Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for an answer to the following problem.

Let $S$ be the multiset $\{1^{d_1},2^{d_2},\dots,m^{d_m}$ $A_{S,k}$ is the number of permutations of $S$ with $k-1$ descents and no descent at the end.

  1. Let $A_S(t)=\sum\limits_k A_{S,k}t^k$ and $d=\sum\limits_{i\leq m} d_i$. I want to show that $$\sum\limits_S \frac{A_S(t)}{(1-t)^{d+1}} \prod\limits_i x_i^{d_i}=(1-t\prod\limits_i(i-x_i)^{-1})^{-1}$$

  2. The Eulerian polynomial $A_d(t)$ is the special case $A_{[d]}$. I want to show that $$\sum\limits_{d=0}^{\infty}\frac{A_d(t)x^d}{(1-t)^{d+1}d!}=\frac{1}{1-te^x}$$

Thanks for your help.

share|improve this question
    
For 1. I somehow have to use that $A_{\emptyset}(t)=1$ and $\frac{A_S(t)}{(1-t)^{d+1}}=\sum\limits_{n=0}^{\infty}t^n\prod\limits_i\binom{n+‌​d_i-1}{d_i}$ –  bronko Dec 13 '12 at 13:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.