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visits | member for | 4 years, 5 months |
seen | Apr 11 at 3:49 | |
stats | profile views | 16,515 |
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. ;)
Apr 2 |
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Simplex in convex polytope, pulling triangulation
How about a star triangulation centered at some vertex $q$ of $P$ then (triangulate each face not containing $q$ and use the corresponding simplexes as bases; it looks like now every simplex has good intersection with the boundary)? |
Apr 2 |
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the validity of a basic statement involving the Hausdorff distance
Is the ambient normed linear space finite-dimensional? |
Mar 30 |
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when does this identity hold?
Look up "Poisson summation formula" and recall that $\frac{\sin y}y$ is the Fourier transform of a certain compactly supported function and that multiplication on the Fourier side is convoluton on the space side. This should answer most of your questions, but if something remains unclear, just let us know :-) |
Mar 30 |
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Absolute convergence of logarithm of polynomial with positive coefficient ($\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} $)
As to homework, the answer is "Yes", if the person took a decent course on Banach algebras recently and "No" otherwise. Our (or, at least, my) memory is short and rusty and our education is patchy, so, Noah, I guess I can kill you with an elementary for me question and, most likely, you can return the favor :-). Let us, hence, assume (unless it is obvious otherwise) that whoever asks a question asks it in good faith and for a good reason ;-). |
Mar 30 |
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Absolute convergence of logarithm of polynomial with positive coefficient ($\ln G(z) = \sum\limits_{i = 0}^\infty {{q_i}{z^i}} $)
Noam has essentially answered it. All we need to notice is that the trick he mentioned allows one to show that the real part of $(1-z)G(z)$ is positive in the closed unit disk except for the point $z=1$, so $G(z)$ cannot be $0$ or negative real (Except, maybe, at $1$? Nah, it equals $1$ there!). Hence, $\log$ has an analytic brunch in some neighborhood ($\mathbb C\setminus(-\infty,0]$) of the compact $G( \text{Clos}\mathbb D )$, so Wiener's theorem finishes the story in no time. |
Mar 30 |
answered | is there any such result about Bernstein polynomials? |
Mar 27 |
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norm of trigonometric polynomials under arbitrary change of signs
Rudin-Shapiro polynomials: coefficients $\pm 1$, all norms about $\sqrt n$. Dirichlet Kernels: all coefficients $1$, $\|...\|_{L^q}\approx n^{1-\frac 1q}$. |
Mar 26 |
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Spearing rolling hula hoops
On the circle the expected time is infinite for $n=2$ (the probability that you have to wait for time $T$ is at least the probability that the initial distance is at least 2 and the speed difference is at most $1/T$, which is about $c/T$, so it shouldn't be hard to prove that for large $n$ (most likely, even for $n\ge 3$) the expectation does not exist. However the question about the distribution is well-posed and, probably, tractable if you want just the order of magnitude rather than an exact number. |
Mar 23 |
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Bound for a certain integral expression
What is $x^{2\alpha}$ for $x<0$ ? |
Mar 22 |
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$L^p$-derivative of Log
Given that the derivative of $\log|\zeta-w|$ with respect to $w$ is $-\frac1{\zeta-w}$, which is clearly not in $L^2(ds)$ when $w\in\gamma=\mathbb[0,1]$, this would be extremely unlikely... |
Mar 22 |
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Tubular neighbourhood which is nowhere piecewise linear
What exactly do you mean by "tubular"? |
Feb 7 |
awarded | Enlightened |
Feb 7 |
awarded | Guru |
Jan 29 |
awarded | Taxonomist |
Jan 20 |
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Simultaneous geometric separator
No. Take one collection of $n$ squares and repeat it an enormous amount $M$ of times. Then any separator of this collection has to have a noticeable size line segment within some fixed compact area. Now for each such line segment draw $n$ squares on that segment and notice that each such set of $n$ squares spoils some neighborhood of that segment as well. So, a finite number $m$ of additions is enough to spoil all such segments and this number does not depend on the number of repetitions, so we can keep $m$ bounded as $M\to\infty$ |
Jan 15 |
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Representing quasianalytic functions in several variables
I'm not sure what exactly you are looking for here. If you restrict your function of several variables to the line connecting the origin with the point at which you want to know the value, you get a function of one variable in pretty much the same class and with all derivatives at the origin known, after which the usual recovery by an appropriate summation method can be made for that value. There is, of course, some trouble for non-convex natural domains but it is there even for the standard analytic continuation and is resolved the same way in both cases. |
Jan 3 |
awarded | Enlightened |
Jan 3 |
awarded | Nice Answer |
Dec 31 |
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Is the Fourier transform of $\frac{1}{\mu+|\xi|^{2\alpha}}$($\mu>0$) a bounded function?
If it were, then so would be $M_t(\xi)=\frac {e^{-t|\xi|^2}}{\mu+|\xi|^{2\alpha}}$ for all $t>0$, and, moreover, the norms would be uniformly bounded (one more convolution with scaled $e^{-|x|^2}$ on the space side). Now the Fourier transform is classical and it becomes obvious that the answer is "No" (value $[\mathcal F^{-1}M_t](0)$ is huge for small $t$). |
Dec 19 |
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A density result for biharmonic functions
The answer is "No" even for 2 balls. The easiest way to see it is to notice that both the value at the center and the integral over a sphere allow one to recover $\sum c_j$ but they are not related. The underlying reason is obvious: it is pretty hard to recover the normal derivative by the boundary difference quotients, which go tangentially. |