alfC
Reputation
258
Top tag
Next privilege 500 Rep.
Access review queues
 Mar 24 comment Is arrow notation for vectors “not mathematically mature”? As physicist, I agree. But I will not relate it with maturity. The arrow implies a lot more than what linear algebra is about. In particular an arrow implies that you have a metric, and you are not going to distinguish between afine and non-affine vectors. So, I agree that mathematicians shouldn't use the arrow, specially when teaching linear algebra. they may use it when teaching other things, like geometry or differential geometry. Mar 13 accepted Piecewise interpolation with derivatives that is also twice differentiable Mar 13 awarded Popular Question Mar 12 comment Is there a classical analog of Bloch's theorem? @1over137: the link doesn't work. Nov 1 comment incomplete gamma function with negative arguments Perhaps, the solution is here arxiv.org/abs/1407.0349. Let us know. Mar 30 revised Real approximation to the maximum using Laplace's method integral fixed typo Mar 30 revised Laplace's method with nontrivial parameter dependency added 12 characters in body Mar 30 revised Laplace's method with nontrivial parameter dependency added 3 characters in body Mar 30 revised Laplace's method with nontrivial parameter dependency added precise form for the error order Mar 28 awarded Revival Mar 28 comment Laplace's method with nontrivial parameter dependency @AntonioVargas, It should say "...still underestimates". My only evidence is numerical, starting from the suspicion that the order must be lower than in the common Laplace Method. Empirically I find that the order of the error is $O(2^{-\lambda}\lambda^{-2}\log^2\lambda)$ and that the prefactor is $11/10$. In summary: $\int = \sqrt{2\pi}\lambda^{-3/2}2^{-\lambda} + \frac{11}{10} 2^{-\lambda}\lambda^{-2}\log^2\lambda + \cdots$. Perhaps the extra order in the error is because you have a infinite series of errors that add up to a higher error (the convergence inside the series may not be uniform) Mar 28 awarded Explainer Mar 27 comment Laplace's method with nontrivial parameter dependency @AntonioVargas, so if it is $O(2^{-\lambda}\dots)$ then it is much better than $O(\lambda^-1 \log\lambda)$, no?. Also, from my numerical experiment I get that $O(2^{-\lambda}\lambda^{-5/2}\log\lambda)$ still overestimates the error. It looks like the error is in practice $O(2^{-\lambda}\lambda^{-2}\log\lambda$. Mar 27 comment Laplace's method with nontrivial parameter dependency @AntonioVargas, Thanks, I think your answer is more formal. I also get to the same approximation and also find that some weaker variant of the Laplace method needs to be used. What I couldn't find is the "order" of the approximation. Did you? Mar 27 revised Laplace's method with nontrivial parameter dependency added 195 characters in body Mar 27 revised Laplace's method with nontrivial parameter dependency added 195 characters in body Mar 27 revised Laplace's method with nontrivial parameter dependency added 1 character in body Mar 27 revised Laplace's method with nontrivial parameter dependency added 1 character in body Mar 27 revised Laplace's method with nontrivial parameter dependency added 75 characters in body Mar 27 revised Laplace's method with nontrivial parameter dependency added 287 characters in body