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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.

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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 $X^3+3X+2$.

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  • $\begingroup$ Great answer already! I will wait for the edit! $\endgroup$ – Vincent Jun 16 at 19:56

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