# Intuition behind the growth condition in the result of Griffin, Ono, Rolen and Zagier on Jensen polynomials

With great pleasure I read the recent paper of Griffin, Ono, Rolen and Zagier proving the surprising result that the Jensen polynomials $$J^{d, n}_\alpha$$ for a sequence $$\alpha = \{\alpha(0), \alpha(1), \ldots \}$$ of real numbers whose growth (?) is controlled in a certain way converges for fixed $$d$$ to a limiting polynomial of the same degree uniformly on compact subsets of $$\mathbb{R}$$.

Their main application is to some sequence of real numbers coming from the Riemann Xi function (see also this other MO question) but I already had lots of fun trying to see how this works out for much simpler sequences such as $$\alpha(n) = 1$$ or $$\alpha(n) = 2^n$$.

My question is however with the conditions in their main non-RH-related result: theorem 8 which reads:

Suppose that $$\{E(n)\}$$ and $$\{\delta(n)\}$$ are positive real sequences with $$\delta(n)$$ tending to $$0$$, and that $$F(t) = \sum_{i =1}^\infty c_i t^i$$ is a formal power series with complex coefficients. For a fixed $$d \geq 1$$, suppose that there are real sequences $$\{C_0(n)\},\ldots,\{C_d(n)\}$$, with $$\lim_{n \to \infty} C_i(n) = c_i$$ for $$0 \leq i \leq d$$, such that for $$0 \leq j \leq d$$, we have

$$\frac{\alpha(n+j)}{\alpha(n)} E(n)^{-j} = \sum_{i =0}^d C_i(n) \delta(n)^i j^i + o(\delta(n)^d) \qquad (*)$$ as $$n \to \infty$$. Then we have:

$$\lim_{n \to \infty} \frac{\delta(n)^{-d}}{\alpha(n)} J^{d, n}_\alpha \left(\frac{\delta(n)X - 1}{E(n)}\right) = H_{F, d}$$

uniformly on compact subsets of $$\mathbb{R}$$ where $$H_{F, d}$$ is defined by the generating function $$F(−t) e^{Xt}=\sum_{m=0}^\infty H_{F,m}(X) \frac{t^m}{m!}$$.

My question is about (*). Hopefully it is clear why I wrote above that I already had fun seeing what this theorem means even for really simple sequences $$\alpha$$: it is a priori not at all clear what $$\delta, E$$ or $$C_i$$ to take and one surprising thing I found is that (unlike their limits $$c_i$$) the sequences $$C_i$$ may depend non-trivially on the choice of the fixed value of $$d$$ even in cases we know a priori that the the limits exist for all $$d$$.

But managing to find sequence $$E, \delta, C_i$$ that 'work' is something quite different from understanding what is going on. My question is: what is, intuitively speaking, the set of conditions (*) trying to convey? Is it saying that the sequence $$\alpha$$ cannot grow too fast? Something else? Is it reasonable to think of finite sum on the right hand side as 'roughly a constant' so that the condition says that $$\alpha$$ grows more or less as $$E^j$$ where $$E$$ is the 'typical' value of $$E(n)$$. Ugh, as soon as I type it it stops making sense.

Any enlightenment is welcome here.

The condition does restrict the rate of growth of the functions considered, and $$E$$, $$C_i$$, and $$\delta$$ do have a certain amount of freedom, however they are meant to encode specific information. The $$E$$ term is meant to account for the exponential part of the growth of $$\alpha(n)$$. Once $$E$$ has been fixed, the right hand side of (*) still has a certain amount of freedom due to the little-oh term, but notice that as $$d$$ becomes large, the $$C_i(n)\delta(n)^i$$ terms are forced to approximate the coefficients of a series expansion in $$j$$ of the left hand side. The $$\delta$$ term should account for the residual decay of the $$C_i(n)$$, so that at least some of the $$C_i(n)$$ have non-zero limits as $$n\to \infty$$. In fact in the example in the paper, $$\delta(n)^2$$ essentially measures the log concavity of the LHS. Once $$\delta(n)$$ and $$E(n)$$ have been fixed, there is still some freedom in choosing the $$C_i(n)$$ functions, however they are always approximations of the series expansion of the LHS, with more freedom allowed for small $$d$$ and less for large $$d$$.
The theorem itself is not very instructive in how to find these numbers, but the proof demonstrates a nice method in the case that the $$\alpha(n)$$ are values of a sufficiently smooth function. The first thing we do is expand $$\log(\frac{\alpha(n+j)}{\alpha(n)})$$ as a power series in $$j$$. I'll walk through a couple examples in a moment, but for now, let's assume we have such an expansion, so that $$\log\left(\frac{\alpha(n+j)}{\alpha(n)}\right)=g_1(n)\cdot j+g_2(n)\cdot j^2+\dots,$$ with $$g_i(n)\to 0$$ as $$n\to \infty$$ for $$i\geq 2$$. We'll take $$E(n)=\exp(g_1(n)),$$ which is the primary exponential contribution. The numbers $$\delta(n)$$ should be chosen to be positive numbers which decay like the slowest of $$\sqrt[i]{|g_i(n)|}$$. In the cases considered in the paper, we took $$\delta(n)=\sqrt{-g_2(n)}$$, for $$n\geq 6$$. The fact that $$g_2(n)<0$$ for large $$n$$ is connected to the log-concavity of these sequences.
The two examples you gave, $$\alpha(n)=1$$ and $$\alpha(n)=2^n$$ will be the same except for the $$E(n)$$ term. In the first case we take $$E(n)=1$$ and in the second we take $$E(n)=2^n.$$ At this point it doesn't matter what we take $$\delta(n)$$ to be. As long as they are non-zero, the $$C_i(n)$$ must all be $$0$$. The function $$F(t)=1$$, and the renormalized polynomial will be (no limit needed) $$X^d$$, which are generated as desired by $$F(-t)e^{Xt}=e^{Xt}.$$
We get a more interesting example when we consider $$\alpha(n)=\frac{k^n}{n}$$ for non-zero $$k$$. Different choices of $$k$$ only result in different $$E$$ terms, so for simplicity, lets assume $$k=1$$, so $$\alpha(n)=1/n$$. As before, in order to find $$E(n)$$, we consider $$\log\left(\frac{\alpha(n+j)}{\alpha(j)}\right)=\log\left(\frac{1}{1+j/n}\right)=-\frac{j}{n}+\frac{j^2}{2n^2}-\frac{j^3}{3n^3}\dots,$$ so we take $$E(n)=e^{-1/n}.$$ For $$\delta(n)$$, we could take $$\delta(n)=\sqrt{|g_2(n)|}=\frac{1}{\sqrt{2}n},$$ but it works just as well (and we get simpler expressions) if we take $$\delta(n)=\frac{1}{n}.$$ Then we have $$F(t)=\frac{1}{1+t}e^{t}.$$ This gives us $$F(-t)\exp(tX)=1+\frac{X}{1!}t+\frac{X^2+1}{2!}t^2+\frac{X^3+3X+2}{3!}t^3+\dots.$$
If we calculate the degree $$d=3$$ re-normalized polynomial for $$n=1000$$ using these choices of $$E(n)$$ and $$\delta(n)$$, we get $$\sim 1.003X^3-.007X^2+2.997X+1.992,$$ which matches with our expected limit of $$X^3+3X+2$$.