Timeline for Sum of exponential functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2012 at 19:24 | vote | accept | John Engbers | ||
| May 17, 2012 at 17:38 | answer | added | Igor Rivin | timeline score: 1 | |
| May 17, 2012 at 17:08 | comment | added | David E Speyer | My suggestion is to show that $\log f$ is convex, not that $f$ is. A plot of $\log f$ looks pretty good wolframalpha.com/input/?i=Plot%20Log%281%2B2^%28x-1%29%29%2Fx&t=mfftb01 | |
| May 17, 2012 at 15:41 | comment | added | John Engbers | Thanks for the suggestions --- I've tried using a H\"{o}lder/Jensen/etc inequality, but haven't found the right thing to use them on yet. @Tom My initial use of the power mean theorem was hurt by the sum of the weights, which also get hit by the $x$th root; I shall keep poking around along these lines. @David Unless I'm missing something here, I don't think that the log convex trick will work, since $f$ isn't necessarily convex (e.g. $a_1=1, a_2=2$). | |
| May 17, 2012 at 15:35 | history | edited | John Engbers | CC BY-SA 3.0 |
added a minor restriction to the problem.
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| May 16, 2012 at 19:14 | comment | added | David E Speyer | I don't have time to think about this carefully either, but one way to prove that a function only changes direction once is to prove that it is log convex. Googling "log convex" and "power mean" turned up files.ele-math.com/articles/jmi-05-24.pdf , which looks like it might have the tools you need. | |
| May 16, 2012 at 16:09 | comment | added | Tom Leinster | I don't have time to think about this properly now, but it seems very likely that it would help to think about power means (a.k.a. generalized means). See e.g. Wikipedia or Hardy, Littlewood and Pólya's text Inequalities. The basic theorem is that the power mean of order $x$ increases with $x$, strictly so unless all the numbers you're taking the mean of are equal. In your case you might want to relate $f(x)$ to the $x$th order power mean of $a_1, \ldots, a_q$, weighted by $a_1^{-1}, \ldots, a_q^{-1}$. | |
| May 16, 2012 at 15:32 | history | asked | John Engbers | CC BY-SA 3.0 |