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Re: A=B example
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32131] Re: [mg32120] A=B example
*From*: BobHanlon at aol.com
*Date*: Sat, 29 Dec 2001 00:46:51 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
In a message dated 12/28/01 3:51:58 AM, erich.neuwirth at univie.ac.at writes:
>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?
Inside the definition of findrecur you use a[]. a is defined globally and
retains its values from the first call. Try
findrecur[f_,ii_,jj_, a_Symbol:a]:=
Module[{yy,zz,ll,tt,uu,r,s,i,j},
Clear[a];
yy= ... ];
Bob Hanlon
Chantilly, VA USA
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