Timeline for A toy model for the t-section problem
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2011 at 1:35 | comment | added | fedja | Yes, that would be another good way to formalize it (provided that you do not start playing with coming very close to $0$ and then lifting off infinitely many times). The bad thing is that I don't know myself what effect I'm looking for. I rather know a long list of trivial effects that I do not care much about. The particular function I wrote would have none of them though, so if you can show that it is impossible, it'll tell me something new. | |
| Jun 5, 2011 at 18:03 | comment | added | Douglas Zare | I wasn't sure if this was supposed to be ruled out by the condition that $H$ is small. One of the things I considered was whether you meant that if $H$ is smooth and vanishes to all orders, then there is some function so that $S = cH$ for some $c \gt 0$. | |
| Jun 5, 2011 at 17:53 | history | edited | Douglas Zare | CC BY-SA 3.0 |
typo
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| Jun 5, 2011 at 17:40 | comment | added | fedja | Thanks a lot! I upvoted it but, unfortunately, it still fits under the "trivial restriction" category. To be more precise, you can assume $H$ to be small in any function space you desire (say, $C^k$ with any $k$). What I mean is that "roughly speaking" $S(x)=\frac f(x)^2$ and cannot change too much when $x$ goes by $f(x)$ to the right or to the left where $f$ is the function whose graph gives $S$, so if you have hard time with the square root or the rate of change is too big, you have trouble for sure. But what if you have $H(x)=10^{-100}[x(1-x)]^{10}$, say? | |
| Jun 5, 2011 at 16:59 | history | answered | Douglas Zare | CC BY-SA 3.0 |