1
$\begingroup$

Let $\mu \in \mathbb{N}_0^n$ be a multi index and set $$P(X_1, \dots, X_n) = X_1^{\mu_1}(X_1 + X_2)^{\mu_2} \cdots (X_1 + \cdots +X_n)^{\mu_n} = \prod_{j=1}^n (X_1 + \cdots + X_j)^{\mu_j}.$$ Since $P$ is a homogeneous polynomial of degree $|\mu| = \mu_1 + \dots + \mu_n$, it is clear that $$ P(X_1, \dots, X_n) = \sum_{|\alpha| = |\mu|} C(\alpha, \mu) \,X_1^{\alpha_1} \cdots X_n^{\alpha_n}$$ for certain coefficients $C(\alpha, \mu)$.

While we expect that there should be a closed expression for $C(\alpha, \mu)$ in terms of binomial coefficients, we were unable to obtain such a formula. Does someone know the answer or a reference to the literature?

$\endgroup$
1
  • $\begingroup$ This will not answer your question, but it might interest you. The number of terms (monomials) in an expansion of $P$, for special values of $\mu$, is given by Corollary 4.5 in the paper: www-math.mit.edu/~rstan/papers/coef.pdf $\endgroup$ Nov 17, 2016 at 16:20

2 Answers 2

4
$\begingroup$

We can express the coefficient as a product of binomial coefficients by expanding by the binomial theorem, starting with the rightmost factor.

For example, with $n=3$ we have $$ \def\m#1{{\mu_{#1}}} \def\i#1{{i_{#1}}} \begin{aligned} X_1^\m1(X_1+X_2)^\m2(X_1+&X_2+X_3)^\m3 = X_1^\m1(X_1+X_2)^\m2\sum_{\i3}\binom{\m3}{\i3} (X_1+X_2)^{\m3-\i3}X_3^\i3\\ &= X_1^\m1\sum_{\i3}\binom{\m3}{\i3}(X_1+X_2)^{\m2+\m3-\i3}X_3^\i3\\ &=\sum_{\i2,\i3}X_1^{\m1+\m2+\m3-\i2-\i3}X_2^{\i2}X_3^{\i3} \binom{\m2+\m3-\i3}{\i2}\binom{\m3}{\i3}. \end{aligned} $$ The formula for general $n$ is similar.

$\endgroup$
3
$\begingroup$

Just for completion, we expand Gessel's suggestion to get: $$\prod_{j=1}^n\left(\sum_{i=1}^jX_i\right)^{\mu_j}= \sum_{i_2,\dots,i_n\geq0}X_1^{\vert\mu\vert-\vert I\vert}\prod_{k=2}^nX_k^{i_k}\prod_{j=2}^n\binom{\mu_n+\cdots+\mu_j-(i_n+\cdots+i_{j+1})}{i_j}$$ where $\vert\mu\vert=\mu_1+\cdots+\mu_n$ and $\vert I\vert=i_2+\dots+i_n$.

Convention: $i_n+\cdots+i_{n+1}=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.