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Im looking for functions $g(z)$ such that

$$g(z) = \sum_{n=0}^{\infty} \frac{g(n)}{(n+1)!} z^n = g(0) + \frac{g(1)}{2} z + \frac{g(2)}{6} z^2 + \frac{g(3)}{24} z^3 + ...$$

and $g(n)$ are all positive reals; $0 \leq g(n)$.

And I wonder how fast $g(n)$ grows.

I got inspired by this :

$f(z) = \sum_{n=0}^{\infty} f(n)^2 z^n$?

But since that did not behave very well - see the edit and the comments there - , I made this variant.

I am not sure this works out nice, Im just wondering.

For clarity with "nice " or behaving "good" I mean being an entire function.

Gary
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mick
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  • I guess fans of fractional calculus know about this since it seems $g$ is close to its own $x$ th derivative of $g(0)$ by taylor's theorem, well at least for positive integers $x$. In fact I could have posted the question that way. – mick Nov 29 '23 at 23:06
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    this is weird again since for $z=1$ you get $g(1)=g(1)+A, A \ge 0$ so you need all $g(k)=0, k \ne 1$ so $g(z)=az$; if you want something like this with $g(k) \ge 0$ you need to pick $a_k$ st $a_k k^k <1$ and $g(z)=\sum a_n g(n)z^n$ – Conrad Nov 30 '23 at 00:31
  • @Conrad yes you are right ! I edited. – mick Nov 30 '23 at 12:05
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    This is impossible because $n^n > (n + 1)!$ for all $n \geq 4$. So if $n \geq 4$, then $$ g(n) = g(0)\frac{n^0}{(0 + 1)!} + \cdots + g(n)\frac{n^n}{(n+1)!} + \cdots \geq g(n)\frac{n^n}{(n+1)!}, $$ which is only true if $g(n) = 0$. but then $g$ is a polynomial, so it cannot be $0$ for all $n \geq 4$. – Polygon Nov 30 '23 at 13:27
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    I expect that without the constraint that $g(n) \geq 0$, there will be an entire solution. This seems related to https://math.stackexchange.com/questions/91855/is-there-a-function-with-the-property-fn-fn0, though I'm not sure if the methods there will be helpful here. – Polygon Nov 30 '23 at 13:47
  • @Polygon I agree that with general complex coefficients one should be able to find a solution just by taking an entire function $g$ for which both the coefficients and the values at $n$ are small enough, perturbing it by $h(z)\sin \pi z$ which doesn't change the values at $n$ and inductively determine the Taylor coefficients of $h$ to get the required expansion for $g+h\sin$ (Match the new Taylor series with the given $g(n)$) and prove those are small enough to determine an entire function – Conrad Nov 30 '23 at 14:09
  • @Polygon yes you are right. maybe $(2n+1)!$ will do the trick ? I am thinking about it. – mick Nov 30 '23 at 22:25
  • I think I can construct a solution that behaves like $g(n) = O(\exp(\pi(n)))$ where $\pi(n)$ is the prime counting function. – mick Nov 30 '23 at 22:28
  • Slightly related : the maximum number of solutions to similar equations : https://math.stackexchange.com/questions/4817807/an-infinite-linear-system-of-equations-with-an-uncountable-number-a-of-equatio – mick Dec 01 '23 at 00:04
  • Notice that for such equations, if $f(x)$ is a solution , then so is $C f(x)$ for constant $C$. So if there is an infinite amount of linear independant solutions then there are an uncountable number of solutions by $ y_1 sol_1 + y_2 sol_2 + ...$ – mick Dec 01 '23 at 00:16

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