Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with zeroes
Search options not deleted
user 59219
5
votes
1
answer
168
views
Existence of Laurent series with zeroes at $𝑒^{2𝑛}$ ($𝑛∈ℕ_0$) and even faster coefficient...
This is an extension of an earlier question of mine which corresponds to the case $A = 1$. Precisely, my question is as follows:
Given $A > 0$ fixed but arbitrary, is there a non-trivial sequence $(c …
- 734
9
votes
1
answer
404
views
Existence of Laurent series with zeroes at $e^{2n}$ ($n \in \Bbb{N}_0$) and extremely fast c...
Real and Complex Analysis, Theorem 15.20]) which shows that if $f : \Bbb{C} \to \Bbb{C}$ is holomorphic, then
$$
M(2r) \geq \prod_{n=1}^{n(2r)} \frac{2r}{|\alpha_n|},
$$
where the $\alpha_i$ are the zeroes … of $f$, and where $|\alpha_1| \leq |\alpha_2| \leq \dots$, and where $n(2r)$ is the number of zeroes of $f$ satisfying $|z| < 2r$. …
- 734