Here is an idea.
If you construct a lower triangular matrix from your original q, you
can do the evaluations with an insertion. For example:
q =: 4 2 1 3 1 2
tri =: 3 : 0
|:(-i.$y)|.!.0"0 1 y
)
m =: tri q
4 0 0 0 0 0
2 4 0 0 0 0
1 2 4 0 0 0
3 1 2 4 0 0
1 3 1 2 4 0
2 1 3 1 2 4
ev =: (+%)/
ev/m
170r39 14r61 48r11 3r13 9r2 1r4
which is correct give or take the odd inversion.
Now, if you'd like to evaluate the continued fraction for q's of
length 1001-1010, you just have to build the appropriate matrix (1010
rows and 10 columns) , do the ev insertion and rectify the inverted
terms.
Does that help?
Mike
On 14-May-08, at 9:37 AM, John Randall wrote:
Suppose q is a truncated continued fraction, say
q=:4 2 1 3 1 2x
J can evaluate these very niftily using (+%)/ :
ev=:(+%)/
ev q
170r39
ev\ q
4 9r2 13r3 48r11 61r14 170r39
What is more difficult is taking as many terms as are needed until
some condition is satisfied: because of J's right to left evaluation,
you cannot just add another term.
One solution is to evaluate the continued fraction backwards and use
various identitites. Another is to use a matrix formulation as in the
Fibonacci sequence (which corresponds to the continued fraction
1 1 1 1....).
In this, the numerator and denominator appear as the bottom row of a
2x2 matrix, and we get the next term by multiplying by the matrix
2 2 $ 0 1 1, an , where an is the next term of the continued fraction.
mp=:+/ .*
init=:2 2 $ 0 1 1 0
ev2=:mp/ @: |. @: (init , 2 2 ($"1) 0 1 1 ,"1 0 ])
ev2 q
61 14
170 39
{: @: ev2\ q
4 1
9 2
13 3
48 11
61 14
170 39
Does anyone have suggestions as to how to improve this code?
Best wishes,
John
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