45
votes
Accepted
A challenging (for me) limit calculation
This limit converges to $\frac{\sqrt3}2$. The idea is that $\sin(x) = x - \frac{x^3}6 + O(x^5)$, so we start with $\frac1{\sqrt n}$ and repeatedly subtract $\frac{x^3}6$. We can approximate this ...
- 2,979
13
votes
Accepted
Asymptotics of a strange oscillatory function
I will show (unless I made a mistake)
$$
f(x) = a\sqrt{x} + O(x^{1/3}),
$$
for some constant $a > 0$.
For each $x > 0$, let $g(t) = \sin(x/t^2)$. Let $\varepsilon > 0$ to be chosen later ...
- 4,633
12
votes
Accepted
Asymptotics for pairs of positive integers whose harmonic (resp. geometric) mean is an integer
[CORRECTED 8/28]
I think the asymptotic ratio is $(4/3)*\ln(2)\approx 0.924196$.
$$G(n) = (6/\pi^2)*n*\ln(n) + O(n)$$
$$H(n) = (8\ln(2)/\pi^2)*n*\ln(n) + O(n)$$
The proof for the geometric mean goes ...
- 316
12
votes
Asymptotic behavior of a certain oscillatory integral
$$I(x):=\int_{0}^{\infty}\frac{e^{i r}}{r^{\frac{1}{2}}}\int_{0}^{\infty}\frac{e^{-s}}{s^{\frac{1}{2}}}\frac{r}{sx+\sqrt{sxr}+r}ds dr.$$
To aid the asymptotic analysis, I regularize $I(x)$ by ...
- 178k
12
votes
Accepted
Lindelöf hypotheses for derivatives of zeta
The Lindelöf hypothesis yields the same bound for each derivative of $\zeta(s)$ via Cauchy's formula. Indeed, let $n\in\mathbb{N}$, $\sigma\in\mathbb{R}$, $T\in(1,\infty)$, $\varepsilon\in(0,1/2)$. ...
- 99.1k
11
votes
Accepted
On an asymptotic integral decay
I'll change $[-1,0]$ to $[0,1]$ (so $t\to -t$), which seems easier on the brain.
$f(t)=e^{-1/t}$ is a counterexample: For $0<t<1/\sqrt{\lambda}$, we have $f\le e^{-\sqrt{\lambda}}$, so this part ...
11
votes
Is there an upper bound on the number of representations as a sum of squares?
It is known since Gauss that
$$r_3(n)=12\dfrac{h(D)}{w(D)/2}(1-(D/2))\sum_{d\mid f}\mu(d)(D/d)\sigma(f/d)\;,$$
where $-n=D(2^vf)^2$, $D$ a fundamental discriminant, $v\ge-1$, $h(D)$ is the class ...
- 11.7k
10
votes
Accepted
Asymptotics of an integral requested
Change variables to $y=n(1-x)$, which gives
$$
I_n=\int\limits_{0}^{n}\left(
1-\frac{y}{n}\right)^{n-1}
\sqrt{1-\frac{\log y}{\log n}}\ {\rm d}y\ ,
$$
then dominated convergence should allow passing ...
- 21.8k
9
votes
Accepted
Does the Riemann hypothesis predict a bound for this prime-counting function?
The Riemann hypothesis is equivalent to the following statement:
$$f(x)=\mathrm{li(x)}-\frac{x}{\log x}+O(\sqrt{x}),\qquad x\geq 2.$$
Note that
$$\mathrm{li(x)}=\mathrm{li(2)}+\frac{x}{\log x}-\frac{2}...
- 99.1k
9
votes
Accepted
On a paper by Dimitrie Pompéiu and on one (in two parts) by Edmund Landau
Below are both references: links then screenshots.
The first reference can be found in full here.
In case permissions etcetera should change, below are screenshots of the pages:
The second reference ...
- 7,741
9
votes
Asymptotics of a strange oscillatory function
Yes, to (1) and (2), but the argument below is much too crude to identify $c$.
The obvious attempt is to use $\sin t\ge 2t/\pi$, $0\le t\le \pi/2$ (or something similar), for the terms with large $n$. ...
8
votes
Accepted
Asymptotic behavior of a certain oscillatory integral
We can evaluate $I(x)$ explicitly, and then asymptotically.
Indeed, using the substitution $s=ru/x$, we get
\begin{equation*}
I(x)=\frac1{\sqrt x}\lim_{R\to\infty}J_R(x), \tag{1}\label{1}
\end{...
- 118k
8
votes
Accepted
Series with the smallest number whose square is divisible by $n$
I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ then $a(k) = ab$, so $$\begin{align*}
\sum_{k\leq x}\frac1{a(k)}
&= \sum_{a b^2 \leq x} \...
- 2,979
7
votes
Probability of large gcd
To get a bound which is worse than the one of GH from MO asymptotically, but which doesn't require any case checking, we can do the following: if $\gcd(t, N) = k$, then $\frac{N}{k} = d$ which is an ...
- 5,061
7
votes
Accepted
Probability of large gcd
The statement is true in the stronger form that
$$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/2}.$$
Indeed, the probability that $\gcd(t,N)$ equals a given $k\mid N$ is at most $1/k$. For $k>N^{3/4}$, ...
- 99.1k
7
votes
Accepted
Randomly removing length 1 intervals in an interval (a fragmentation process)
This is exactly the so-called "Renyi parking process." Renyi proved
$$\varphi(x) = cx + c - 1 + o(1)$$
where $$c = \int_0^\infty \exp\left( -2 \int_0^x \frac{1 - e^{-y}}{y}\,dy \right)\,dx \...
- 909
6
votes
Accepted
5
votes
Accepted
Asymptotic behavior of the integral that contains $\delta$ function
$$P(s)=2\int_{-\infty}^{\infty}{\rm d}p\int_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}$$
$$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{\...
- 178k
5
votes
Accepted
On the asymptotic behaviour of the series $\sum_{n=1}^{\infty} \left ( \frac{\zeta(ns)}{n}+\frac{s}{1-ns}\right )$ near $s=0$
Proposition. The following approximation holds true:
$$f(s)=\frac{1}{2}\log s+O(1),\qquad s\in(0,1).$$
Proof. Following Carlo Benakker, we remark that
$$\zeta(s)+\frac{s}{1-s}=-\frac{1}{2}+O(s),\qquad ...
- 99.1k
4
votes
Accepted
The asymptotic behavior of $F(\lambda):=\sum_{k=0}^{\infty}\frac{\Gamma{(a k)}}{\Gamma{(b k)}}{\lambda}^{-k}$, $b>a>0$
$\newcommand{\Ga}{\Gamma}$The value of
\begin{equation*}
r_k:=\frac{\Ga(ak)}{\Ga(bk)}
\end{equation*}
is undefined at $k=0$. So, let
\begin{equation*}
r_0:=\lim_{k\to0}\frac{\Ga(ak)}{\Ga(bk)}=\...
- 118k
4
votes
Slick proof of Stirling's Formula?
I've played around with this a bit. I have a slick lower bound, but not a slick upper bound.
We start with the $\Gamma$-integral:
$$n! = \int_{x=0}^\infty x^n e^{-x} dx = \int_{y=-n}^\infty (n+y)^n e^{...
- 151k
4
votes
Accepted
A limit arising from Mellin Inversion: How to compute a specific term of an asymptotic series?
As $s\to0^+$, we have $s\Gamma(s)=\Gamma(s+1)\to1$, $x^s\to1$ unless $x=0$, and $e^{-s f(n)}\to1$ for each $n$. So, by (say) the Fatou lemma,
$$ L:=\lim_{s\to0^+} \left[s \sum_{n=0}^{\infty} e^{-s f(...
- 118k
4
votes
Accepted
How fast does this summation grow?
Let us obtain the asymptotic of $\log S_n$, where $\log$ denotes the base-$2$ logarithm,
\begin{equation*}
S_n:=\sum_{k=1}^n\prod_{l=1}^k\binom{2^n}{2^l}^i
=\sum_{k=1}^n a_{n,k},
\end{...
- 118k
4
votes
Accepted
$L^1$ error between indicator function and smoothed out version
Yes, this works, and the only ingredient we need is the estimate $\int_r^{\infty} e^{-t^2}\, dt\lesssim e^{-r^2}$.
We then have (for example)
\begin{align*}
\int_r^{\infty} |f_r(x)|\, dx &=\frac{1}...
4
votes
Accepted
Asymptotic solution of a system of ODEs
This is my comment as an answer.
Using Mathematica's ParametricNDSolve[] with 50 digits precision as well as a series expansion around $u=\infty$, I get
\begin{align}
a(u) &= -1.02418609585 u^{-1} ...
- 2,705
4
votes
On the asymptotic behaviour of the series $\sum_{n=1}^{\infty} \left ( \frac{\zeta(ns)}{n}+\frac{s}{1-ns}\right )$ near $s=0$
We seek the small-$s$ asymptotics of the sum
$$f(s)=s\sum_{n=1}^\infty g(ns),\;\;g(x)=\frac{\zeta(x)}{x}+\frac{1}{1-x}.$$
First note that
$$\lim_{s\rightarrow 0}\sum_{n=\lfloor{2/s}\rfloor+1}^\infty ...
- 178k
3
votes
Asymptotic scaling of mean and variance for non-central chi distribution
For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward, $k(\mu/\sigma)^2\mapsto\sum_{i=1}^k(\mu_i/\sigma_i)^2={\cal O}(k).$
...
- 178k
3
votes
Accepted
Extreme case bounds on Diophantine approximation
If $\alpha$ is a real irrational number, then there are infinitely many coprime integers $p,q$ with $q > 0$ such that
$$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^...
- 25.5k
3
votes
Accepted
Estimating the bound of the integral over whole $\mathbb{R}$ of the Taylor remainder term?
$\newcommand\R{\mathbb R}$We want to bound the ratio
\begin{equation*}
R:=\frac ND,
\end{equation*}
where
\begin{equation*}
N:=\iint_{\R^2}(F(x,w)-F(y,w))^2 \mu(dx)\mu(dy),\quad
D:=\...
- 118k
3
votes
How fast does this summation grow?
For $$A_n=\sum_{k=1}^n\prod_{l=1}^k\binom{2^n}{2^l}$$ it seems that
$$\log(\log(A_n))\sim \alpha +n \,\log(2) \qquad \qquad \alpha \sim -0.383$$
- 1,422
Only top scored, non community-wiki answers of a minimum length are eligible