Timeline for Quotient of two Laplace integrals
Current License: CC BY-SA 2.5
16 events
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| Dec 22, 2010 at 14:06 | vote | accept | LI Yutian | ||
| Dec 6, 2010 at 22:19 | history | edited | LI Yutian | CC BY-SA 2.5 |
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| Dec 6, 2010 at 22:05 | history | edited | LI Yutian | CC BY-SA 2.5 |
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| Dec 6, 2010 at 21:56 | history | edited | LI Yutian | CC BY-SA 2.5 |
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| Dec 5, 2010 at 14:35 | comment | added | Willie Wong | @Theo: I should note that even in the case $\theta$ is $C^1$, I don't have a good argument to rule out oscillations. (In fact, seeing now Michael's answer below, I have to re-evaluate even the original statement about Laplace's method by the OP.) | |
| Dec 5, 2010 at 14:26 | comment | added | Willie Wong | @Theo: there can be oscillations. If $f = r e^{i\theta}$ and $\theta$ has a cusp at $\xi$, so that $\theta'(\xi) = \infty$ (by abuse of notation), when sending $n\to\infty$, in a small neighorhood of $\xi$ the integral may go to zero faster then elsewhere: basically one can imagine that Riemann-Lebesgue lemma giving you additional decay. Without the complex phase part the estimate of-course holds for continuous, or even weaker functions $f$. | |
| Dec 5, 2010 at 6:05 | answer | added | Michael Greenblatt | timeline score: 7 | |
| Dec 5, 2010 at 4:48 | comment | added | Theo Johnson-Freyd | (continuation) I think this completes the proof when $\phi(\xi)\neq0\neq\psi(\xi)$. If both $\phi(\xi),\psi(\xi)$ vanish, I will take the problem as vacuous, and by reciprocating if necessary the only other possibility is $\phi(\xi)=0$, in which case I win by linearity. But I haven't thought carefully enough, so there might be more nuance that I'm not seeing. It's just, it seems that even $C^1$ isn't necessary, only continuity is. | |
| Dec 5, 2010 at 4:46 | comment | added | Theo Johnson-Freyd | Also, can someone explain to me why this problem is difficult? I can rescale $f$ by $f(\xi)$ to assume that $f(\xi)=1$ and $|f(x)|<1$ for $x\neq\xi$. Pick $\epsilon$, and find an interval around $\xi$ in which $\phi,\psi,f$ don't vary more than $\times(1\pm\epsilon)$ (I guess I'm assuming $\phi(\xi)\neq0\neq\psi(\xi)$). Moreover, outside this small interval $f(x)\leq c<1$. So in the limit only this interval contributes to the integrals. But then I approximate the integrals by something with error $\times(1\pm \epsilon)^4 \leq \times(1 \pm 5\epsilon)$. (continued) | |
| Dec 5, 2010 at 4:38 | comment | added | Theo Johnson-Freyd | I would like to echo Yemon's comment. I'm not a fan of "questions" without question marks: this one is written like a homework problem: "prove or disprove". Much better would be for you to provide some motivation --- why did you think about this question? how does it relate to your research? --- or at least some more background --- what other results are related to this? Remember that this site is for questions (and "research" questions at that), so demands like "prove or disprove" come across (usually unintentionally) as rude. | |
| Dec 5, 2010 at 0:26 | history | edited | LI Yutian | CC BY-SA 2.5 |
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| Dec 5, 2010 at 0:18 | comment | added | LI Yutian | This problem comes from an attempt of proving a probability theorem (of K.L. Chung and P. Erd\"os, 1951) using analytic argument. | |
| Dec 4, 2010 at 23:53 | history | edited | LI Yutian |
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| Dec 4, 2010 at 22:37 | comment | added | Willie Wong | Hum, I voted to close, but after looking closer at the question and noticing that it asks for $[f(x)]^n$ rather than $|f(x)|^n$ as I initially misread, I feel I may have been a bit too hasty with the vote: that $f(x)$ is required to be $C^1$ is a tiny bit subtle. Since I don't think I can undo my vote: can the next person who wants to vote to close instead leave a comment "cancelling" this request? Thanks. | |
| Dec 4, 2010 at 19:49 | comment | added | Yemon Choi | The way you have phrased this question makes it seem like an exercise or coursework. If this is not the case, could you please give some more background, e.g. where did you come across this problem? | |
| Dec 4, 2010 at 18:51 | history | asked | LI Yutian | CC BY-SA 2.5 |