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MathGroup Archive 1997

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Fwd: Permutation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg9189] Fwd: [mg9161] Permutation
  • From: BobHanlon at aol.com
  • Date: Tue, 21 Oct 1997 02:03:02 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

In the expansion, Sum cannot be used since it would evaluate to the 
original expression.  Consequently, a new function "summ" is used to
hold the

evaluation.  Evaluate the following:

summ[x_ * summ[y_, iterInner__], iterOuter__] := 
	summ[x y, iterOuter, iterInner]; (* nested summation *)

binomialExpand[m_] := 
	(x_ + y_)^k_ -> summ[Binomial[k, m] x^m * y^(k-m), {m, 0, k}];

((a+y) + (b+c*x+d*x^2))^k /. binomialExpand[m1] % /. binomialExpand[m2]
% /. binomialExpand[m3]
% /. binomialExpand[m4]
ans1 = % // PowerExpand

Check:

ans1 /. summ -> Sum
% // PowerExpand

Looking at an alternate representation:

ans1 // FunctionExpand
% /. Gamma[x_ + 1] -> x!
ans2 = % /. k! -> Multinomial[m1, m2, m3, m4, k-m1-m2-m3-m4] * 
	m1! m2! m3! m4! (k-m1-m2-m3-m4)!

Check:

ans2 /. summ -> Sum
% // FunctionExpand
% // PowerExpand


Bob Hanlon
---------------------
Forwarded message:
From:	dd4b at virginia.edu (David Djajaputra) To: mathgroup at smc.vnet.net
To:	mathgroup at smc.vnet.net

Hello,

Is it possible to ask Mathematica to expand an abstract algebraic
formula? I mean, to do
something like

(1+x)^n = Sum[Binomial[n,m] x^m, {m,0,n}]

I have in mind problems like determining, in closed form, the
coefficient of (x^m y^n) in

((a+y) + (b+cx+dx^2))^k

for example, which then can help in proving formulas, etc.


Thanks a lot.

David Djajaputra




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