# Tag Info

## New answers tagged asymptotics

### On asymptotic of logarithm of modulus of a function

It will be better use another integral representation of $\zeta(s)$: $$\zeta(s)={s\over s-1}-s\int_1^\infty{x-\lfloor x\rfloor\over x^{s+1}}\mathrm dx$$ Set $s=\frac12+iT$, then we have for $T\ge2$ ...
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### How to obtain logarithmic asymptotic behavior for this integral?

Using a CAS, there is an exact result $$I=\int_0^t \int_0^t \frac{ dt_1\, dt_2}{\sinh^2(t_1-t_2-i\delta)}\,$$ $$I=2 \log (-i \sin (\delta ))-\log (\sinh (t-i \delta ))-\log (-\sinh (t+i \delta ))$$ ...
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### Difference between the usage of Big-Omega notation as used by Computer Scientists and Mathematicians.

The math definition is that there exists some infinite subsequence $I=\{n_i:i\geq 1\}$of the natural numbers and a positive constant $c$ such that for all $n\in I,$ $$f(n)\geq c g(n).$$ One could ...
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### Asymptotics of a two dimensional integral

It turns out that @Gary 's suggestion to reverse the order of integration is a fruitful first step because the integral over $\epsilon$ is analytically tractable. Exchanging the order results in the ...
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### Expansion of logarithm with little o

When you have on $o(z)$ in a development it means that anything smaller than order $z$ is simply ignored. When $z\to 0$ we have to ignore any terms in $z^a$ where $a>1$, in particular $z^2$ and of ...
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### Big O notation problem

Thanks for comments I fix my problem. The key point is $ln(n)$ is $O(ln(n))$ $∃ C, k:$ any $x >k (ln(n) <= Cln(n))$ k = 1, C can be any number greater than 1. So $ln(n)$ is $O(ln(n))$

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### maximise $(1-x)^{n-1}x - x^n$ subject to $0\le x\le 1/n$, where n is a natural number

I don't know if this would help but ... ...
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### maximise $(1-x)^{n-1}x - x^n$ subject to $0\le x\le 1/n$, where n is a natural number

$$f(x) = (1-x)^{n-1}x - x^n \implies f'(x)=(1-x)^{n-2} (1-n x)-n x^{n-1}$$ Expanding $f'(x)$ as a series around $x=\frac{1}{n}$ gives f'(x)=-n^{2-n}-\frac{\left((n-1)^n+(n-1)^3\right) n^{3-n}}{(n-1)^...
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