Timeline for Inequality for an integral [closed]
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2014 at 23:02 | history | closed |
Will Jagy Stefan Kohl♦ Tilman Ryan Budney S. Carnahan♦ |
Not suitable for this site | |
| Jul 16, 2014 at 21:41 | comment | added | Marjo | @ Yemon Choi By using Cauchy formula, you obtain the function $f(r)$, and mathematica 8 software tells us that the maximum of $f$ is in $0$. | |
| Jul 16, 2014 at 21:39 | comment | added | Yemon Choi | When you say you guess this is true, how did you guess these exact constants? | |
| Jul 16, 2014 at 20:45 | comment | added | Marjo | It should be $|f'(z)-f'(0)|\le \frac{4|z|}{\pi(1-|z|^2)} \max|f(z)|$ instead. | |
| Jul 16, 2014 at 20:29 | comment | added | Christian Remling | If that's (= your comment) what you're after, then Cauchy's formula seems to give a better estimate quite easily. | |
| Jul 16, 2014 at 20:23 | comment | added | Marjo | Not an exercise, I am guessing that this is true but probably a problem which no one can solve it. | |
| Jul 16, 2014 at 20:00 | comment | added | Yemon Choi | OK, so this looks like an exercise from a book. If so, then I think math.stackexchange.com would be a more suitable site for this question. | |
| Jul 16, 2014 at 19:54 | comment | added | Marjo | A step on the proof of the following inequality $|f'(z)-f'(0)|\le \frac{8|z|}{\pi(1-|z|^2)} \max|f(z)|$, where $f$ is analytic in the unit disk. | |
| Jul 16, 2014 at 19:51 | comment | added | Yemon Choi | Could you give some more background to this problem? Surely this is not a randomly chosen integral; is this an exercise, or a step in the proof of a larger result? | |
| Jul 16, 2014 at 19:09 | history | edited | Marjo | CC BY-SA 3.0 |
added 9 characters in body
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| Jul 16, 2014 at 18:53 | review | Close votes | |||
| Jul 18, 2014 at 23:02 | |||||
| Jul 16, 2014 at 18:36 | review | First posts | |||
| Jul 16, 2014 at 19:01 | |||||
| Jul 16, 2014 at 18:32 | history | asked | Marjo | CC BY-SA 3.0 |