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When doing the expansion of the following polynomial expression, the result is zero. \begin{align*} &- 1 (-1 - 2 x) (0 - 2 x) (1 - 2 x) (2 - 2 x) \\ &+5 (-1 - 1 x) (0 - 1 x) (1 - 1 x) (2 - 1 x) \\ &-10 (-1 - 0 x) (0 - 0 x) (1 - 0 x) (2 - 0 x) \\ &+10 (-1 + 1 x) (0 + 1 x ) (1 + 1 x) (2 + 1 x) \\ &-5 (-1 + 2 x) (0 + 2 x) (1 + 2 x) (2 + 2 x) \\ &+1 (-1 + 3 x) (0 + 3 x ) (1 + 3 x) (2 + 3 x)\\ &=0. \end{align*}

More generally, for $1 \leq k <n$, it seems that $$\tag 1 \sum_{j=0}^n (-1)^j{n \choose j}{k-1+x(j-k) \choose n-1}=0.$$

I have tried, but failed so far, to prove it. Any idea ?

For the context, a proof of (1) would implies that $S(k,n,m)=S(n-k,n,m)$, where $S(k,n,m)$ is defined in this question.

René Gy
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2 Answers2

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I'm the author of the other question. I will give a more or less detailed proof of the question you made. I will skip the long calculation details, but they are not complicated. I found an approach slightly clearer than what I commented in the original question (which I encourage interested people to think of!)

Let's fix $k,n,t$. The coefficient of $x^{kt}$ in the polynomial $(1+x+x^2+\ldots+x^t)^n$ is clearly equal to the coefficient of $x^{(n-k)t}$. We can write that polynomial simply as $\left(\frac{1-x^{t+1}}{1-x}\right)^n$, and using generating functions of both the numerator and denominator, we can find the following formula:

$$ E_{k,n}(t) = \sum_{j=0}^{k-1} (-1)^j \binom{n}{j} \binom{(k-j)t+n-1-j}{n-1} .$$

So the coefficient of $x^{kt}$ in $(1+x+\ldots+x^t)^n$ is exactly the last expression (remarkably, it is a polynomial in $t$).

If you try to isolate the coefficient of $t^m$ in the last expression what you get is simply the expression of $S(k,n,m)$ of the question you linked. Essentialy, since $E_{k,n}(t) = E_{n-k,n}(t)$ then it is evident that $S(k,n,m)=S(n-k,n,m)$.

With that language, the difference $S(k,n,m)-S(n-k,n,m)$ is exactly the sum you are asking in this post, and then it is clearly 0.

As I said in the other post, it of course can be proven only working with some recurrences and induction, but this is a more transparent proof. Of course, the other question may be restated as proving whether the polynomials $E_{k,n}$ have positive coefficients.

Luis Ferroni
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  • That is a very nice proof! I believe that $(-1)^j$ is missing in your expression for $E_{k,n}(t)$, though. I would accept it as an answer. My original question is settled since in can be re-phrased as $E_{k,n}(x-1)-E_{n-k,n}(x-1)=0$. – René Gy Nov 03 '19 at 12:17
  • It seems that $F_{n,k}(t):=E_{n,k}(t-1)$ is also a polynomial with positive coefficients. This is the coef. of $x^{kt}$ in $(1+x+\cdot\cdot +x^{t-1})^n$. – René Gy Nov 03 '19 at 15:31
  • No, that's not true. For $k=2$ and $n=9$ the cubic coefficient is negative. – Luis Ferroni Nov 03 '19 at 16:04
  • ah, good catch! I did'nt check that far. But the coef. for $F_{n,k}$ also seem worth investigating: they have an absolute value which is smaller than those of $E_{n,k}$. – René Gy Nov 03 '19 at 16:17
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Idea (for the first polynomial):

Since it is of the degree $4$ you only need to find $5$ different roots. The $0$ is obviously one, then try with $\pm 1$ and $\pm2$. Then it must be constantly $0$.

Say: $$p(1) = 0+0+0+0-120+120 =0$$

nonuser
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