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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
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Mar
30
revised Laplace's method with nontrivial parameter dependency
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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
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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
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Mar
27
revised Laplace's method with nontrivial parameter dependency
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Mar
27
revised Laplace's method with nontrivial parameter dependency
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Mar
27
revised Laplace's method with nontrivial parameter dependency
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