Re: Fwd: Permutation
- To: mathgroup at smc.vnet.net
- Subject: [mg9321] Re: Fwd: [mg9161] Permutation
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Mon, 27 Oct 1997 02:47:38 -0500
- Organization: University of Western Australia
- Sender: owner-wri-mathgroup at wolfram.com
David Djajaputra wrote:
> 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
One way is to omit the Sum and use the Einstein summation convention
which sums over repeated indicies. (This is, implicitly, what a human
really does). For your example (the following can be improved):
In[1]:= (d*x^2 + c*x + b + (a + y))^k /.
(a + y + (b_))^k -> Binomial[k, m]*(a + y)^m*b^(k - m)
Out[1]=
2 k - m m
(d x + c x + b) (a + y) Binomial[k, m]
In[2]:= % /. ((a_) + (b_))^m -> Binomial[m, p]*b^p*a^(m - p)
Out[2]=
m - p 2 k - m p
a (d x + c x + b) y Binomial[k, m] Binomial[m, p]
In[3]:= PowerExpand[Simplify[% /.
((a_) + b)^(q_) -> Binomial[q, r]*b^r*a^(q - r)]]
Out[3]=
m - p r k - m - r k - m - r p a b x (c + d
x) y Binomial[k, m]
Binomial[k - m, r] Binomial[m, p]
In[4]:= PowerExpand[% /. ((a_) + (b_))^(q_) -> Binomial[s, t]*b^s*a^(t -
s)]
Out[4]=
m - p r t - s s k - m - r + s p a b c d x
y Binomial[k, m]
Binomial[k - m, r] Binomial[m, p] Binomial[s, t]
Cheers,
Paul
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia Nedlands WA 6907
mailto:paul at physics.uwa.edu.au AUSTRALIA
http://www.pd.uwa.edu.au/~paul
God IS a weakly left-handed dice player
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