Timeline for Possible to find a set of log-concave functions with log-concave sums?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 2, 2013 at 9:27 | comment | added | Fred B | Just checked: the criterion holds for sums of the form $\left(\frac{x}{A}\right)^\alpha + \left(\frac{x}{B}\right)^\beta$, with $A,B>0$ constants. | |
| Apr 2, 2013 at 9:06 | comment | added | Fred B | Intermediate step: polynomial discriminant can be factored into $(\alpha-\beta)^2 (\alpha^2 + \beta^2 -2 \alpha \beta - 2 \alpha - 2 \beta +1)$. | |
| Mar 26, 2013 at 15:00 | comment | added | Fred B | Thanks! The condition for log-concavity is very neat. Assuming $\beta > \alpha \geq 1$, $\gamma=\beta-\alpha$, the condition further simplify into $\gamma<1+2\sqrt{\alpha}$. In the general case, maybe I should seek similar relationships for elasticities ($\frac{x}{f}\frac{df}{dx}$). Writing down $\frac{d^2}{dx^2}\ln(x^\alpha+x^\beta)$, I obtained $\frac{x^{2\alpha-2}}{x^\alpha+x^\beta} (-\beta x^{(\beta-\alpha)2}+(\alpha^2+\beta^2-\alpha-\beta-2\alpha \beta)x^{(\beta-\alpha)}-\alpha)$ I suppose you used $t=x^{\beta-\alpha}$ and factored the discriminant as in the case $\alpha=2$? | |
| Mar 22, 2013 at 19:14 | history | edited | Robert Israel | CC BY-SA 3.0 |
added 156 characters in body
|
| Mar 22, 2013 at 19:01 | history | answered | Robert Israel | CC BY-SA 3.0 |