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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3$\begingroup$ The title and the body ask different questions. A simple example for the body question is $2 \pi i$. $\endgroup$– Qiaochu YuanCommented May 29, 2013 at 5:00
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$\begingroup$ @: Qiaochu Yuan : Thank you. It has been fixed. $\endgroup$– MarkCommented May 29, 2013 at 5:02
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$\begingroup$ A related question is mathoverflow.net/questions/42449. $\endgroup$– Richard StanleyCommented May 29, 2013 at 13:24
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$\begingroup$ The question was asked at XI St.Petersburg Summer Meeting in Mathematical Analysis in 2002, but not answered there. $\endgroup$– MarkCommented May 29, 2013 at 13:38
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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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$\begingroup$ @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. $\endgroup$– MarkCommented May 29, 2013 at 7:37
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2$\begingroup$ @Mark : the $n$-th coeff of $g$ is $a_n-a_{n-1}/z$. $\endgroup$ Commented May 29, 2013 at 11:05
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1$\begingroup$ One should also require $a_0$ to be rational. $\endgroup$– yuriusCommented May 29, 2013 at 11:06
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1$\begingroup$ 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. $\endgroup$– S. Carnahan ♦Commented May 29, 2013 at 11:34