Doubts in some inequalities used in asymptotic analysis

I was going through this recent paper, and I had a couple of doubts in the final analysis of complexity of the scheme (you don’t need to go through the whole paper). I included three questions here, as I thought they seem to simple questions.

  1. In Pg 17 (last four lines, see after equation 7), there is this inequality that is used (here, $ k(n) = \sqrt{\frac{n\delta(n)}{\log n}}$ and $ \delta(n) = o(n / \log n)$ ):

$ \frac{\binom{n}{a}}{\binom{^n/_k}{^a/_k}^k} \leq (^n/_k)^k$

Can I know a proof for it?

  1. Similarly, in the beginning of Pg 18, how is this possible? (the above inequality is used to get here though, $ (n / k)^{k / 2}$ thing, and don’t worry about the meaning of $ \mathtt{ss}$ )

$ \mathtt{ss} \leq \sqrt{\binom{n}{\frac{n}{2}-\delta(n)}}\cdot (n / k)^{k / 2} \cdot \binom{k}{2\delta(n)} \cdot 2^{(2\delta(n)+1)n / k} \leq 2^{n / 2 + o(n)}$

  1. Also, this one might be a bit trivial, in Pg 17, one inequality above equation 7, the $ O(n)$ term is dropped, isn’t that relevant?

Proof involving asymptotic complexity

The question in Proof of big-o propositions asked to prove:

$ O(f(n))=O(g(n))\iff\Omega(f(n))=\Omega(g(n))\iff\Theta(f(n))=\Theta(g(n))$

The accepted answer starts the proof with:

Suppose that $ O(f(n))=O(g(n))$ . It is easy to check that $ g(n)=O(g(n))$ … and so $ O(f(n))=O(g(n))$ implies that $ g(n)=O(f(n))$ .

I believe that there is a mistake in quoted part of the answer above.

It claims that $ O(f(n))=O(g(n))\tag{1}$

and $ g(n)=O(g(n))\tag{2}$

implies

$ g(n)=O(f(n))\tag{3}$

Using the interpretation given in “Introduction to Algorithms (CLRS) Edition 3, page 50”, (1) is interpreted as $ \forall\Phi(n)\in O(f(n)), \exists\Psi(n)\in O(g(n))$ such that $ \Phi(n)=\Psi(n)$ .

Additionally, the definition of $ O$ -notation given in “Introduction to Algorithms (CLRS) Edition 3, page 47” states:

$ O(g(n))=\{f(n):$ there exist positive constants $ c$ and $ n_0$ such that $ 0 \leq f(n)\leq c\cdot g(n)$ for all $ n\geq n_0\}$

Considering the counter example whereby $ f(n)=n$ and $ g(n)=n^2$ . Then $ O(n)=O(n^2)$ and $ n^2=O(n^2)$ hold, but $ n^2\neq O(n)$ , thereby contradicting the assertion given in the accepted answer. Note: “$ \neq$ ” in this case refers to “$ \not \in$ “.

Therefore, I would like to ask if the answer provided is wrong or my interpretation and/or reasoning is wrong.

Asymptotic notation and random variables

I have two random variables $ X$ and $ Y$ and I want to bound the value of one in terms of the other (for now, I don’t care about the actual distribution of their values).

Suppose that the two variables can have different distributions with values chosen from $ [1, n]$ . But $ X$ is always upper bounded by $ Y \cdot \log{n}$ . Can I write this as $ X = O(Y\log{n})$ (if I care about the behavior for large $ n$ ). I’m not sure what is the convention wrt to random variables and asymptotic notation.

The role of asymptotic notation in $e^x=1+𝑥+Θ(𝑥^2)$?

I’m reading CLRS and there is the following:

When x→0, the approximation of $ e^x$ by $ 1+x$ is quite good: $ $ e^x=1+𝑥+Θ(𝑥^2)$ $

I suppose I understand what means this equation from math perspective and, also, there is an answer in another cs question. But I don’t understand some things, so have a few questions.

  1. Why do they use $ Θ$ here and why do they use $ =$ sign?
  2. Is it possible to explain how the notation here is related to the author’s conclusion that $ e^x$ is very close to $ 1 + x$ when $ x \rightarrow 0 $ ?
  3. Also, how is it important here that $ x$ tends to $ 0$ rather than to $ \infty$ as we usually use asymptotic notations?

I’m sorry if there are a lot of questions and if they are stupid, I’m just trying to master this topic.

Asymptotic formula for the number of path-connected graphs

It can be shown that the set of graphs with $ N$ vertices $ G_N$ has cardinality:

\begin{equation} \lvert G_N \rvert = 2^{N \choose 2} \tag{1} \end{equation}

Recently, I wondered how much bigger $ \lvert G_N \rvert$ is compared to the number of graphs with $ N$ vertices that are path-connected, $ PC_N$ .

If we denote the set of Hamiltonian paths by $ H_N$ we can easily show that:

\begin{equation} H_N \subset PC_N \tag{2} \end{equation}

and by Stirling’s approximation:

\begin{equation} \lvert H_N \rvert = N! \sim \sqrt{2\pi N} \big(\frac{N}{e}\big)^N \tag{3} \end{equation}

and therefore $ \lvert PC_N \rvert$ must grow exponentially fast.

Now, I’m curious about asymptotic formulas for $ \lvert PC_N \rvert$ and I’m fairly confident that:

\begin{equation} \frac{\lvert PC_N \rvert}{\lvert G_N \rvert} \leq e^{-N} \tag{4} \end{equation}

but I suspect that a proof for this statement would be fairly subtle.

Find an Asymptotic Upper Bound using a Recursion Tree

The problem is this: Use the recursion-tree method to give a good asymptotic upper bound on T(n) = 9T(n^(1/3)) + Big-Theta(1). I am able to get the tree started and find a pattern with the sub-problems, but I am having difficulty finding the total cost of the running times throughout the tree. I can not figure out how to get the number of sub-problems at depth i when n=1. I have a feeling the answer is O(log3(n)), but I can not verify that at the moment. Any help would be appreciated.

T(n) = 9T(n^(1/3)) + Big-Theta(1) can be written as: T(n) = 9T(n^(1/3)) + C, where C is some constant since any constant will always be treated as 1 asymptotically. My recursion tree is explained by each level below: Level 0: This is the constant C

Level 1: T(n^(1/3)) is written 9 times which represent the sub-problems of C. This adds up to 9cn^(1/3).

Level 2: Each of the 9 sub-problems from level 1 gets divided into 9 more sub-problems, which are each written as T(n^(1/9)). All of these add up to 81cn^(1/9).

Sub-Problem Sizes and Nodes: The number of nodes at depth i is 9^i We know that the sub-problem size for a node at depth i is n^(1/(3^i)). The problem size hits n=1 when this size equals 1. Solving for i yields:

(n^(1/(3^i)))^(3i) = 1^(3i) n = 1^(3i). This results in n being 1 which doesn’t give a logarithmic form!

The optimal asymptotic behavior of the coefficient in the Hardy-Littlewood maximal inequality

It is well-known that for $ f \in L^1(\mathbb{R^n})$ ,$ \mu(x \in \mathbb{R^n} | Mf(x) > \lambda) \le \frac{C_n}{\lambda} \int_{\mathbb{R^n}} |f| \mathrm{d\mu}$ , where $ C_n$ is a constant only depends on $ n$ .

It is easy to see $ C_n \le 2^n$ , but how to determine its optimal asymptotic behavior? For example, does $ C_n$ bounded in $ n$ ? Is $ C_n$ bounded by polynomial in $ n$ ?

How can to find asymptotic form of power series?

The Power series is $ $ R_n(x)=\Sigma_{p=2}^{n+1}\frac{(p-1)(2n-p)!}{n!(n+1-p)!}x^p$ $

It says the following:

For large $ n$ , the asymptotic form of $ R_n(1/\beta)$ can be obtained by looking at the values of $ p$ which dominate the sum

  • $ p$ of order $ 1$ for $ \beta > 1/2$ ,
  • $ p$ of order $ \sqrt n$ for $ \beta = 1/2$ , and
  • $ p \propto (1 – 2\beta)/(1-\beta)$ for $ \beta < i$

See This Picture

Asymptotic Bound on Minimum Epsilon Cover of Arbitrary Manifolds

Let $ M \subset \mathbb{R}^d$ be a compact smooth $ k$ -dimensional manifold embedded in $ \mathbb{R}^d$ . Let $ \mathcal{N}(\epsilon)$ denote the size of the minimum $ \epsilon$ cover $ P$ of $ M$ ; that is for every point $ x \in M$ there exists a $ p \in P$ such that $ \| x – p\|_{2}$ .

Is it the case that $ \mathcal{N}(\epsilon) \in \Theta\left(\frac{1}{\epsilon^k}\right)$ ? If so, is there a reference?

Always exists an asymptotic component with large size in a minimal subshift?

Let $ (X, T)$ be a minimal subshift, i.e. $ X$ is a closed $ T$ -invariant subset of $ A^\mathbb{Z}$ , where $ T$ is the shift. A pair $ x,y\in X$ is asymptotic if $ d(T^nx, T^ny)$ goes to zero as $ n\to\infty$ . Always exists such a pair when $ X$ is infinite: for every $ n\geq1$ there exists $ x^{(n)}, y^{(n)} \in X$ such that $ x^{(n)}_0\not= y^{(n)}_0$ and $ x^{(n)}_{[1,N]}= y^{(n)}_{[1,N]}$ (if not, for some $ n$ , $ x_{[1,n]} = y_{[1,n]}$ implies $ x_0=y_0$ , i.e., $ x_{[1,n]}$ determines $ x_0$ , and this forces $ X$ to be periodic), and any pair of convergent subsequences of $ x^{(n)}$ and $ y^{(n)}$ will do the trick.

My question is: do exist $ k$ -tuples of asymptotic points, for every $ k\geq1$ ? More precisely, is it true that for every $ k\geq1$ there exists $ x_1,\dots,x_k\in X$ such that $ $ \lim_{n\to\infty}d(T^nx_i, T^nx_j) = 0\ \forall i,j$ $