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 ...
- 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) \\ &...
- 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^{ - ...
- 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 ...
- 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 ...
- 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)
$$ \left| \sum_{i=1}^{n} f_n(i) - \int_{1}^{n} f_n(x) \, \mathrm{d}x \right| \leq \max_{1\leq x \leq n} f_n(x). $$
...
- 143k
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 ...
- 340k
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 ...
- 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
$$
T(n) = a_1 T(b_1 n) + a_2 T(b_2 n ) + g(n) =...
- 27.4k
4
votes
Accepted
maximise $(1-x)^{n-1}x - x^n$ subject to $0\le x\le 1/n$, where n is a natural number
$$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)^...
- 217k
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
$$\...
- 51.9k
4
votes
Accepted
Asymptotics of Laplace like integral with shrinking integration intervals
With the change of variables $x=e^{-t}$, we have
$$
I(M) = (\log M)^{M + 1} \int_{\log \log M}^{ + \infty } {e^{ - Mg(t)} e^{ - t} dt} ,
$$
where $g(t) = t+ f(e^{ - t} )$. This further equals to
\...
- 20.4k
3
votes
What is $\sum_{k = 1}^n (k \log k)\binom{n}{k}$? If the exact answer is difficult to find, what is the tightest asymptotic upper bound?
a) rewriting the sum
First of all we have better to rewrite the sum as
$$
\eqalign{
& S(n) = \sum\limits_{k = 1}^n {k\ln k\left( \matrix{
n \cr k \cr} \right)} = \sum\limits_{k = 1}^n {\ln ...
- 33.5k
3
votes
Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?
$$\begin{align*}\lim_{n\to\infty}\big(\frac{\ln(n!)}{n}-\ln(n)\big)&=\lim_{n\to\infty}\frac1n\big(\ln(\frac1n)+\ln(\frac2n)+\dots+\ln(\frac nn)\big)\\
&=\int_0^1\ln(x)dx\\
&=[x\ln x-x]_0^1=...
- 9,970
3
votes
How to obtain logarithmic asymptotic behavior for this integral?
Using a CAS, there is an exact result
$$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 ))$$
...
- 217k
2
votes
Asymptotic expansion of inhomogenous differential equation
I'm not 100% sure what you are looking for but let me try something. First of all, identifying the constant term in the asymptotic expansion without giving an initial data seems impossible. The reason ...
- 91
2
votes
Accepted
Asymptotics of an integral with singular derivation
While it is true that the integrand is singular at one of the endpoints of integration, this is not the biggest problem about the question you're asking. In fact, it is the requirements for a series ...
- 7,978
2
votes
Accepted
Possible growth rates of a matrix entry with respect to exponentiation
Due to the Cayley-Hamilton theorem, matrix elements with respect to exponentiation adhere to the linear recurrence defined by the characteristic polynomial of a matrix. And a linear recurrence $A_0, ...
- 231
2
votes
Possible growth rates of a matrix entry with respect to exponentiation
Consider $A=I+S,$ where $S$ a matrix with entries $s_{i,i+1}=1$ and $0$ otherwise. If the dimension is equal $m,$ then $$A^n=I+ \sum_{k=1}^{m-1} {n\choose k}S^k$$ Thus $$(A^n)_{1,1+j}={n\choose j}\...
- 6,340
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 $s=nx$
$$=\frac{1}{n^{5/4}}\int_0^ns^{1/4}ds\int_0^1dy\sqrt{1-\sqrt y}\sqrt{2-\sqrt\frac{s}{...
- 6,324
2
votes
Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?
Suppose that $a_n\sim b_n$ as $n\to +\infty$. This means, by definition, that, in particular,
$$
\frac{1}{2}<\frac{a_n}{b_n}<\frac{3}{2}
$$
for all sufficiently large $n$. Taking $n$th roots and ...
- 20.4k
2
votes
Meaning of asymptotically equal
Some definitions first:
$$
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),
...
- 231
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
$$ \color{red}{K}=\lim_{n\to +\...
- 340k
2
votes
Accepted
Difference between the usage of Big-Omega notation as used by Computer Scientists and Mathematicians.
The math definition is that there exists some infinite subsequence $I=\{n_i:i\geq 1\}$of the natural numbers and a positive constant $c$ such that for all $n\in I,$
$$
f(n)\geq c g(n).
$$
One could ...
- 6,378
1
vote
Accepted
Asymptotics of Bessel function $J_n(z)$ for $|z| \gg n$ but $|z|\not\gg n^2$
In terms of the modulus function $M_\nu (z)$ and phase function $\theta _\nu (z)$, we can write
$$
H_\nu ^{(1)} (z) = M_\nu (z)e^{i\theta _\nu (z)}, \quad
J_\nu (z) = M_\nu (z)\cos \theta _\nu (...
- 20.4k
1
vote
Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$
If you use Pochhammer symbols and $I_0=\frac{\pi}{2\sqrt2}$
$$I_n=\frac{4n-3}{4n}I_{n-1}\quad \implies I_n=\frac{\pi }{8 \sqrt{2}}\frac{\left(\frac{5}{4}\right)_{n-1}}{(2)_{n-1}}=\frac{\pi }{8 \sqrt{2}...
- 217k
1
vote
Accepted
Computing the asymptotics of a principal value integral
We start with $(8.6.4)$:
$$\tag{1}
\Gamma (\sigma ,z) = \frac{{z^\sigma e^{ - z} }}{{\Gamma (1 - \sigma )}}\int_0^{ + \infty } {\frac{{t^{ - \sigma } e^{ - t} }}{{z + t}}dt}
$$
valid for $|\arg z|&...
- 20.4k
1
vote
Accepted
Find mle of theta from some mixed density
Well, yes, your calculations are correct.
Consistency follows from the compactness of the parameter space: $\theta \in [-1, 1] $
Hint:
$$
P( |\theta_*-\theta| < \epsilon)=P\Big( \frac{n}{2}(1 + \...
- 1,110
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 ...
- 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 ...
- 10.6k
Only top scored, non community-wiki answers of a minimum length are eligible