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        riddles >> easy >> polynomials
        
(Message started by: fatball on Dec 14th, 2010, 7:40am)
Title: polynomials
Post by fatball on Dec 14th, 2010, 7:40am
can you tellme how to expand the following or which theorem to use for that pleae?

(a1+a2+...+a12)^4

urgent please!  Thanks!
Title: Re: polynomials
Post by SMQ on Dec 14th, 2010, 8:31am
(a1 + a2 + ... + an)4
= a14 + a24 + ... + a124
   + 4a13a2 + 4a13a3 + ... + 4an-13an
   + 6a12a22 + 6a12a32 + ... + 6an-12an2
   + 4a1a23 + 4a1a33 + ... + 4an-1an3
   + 12a12a2a3 + 12a12a2a4 + ... + 12an-22an-1an
   + 12a1a22a3 + 12a1a22a4 + ... + 12an-2an-12an
   + 12a1a2a32 + 12a1a2a42 + ... + 12an-2an-1an2
   + 24a1a2a3a4 + 24a1a2a3a5 + ... + 24an-3an-2an-1an

Edit: got it right the second time, as usual...

--SMQ
Title: Re: polynomials
Post by fatball on Dec 14th, 2010, 8:51am
general formula or theorem to use?
Title: Re: polynomials
Post by towr on Dec 14th, 2010, 9:58am
You can do it recursively with the binomial theorem, if you're so inclined.

(a+b)n = sum k=0..n C(n,k) an-k * bk

So  (a1 + (a2+..+am) ) =
1* a14 +  
4* a13*(a2+..+am) +  
6* a12*(a2+..+am)2  +  
4* a1*(a2+..+am)3 +  
1* (a2+..+am)4



You could also simply expand the formula

Sum(ai)n =
Sum(ai * Sum(ai)n-1)


[edit]
I think you're missing a few terms, SMQ. You seem to only have terms with at most two different factors, but there can be up to 4.
[/edit]
Title: Re: polynomials
Post by SMQ on Dec 14th, 2010, 11:16am

on 12/14/10 at 09:58:52, towr wrote:
I think you're missing a few terms, SMQ. You seem to only have terms with at most two different factors, but there can be up to 4.

Right you are -- fixed now.  That's what I get for generalizing from the binomial case in my head... :-[

--SMQ
Title: Re: polynomials
Post by pex on Dec 14th, 2010, 11:18am

on 12/14/10 at 08:51:44, fatball wrote:
general formula or theorem to use?

It's called a multinomial series (http://mathworld.wolfram.com/MultinomialSeries.html).


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