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seen | Aug 18 at 10:03 | |
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If I like your problem, I'll think of it in honest. However, if the question is badly posed or the information is incomplete, I'll vote to close it when in a bad mood and waste several hours of your and my time asking for clarification, pointing out trivial counterexamples, etc., when in a good mood. In both cases, the most likely outcome is that I'll finally switch my attention to something else. ;)
Aug 15 |
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A topologist is not a mathematician - a small question
@DouglasZare On the other hand, I should confess that I just left a note on the door to the janitor who was supposed to clean my office before the new semester that read "Everything on the floor except furniture can be thrown away" and, when I had to catch a car ride home during a heavy rain next day added "or bicycle" before departing. :-) |
Aug 15 |
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Distribution of zeroes of lacunary functions
I have no time to post an answer, but read the fifth from the top paper on math.msu.edu/~fedja/pubpap.html and third from the top paper on math.msu.edu/~fedja/prepr.html. Together they'll give you all you need. |
Aug 11 |
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A topologist is not a mathematician - a small question
@TomGoodwillie So was he overly humble or overly arrogant? |
Aug 9 |
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Gaps between roots of trigonometric polynomials
It is a little bit unclear what exactly you are asking because the word "general" may mean about 100 different things. The most trivial interpretation is that $c_k$ are standard i.i.d. complex Gaussians ($k>0$); show that, with probability close to $1$, .... This certainly has been done but I have no idea if this is even remotely close to what you are looking for except your last sentence shows that the answer is somewhat more likely to be "no" than "yes". Can you elaborate a bit on what you would consider a useful characterization? |
Aug 9 |
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Newtonian potential for continuous $f$
It is always true in the sense of generalized functions but then it is trivial because you can pass all the convolutions and differentiations on the $C_0^\infty$ mollifier. If you want it pointwise, you have to tell in exactly what sense you want to understand the Laplacian in this equation. |
Aug 8 |
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A topologist is not a mathematician - a small question
@Adam Przeździecki That depends on the country. In Israel a lot of customs staff are Tel Aviv university students working part time. So, not only do they ask for specific field, but if I'm careless enough to say that I came to work with some faculty at TAU, they ask "Who?", "How does he look?", etc. |
Aug 8 |
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Distribution of dropped objects
"Perfectly elastic"??? I guess they are still bouncing then happily hopping away to infinity... |
Jul 11 |
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Quadratic PDE Systems
Well, you certainly cannot make it perfectly linear. Also, from the viewpoint of numerics a fractional power is just as bad as the function in front of derivatives: once $w$ hits $0$, you are screwed either way. :-) |
Jul 9 |
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Simultaneous multiple perturbations in Markov chain Monte Carlo
With constraints, all you need to do is to use conditional probabilities instead of full ones, so the "normal proposal" just gets divided by the total probability that the constraint is satisfied in the unconstrained model (I agree that the computation of this one may be ugly...). |
Jul 9 |
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Simultaneous multiple perturbations in Markov chain Monte Carlo
What's wrong with using vectors and density matrices/their determinants (inverse to correlation matrices) instead of numbers in the same very formula for $q(v'|v)$ you wrote? It does not look like the multivariate normal distribution is any different from the univariate one for your question unless you swept something subtle under the rug. |
Jul 9 |
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Asymptotic formula for restricted partition function
The moral is, of course, not that you cannot compute the number of partition with small relative error quickly, just that you need to consider some quantities other than density when doing so. So, the answer to your last question is still "most likely, yes" though I have no time now to attempt a computation. I put this remark separately because I thought it was obvious, but then I looked at my second one, at your question, and thought again :-) |
Jul 9 |
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Asymptotic formula for restricted partition function
To get the last statement immediately and without any computation, just look at it by the eyes of asymptotics (that has to kick in somewhere) and ask yourself how you can possibly know by the moment you decide to board the train and start making accurate predictions whether the density is not going to change somewhere far beyond that point. |
Jul 9 |
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Asymptotic formula for restricted partition function
The existence of density is an extremely weak assumption. Indeed, suppose that something like you said were true. Add just one number $a$ to $A$. Then the number of new partitions of $n$ is $Q(n)=P(n)+P(n-a)+P(n-2a)+\dots$ and you can see that $C_A$ (if it was there for $A$) now becomes a growing function in $n$ (any finite number of terms would be of the same order of magnitude if your conjecture were anywhere close to the truth). Adding a sequence of numbers of $0$ uniform density results in a total havoc showing that you do not have any chance to change $o(1)$ to anything explicit. |
Jul 8 |
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Quadratic PDE Systems
$w^2$ changes nothing. Of course, there is no reason to accept until you get sure the argument works. |
Jul 8 |
revised |
Quadratic PDE Systems
added 4 characters in body |
Jul 8 |
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Quadratic PDE Systems
>>>>Please avoid extended discussions in comments. Would you like to automatically move this discussion to chat?<<<< No, it is not an individual/private discussion. |
Jul 8 |
answered | Quadratic PDE Systems |
Jul 8 |
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Quadratic PDE Systems
Also, what is the exact range of $t$: the whole line, a ray (up or down), an interval? |
Jul 8 |
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Quadratic PDE Systems
>>> also $c_n$ non zero <<< any chance to know which ones are positive and which ones are negative? |
Jul 8 |
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Quadratic PDE Systems
OK, but it is still unclear if you want to find just one solution (which is trivial: look for functions that depend on $t$ only) or you want to show that nothing else is there, or something else. |