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In this question I address the problem of proving the nonnegativity of a numerical sequence $a_0,a_1,a_2,\dots$ via generating series technique. In the notation $A(x)=\sum_{n=0}^\infty a_nx^n\ge0$ meaning that all coefficients of the power series are nonnegative (and this extends to formal power series in as many variables as needed), $$ A(x)\ge0 \iff \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)\ge0. $$ Note that $$ \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr) =\sum_{n=0}^\infty a_n\sum_{k,m=0}^\infty\binom{n+k}n\binom{n+m}nx^{n+k}y^{n+m}. $$

I would like to have a similar criterion for a sequence $a_n(q)$ of polynomials in one variable $q$, that is, something to prove that their coefficients are nonnegative, $a_n(q)\ge0$. For that I replace the binomial coefficients by their $q$-analogue, the Gaussian polynomials $$ \left[\begin{matrix} a \cr b\end{matrix}\right] _q =\frac{(q;q) _a}{(q;q) _b(q;q) _{a-b}}, $$ where $$ (x ; q)_n = \prod _ {j = 1} ^n (1-q^{j-1} x) $$ (the empty product for $n=0$ has to be interpreted as $1$). Note that the coefficients of the Gaussian polynomials are all nonnegative.

Question. Is it true, for a given sequence $a_n(q)\in \mathbb R[q]$, that $$ \sum_{n=0}^\infty a_n(q)x^n\ge0 \iff \sum_{n=0}^\infty a_n(q) \sum_{k,m=0}^\infty \left[\begin{matrix} n+k \cr n\end{matrix}\right] _q \left[\begin{matrix} n+m \cr n\end{matrix}\right] _q x^{n+k}y^{n+m} \ge0 \ ? $$

With the help of the $q$-binomial theorem this can be stated as $$ \sum_{n=0}^\infty a_n(q)x^n\ge0 \iff \sum_{n=0}^\infty\frac{a _ n (q) }{{ ( x ; q ) _ {{n+1} } } {( y ; q ) _ { {n+1} } }} (xy)^n \ge0 . $$

Remark. The polynomiality condition cannot be relaxed to $a_n(q)\in \mathbb R[[q]]$ (power series), as the example $$ a_0(q)=a_2(q)=a_3(q)=a_4(q)=a_5(q)=\sum_{k=0}^\infty q^k\ge0 \quad\text{and}\quad a_1(q)=-1 $$ shows. In this case, we have $$ \sum_{n=0}^5\frac{a _ n (q) }{{ ( x ; q ) _ {{n+1} } } {( y ; q ) _ { {n+1} } }} (xy)^n \ge0 . $$ The trouble with formal power series seems to be related to the fact that they do not possess, in general, the limiting $q\to1$ case.

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