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A complex number is said to be hypertranscendental if the one is not a zero of any entire function with all rational Maclaurin coefficients. Does there exist a transcendental number which is not hypertranscendental?

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The title and the body ask different questions. A simple example for the body question is $2 \pi i$. – Qiaochu Yuan May 29 '13 at 5:00
@: Qiaochu Yuan : Thank you. It has been fixed. – Mark May 29 '13 at 5:02
A related question is – Richard Stanley May 29 '13 at 13:24
The question was asked at XI St.Petersburg Summer Meeting in Mathematical Analysis in 2002, but not answered there. – Mark May 29 '13 at 13:38
up vote 11 down vote accepted

Let $z$ be an arbitrary complex number. Since $\mathbb{Q}[i]$ is dense in $\mathbb{C}$, we can choose a sequence of complex numbers $a_n$ such that $|a_n|< \frac{1}{n!}$ and $a_{n+1} - \frac{a_n}{z} \in \mathbb{Q}[i]$. Define an entire function

$ f(t) = \sum_{n=0}^{\infty} a_n t^n $

Then the function $g(t) = (1-t/z)f(t)$ is also entire, has coefficients in $\mathbb{Q}[i]$ and vanishes at $t = z$. Now let $h(t)$ be the function whose power series coefficients are the complex conjugates of the coefficients of $g(t)$. Then $g(t)h(t)$ is a power series with rational coefficients that vanishes at $z$. Hence there are no hypertranscendental numbers.

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@Lucas Culler: Could you explain in detail why $g(t)$ and $g(t)h(t)$ have rational coefficients? As far as I understand it, $z$ is not necessarily rational. – Mark May 29 '13 at 7:37
@Mark : the $n$-th coeff of $g$ is $a_n-a_{n-1}/z$. – François Brunault May 29 '13 at 11:05
One should also require $a_0$ to be rational. – Yurii Savchuk May 29 '13 at 11:06
Here is a comment by CofWsug in response to Mark (from a deleted answer): Since $g$ has coefficients in $\mathbb{Q}[i]$, it suffices to write $g=g_1+ig_2$ where $g_1$, $g_2$ are entire functions with rational coefficients, then you can see that $h=g_1−ig_2$. It follows that $gh=g^2_1+g^2_2$ which is obviously an entire function with rational coefficients. – S. Carnahan May 29 '13 at 11:34

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