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If i have the polynomial expression $(a_1x+b_1y+c_1)^p. (a_2x+a_2y+c_2)^d$, and with assumptions $a_1+b_1<<c_1$ , $a_2+b_2<<c_2$, can i expand this as a product of binomials using the multi binomial theorem? http://en.wikipedia.org/wiki/Multi-binomial_theorem#The_multi-binomial_theorem. Could someone explain the Multi Binomial theorem in more detail? How is this summation with "v" to be done?

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isnt this the multinomial theorem, because i dont think those are binomials –  Jorge Fernández Jul 23 '12 at 13:39
    
Thats what i m not sure. Under the assumptions that i have given, we can approximate each of the terms above as (a1/c1)x+(b1/c1)y+1. So should i expand it using multi multinomial theorem? –  hAcKnRoCk Jul 23 '12 at 13:43

1 Answer 1

I presume that you mean the term $\nu \leq \alpha$? It means the the inequality holds componentwise. So, if $\alpha = (\alpha_1,\cdots,\alpha_d)$, $\nu = (\nu_1,\cdots, \nu_n)$, then $\nu \leq \alpha$ is equivalent to $\nu_1 \leq \alpha_1, \cdots, \nu_n \leq \alpha_n$.

Of course, implicitly the indices $\nu_k$ are also taken to be integers, and each $\nu_k \geq 0$, as well.

A slightly clearer, but less concise, notation would be $0 \leq \nu \leq \alpha$, $\nu \in \mathbb{Z}^d$.

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yes. I did mean that $v<=alpha$. term. thanks. I got it. But still not sure of how to proceed with the expansion. –  hAcKnRoCk Jul 23 '12 at 13:52

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