       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
>
>
> 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