Positivity of sequences via generating series - MathOverflow

Positivity of sequences via generating series

2010-07-16T00:47:54Z

There are different ways of showing that a given sequence $a_0,a_1,a_2,\dots$ of integers, say, is nonnegative. For example, one can show that $a_n$ count something, or express $a_n$ as a (multiple) sum of obviously positive numbers. Another recipe is manipulating with the corresponding generating series $A(x)=\sum_{n=0}^\infty a_nx^n$ and showing that $A(x)\ge0$ (this is the notation for a series which has all coefficients nonnegative, and this extends to formal power series in as many variables as needed).

An example of criterion in this direction is $$ (*) \qquad A(xy)\ge0 \iff \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)\ge0 $$ (the multiple $1/(1-x)(1-y)$ is introduced for cosmetic purposes only and, of course, both $A(x)\ge0$ and $A(xy)\ge0$ are by definition equivalent to the nonnegativity of the sequence $a_n$). The latter can be verified by comparing the corresponding coefficients in the power series expansion $$ \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr) =\sum_{n=0}^\infty a_n\sum_{k,m=0}^\infty\binom{n+k}n\binom{n+m}nx^{n+k}y^{n+m}. $$

On the other hand, the one-dimensional version of $( * )$, $$ A(x)\ge0 \iff \frac1{1-x}A\biggl(\frac{x}{1-x}\biggr)\ge0, $$ is simply false.

My question is whether it is possible to find two nontrivial rational functions $p(x)\in\mathbb Q[[x]]$ and $r(x)\in x\mathbb Q[[x]]$ in one variable $x$ such that $$ A(x)\ge0 \iff p(x)A\bigl(r(x)\bigr)\ge0. $$ Although I am not supposed to put several problems in one question, I would also ask about a more direct proof of $( * )$ and about general ways of constructing such $p$ and $r$ in more than one variable.

Motivation. Basically I am interested in proving nonnegativity of certain $q$-series sequences $a_0(q),a_1(q),a_2(q),\dots$ by manipulating with the corresponding generating series $A_q(x)=\sum_{n=0}^\infty a_n(q)x^n$. Some of them can be "guessed" from non-$q$-versions, for example there is a neat $q$-analogue of the criterion $( * )$.

Answer by Qiaochu Yuan for Positivity of sequences via generating series

2010-07-17T22:06:27Z

Just a quick observation. It is not hard to see that this is impossible if there exists some $i$ such that for all $j \ge i$, the coefficients of $p(x) r(x)^j$ are all positive after the first positive coefficient. (This occurs, for example, whenever $p, r$ themselves have this property.) This is because by induction one can take $A(x) = a_i x^i - x^{i+1} + a_{i+2} x^{i+2} + ...$ where $a_i, a_{i+2}, ... $ can be chosen to be large enough so that $p(x) A(r(x)) \ge 0$. In general some weird things happen that you might be able to fix with the Skolem-Mahler-Lech theorem, and I suspect that when $r$ is a polynomial it should always be possible to find a counterexample.

Answer by Ryan Reich for Positivity of sequences via generating series

2010-07-22T05:09:18Z

It is impossible, and not just for rational functions. To see this, let's consider the coefficients $b_n$ of $p(x) A(r(x))$ as functions of the $a_n$, the coefficients of $A(x)$. Since $r(0) = 0$ (as it must be) we see that:

Each $b_n$ is a linear combination of $a_0, \dots, a_n$; i.e. we have an upper-triangular infinite matrix $F$, not depending on $a_n$, such that (writing $\vec{a} = (a_n), \vec{b} = (b_n)$) $\vec{b} = F \vec{a}$. Suppose for now that $p(0), r'(0) \neq 0$; then $F$ has a nonzero diagonal, and so is invertible.
If $\vec{b} \geq 0$ for all $\vec{a} \geq 0$, then in particular this is true of the columns of $F$, taking $\vec{a}$ to be infinite "basis" vectors. Conversely, we have equivalently that $\vec{a} = F^{-1} \vec{b}$, so if $\vec{a} \geq 0$ for all $\vec{b} \geq 0$ this must be true of the columns of $F^{-1}$. We conclude that both $F$ and $F^{-1}$ have nonnegative entries.
Lemma in linear algebra: if $F$ is upper-triangular and both it and $F^{-1}$ have nonnegative real entries, then $F$ is diagonal. Proof by induction: true for $1 \times 1$ matrices vacuously. In general, by induction we may assume that the upper-left and lower-right $(n - 1) \times (n - 1)$ blocks of $F$ are diagonal, so only the $(1,n)$ entry of $F$ is nonzero off the diagonal. Then we have $(F^{-1})_{1n} = -F_{1n} F_{nn}/F_{11}$ . Since $F_{11}$ and $F_{nn}$ are both positive, $F_{1n}, (F^{-1})_{1n} \geq 0$ implies $F_{1n} = 0$.
This is also true of infinite matrices, since we can compute the finite upper-left blocks independently of the rest of the matrix.
However, if $F$ is diagonal then we see that $p(x)A(r(x)) = \sum a_n p(x) r(x)^n = \sum F_{nn} a_n x^n$ for all choices of $a_n$, so (for example, taking $a_n = t^n$ for a new variable $t$ and rewriting both sides as power series in $t$) we have $p(x) r(x)^n = F_{nn} x^n$ for all $n$ (and some $F_{nn} > 0$). That is, $(r(x)/x)^n = F_{nn} p(x)^{-1}$ for all $n$, so in fact $r(x)/x = F_{11}/F_{00}$ is constant, and finally, $p(x)$ is constant as well.
Now we remove the assumptions that $p(0), r'(0) \neq 0$. If $x^k$ divides $p(x)$, then replacing $p(x)$ by $p(x)/x^k$ does not change positivity of the coefficients. Now suppose the bottom exponent of $r(x)$ is $x^m$ with positive coefficient; then in $\mathbb{R}[[x]]$ we can write $r(x) = s(x)^m$, and if we denote $A_m(x) = A(x^m)$, we have $A(r(x)) = A_m(s(x))$. Clearly, $A_m$ has nonnegative coefficients if and only if $A$ does, and $s'(0) \neq 0$, so the previous proof applies and $s(x)$ is a multiple of $x$, i.e. $r(x)$ is a multiple of $x^m$, and $p(x)$ is a multiple of $x^k$.