sfu.ca/~cbm/aands/page_68.htm
>
>
> --Kip Murray
>
> Sent from my iPad
>
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