Timeline for Exponential Convexity
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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Feb 5, 2014 at 19:24 | vote | accept | Shinning Star | ||
Feb 5, 2014 at 17:22 | review | Close votes | |||
Feb 6, 2014 at 10:46 | |||||
Feb 5, 2014 at 13:48 | comment | added | Mark Meckes | @RSG: Yes, you're missing the square in $\zeta^2 h(x)$. | |
Feb 5, 2014 at 13:05 | answer | added | UwF | timeline score: 2 | |
S Feb 5, 2014 at 12:53 | history | suggested | UwF | CC BY-SA 3.0 |
the x_i are also chosen from R
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Feb 5, 2014 at 12:50 | review | Suggested edits | |||
S Feb 5, 2014 at 12:53 | |||||
Feb 5, 2014 at 12:00 | comment | added | RSG | I am slightly confused. if $n=1$ (i.e. there is only $x$), we can always choose a $\zeta\in\mathbb{R}$ to make the product $\zeta h(x)$ not positive, whatever the function $h$ may be. Is it that, I am missing something? You have mentioned for all $n$. (Even for $n=2$, we can choose accordingly). | |
Feb 5, 2014 at 9:30 | history | edited | Shinning Star |
edited tags
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Feb 5, 2014 at 9:14 | comment | added | Shinning Star | Yes, it was mistyping. Its fine now. | |
Feb 5, 2014 at 9:13 | history | edited | Shinning Star | CC BY-SA 3.0 |
deleted 8 characters in body
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Feb 5, 2014 at 9:07 | comment | added | Mark Meckes | In any case, something has to be wrong or missing here. In (ii) of the Proposition, $(x_i + x_j)/2$ isn't necessarily in the domain of $h$. | |
Feb 5, 2014 at 9:05 | comment | added | Mark Meckes | Okay, your phrasing is confusing because you refer to "all choices of $\xi_i$… such that" something is true about the $x_i$. | |
Feb 5, 2014 at 9:02 | comment | added | Shinning Star | its mentioned in definition. there is not any relationship between these two. $x_i\in(a,b)\subset\mathbb{R}$ and $\xi_i\in\mathbb{R}$. for all $1\leq i\leq n$. | |
Feb 5, 2014 at 8:48 | comment | added | Mark Meckes | What is the relationship between $\xi_i$ and $x_i$? | |
Feb 5, 2014 at 8:37 | history | asked | Shinning Star | CC BY-SA 3.0 |