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| visits | member for | 1 year, 3 months |
| seen | 2 days ago | |
| stats | profile views | 457 |
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May 16 |
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Convergence of $\sum \limits_{n=1}^{\infty}\sin(n^k)/n$ @curious I don't think so. Suppose $f(x)=\sin(x^2)/x$, we have $f'(x)=2\cos(x^2)-\sin(x^2)/x^2$. If $\int f'(x)$ converges absolutely, then $\int \cos(x^2)$ converges absolutely, contradiction! I think integral test wouldn't work if the integrand is oscillating near $\infty$. |
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May 11 |
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how to prove $f$ is an arithmetic function with this property $\sum_{d\mid n} f(d)=n^2$ @draks... Oh, sorry, I misunderstood. Now it seems that the propose of OP is unclear. |
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May 11 |
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how to prove $f$ is an arithmetic function with this property $\sum_{d\mid n} f(d)=n^2$ I think it's a direct result of Möbius inversion formula |
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May 11 |
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real analysis: continous I think the condition that $\partial f/\partial t$ continuous could be replaced with a weaker one: $\partial f/\partial t$ is uniformly bounded and the integrand in the second integral is Riemann-Stieltjes integrable. |
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May 10 |
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Absolute values in $\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$ Another approach: Let $u=x+2$, we need to integrate $du/u\sqrt{u^2-1}$. Set $v=\sqrt{u^2-1}$, we have $du/u=vdv/(v^2+1)$ and $dx/(x+2)\sqrt{(x+1)(x+3)}=dv/(v^2+1)$, therefore the answer is $\arctan v$ where $v=\sqrt{(x+1)(x+3)}$. |
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May 10 |
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Conditions that allow Integration by Substitution By minimum, you mean the necessary and sufficient condition of $\phi$ such that for each $f$ is (Riemann/Lebesgue)-integrable on $[a,b]$, we have $\int_a^b f(\phi(x))\phi^\prime(x)dx=\int_{\phi(a)}^{\phi(b)}f(x)dx$. It's quite hard. I know that it's true when $\phi$ is monotone. |
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May 10 |
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Conditions that allow Integration by Substitution $\phi$ needn't to be continuously differentiable. For example, if $\phi$ is monotone and differentiable, where $\phi^\prime$ is integrable, the theorem is also correct. @ήλιος |
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May 10 |
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Asymptotic related to the infinite product of sine @AntonioVargas I think I've got the key to attack the problem. Substract $S_n$ with $\sum_{k=1}^n\ln(1-x^2/k^2\pi^2)$, and estimate the summation through Euler-Maclaurin formula. You can plot a graph for the summation, and find that the distribution is very flat, which case is suitable for Euler-Maclaurin! Detailed calculation is not performed since the calculation is tedious. |
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May 5 |
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How prove this linear algebra $AB=BA$? It is also related, Motzkin & Taussky's original proof. |
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May 5 |
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How prove this linear algebra $AB=BA$? I hope if there's some pure elementary approach. |
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May 5 |
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How prove this linear algebra $AB=BA$? The second link seems unavailable. |
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May 5 |
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Snags when discovering the asymptotic behavior of an integral Although it's not a complete answer, it's well-informed and informative. |
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May 4 |
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Laplace integration after the first term Bibliography: de Bruijn's Asymptotic methods in analysis |
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May 4 |
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Laplace integration after the first term The substitution $t=h(s)$ does work; however, the integral become an improper integral. Note that the improper integral $\Gamma(1+\alpha)=\int_0^\infty e^{-q}q^\alpha dq$ converges even when $-1<\alpha<0$. |
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May 3 |
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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 |
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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. |
