New answers tagged asymptotics
0
votes
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 $T\ge2$ ...
- 3,635
0
votes
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
$$ \log(a_n)=\log (n)-1+\frac{\log (2 \pi n)}{2 n}+\frac{1}{12 n^2}-\frac{1}{360
n^4}+O\...
- 216k
0
votes
Follow up question: asymptotics of a two dimensional integral
First, let's simplify the given integral using the same $\int_0^1 d\epsilon\int_{-\epsilon}^\epsilon dt=\int_{-1}^1 dt\int_{|t|}^1 d\epsilon$. Denote $$g(\rho,t)=\sin\big[({\textstyle\sqrt{1-(\rho+t)^...
- 32.1k
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
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
0
votes
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 $...
- 17.5k
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
\...
- 20.4k
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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0
votes
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
Accepted
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
$$ \int_{1}^{n}\ln(x)\,dx \geq \ln(2)+\ln(3)+\ldots+\ln(n-1)+\frac{1}{...
- 340k
0
votes
Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes).
Informally, if you're going to let $n$ approach infinity, then all but the highest power of a polynomial can be ignored. In your example, as $x$ approaches infinity
$\sqrt{x^2+9} \rightarrow \sqrt{x^2}...
- 25.8k
4
votes
Accepted
Exponential sum behaves like linear term for large $t$
The Fourier transform of $f(x) = \exp\left(-\tfrac{\pi x^2}{t^2}\right)$ is $\widehat{f}(\omega) = t\exp\left(-\tfrac{t^2\omega^2}{4\pi}\right)$. So by using the Poisson summation formula, we have
$$\...
- 51.9k
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
$$\|L_{K_t}\|=t^{-1/2},\qquad {\...
- 6,307
1
vote
sequence such that $x_{n+1}=n(x_n-n)$
Your question is about a sequence that satisfies
$$ 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} $$
where $\...
- 29.6k
5
votes
Accepted
sequence such that $x_{n+1}=n(x_n-n)$
By some experimentation from the starting point $x_n=an+b+y_n$ and reporting in the equation I quickly found that $x_n=n+1+2y_n$ lead to the following simplification
$$y_{n+1}=ny_n-1$$
So we have ...
- 24.6k
1
vote
Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$
If you use Pochhammer symbols and $I_0=\frac{\pi}{2\sqrt2}$
$$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}...
- 216k
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
$$
H_\nu ^{(1)} (z) = M_\nu (z)e^{i\theta _\nu (z)}, \quad
J_\nu (z) = M_\nu (z)\cos \theta _\nu (...
- 20.4k
0
votes
Accepted
How to show that $\forall n \geq 1$, $\frac{n^{2n}}{n!^2} \geq (\frac{n+1}{n})^{n^2-n}$?
We can proceed by induction on $n$. The case $n=1$ is easy to check. Suppose that the inequality holds for $1,2,\ldots,n$. Then
\begin{align*}
\frac{{(n + 1)^{2n + 2} }}{{(n + 1)!^2 }} = \frac{{(n + 1)...
- 20.4k
2
votes
Equivalent of a recurrence sequence $I_n=\frac{4n-3}{4n}I_{n-1}$
By Euler's Beta function
$$ 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
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 ...
- 331
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
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
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:
$$
P( |\theta_*-\theta| < \epsilon)=P\Big( \frac{n}{2}(1 + \...
- 1,110
6
votes
Accepted
Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?
We have
\begin{align*}
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ &...
- 20.4k
4
votes
Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?
Asymptotically,
$$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}$$
And presumably your expression is simply the ...
- 1,385
2
votes
Accepted
Asymptotics of an integral with singular derivation
While it is true that the integrand is singular at one of the endpoints of integration, this is not the biggest problem about the question you're asking. In fact, it is the requirements for a series ...
- 7,978
0
votes
Asymptotics of an integral with singular derivation
Try to rewrite $(\sqrt{1-(\rho+t)^2}-\sqrt{1-\rho^2})$ as
$$(\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 -...
- 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 ...
- 24.6k
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 $$(A^n)_{1,1+j}={n\choose j}\...
- 6,307
1
vote
Expansion of logarithm with little o
You were too aggressive with your $o$s, and confused an $O(|z+o(z)|^2)$ term with $o(|z+o(z)|^2)$. You could have instead expanded as: $$z+o(z)-\frac{1}{2}(z+o(z))^2+o(|z+o(z)|^2)$$
We can clean this ...
- 10.5k
1
vote
Accepted
Bounding the difference between two points of the logarithm
$$\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}\\&...
- 10.5k
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
0
votes
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 ...
- 31
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
$$
f(x) - g(x) = g(x)t(x),
...
- 231
0
votes
Big-O of Equations: How to understand it.
Since
$$\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}{...
- 6,314
0
votes
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
$$\int_0^1\int_0^v \left(1-(1-\sqrt{v}-\sqrt{u})^2\right)^{1/2}e^{-nu}\frac{1}{...
- 4,351
4
votes
Accepted
Bounding a convergent sum
For each $n$, let $f_n(x) = \frac{\exp(-n/2^x)}{2^x x}$. Then we can show that(1),(2)
$$ \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). $$
...
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4
votes
Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?
You may just use the following
Lemma if $\{a_n\}_{n\geq 1}$ is a sequence of positive numbers such that $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to +\infty}\sqrt[n]{a_n}=L$.
Apply ...
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7
votes
Accepted
asymptotic of a determinant
The matrix $M_n$ is the Hilbert matrix $H_n = \left(\dfrac{1}{i+j+1}\right)_{0 \le i,j \le n-1}$ but flipped horizontally. Hence, $|\det(M_n)| = |\det(H_n)|$ for all $n$.
From the Wikipedia article on ...
- 51.9k
2
votes
Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?
Suppose that $a_n\sim b_n$ as $n\to +\infty$. This means, by definition, that, in particular,
$$
\frac{1}{2}<\frac{a_n}{b_n}<\frac{3}{2}
$$
for all sufficiently large $n$. Taking $n$th roots and ...
- 20.4k
3
votes
Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?
$$\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=...
- 9,925
0
votes
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 \...
- 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 $O(ln(n))$
0
votes
How can I find an asymptotic solution to this recurrence?
Hint
If you let $T(n)=U(n)-\frac c5$, you end with
$$U(n)=4 U\left(\frac{n}{4}\right)+2 U\left(\frac{n}{2}\right)$$
$$U(n)=c_1 \,\left(1-\sqrt{5}\right)^{\log_2 (n)}+c_2\, \left(1+\sqrt{5}\right)^{\...
- 216k
4
votes
Accepted
How can I find an asymptotic solution to this recurrence?
You can use a sharper form of the master theorem called the Akra-Bazzi Method.
You can find details on the linked wikipedia page, but broadly we see that
$$
T(n) = a_1 T(b_1 n) + a_2 T(b_2 n ) + g(n) =...
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0
votes
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 ...
...
- 7
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)^...
- 216k
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