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Re: A=B example
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32133] Re: [mg32120] A=B example
*From*: Rob Pratt <rpratt at email.unc.edu>
*Date*: Sat, 29 Dec 2001 00:46:53 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
I don't have access to Mathematica at the moment, but here's a thought.
Note that a is not a local variable. Try Clear[]-ing a before repeating
the evaluation.
Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill
rpratt at email.unc.edu
http://www.unc.edu/~rpratt/
On Fri, 28 Dec 2001, Erich Neuwirth wrote:
> this has probably been asked before.
> being new to the list i would be happy with a pointer to previous
> answers.
>
>
> in A=B by Wilf and Zeilberger,
> on page 61 there is the following program for deriving recursions
> explicitly
> (i had to change FactorialSimplify to FullSimplify
> because the original version was for mathematica 2.2):
>
>
> findrecur[f_, ii_, jj_] :=
> Module[{yy, zz, ll, tt, uu, r, s, i, j},
> yy = Sum[
> Sum[a[i, j] *FullSimplify[f[n - j, k - i]/f[n, k]], {i, 0, ii}],
> {j,
> 0, jj}];
> zz = Collect[Numerator[Together[yy]], k];
> ll = CoefficientList[zz, k];
> tt = Flatten[Table[a[i, j], {i, 0, ii}, {j, 0, jj}]];
> uu = Flatten[Simplify[Solve[ll == 0, tt]]];
> For[r = 0, r <= ii, r++,
> For[s = 0, s <= jj, s++,
> a[r, s] = Replace[a[r, s], uu]]];
> Sum[Sum[a[i, j] F[n - j, k - i], {i, 0, ii}], {j, 0, jj}] == 0]
>
> defining
> f[n_, k_] := n!/(n - k)!
>
> and executing
> findrecur[f,1,1]
> works,
> but trying to run exactly the same statement a second time
> produces a lot of errors and effectively hangs mathematica.
>
> is there a solution?
>
>
>
> --
> Erich Neuwirth, Computer Supported Didactics Working Group
> Visit our SunSITE at http://sunsite.univie.ac.at
> Phone: +43-1-4277-38624 Fax: +43-1-4277-9386
>
>
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