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Given a pair of strictly increasing functions $f,g:\mathbb{N}\to \mathbb{N}$ define:

$P_N(f,g)\doteq \left(z\in \mathbb{C}\mapsto \prod_{i=1}^{f(N)}\left(1+\frac{z}{v_i(N)}\right)\in \mathbb{C}\right),$

where $v_i:\mathbb{N}\to \mathbb{C}, |v_i(N)|\geq g(N) \mbox{ for } i=1,2,\ldots,f(N);$

and such that $\lim_{N\to\infty} \sum_{i=1}^{f(N)}\frac{1}{v_i(N)}=1.$

Define $S$ to be the set of functions $\varphi:\mathbb{C}\to \mathbb{C}$ such that there exists a sequence $(P_N(f,g))_{N\in\mathbb{N}}$ with some $f,g:\mathbb{N}\to \mathbb{N}$ strictly increasing functions such that $P_N(f,g)$ converges compactly to $\varphi$ as $N\to\infty.$

Question: Can you describe $S$?

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  • $\begingroup$ Looks to me like a collection of entire complex functions with infinitely many roots, whose value at zero is one, with derivative having value 1 at zero, and maybe even higher order derivatives having absolute value at most one when evaluated at zero. I'm not a complex analyst though: there are likely better descriptions. $\endgroup$
    – The Masked Avenger
    Commented May 29, 2014 at 15:33
  • $\begingroup$ The answer depends substantially on what limits are allowed (pointwise, uniform on compacts, or something else), so it would be nice to specify it. $\endgroup$
    – fedja
    Commented Jun 1, 2014 at 2:34
  • $\begingroup$ I have specified the limit that it is allowed. Thanks. $\endgroup$
    – user39115
    Commented Jun 2, 2014 at 9:29
  • $\begingroup$ My guess is that $S$ is the set of zero-free entire functions of finite order. $\endgroup$
    – user39115
    Commented Jun 9, 2014 at 18:38

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$S$ is the set of all entire zero-free functions $F$ with $F(0)=F'(0)=1$. To approximate such an $F$, just cut off its Taylor series $F(z)=1+z+\sum_{n\ge 2} a_n z^n$ at high enough degree $N_1=f(1)$. Make sure this polynomial $p$ approximates $F$ well enough on $|z|\le 1=g(1)$ (say) so that it will be zero-free there. Since $p(0)=p'(0)=1$ also, its factorization is of the required form. Continue in this way, with $f(2)>f(1)$ and $g(2)>g(1)$ etc.

Conversely, any locally uniform limit of $P_n(f,g)$'s is of course zero-free and satisfies $F(0)=F'(0)=1$.

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    $\begingroup$ I like your answers all of them are short and natural, but give me a short time to convince myself. $\endgroup$
    – user39115
    Commented Jun 10, 2014 at 9:27
  • $\begingroup$ Is it true then that $S=\{e^{z+\sum_{i=2}^N a_i z^i}: a_i\in\mathbb{C},N\in\mathbb{N}\}$ ? $\endgroup$
    – user39115
    Commented Jun 10, 2014 at 16:34
  • $\begingroup$ No, you need a general entire function in the exponent, not just a polynomial: $S=\{ e^{z+\sum_{n=2}^{\infty} a_nz^n}\}$ (and of course we only consider $R=\infty$ power series here) $\endgroup$
    – Christian Remling
    Commented Jun 10, 2014 at 16:54

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