This is a solution
etc=.[: T [ {. [: T ] NB. extend polynomial sequence
T NB. Transformation from values to coefficients, or back
3 : 0 NB. (This should be done more elegantly)
a=.0$0
while. #y do.
a=. a,{.y
y=.(}:-}.)y
end.
a
)
20 F 1 4 NB. extend sequence se0 from 2 to 20 elements
1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58
20 F 0 4 20 54 NB. extend sequence se1 from 4 to 20 elements
0 4 20 54 112 200 324 490 704 972 1300 1694 2160 2704 3332 4050 4864 5780 6804
7942
b= T T b=. ? 20$100 NB. test (T T) = ]
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Bo
--- Den tors 20/8/09 skrev Kip Murray <[email protected]>:
> Fra: Kip Murray <[email protected]>
> Emne: Re: [Jprogramming] Takes from a mystery sequence
> Til: "Programming forum" <[email protected]>
> Dato: torsdag 20. august 2009 16.52
> Thank you, Ambrus, I will study your
> verbs. Of course I would like the lowest
> degree that fits the data. I'm assuming data that has
> an exact polynomial fit.
>
> Kip
>
> Zsbán Ambrus wrote:
> > On Tue, Aug 18, 2009 at 11:27 PM, Kip Murray<[email protected]>
> wrote:
> >> Have you seen a polynomial-fitting verb for data
> such as iii0 10 and iii1 10 ?
> >
> > Sure, and it's quite simple:
> >
> > se0=: 1 4 7 10 13
> 16 19 22 25 28
> > se1=: 0 4 20 54 112 200 324 490 704 972
> > ]p0 =: (%.[:^/~...@#) se0
> > 1 3 7.97184e_8 _7.964e_8 4.15017e_8 _1.25439e_8
> 2.27913e_9
> > _2.45331e_10 1.44069e_11 _3.55401e_13
> > p0 p. i. 10
> > 1 4 7 10 13 16 19 22 25 28
> > ]p1 =: (%.[:^/~...@#) se1
> > _7.96167e_11 2.1526e_7 3 1 _2.66006e_7 7.83796e_8
> _1.39068e_8
> > 1.46474e_9 _8.43371e_11 2.04392e_12
> > p1 p. i. 10
> > _7.96167e_11 4 20 54 112 200 324 490 704 972
> > NB. or, if you want lower degree
> polynomyals
> > ]p0d2 =: (%.2^/&i.~#) se0
> > 1 3
> > p0d2 p. i. 10
> > 1 4 7 10 13 16 19 22 25 28
> > ]p1d4 =: (%.4^/&i.~#) se1
> > 1.3074e_12 _2.50111e_12 3 1
> > p1d4 p. i. 10
> > 1.3074e_12 4 20 54 112 200 324 490 704 972
> >
> > Ambrus
> >
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> ----------------------------------------------------------------------
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>
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