Timeline for Is this entire function a square?
Current License: CC BY-SA 4.0
15 events
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| Jul 16, 2023 at 13:38 | history | edited | Mikhail Katz | CC BY-SA 4.0 |
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| Jul 13, 2023 at 17:49 | comment | added | Emil Jeřábek | I should have mentioned that the last thing in my previous comment is an application of the Borel–Carathéodory theorem. | |
| Jul 12, 2023 at 15:06 | history | edited | Mikhail Katz | CC BY-SA 4.0 |
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| Jul 12, 2023 at 14:37 | history | edited | Mikhail Katz | CC BY-SA 4.0 |
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| Jul 12, 2023 at 13:08 | comment | added | Emil Jeřábek | I don’t see how you’d do that, but then I’m no expert in complex analysis. However, Christian Remling’s argument is not really complicated. You don’t need any results on the existence of Hadamard factorization: the assumption that $f(z)$ has no zero besides the double zero at $z=0$ already gives that $f(z)=z^2e^{g(z)}$ for some entire $g$, and then you just need to argue that since $\operatorname{Re}g(z)=O(|z|)$, $g$ must be a linear polynomial. | |
| Jul 12, 2023 at 12:46 | comment | added | Mikhail Katz | @EmilJeřábek: Good point. It seems to me that Remling's argument with Hadamard representation is a little too convoluted. Couldn't one use the transcendental nature of the function a bit more directly? | |
| Jul 12, 2023 at 12:41 | comment | added | Emil Jeřábek | No, that’s the point. I just gave you a counterexample: $ze^z$. Or more generally, $p(z)e^z$ for a nonconstant polynomial $p(z)$. This attains all values, but it attains zero only finitely many times. | |
| Jul 12, 2023 at 12:01 | history | edited | Mikhail Katz | CC BY-SA 4.0 |
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| Jul 12, 2023 at 12:01 | comment | added | Mikhail Katz | @EmilJeřábek, but the values it does attain (all except possibly one value), it attains infinitely many times, doesn't it? | |
| Jul 12, 2023 at 10:54 | comment | added | Emil Jeřábek | Can you elaborate on since $f$ is a transcendental function, it must hit every value infinitely many times? This is not true in general: e.g., the transcendental function $ze^z$ has exactly one zero. What follows from Great Picard is that a meromorphic function that is not a rational function attains all but possibly one values infinitely many times. | |
| Jul 12, 2023 at 6:14 | history | edited | Mikhail Katz | CC BY-SA 4.0 |
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| Jul 11, 2023 at 19:42 | comment | added | LSpice | @DavidESpeyer, that argument was given by @ChristianRemling. | |
| Jul 11, 2023 at 19:32 | comment | added | David E Speyer | If $\sin z = z$ and $1=\cos z$, then $z^2+1=\sin^2 z + \cos^2 z = 1$ and thus $z=0$. So $z=0$ is the only double zero of $\sin z - z$. | |
| Jul 11, 2023 at 14:23 | comment | added | Bazin | Kats Thanks for your interest in my question. If $1=\cos z$ and $w=e^{iz}$, you get $w+\frac{1}{w}=2$ (not $2i$). | |
| Jul 11, 2023 at 12:49 | history | answered | Mikhail Katz | CC BY-SA 4.0 |