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Question : Letting $k,n$ be positive integers, let's define a sequence $\{a_i\}\ (i=0,1,\cdots, kn)$ as $$(1+x+\cdots+x^k)^n=\sum_{i=0}^{kn}a_ix^i.$$ Then, is the 'special' difference-sequence $\{d^Na_i\}$ a unimodal sequence for every non-negative integer $N$? If the answer is yes, then prove that. If the answer is no, then find a counterexample.

In the following, I'm going to define some words.

1. Let's call a sequence $\{a_i\}\ (i=0,1,\cdots, kn)$ which satisfies the following condition 'a unimodal sequence '.

Condition : $\ $There exists an integer $t$ such that $$a_0\le a_1\le \cdots\le a_t\ge a_{t+1}\ge a_{t+2}\ge\cdots.$$

2. Let's define $\{d^Na_i\}\ (N\in\mathbb N)$ as the following:

$$d^Na_i=\max(d^{N-1}a_i-d^{N-1}a_{i-1},0)\ \ \ (i=0,1,\cdots, kn),$$ $$d^0a_i=a_i, \ a_{-1}=0.$$

Please note that this is not an usual difference-sequence. (we may call this 'a special difference-sequence')

In the following, I'm going to write why I'm interested in this question.

When I saw Pascal's triangle, I found the following property about $a_i=\ _nC_i$. Note that this is the $k=1$ case of the above question.

For example, let's see the $n=8$ case. $$a_i(=d^0a_i) : 1\ \ 8\ \ 28\ \ 56\ \ 70\ \ 56\ \ 28\ \ 8\ \ 1$$

$$d^1a_i : 1\ \ 7\ \ 20\ \ 28\ \ 14\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^2a_i : 1\ \ 6\ \ 13\ \ 8\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^3a_i : 1\ \ 5\ \ 7\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^4a_i : 1\ \ 4\ \ 2\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^5a_i : 1\ \ 3\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^6a_i : 1\ \ 2\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^7a_i : 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^Na_i (N\ge 8): 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

Hence, we can see that $\{d^N{_8C_i}\}\ (i=0,1,\cdots,8)$ is a unimodal sequence for every non-negative integer $N$.

Then, I reached an expectation, which is the above question. The answer seems yes, but I'm facing difficulty. I need your help.

Update : I crossposted to MO.

http://mathoverflow.net/questions/142565/about-the-unimodality-of-the-coefficients-sequence-of-1x-cdotsxkn

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What you call single-peak sequence is called unimodal sequence in the literature. Using that term might attract a proper audience (and help find information about it). –  Marc van Leeuwen Sep 13 '13 at 14:43
    
@MarcvanLeeuwen: Thank you. I edited them. –  mathlove Sep 13 '13 at 14:47
    
I suspect that there would exist some papers which are helpful. Could anyone please tell me whether the following papers are helpful? sciencedirect.com/science/article/pii/S0097316504001153 maa.org/sites/default/files/pdf/upload_library/22/Ford/… kurims.kyoto-u.ac.jp/EMIS/journals/JACO/Volume7_1/… math.miami.edu/~wachs/papers/unimodal.pdf –  mathlove Sep 16 '13 at 7:04
    
I think your main difficulty is that your difference operator is quite unusual (the usual one does not cut off negative values) and the property you want to prove clearly fails if you take the usual definition of the difference operation. This means that most theory you will find will not apply to your problem statement. –  Marc van Leeuwen Sep 16 '13 at 7:07
    
@MarcvanLeeuwen: I know this is unusual, but I would like to know this. Anyway, thanks always. –  mathlove Sep 16 '13 at 7:17

1 Answer 1

up vote 0 down vote accepted

I'm posting an answer just to inform that the question has received an answer by David Speyer on MO.

http://mathoverflow.net/questions/142565/defining-a-i-as-1xxkn-sum-i-0kna-ixi-then-is-the-speci

He uses Descartes' rule of signs.

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