With great pleasure I read the [recent paper of Griffin, Ono, Rolen and Zagier](https://www.pnas.org/content/early/2019/05/20/1902572116) 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](https://mathoverflow.net/q/323781/41139)) 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. 



Intuition behind the growth condition in the result of Griffin, Ono, Rolen and Zagier on Jensen polynomials
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 $X^3+3X+2$.


