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| bio | website | |
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| age | ||
| visits | member for | 1 year, 3 months |
| seen | 2 days ago | |
| stats | profile views | 457 |
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May 4 |
answered | Prove an integral limit |
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May 4 |
revised |
Asymptotic related to the infinite product of sine added 273 characters in body |
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May 3 |
revised |
Asymptotic related to the infinite product of sine added 356 characters in body |
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May 3 |
comment |
Asymptotic related to the infinite product of sine @AntonioVargas $x$ is a constant, as I've said. What I really need, is methods, discipline to deal with such a summand. Comparing is just yielding by-products. |
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May 2 |
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Snags when discovering the asymptotic behavior of an integral Huh, since these days I read part of de Bruijn's, I picked some stuff out from my mathematical analysis textbook and tried to determine the asymptotic behavior. Unfortunately, I found the road is extremely distorted. This one is just one of these problems. I've posted a new question related to the proof of the infinite product of sine. I will be pleased if there's some asymptotic analyst (professor) around me. |
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May 2 |
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Snags when discovering the asymptotic behavior of an integral I doubt that it could be expanded. I guess the term after $Cn^{-2}$ is $O(n^{-3}\log n)$, not $O(n^{-3})$. |
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May 2 |
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Snags when discovering the asymptotic behavior of an integral Typo, the remainder is $\displaystyle\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{x\sin xdx}{(\sin x+1/n\pi)(\sin x+(1+x\sin x)/n\pi)}$. |
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May 2 |
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Snags when discovering the asymptotic behavior of an integral The remainder is $\displaystyle\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{x\sin xdx}{\sin x+1/n\pi}{\sin x+(1+x\sin x)/n\pi}\asymp\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{xdx}{\sin x}$, where the absolute error is $\displaystyle-\frac1{n^3\pi^3}\int_0^{\pi/2}\frac{(2\sin x+x\sin^2x+(1+x\sin x)/n\pi)xdx}{(\sin x+1/n\pi)(\sin x+(1+x\sin x)/n\pi)\sin x}$, which is not easy to determine the major part (it seems that Lebesgue's dominated convergence theorem doesn't work well). |
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May 2 |
comment |
Snags when discovering the asymptotic behavior of an integral Well, I'll compute it soon. Incidentally, does some method called saddle point method work for this integral? I have no idea about complex analysis and Cauchy integral theorem. It is said that such a method is very powerful in de Bruijn's Asymptotic methods in analysis. |
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May 1 |
asked | Asymptotic related to the infinite product of sine |
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Apr 30 |
comment |
Snags when discovering the asymptotic behavior of an integral I wonder whether we could look into $O(1/n)$. The $\sin$ function is very nasty. The major distribution of the integrand is near $0$, but we couldn't decompose the interval $[0,\pi/2]$ easily to obtain the result. |
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Apr 30 |
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How to interpret little-o notation in an exponent. A systematical way to interpret these phenomena is to consider $o$ and $O$ sets of functions. For example, $o(g(n))=\{f(n):\forall\epsilon>0,\exists N,\forall n\ge N,\lvert f(n)\rvert<\epsilon\lvert g(n)\rvert\}$, and $O(g(n))=\{f(n):\exists M,\forall n,\lvert f(n)\rvert\le M\lvert g(n)\rvert\}$, and $f(n)=o(g(n))$ is conventional abbreviation for $f(n)\in o(g(n))$. |
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Apr 29 |
comment |
Asymptotic for the integral involving exponential Since $\exp(x^n)=\sum_0^\infty x^{nk}/k!$ converges absolutely on $[0,1]$, the interchange of summation and integration is justified. The interchange of summation is valid for the absolute convergence of the double sum. Am I right? |
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Apr 29 |
accepted | Asymptotic for the integral involving exponential |
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Apr 29 |
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Asymptotic for the integral involving exponential Very good, thanks! |
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Apr 29 |
revised |
Asymptotic for the integral involving exponential added 70 characters in body |
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Apr 29 |
comment |
Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral? One could calculate $\int_0^\infty\sin xdx/x$ elementarily. It's equivalent to $\lim_{n\to\infty}\int_0^{\pi/2}\sin2nxdx/x$. We're pleased to see that $g(x)=1/x-1/\sin x=O(1/x)$ and so $\lim_{n\to\infty}\int_0^{\pi/2}g(x)\sin2nxdx=0$ (Generally, Riemann-Lebesgue lemma; however, since $g$ is of $C^\infty$, we could integrate it by part and obtain the result). Since $\sin2nx/\sin x=\sum_{k=1}^n(\sin2kx-\sin2(k-1)x)$, we could easily determine $\int_0^{\pi/2}\sin2nxdx/\sin x$. |
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Apr 29 |
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Compactness proof of $\mathbb{R}^2$ It could be rather general if we come into the situation of topology spaces. The proof is similar in such a way: if $Y$ is compact and a open set of $X\times Y$ contains $x_0\times Y$, then there's a neighborhood $U$ of $x_0$ in $X$ such that $V$ covers $U\times Y$. For detailed information, see wiki or Munkres' topology. |
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Apr 29 |
asked | Asymptotic for the integral involving exponential |
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Apr 27 |
asked | Snags when discovering the asymptotic behavior of an integral |
