7 votes
Accepted

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 ...
user avatar
  • 51.9k
6 votes
Accepted

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) \\ &...
user avatar
  • 20.4k
5 votes
Accepted

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^{ - ...
user avatar
  • 20.4k
5 votes
Accepted

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 ...
user avatar
  • 24.6k
4 votes
Accepted

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 ...
user avatar
  • 7,978
4 votes
Accepted

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) $$ \left| \sum_{i=1}^{n} f_n(i) - \int_{1}^{n} f_n(x) \, \mathrm{d}x \right| \leq \max_{1\leq x \leq n} f_n(x). $$ ...
user avatar
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 ...
user avatar
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 ...
user avatar
  • 1,385
4 votes
Accepted

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) =...
user avatar
4 votes
Accepted

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)^...
user avatar
4 votes
Accepted

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 $$\...
user avatar
  • 51.9k
4 votes
Accepted

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 \...
user avatar
  • 20.4k
3 votes

What is $\sum_{k = 1}^n (k \log k)\binom{n}{k}$? If the exact answer is difficult to find, what is the tightest asymptotic upper bound?

a) rewriting the sum First of all we have better to rewrite the sum as $$ \eqalign{ & S(n) = \sum\limits_{k = 1}^n {k\ln k\left( \matrix{ n \cr k \cr} \right)} = \sum\limits_{k = 1}^n {\ln ...
user avatar
  • 33.5k
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=...
user avatar
  • 9,970
3 votes

How to obtain logarithmic asymptotic behavior for this integral?

Using a CAS, there is an exact result $$I=\int_0^t \int_0^t \frac{ dt_1\, dt_2}{\sinh^2(t_1-t_2-i\delta)}\,$$ $$I=2 \log (-i \sin (\delta ))-\log (\sinh (t-i \delta ))-\log (-\sinh (t+i \delta ))$$ ...
user avatar
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 ...
user avatar
2 votes
Accepted

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 ...
user avatar
  • 7,978
2 votes
Accepted

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, ...
user avatar
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}\...
user avatar
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}{...
user avatar
  • 6,324
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 ...
user avatar
  • 20.4k
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), ...
user avatar
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 +\...
user avatar
2 votes
Accepted

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 ...
user avatar
  • 6,378
1 vote
Accepted

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 (...
user avatar
  • 20.4k
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}...
user avatar
1 vote
Accepted

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|&...
user avatar
  • 20.4k
1 vote
Accepted

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 + \...
user avatar
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 ...
user avatar
  • 24.6k
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 ...
user avatar
  • 10.6k

Only top scored, non community-wiki answers of a minimum length are eligible