How to use involutoric newtontransform to extrapolate a list of numbers.
]data =. 3 ^~ i. 4 NB. test data
0 1 8 27
T data NB. transform once
0 _1 6 _6
T T data NB. transform twice, reproducing data
0 1 8 27
T NB. transform using only '-', neither '*' nor '%' nor even '+'.
3 : 0
a =. 0 $ 0
while. # y do.
a =. a , {. y
y =. ( }: - }. ) y
end.
a
)
etc NB. extend transform with zeroes and transform back.
[: T [ {. [: T ]
11 etc data
0 1 8 27 64 125 216 343 512 729 1000
- Bo
--- Den fre 21/8/09 skrev Kip Murray <[email protected]>:
> Fra: Kip Murray <[email protected]>
> Emne: Re: [Jprogramming] A better conjunction
> Til: "Programming forum" <[email protected]>
> Dato: fredag 21. august 2009 14.41
> And, of course,
>
> diff =: 2&(-~/\) NB. differences of a vector
>
> (For the notes at the end)
>
> Kip Murray wrote:
> > 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
> >
> > ----------
> >
> >
> > [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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>
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