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A=B example
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32120] A=B example
*From*: Erich Neuwirth <erich.neuwirth at univie.ac.at>
*Date*: Fri, 28 Dec 2001 02:41:35 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
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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