A uniform upper bound for Fredholm index of quasi Laplace operators on a compact parallelizable manifold

Assume that $ M$ is a compact parallelizable manifold. Is there an upper bound for the absolute value of Fredholm index of all operators in the form $ D=\sum_{i=1}^n \partial^2/\partial{X_i^2}$ where $ \{X_1,X_2,\ldots,X_n\}$ is a global smooth frame?

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!

Change an entire google sheets to UPPER case, not with an add-on or onEdit(e) [on hold]

I need to change all the text in Google Sheets to upper case, the problem is it’s filled in by Zapier which doesn’t register as an edit for onEdit functions. It’s also automatically converted to .xlsx and emailed so none of the add-ons will work because they need to manually run. After the email sends a script wipes it and we start over as a daily report. I’ve looked everywhere for a script that will work but haven’t been able to find anything.

I currently have it transferring to other sheets using importrange wrapped in arrayformula(upper, which works but then my xlsx cell contents is all iferror functions.

Lower and upper bounds of the distance between two Frobenius numbers

I consider two sequences of numbers: $ A=\{a_1,…,a_{m-1},n\}$ and $ B=\{n-a_{m-1},…,n-a_1,n\}$ , where $ a_1 < a_2 < … < a_{m-1} < n$ and $ \gcd(A) = \gcd(B) = 1$ .


I investigate the lower and upper bounds of the distance between two Frobenius Numbers:

$ Dist Lower Bound \le \left| F(A) – F(B) \right| \le Dist Upper Bound $

I found only one trivial estimate for lower bound: It is known that $ F(a_1,…,a_m) \ge a_1 – 1$ if $ a_1 \ge 2$ . Using this property, we obtain that:

$ \left| F(A) – F(B) \right| \ge \left| n-a_{m-1}-a_1 \right|$ .


Also, I investigate the lower and upper bounds of the sum of two Frobenius Numbers:

$ Sum Lower Bound \le F(A) + F(B) \le Sum Upper Bound $

I found only one trivial estimate for upper bound: I am using simple way. I took well-known estimates and combine it because I having knowledge about $ a_{m-1}$ .

$ 1. F(a_1,…,a_m) \le 2a_{m-1} \lfloor{\frac{a_m}{m}}\rfloor – a_m = 2a_{m-1} \lfloor{\frac{n}{m}}\rfloor – n$ .

$ 2. F(b_1,…,b_m) \le 2b_m \lfloor{\frac{b_1}{m}}\rfloor – b_1 = 2n \lfloor{\frac{n-a_{m-1}}{m}}\rfloor – n + a_{m-1}$ .


I am convinced that there are other nearer solutions. I will be grateful for any help in search $ Sum Lower Bound $ and $ Dist Upper Bound $ as well as improving my estimates.

About Frobenius Number can be read here: link 1, link 2.

How do you empirically estimate the most popular seat and get an upper bound on total variation?

Say there are $ n$ seats $ \{s_1, …, s_n\}$ in a theater and the theater wants to know which seat is the most popular. They allow $ 1$ person in for $ m$ nights in a row. For all $ m$ nights, they record which seat is occupied.

They are able to calculate probabilities for whether or not a seat will be occupied using empirical estimation: $ P(s_i ~\text{is occuped})= \frac{\# ~\text{of times} ~s_i~ \text{is occupied }}{m}$ . With this, we have an empirical distribution $ \hat{\mathcal{D}}$ which maximizes the likelihood of our observed data drawn from the true distribution $ \mathcal{D}$ . This much I understand! But, I’m totally lost trying to make this more rigorous.

  • What is the upper bound on $ \text{E} ~[d_{TV}(\hat{\mathcal{D}}, \mathcal{D})]$ ? Why? Note: $ d_{TV}(\mathcal{P}, \mathcal{Q})$ is the total variation distance between distributions $ \mathcal{P}$ and $ \mathcal{Q}$ .
  • What does $ m$ need to be such that $ \hat{\mathcal{D}}$ is accurate to some $ \epsilon$ ? Why?
  • How does this generalize if the theater allows $ k$ people in each night (instead of $ 1$ person)?
  • Is empirical estimation the best approach? If not, what is?

If this is too much to ask in a question, let me know. Any reference to a textbook which will help answer these questions will happily be accepted as well.

Upper bound on $\sum_{k=1}^T \frac{1}{k (1+a)^{T-k}}$

Is there any reasonable upper bound for the following quantity $ $ \sum_{k=1}^T \frac{1}{k (1+a)^{T-k}} $ $

where $ a>0$ with respect to $ T$ and $ a$ (something like $ \mathcal{O}(\frac{\log (T)}{aT}$ )? I tried to compute integral $ $ \int_{0}^T \frac{1}{x (1+a)^{T-x}}dx, $ $ which should be upper bound on this sum as $ f(x) = \frac{1}{x (1+a)^{T-x}}$ is decreasing on $ (0, T)$ , but I did not achieve to get reasonable expression.

Prove that the upper bound in the Noiseless-coding theorem is strict

Given a probability distribution $ p$ across an alphabet, we define redundancy as:

Expected Length of codewords – entropy of p = $ \ E(S) – h(p)$

Prove that for each $ \epsilon$ with $ 0 \le \epsilon \lt 1$ there exists a $ p$ such that the optimal encoding has redundancy $ \epsilon$ .

Attempts

I have tried constructing a probability distribution like $ p_o = \epsilon, p_1 = 1 – \epsilon $ based on a previous answer, but I can’t get it to work.

Any help would be much appreciated.