Timeline for Proving that a complicated function is eventually concave
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2012 at 22:28 | comment | added | Thierry Zell | I'm sorry that I didn't have the time to come back to this. I have been in email contact with the OP. Emil raises some good objections, and I am not sure how to fix them yet or when I'll have the time to tackle this. Sorry about that! | |
| Oct 26, 2012 at 12:48 | comment | added | Yair Carmon | @Thierry, can you elaborate on why is $\Phi$ in the Pfaffian closure? We might very well end up with the most mathematically advanced proof ever made in Information Theory :) | |
| Oct 26, 2012 at 10:55 | comment | added | Emil Jeřábek | $\partial/\partial y$ is clear, but what about $\partial/\partial\gamma$? The definition of a Pfaffian chain requires all partial derivatives. The Pfaffian closure actually allows more than just Pfaffian chains, but the definition using Rolle leaves is a bit over my head. | |
| Oct 25, 2012 at 18:52 | comment | added | Yair Carmon | In this case, doesn't the fact that $\frac{\partial}{\partial y}\Phi(\gamma,y)=\phi(\gamma,y)$ imply that $\Phi$ is in the Pfaffian closure of a set containing $\phi$? | |
| Oct 25, 2012 at 17:41 | comment | added | Emil Jeřábek | A function $f(x_1,\dots,x_n)\colon\mathbb R^n\to\mathbb R$ is definable if there is a first-order formula $A(x_1,\dots,x_n,y)$ in the language of $\mathbb R_{\exp}$ which describes the graph $\{(x_1,\dots,x_n,y)\in\mathbb R^{n+1}:f(x_1,\dots,x_n)=y\}$. The function does not have to be unary (and it does not have to be total, as I wrote here for simplicity). $\mathbb R_{\exp}$ even has binary functions in its signature: $x+y$ and $x\cdot y$. Here, it suffices to show that $\log x$ and $x/y$ are definable (e.g., $x/y=z$ iff $x=yz\land y\ne0$), and then $\phi$ is constructed by composition. | |
| Oct 25, 2012 at 17:10 | comment | added | Yair Carmon | @Thierry, when you say $\phi$ is definable in the exponential field, do you mean that for every $t$, $\phi(\cdot,t)$ is definable? If so I also don't understand why $\Phi$ is in the Pfaffian closure of $\mathbb{R}_{\exp}$. Otherwise, I don't understand in what sense is $\phi$ definable - is there an extension of the exponential field to functions of more than one variable? Thanks! | |
| Oct 25, 2012 at 15:45 | comment | added | Emil Jeřábek | Sorry for my ignorance, but is there a simple reason why the Pfaffian closure is closed under parametric integration like this? (By the way, there is a confusion of $x$ and $\gamma$ in the definition of $\phi$.) | |
| Oct 25, 2012 at 15:26 | history | answered | Thierry Zell | CC BY-SA 3.0 |