2
$\begingroup$

Recall the following:

Multinomial Theorem. For all finite sets $X$, we have:

$$\left(\sum_{x \in X} x\right)^n = \sum_{a}[a](X)^a$$

where $a$ ranges over the set of partitions of $n$ into $|X|$-many parts (with each part possibly $0$), $[a]$ is the relevant multinomial coefficient, and $(X)^a$ is the relevant monomial symmetric polynomial.

For example:

$$\left(x+y+z\right)^3 = [3,0,0](x,y,z)^{3,0,0}+[2,1,0](x,y,z)^{2,1,0}+[1,1,1](x,y,z)^{1,1,1}$$

$$= (x^3+y^3+z^3)+3(x^2y+x^2z+y^2z)+6xyz$$

Anyway, I don't particularly like this business of leaving out the domain of quantification for $a$ and just putting it in words.

Question. It would be good if there were a moderately standard notation for the set of all ways of partitioning a natural number $a$ into $b$ parts, so the multinomial theorem can be stated properly. Is there?

$\endgroup$
  • $\begingroup$ Related, but not what you want: $\lambda \vdash n$ means that $\lambda$ is an (unordered) partition of $n$ into arbitrarily many parts. $\endgroup$ – Misha Lavrov Mar 24 '17 at 1:21
  • $\begingroup$ @Misha, that's not too bad. Worst case scenario we can write something like $\lambda \vdash a \rightarrow b$ or something like that. $\endgroup$ – goblin Mar 24 '17 at 1:22
  • $\begingroup$ @Misha, it creates a problem, though, namely that $\lambda \vdash n$ means non-zero parts, but $\lambda \vdash a \rightarrow b$ means possibly-empty parts. $\endgroup$ – goblin Mar 24 '17 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.