Timeline for Do there exist transcendental numbers which are not hypertranscendental?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 29, 2013 at 11:34 | comment | added | S. Carnahan♦ | 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. | |
| May 29, 2013 at 11:25 | vote | accept | Mark | ||
| May 29, 2013 at 11:06 | comment | added | yurius | One should also require $a_0$ to be rational. | |
| May 29, 2013 at 11:05 | comment | added | François Brunault | @Mark : the $n$-th coeff of $g$ is $a_n-a_{n-1}/z$. | |
| May 29, 2013 at 7:37 | comment | added | Mark | @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. | |
| May 29, 2013 at 7:22 | history | answered | Lucas Culler | CC BY-SA 3.0 |