Timeline for smooth functional to detect whether a function has a zero
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2016 at 15:49 | history | edited | Dan Christensen | CC BY-SA 3.0 |
Remove unneeded steps, add more details, and clarify phrasing throughout.
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| Oct 17, 2016 at 13:08 | comment | added | Willie Wong | @DanChristensen: you are right. The discussion of $g$ was a vestigial paragraph from an earlier version of a "proof" that didn't quite work, I just failed to completely erase its presence. Sorry about that! You can go ahead and edit: after all, I am the interloper and it is your answer :-) | |
| Oct 17, 2016 at 0:18 | history | edited | Dan Christensen | CC BY-SA 3.0 |
Replace $x$ with $x_0$ in two places.
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| Oct 16, 2016 at 22:56 | comment | added | Dan Christensen | @WillieWong If $t=0$ is in the boundary of the closed set $g^{-1}(0)$, I don't see why there is an interval $(0,t_0)$ on which $g(t) > 0$. Instead, I would drop the discussion of $g$ completely. All of the inequalities should still hold for each $t$ such that $f(x,t)$ has no zeros for $x \in [0,1]$. And for $t$ such that $f(x,t)$ does have such a zero, $G(f) = 0$, so there is nothing to prove. I think this fixes the mistake and simplifies the argument at the same time. If you agree with this reasoning, I'm happy to make the changes. | |
| Oct 13, 2016 at 22:26 | comment | added | Dan Christensen | @NawafBou-Rabee I don't intend to find zeros by practically computing this functional. The use would be to find a local trivialization of a principle $G$-bundle with an associated smooth partition of unity, following Dold's argument in the topological case, which uses the function $g$ that arose in Willie Wong's proof of (*). | |
| Oct 13, 2016 at 22:23 | comment | added | Dan Christensen | @WillieWong Thanks for filling in the proof of (*). It's essentially the same as my proof, although I didn't need to treat interior points of $g^{-1}(0)$ specially. However, Gord Sinnamon pointed out an error in the last sentence of the argument I gave above; it only handles the first derivative, giving the same result I had already proved for $(\int 1/f)^{-2}$. Any ideas for how to handle higher derivatives? | |
| Oct 13, 2016 at 22:21 | history | edited | Dan Christensen | CC BY-SA 3.0 |
point out my mistake
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| Oct 13, 2016 at 21:22 | comment | added | Willie Wong | @NawafBou-Rabee: since the answerer is the OP, I can only assume he knows how it answered the question. :-) | |
| Oct 13, 2016 at 21:02 | comment | added | Nawaf Bou-Rabee | BTW, I don't see how this functional is practical as a means to detect a zero of f, because its evaluation seems to rely on another routine to detect this zero. So I don't see how this functional answers the OPs question. | |
| Oct 13, 2016 at 20:52 | comment | added | Nawaf Bou-Rabee | @WillieWong sorry for the delay: I had to teach. It was in relation to $x$, but it looks like your argument fills in the details. I was concerned that the Taylor series argument may have assumed that $f(\cdot,t)$ had a discrete number of zeros. For the purpose of detecting a zero of $f$, its clear that one can take $f$ to be non-negative WLOG. | |
| Oct 13, 2016 at 19:17 | history | edited | Willie Wong | CC BY-SA 3.0 |
added 1036 characters in body
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| Oct 13, 2016 at 19:14 | comment | added | Willie Wong | @NawafBou-Rabee: I edited the answer and filled in the proof. | |
| Oct 13, 2016 at 19:11 | history | edited | Willie Wong | CC BY-SA 3.0 |
added 1036 characters in body
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| Oct 13, 2016 at 18:48 | comment | added | Willie Wong | @NawafBou-Rabee: I am not entirely sure I understand your objection, in part because $f$ depends on two variables and your last comment only talks about support in [1/4,3/4]. Is that in relation to $x$ or to $t$? | |
| Oct 13, 2016 at 17:12 | comment | added | Nawaf Bou-Rabee | I still don't understand that sentence. What if the support of f is in [1/4,3/4] then there are a continuum of zeros. How is your taylor series argument valid? | |
| Oct 13, 2016 at 16:56 | comment | added | Dan Christensen | @NawafBou-Rabee Thanks for pointing out the typo about the case where $t$ might be zero. I've added "nonzero" to that sentence. Regarding your second question, for each fixed $t$, either $f(x,t)$ has a zero for some $x \in [0,1]$, or it doesn't. Which of those is true determines which formula is used to define $G(f)$ for that $t$. | |
| Oct 13, 2016 at 16:56 | history | edited | Dan Christensen | CC BY-SA 3.0 |
add "all" to clarify quantification
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| Oct 13, 2016 at 16:36 | comment | added | Nawaf Bou-Rabee | The inequality ($*$) makes no sense at $t=0$, so how can it hold for all $t \in [-1,1]$? Also I do not understand the assumptions on $f$ in the sentence If $f(x,0)$ has a zero in $[0,1]$, then ... one can show that ... such that $f(x,t)$ is nonzero for $x \in [0,1]$. | |
| Oct 13, 2016 at 15:02 | history | edited | Dan Christensen | CC BY-SA 3.0 |
Improve phrasing
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| Oct 13, 2016 at 14:54 | history | answered | Dan Christensen | CC BY-SA 3.0 |