Timeline for Hermite-Kakeya Theorem for entire functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Jan 3, 2016 at 7:16 | vote | accept | mike | ||
| Jan 3, 2016 at 4:58 | answer | added | Alexandre Eremenko | timeline score: 3 | |
| Jan 3, 2016 at 4:32 | comment | added | Alexandre Eremenko | Where is $r$ in your statement for entire functions? | |
| Jan 3, 2016 at 2:42 | comment | added | mike | @ChristianRemling. You are right. I only need to prove that $h(z)=f(z)-g(z)$ has only real zeros. Let $f_m(z),g_m(z)$ be the cut off version of the products up to $m$ terms. I think that I might be able to prove that $f'(x)g(x)-g'(x)f(x)>0,x\in\mathbb{R}$. But it might be hard for me to prove that $f'_m(x)g_m(x)-g'_m(x)f_m(x)>0$. Thanks again for the advice. | |
| Jan 3, 2016 at 2:25 | comment | added | Christian Remling | I'm not sure I understand all the details fully (you want to assume (1), (2), so these are not general entire functions?), but I think such a version follows immediately from the one you quoted. If $f-g$ has a non-real zero, then the same holds for the approximating polynomials when you cut off the products. | |
| Jan 3, 2016 at 2:17 | history | edited | mike | CC BY-SA 3.0 |
added 9 characters in body
|
| Jan 3, 2016 at 2:03 | history | asked | mike | CC BY-SA 3.0 |