In (long sentence here, watch for the period)
]data =: 3 ^~ i. 11
0 1 8 27 64 125 216 343 512 729 1000
p =: 3 interp data
p NB. conjunction interp embedded
(3) (2 : '(y !~ i. >: m) +/ . * {. (>: m) difftb n') 0 1 8 27 64 125 216 343
512
729 1000
the conjunction interp is supposed to look at the difference table
4 difftb data NB. successive differences of first column, see note [1]
0 1 6 6
1 7 12 0
8 19 0 0
27 0 0 0
and, reading the first line, produce the equivalent of
q =: 0 1 6 6 +/ . * (i.4) ! ] NB. A Newton series, see note [2]
.
Conjunction interp _does_ produce an equivalent:
(p"0 ,: q"0) i. 11
0 1 8 27 64 125 216 343 512 729 1000
0 1 8 27 64 125 216 343 512 729 1000
But I don't want p to recreate the difference table every time it is called.
Is
there a way to rewrite conjunction interp so that
q =: 3 interp1 data
q
0 1 6 6 +/ .* 0 1 2 3 ! ]
?
Kip
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[1] x difftb y is an x by x table whose kth column has
kth differences for vector x {. y (y is a vector and x <: #y)
difftb =: 4 : '|: diff^:(i.x) x {. y'
[2] In algebra notation q is the third degree Newton series
q(x) = 0 + 1 x/1 + 6 (x*(x-1))/(1*2) + 6 (x*(x-1)*(x-2))/(1*2*3)
This one simplifies to q(x) = x^3 .
For Newton series, see pages 190-191 of Graham, Knuth, and Patashnik, Concrete
Mathematics Second Edition.
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