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The usual binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!} $ can be generalized to real upper argument, lower argument still a nonnegative integer, by the definition $\binom{\alpha}{k} = \frac{\alpha (\alpha-1)\dots (\alpha-k+1)}{k!}$.

In the same way we could generalize the multinomial coefficient $ \binom{k_1 + \dots + k_s}{k_1 k_2 \dots k_s} = \frac{(k_1+\dots+k_s)!}{k_1! k_2! \dots k_s!}$ to a real upper argument by $ \binom{\alpha}{k_1 k_2 \dots k_s} = \frac{\alpha (\alpha-1) \dots (\alpha -k_1 -\dots -k_s+1)}{k_1! \dots k_s!}$ .

This definition, for instance, will allow us to generalize the Newton Binomial Theorem, to "the Newton Multinomial Theorem", in the following way (trinomial case):

\begin{equation} (1+y+z)^{\alpha} = \sum_{r=0}^\infty \sum_{s=0}^\infty \binom{\alpha}{r s} y^r z^s \end{equation}

My question is simple: I did a lot of googling, but I cannot find any reference for this!

¿Any comments? ¿Any references?

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We have $$(1+y+z)^{\alpha} = \sum_{t=0}^{\infty} \binom{\alpha}{t}(y+z)^t$$ The trinomial identity is just a rearrangement of the above by $$\binom{\alpha}{r,s}=\binom{\alpha}{r+s}\binom{r+s}{s}$$

To have convergence, we assume $|y+z|<1$, $|y|<1$, and $|z|<1$. The generalization follows similarly.

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  • $\begingroup$ Yes, but the question was to references (published! or web-pages) for the generalization of the multinomial coefficient. $\endgroup$ May 4, 2012 at 4:13
  • $\begingroup$ I don't see any reason that anyone would want to publish this. I think it will be okay to just use this fact without proof. $\endgroup$ May 4, 2012 at 4:29
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I don't see a reference but the point of the answer above is that your generalized multinomial coefficient is always the product of a generalized binomial coefficient and an ordinary multinomial coefficient. $$\binom{\alpha}{k_1 k_2 \dots k_s} =\binom{\alpha}{k}\binom{k}{k_1 k_2 \dots k_s}$$ for $k=k_1+k_2+\cdots k_s.$

The general binomial theorem is usualy derived as a very special Taylor series. In the same way the multivariate Taylor series for $(x_1+x_2+\cdots+x_s)^{\alpha}$ comes out to be just what you want.

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