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Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly. I think maybe someone can give me a hand on this.

Synthetically, what I want to prove is that the following sum is positive:

$$S(k,n,m)=\sum_{i=0}^{n-m-1} \sum_{j=0}^{k-1} (-1)^{i+j} \binom{n}{j}(k-j)^m {j \brack {j-i}} {{n-j}\brack {m+1+i-j}}$$

Where the symbol ${x \brack y}$ stands for the Stirling numbers of the first kind (without sign).

I'm interested in the case $1\leq m,k\leq n-1$.

I have already proven the following:

1) If in the sum we set $m=n-1$, we get just the well known recurrence for Eulerian numbers, so it is positive. For $m=n-2$, the result is a sum of two Eulerian numbers.

2) If we replace $k$ by $n-k$, the sum remains the same.

3) With $k=1$, we get simply the Stirling numbers of the first kind.

4) With $m=1$ the sum is always positive.

Probably someone with more experience working on this kind of sums can give me a hand. I bet that there may be even a way to understand this sum combinatorially.

Luis Ferroni
  • 733

1 Answers1

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For the moment at least, I can just individuate the first step of an approach which might be possibly interesting.

The sum can be rewritten as $$ \eqalign{ & S(q,n,m) = \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( { \le \,n - m - 1} \right)} {\;\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^{\,i + j} \left( \matrix{ n \cr j \cr} \right)\left( {q - j} \right)^{\,m} \left[ \matrix{ j \cr j - i \cr} \right]\left[ \matrix{ n - j \cr m + 1 + i - j \cr} \right]} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)} {\;\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^{\,k} \left( \matrix{ n \cr j \cr} \right)\left( {q - j} \right)^{\,m} \left[ \matrix{ j \cr k \cr} \right]\left[ \matrix{ n - j \cr m + 1 - k \cr} \right]} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( \matrix{ n \cr j \cr} \right)\left( {q - j} \right)^{\,m} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)} {\left( { - 1} \right)^{\,k} \left[ \matrix{ j \cr k \cr} \right]\left[ \matrix{ n - j \cr m + 1 - k \cr} \right]} } \cr} $$ where putting the bounds in parentheses is meant to underline that they are implicit in the binomial / Stirling n. , which is a useful indication for dealing with convolutions.

Since $$ x^{\,\overline {\,n\,} } x^{\,\overline {\,m\,} } = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n + m} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left[ \matrix{ n \cr j \cr} \right]\left[ \matrix{ m \cr k - j \cr} \right]x^{\,k} } } $$ where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial
then the inner sum above can be written as $$ \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)} {\left( { - 1} \right)^{\,k} \left[ \matrix{ j \cr k \cr} \right]\left[ \matrix{ n - j \cr m + 1 - k \cr} \right]} = \left[ {x^{\,m + 1} } \right]\left( {\left( { - x} \right)^{\,\overline {\,j\,} } x^{\,\overline {\,n - j\,} } } \right) = \cr & = \left[ {x^{\,m + 1} } \right]\left( {\left( { - 1} \right)^j x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } } \right) = \left[ {x^{\,m + 1} } \right]\left( {\left( { - 1} \right)^j x^{\,\underline {\,j\,} } \left( {x + n - 1 - j} \right)^{\,\underline {\,n - j\,} } } \right) \quad \left| \matrix{ \;1 \le n \hfill \cr \;j \le n \hfill \cr} \right. \cr} $$ thus giving $$ \bbox[lightyellow] { S(q,n,m) = \left[ {x^{\,m + 1} } \right]\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^j \left( \matrix{ n \cr j \cr} \right) \left( {q - j} \right)^{\,m} x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } } \quad \left| {\;1 \le n} \right. }$$

The function on RHS can be further rewritten as $$ \eqalign{ & F(q,n,m,x) = \sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^j \left( \matrix{ n \cr j \cr} \right)\left( {q - j} \right)^{\,m} x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } } = \cr & = n!\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^j \left( {q - j} \right)^{\,m} \left( \matrix{ x \cr j \cr} \right)\left( \matrix{ x + n - 1 - j \cr n - j \cr} \right)} = \cr & = n!\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( {q - j} \right)^{\,m} \left( \matrix{ j - x - 1 \cr j \cr} \right)\left( \matrix{ x + n - 1 - j \cr n - j \cr} \right)} \cr} $$

G Cab
  • 35,964
  • Thank you very much! I can only say that your conclusion was where I came from. That was exactly my original polynomial. I individuated every coefficient making the same steps (backwards, obviously), and expressed it as in the original question. – Luis Ferroni Oct 30 '19 at 18:35
  • Uh-Oh ! Anyway, I don't know from which end is better to start (!?) – G Cab Oct 30 '19 at 18:51
  • shouldn't the exact upper bound for the sum over $k$ be min$(m+1,q-1)$ ? then, for the upper bound to be exactly $m+1$ we would need $m \leq q-2 $ , but this is not specified in the OP. Does this matter? – René Gy Oct 30 '19 at 19:45
  • @RenéGy: the two Stirlings are implicitly providing that $k \le m+1 $ and $0 \le k$ and $k \le j$. The bound on $j$ is instead kept explicitly. So we do not need bounds on $k$: for computation you can take $m+1$ or $j$ or $q-1$. – G Cab Oct 30 '19 at 20:05
  • @GCab: ok, thanks for the explanation. I find that the index $k$ for the inner sum can be more precisely bounded between max$(0,j-(n-q-1))$ and $j$. This would mean that the number of terms in the inner sum would be $j+1$ when $q \leq n-m$ and $j+1 +n-m$ when $q>n-m$. Is that correct? This might indicate that for studying further $S(q,n,m)$ we would need to study separately two cases depending on whether $q \leq n-m$ or not. Unless it can be found a kind of symetry relation between $S(q,n,m)$ and $S(n-q,n,n-m)$? – René Gy Oct 30 '19 at 22:31
  • @RenéGy: to that aim, yes, you have to fix the actual bounds. For that just consider that the Stirling N. (1st and 2nd) are bounded to be non null same as the binomial, plus the fact that $St1(n,0)$ is non null only for $n=0$. – G Cab Oct 30 '19 at 23:01
  • @RenéGy, concerning the symmetry, as the OP said, $S(q,n,m)=S(n-q,n,m)$ for $1 \le m$ and the computation for small values confirms that, but I could not reach to demonstrate it. – G Cab Oct 30 '19 at 23:15
  • @GCab that is good, because then it suffices to consider the case where $q \leq n-m$. I will now try to prove $S(q,n,m)=S(n-q,n,m)$. – René Gy Oct 31 '19 at 15:35
  • The fact that $S(q,n,m)=S(n-q,n,m)$ may be assumed as truth. I have a proof but it's a bit lengthy. Personally I don't find my proof very instructive for proving the positivity of $S(q,n,m)$. – Luis Ferroni Oct 31 '19 at 16:36
  • @Vladislao: if your proof is too long better not to insert it here. However knowing the basics of it (and from where the problem originates) could cast some light on the resolution approach. – G Cab Oct 31 '19 at 18:59
  • The key idea is to define $f_{j,n,m} \doteq \sum_{i=0}^{n-m-1} (-1)^i {j \brack {j-i}} {{n-j}\brack{m+1+i-j}}$. After that, one may prove that $f_{j,n,m} = (-1)^{n-m-1} f_{n-j,n,m}$, and from there, playing a little bit with the expression of $S(n-q,n,m)$ and this symmetry formula for the $f_{j,n,m}$, one gets the desired result after substracting $S(q,n,m)$ and $S(n-q,n,m)$. – Luis Ferroni Oct 31 '19 at 19:10
  • About the symetry of $q$ with $n-q$, by playing with the indices,I find that $S(q,n,m)-S(n-q,n,m)= \sum_{j,h}(q-j)^m {n\choose j}{j\brack h}{n-j\brack m+1-h}$ where the indices of the double sum run over all non-negative integers. Still the sum is finite, because the numbers of non-zero summands is finite. If I am not mistaken, this is $(-1)^n n!$ times the coefficient of $x^{m+1}$ in the polynomial $F_{n,q}(x)=\sum_{j=0, j\neq q}^n\frac{1}{q-j}{x(q-j)\choose j}{-x(q-j)\choose n-j}$. $F_{n,q}$ seems to be the zero-polynomial indeed, but I could not demonstrate it. – René Gy Nov 01 '19 at 13:29
  • We want to compute the sum $\sum_{j=0}^n (-1)^j \binom{n}{j} (k-j)^m f_{j,n,m}$. If you write it as a coefficient of a product of series, it is sufficient to show that $\sum_{j=0}^n (-1)^j \binom{n}{j} f_{j,n,m}=0$ which can be proven by induction in $n-m$ for example (again this is a coefficient in a product, $(1-x)^n$ and the generating function of $f_{j,n,m}$ for $j$), so using some easy recurrences on $f_{j,n,m}$, this gets you to the result. Notice that, as I said, the proof is non instructive. Look at the EDIT in the original post, I reformulated the problem by reducing it. – Luis Ferroni Nov 02 '19 at 13:59
  • I think I have solved both the original question and the one I added in the edit part. The proof is too long to post it here. I gave a combinatorial (i.e. positive integer) but more or less complicated interpretation for all of these numbers and it seems to work perfectly. – Luis Ferroni Nov 04 '19 at 00:18
  • @Vladislao: congratulations, it is quite a hard problem ! pity that you cannot post your solution. – G Cab Nov 04 '19 at 01:19