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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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Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?

$$a_n=\sqrt[n]{n!} \qquad \implies \qquad \log(a_n)=\frac 1 n \log(n!)$$ Now, using Stirling approximation $$ \log(a_n)=\log (n)-1+\frac{\log (2 \pi n)}{2 n}+\frac{1}{12 n^2}-\frac{1}{360 n^4}+O\...
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Follow up question: asymptotics of a two dimensional integral

First, let's simplify the given integral using the same $\int_0^1 d\epsilon\int_{-\epsilon}^\epsilon dt=\int_{-1}^1 dt\int_{|t|}^1 d\epsilon$. Denote $$g(\rho,t)=\sin\big[({\textstyle\sqrt{1-(\rho+t)^...
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1 vote

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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2 votes
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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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Illegal transformation of condition for asymptotes?

I'm assuming that your limits are for $x \to \infty$, but the essential arguments still hold with a bit of tweaking if you intend $x \to a$ for some $a \in \mathbb{R}$. If $g(x)$ is bounded away from $...
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4 votes
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Asymptotics of Laplace like integral with shrinking integration intervals

With the change of variables $x=e^{-t}$, we have $$ I(M) = (\log M)^{M + 1} \int_{\log \log M}^{ + \infty } {e^{ - Mg(t)} e^{ - t} dt} , $$ where $g(t) = t+ f(e^{ - t} )$. This further equals to \...
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1 vote

Evaluate/asymptotic of the sum $\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$

This is a partial answer that covers the case when $L \left({N}\right) = \lfloor{\frac{N}{6} - \frac{1}{2}}\rfloor$. First a correction, the above initial solution involving $m$ was based on the ...
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Exponential sum behaves like linear term for large $t$

Thanks to the remark from Steven Clark I came up with a nice solution using the modular theta function $$ \theta : \mathbb H \to \mathbb C,\quad \theta(z) := \sum_{n \in \mathbb Z }e^{\pi i n^2 z}. $$ ...
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2 votes

Asymptotic expansion of inhomogenous differential equation

I'm not 100% sure what you are looking for but let me try something. First of all, identifying the constant term in the asymptotic expansion without giving an initial data seems impossible. The reason ...
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1 vote
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Comparing $n \log_2 (n) /2$ and $\log_2 n!$ asymptotically

Let us consider $f(x)=\ln(x)$ over the interval $[1,n]$. $\log(x)$ is a concave function, hence by the Hermite-Hadamard inequality $$ \int_{1}^{n}\ln(x)\,dx \geq \ln(2)+\ln(3)+\ldots+\ln(n-1)+\frac{1}{...
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Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes).

Informally, if you're going to let $n$ approach infinity, then all but the highest power of a polynomial can be ignored. In your example, as $x$ approaches infinity $\sqrt{x^2+9} \rightarrow \sqrt{x^2}...
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4 votes
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Exponential sum behaves like linear term for large $t$

The Fourier transform of $f(x) = \exp\left(-\tfrac{\pi x^2}{t^2}\right)$ is $\widehat{f}(\omega) = t\exp\left(-\tfrac{t^2\omega^2}{4\pi}\right)$. So by using the Poisson summation formula, we have $$\...
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1 vote

Relationship between the trace distance and the operator norm for some time dependent integral operator with kernel $K(x,y,t)$.

The conclusion is not true. Consider $\ell^2(\mathbb{N})$ and $t\ge 1.$ $$K(i,j,t)= \begin{cases} t^{-1/2} & i=j\le t \\ 0 & {\rm otherwise}\end{cases} $$ Then $$\|L_{K_t}\|=t^{-1/2},\qquad {\...
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1 vote

sequence such that $x_{n+1}=n(x_n-n)$

Your question is about a sequence that satisfies $$ x_{n+1} = n(x_n-n) \tag{1} $$ where initial value $\,x_1\,$ is given. This is a linear recursion so that $$ x_{n+1} = n!\,x_1-a_n \tag{2} $$ where $\...
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5 votes
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sequence such that $x_{n+1}=n(x_n-n)$

By some experimentation from the starting point $x_n=an+b+y_n$ and reporting in the equation I quickly found that $x_n=n+1+2y_n$ lead to the following simplification $$y_{n+1}=ny_n-1$$ So we have ...
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1 vote

Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$

If you use Pochhammer symbols and $I_0=\frac{\pi}{2\sqrt2}$ $$I_n=\frac{4n-3}{4n}I_{n-1}\quad \implies I_n=\frac{\pi }{8 \sqrt{2}}\frac{\left(\frac{5}{4}\right)_{n-1}}{(2)_{n-1}}=\frac{\pi }{8 \sqrt{2}...
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1 vote
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Asymptotics of Bessel function $J_n(z)$ for $|z| \gg n$ but $|z|\not\gg n^2$

In terms of the modulus function $M_\nu (z)$ and phase function $\theta _\nu (z)$, we can write $$ H_\nu ^{(1)} (z) = M_\nu (z)e^{i\theta _\nu (z)}, \quad J_\nu (z) = M_\nu (z)\cos \theta _\nu (...
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How to show that $\forall n \geq 1$, $\frac{n^{2n}}{n!^2} \geq (\frac{n+1}{n})^{n^2-n}$?

We can proceed by induction on $n$. The case $n=1$ is easy to check. Suppose that the inequality holds for $1,2,\ldots,n$. Then \begin{align*} \frac{{(n + 1)^{2n + 2} }}{{(n + 1)!^2 }} = \frac{{(n + 1)...
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2 votes

Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$

By Euler's Beta function $$ I_n = \int_{0}^{1} t^{n-3/4} (1-t)^{-1/4}\,dt = \frac{\Gamma\left(n+\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma(n+1)} $$ hence $$ \color{red}{K}=\lim_{n\to +\...
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1 vote

von Mangoldt's formula for Chebyshev $\psi$ function

The answer to your first question, in a nutshell, is no. Depending on whether the Riemann Hypothesis is true or not, $\psi$ will take certain or other values. In particular, if the Riemann Hypothesis ...
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1 vote
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Computing the asymptotics of a principal value integral

We start with $(8.6.4)$: $$\tag{1} \Gamma (\sigma ,z) = \frac{{z^\sigma e^{ - z} }}{{\Gamma (1 - \sigma )}}\int_0^{ + \infty } {\frac{{t^{ - \sigma } e^{ - t} }}{{z + t}}dt} $$ valid for $|\arg z|&...
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4 votes
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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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1 vote
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Find mle of theta from some mixed density

Well, yes, your calculations are correct. Consistency follows from the compactness of the parameter space: $\theta \in [-1, 1] $ Hint: $$ P( |\theta_*-\theta| < \epsilon)=P\Big( \frac{n}{2}(1 + \...
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6 votes
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Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

We have \begin{align*} \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ &...
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4 votes

Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

Asymptotically, $$g(n) = \prod_{2 < p < n}^{p \text{ prime}} 1-\frac2p \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$ And presumably your expression is simply the ...
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2 votes
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Asymptotics of an integral with singular derivation

While it is true that the integrand is singular at one of the endpoints of integration, this is not the biggest problem about the question you're asking. In fact, it is the requirements for a series ...
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Asymptotics of an integral with singular derivation

Try to rewrite $(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})$ as $$(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})\dfrac{\sqrt{1-(\rho+t)^2}+\sqrt{1-\rho^2}}{\sqrt{1-(\rho+t)^2}+\sqrt{1-\rho^2}} = \dfrac{1-(\rho+t)^2 -...
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  • 2,468
1 vote

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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2 votes
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Possible growth rates of a matrix entry with respect to exponentiation

Due to the Cayley-Hamilton theorem, matrix elements with respect to exponentiation adhere to the linear recurrence defined by the characteristic polynomial of a matrix. And a linear recurrence $A_0, ...
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2 votes

Possible growth rates of a matrix entry with respect to exponentiation

Consider $A=I+S,$ where $S$ a matrix with entries $s_{i,i+1}=1$ and $0$ otherwise. If the dimension is equal $m,$ then $$A^n=I+ \sum_{k=1}^{m-1} {n\choose k}S^k$$ Thus $$(A^n)_{1,1+j}={n\choose j}\...
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1 vote

Expansion of logarithm with little o

You were too aggressive with your $o$s, and confused an $O(|z+o(z)|^2)$ term with $o(|z+o(z)|^2)$. You could have instead expanded as: $$z+o(z)-\frac{1}{2}(z+o(z))^2+o(|z+o(z)|^2)$$ We can clean this ...
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1 vote
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Bounding the difference between two points of the logarithm

$$\begin{align}\log(n)-\log(n-k)&=\log\left(1+\frac{k}{n-k}\right)\\&\lt\log\left(1+\frac{1}{n^{1-\varepsilon}-1}\right)\\&\lt\frac{1}{n^{1-\varepsilon}-1}\\&\lt\frac{1}{\sqrt{n}-1}\\&...
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5 votes
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Asymptotic behavior of integral with Laplace's method

Elaborating on Maxim's comment. If $v=xT$, then \begin{align*} I(n)&=\int_0^1\!\! {\int_0^1 {e^{ - nxT} \sqrt {1 - (1 - \sqrt x + \sqrt {xT} )^2 } dT} dx} \\& = \int_0^1\!\! {\int_0^x {e^{ - ...
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What does it mean for a function to be polynomially bounded

It depends on the context. In general, $f$ is said polinomially bounded if $|f(x)|\leqslant p(x)$ for some real polynomial $p$. But for example, in the context of the theory of distributions, $f$ is ...
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2 votes

Meaning of asymptotically equal

Some definitions first: $$ f(x) \in o(g(x)) \iff \lim\limits_{x \to \infty} \frac{f(x)}{g(x)} = 0. $$ As the notion goes, I assume that by $f(x) = g(x)(1+o(1))$ we mean that $$ f(x) - g(x) = g(x)t(x), ...
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Big-O of Equations: How to understand it.

Since $$\frac{f(n)}{g(n)}=\frac{3n^3+20n^2+5}{n^3}=3+\frac{20}{n}+\frac{5}{n^3}\rightarrow 3$$ as $n\rightarrow\infty$, $c$ can be any number larger then $3$. Fix $c>3$. We want to find $n_0$ so ...
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2 votes

Asymptotic behavior of integral with Laplace's method

Let's denote $$I=\int_0^1dx\int_0^1dy\sqrt{1-(1-\sqrt x+\sqrt x\sqrt y)^2}e^{-nxy}$$ Making the substitution $s=nx$ $$=\frac{1}{n^{5/4}}\int_0^ns^{1/4}ds\int_0^1dy\sqrt{1-\sqrt y}\sqrt{2-\sqrt\frac{s}{...
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Asymptotic behavior of integral with Laplace's method

As suggested in the comments by Ian, make the change of variables $u=xT$, $v=x$, after which your integral takes the form $$\int_0^1\int_0^v \left(1-(1-\sqrt{v}-\sqrt{u})^2\right)^{1/2}e^{-nu}\frac{1}{...
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  • 4,351
4 votes
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Bounding a convergent sum

For each $n$, let $f_n(x) = \frac{\exp(-n/2^x)}{2^x x}$. Then we can show that(1),(2) $$ \left| \sum_{i=1}^{n} f_n(i) - \int_{1}^{n} f_n(x) \, \mathrm{d}x \right| \leq \max_{1\leq x \leq n} f_n(x). $$ ...
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4 votes

Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?

You may just use the following Lemma if $\{a_n\}_{n\geq 1}$ is a sequence of positive numbers such that $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to +\infty}\sqrt[n]{a_n}=L$. Apply ...
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7 votes
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asymptotic of a determinant

The matrix $M_n$ is the Hilbert matrix $H_n = \left(\dfrac{1}{i+j+1}\right)_{0 \le i,j \le n-1}$ but flipped horizontally. Hence, $|\det(M_n)| = |\det(H_n)|$ for all $n$. From the Wikipedia article on ...
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2 votes

Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?

Suppose that $a_n\sim b_n$ as $n\to +\infty$. This means, by definition, that, in particular, $$ \frac{1}{2}<\frac{a_n}{b_n}<\frac{3}{2} $$ for all sufficiently large $n$. Taking $n$th roots and ...
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3 votes

Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?

$$\begin{align*}\lim_{n\to\infty}\big(\frac{\ln(n!)}{n}-\ln(n)\big)&=\lim_{n\to\infty}\frac1n\big(\ln(\frac1n)+\ln(\frac2n)+\dots+\ln(\frac nn)\big)\\ &=\int_0^1\ln(x)dx\\ &=[x\ln x-x]_0^1=...
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  • 9,925
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Little o notation within another little o

It seems like you want to justify that $f(x) = o(x + o(x)) \implies f(x) = o(x)$, as $x \to 0$. By definition $\lim_{x \to 0} \frac{o(x)}{x} = 0 \land \lim_{x \to 0} \frac{x}{x} = 1 \implies \lim_{x \...
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  • 370
1 vote

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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How can I find an asymptotic solution to this recurrence?

Hint If you let $T(n)=U(n)-\frac c5$, you end with $$U(n)=4 U\left(\frac{n}{4}\right)+2 U\left(\frac{n}{2}\right)$$ $$U(n)=c_1 \,\left(1-\sqrt{5}\right)^{\log_2 (n)}+c_2\, \left(1+\sqrt{5}\right)^{\...
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4 votes
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How can I find an asymptotic solution to this recurrence?

You can use a sharper form of the master theorem called the Akra-Bazzi Method. You can find details on the linked wikipedia page, but broadly we see that $$ T(n) = a_1 T(b_1 n) + a_2 T(b_2 n ) + g(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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4 votes
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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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