Let $i$, $k$ be integers such that $2 \leq i \leq k$. I would like to show that the sum $$ \sum_{j=1}^{i-1} \frac{(-1)^{j-1}(i-j)^k}{(i-j)! (j-1)!} $$ is positive. I have carried out extensive numerical experiments to check this for small values of $k$. In fact, much more should be true. Define polynomials $$ U(x)=(x+i-1)^k $$ and $$ V(x)=x(x+1)\cdots(x+i-1). $$ Let $Q$ and $R$ be the quotient and remainder on dividing $U$ by $V$. The above sum is the leading coefficient of $R$. It seems that all the coefficients of $Q$ and $R$ are always positive, and it would be nice to prove this, but I only need the positivity of the above sum. This question has applications for proving the irrationality of certain series.
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ This is a derivative of the binomial $(1-x)^N$ for a suitable $N$. $\endgroup$– Wadim ZudilinCommented Jul 16, 2010 at 11:27
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
These are Stirling numbers of the second kind. More precisely your sum is S(k,i-1) where $S$ denotes Stirling number of the second kind.
answered Jul 16, 2010 at 12:01
-
3$\begingroup$ ... in particular, they have a combinatorial interpretation in terms of counting partitions that makes them manifestly positive. $\endgroup$– S. Carnahan ♦Commented Jul 16, 2010 at 14:28
$\begingroup$
$\endgroup$
0
More general result is the following:
if $R$ is remainder of $x^k$ modulo $(x-c_1)(x-c_2)\dots (x-c_i)$ with non-negative $c_i$'s, then leading coefficient of $R$ is positive.
Indeed, let $R=ax^{i-1}+\dots$, then $f(x):=x^k-ax^{i-1}-\dots$ has roots in $c_i$'s, then by Rolle theorem $f^{(i-1)}$ has at least one positive root, which is true iff $a>0$.
answered Jul 16, 2010 at 21:47