A solution, found this evening, follows the problem statement. /Kip
Kip Murray wrote:
> Newton polynomial p below is based on table au whose first row contains
> coefficients, second row, special values of the argument.
>
> ]au =: a ,: u
> 0 1 4 1
> 0 1 3 6
>
> p =: 3 : '+/ 0 1 4 1 * 1,(y-0),((y-0)*(y-1)),((y-0)*(y-1)*(y-3))'
>
> ----------------------------------------------------------------------
> Can you write an adverb Np (Newton polynomial) such that verb au Np is
> equivalent to the Newton polynomial specified in table au?
> ----------------------------------------------------------------------
Np =: 1 : '(({.m) +/ . * [: */\ 1 , (}:{:m) -~ ])"0'
(p ,: au Np) i. 5
0 1 8 27 64
0 1 8 27 64
au
0 1 4 1
0 1 3 6
p
3 : '+/ 0 1 4 1 * 1,(y-0),((y-0)*(y-1)),((y-0)*(y-1)*(y-3))'"0
au Np
(0 1 4 1 +/ .* [: */\ 1 , 0 1 3 -~ ])"0
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