Timeline for Maximum and concavity of function
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8 at 2:58 | vote | accept | nervxxx | ||
| Jul 31 at 12:25 | answer | added | fedja | timeline score: 4 | |
| S Jul 24 at 14:01 | history | bounty ended | CommunityBot | ||
| S Jul 24 at 14:01 | history | notice removed | CommunityBot | ||
| Jul 22 at 11:46 | comment | added | ResearchMath | Finally, to compute the value at the symmetric point I would use the formula $\log(x)=2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{x-1}{x+1}\right)^{2k+1}$ and similar computational methods as above. | |
| Jul 22 at 11:43 | comment | added | ResearchMath | Have you tried it? I did try it. You said "even a sketch of the steps would be much appreciated!", but you don't even try what I say... | |
| Jul 22 at 11:41 | comment | added | nervxxx | @ResearchMath well, I threw my integral into Mathematica and it could not find a closed form expression for it (only numerically integrate it). I doubt the website you linked is more computational powerful.. | |
| Jul 22 at 11:33 | comment | added | ResearchMath | You just compute one integral at a time. I am pretty sure the antiderivatives exist. | |
| Jul 22 at 11:30 | comment | added | nervxxx | @ResearchMath this sounds promising, but I think without some steps from you it is hard to piece together your argument... The integrals I get that determine whether I have a local minimum or not, are 3-dimensional integrals (over the 3-torus) --- the website you linked only does 1 dimensional integrals as far as I can see, so maybe you have different objects you are working with than what I have in mind? | |
| Jul 22 at 11:22 | comment | added | ResearchMath | The integrals which express the conditions for a local maximum can be handled by first integrating by parts on the cosine to get rid of the logarithm, then using the addition formulas for the cosine and the sine to get a rational function of trigonometric functions and then using the tangent half-angle substitution. Subsequent integrals also seem computable, just try them on integral-calculator.com to see which steps are involved. | |
| Jul 22 at 11:15 | comment | added | ResearchMath | I would write $x_i=y_i^2$ and use the method used here mathoverflow.net/questions/471487/… to translate the problem to the Lie group $SO(3)$. | |
| Jul 22 at 11:06 | comment | added | nervxxx | @ResearchMath even a sketch of the steps would be much appreciated! | |
| Jul 22 at 10:15 | comment | added | ResearchMath | It even seems possible to express the value at the symmetric point as an infinite series, but I don't have time to carry this computation out in full detail. | |
| Jul 22 at 10:12 | comment | added | ResearchMath | It is quite a lengthy computation, but doable I think. | |
| Jul 22 at 10:10 | comment | added | ResearchMath | I think I could prove it is a local maximum. | |
| Jul 22 at 9:49 | comment | added | nervxxx | @ResearchMath I agree, the question is how :) I tried computing the Hessian and reduced what I need to show to negativity of some integral. Mathematica tells me it is but I'd like to understand why it is negative (hence generalizing the method to solve the problem from 3 angles and 3 simplex to general n).. | |
| Jul 22 at 8:12 | comment | added | ResearchMath | Showing that it is a local maximum might be doable. | |
| Jul 21 at 11:41 | comment | added | nervxxx | @AlexRavsky thanks. At this stage I'm just happy showing the symmetric point is the maximum. It is natural but resists a proof so far... | |
| Jul 21 at 8:44 | comment | added | Alex Ravsky | Concerning a simple closed-form expression for the maximum value, there is a weak hope that when you calculate it numerically up to a dozen of digits, you will be able to find it via Google. | |
| Jul 20 at 22:48 | comment | added | nervxxx | @Steve ah good catch! It should not be, sorry for the typo. | |
| Jul 20 at 22:47 | history | edited | nervxxx | CC BY-SA 4.0 |
deleted 3 characters in body
|
| Jul 20 at 13:52 | comment | added | Steve | Just to double check: The case $i=j$ should really be included in the sum in the definition of $p$? | |
| Jul 16 at 18:28 | history | edited | LSpice | CC BY-SA 4.0 |
Slightly less squashed `\sqrt`; `_{max}` -> `_\text{max}`
|
| Jul 16 at 12:26 | history | edited | nervxxx | CC BY-SA 4.0 |
added 23 characters in body
|
| S Jul 16 at 12:25 | history | bounty started | nervxxx | ||
| S Jul 16 at 12:25 | history | notice added | nervxxx | Draw attention | |
| S Jul 13 at 10:16 | history | suggested | George Giapitzakis | CC BY-SA 4.0 |
Subscripts in x_i's
|
| Jul 13 at 9:51 | review | Suggested edits | |||
| S Jul 13 at 10:16 | |||||
| Jul 13 at 4:40 | history | asked | nervxxx | CC BY-SA 4.0 |