## What are the points about representation of groups?

For a fixed (let say finite-, or Lie-, to respect the historical motivations) group, why does the study of all its linear representations over a fixed field, leads to some knowledge about its structure that wouldn’t be posible to reach without this tool ? What are the fundamental key points to understand here ?

## 3 open conjectures about simple theories?

To my knowledge, there are still 3 open conjectures concerning simple theories : elimination of hyperimaginaries, Lascar str type = srt type, and stable forking conjecture (if a type forks over some parameters, this is due to an instance of a stable forumula). I haven’t think of it very hard, but intuitively elimination of hyperimaginaries implies Lstr type = srt type, isn’t it ? Are there more implications between those 3 conjectures ? On the other hand, do you think the resolution of one of them would be an important breakthrough in model theory ? Or at the contrary would be quite anecdotal ?

## system becomes unresponsive as soon as any process uses Disk I/O at about 300 MB/s speed or more

When any process uses Disk I/O at about 300 MB/s speed or more as seen in htop, the system becomes unresponsive (it often may take longer than a second to react to mouse movement or key presses) which IMO should never happen. Is this something unavoidable on Ubuntu, or are there any things to try which may solve this?

Also I use a SSD, a relatively fast one I think. No HDDs are even connected to the PC.

This is, for example, one of the issues that make it near impossible to use Android Emulator.

## A proof about Automorphism in congruence class [on hold]

Suppose gcd(m,n)=1, and let $$F :Z_n→Z_n$$ be defined by $$F([a])=m[a]$$. Prove that $$F$$ is an automorphism of the additive group $$Z_n$$. I find it is diffcult to prove $$F$$ is injective and surjective. Could you please to help me proof it with all the details. I type it roughly, and i am sorry and sincerely looking for a result.

## Conflicting info about visa-free South Korea countries?

I was wondering which countries are OK for visa-free tourism to South Korea. Particularly just USA and UK.

Strangely, I found it hard to find decisive information!

https://english.visitkorea.or.kr/enu/TRV/TV_ENG_2_1.jsp

Notice it does not (!) list United Kingdom in the “Designated Countries..” box.

https://www.hikorea.go.kr/pt/en/info/popup/icis/VisaWaiver_pop.htm

Notice it does not list the USA (!)

Further: in fact it’s really unclear to me if thes are official Korean government -supplied web sites?

Specific Questions

1. Where exactly is the most official web page from some part of the Sth Korean government, showing the list of visa-free tourism countries? As of April 2019 specifically.1

2. Is it absolutely a fact that both USA and UK are visa-free? (I guess, we’d have to know (1) to have the answer to (2), unless someone has another absolutely-certain information route.)

3. Ideally as well as the USA/UK facts, I’d like the exact complete list (per April 2019), but I guess that would come from (1) also.

My googlefu is pretty good but I’ve completely drawn a blank on this basic quest! :O

1 I guess there could conceivably be some hoo-ha in relation to Brexit?

## Series solution about $x=0$ of $xy”-y’+4xy=0$.

I want to find at least one solution of the differential equation $$xy”-y’+4xy=0$$ about the point $$x=0$$. I identified that $$x=0$$ is a regular singular point and thus Frobenius Theorem is applicable.

Now, assuming a solution of the form $$y(x)={x^r}\sum_{n=0}^\infty {c_n x^n}=\sum_{n=0}^\infty {c_n x^{n+r}}$$

I get $$y’=\sum_{n=0}^\infty {(n+r)c_n x^{n+r-1}}$$ and $$y”=\sum_{n=0}^\infty {(n+r-1)c_n x^{n+r-2}}$$.

Substituting, and simplifying, I end up with the equation: $$\sum_{n=0}^\infty {(n+r)(n+r-1)c_n x^{n+r-1}} – \sum_{n=0}^\infty {(n+r)c_n x^{n+r-1}} + \sum_{n=2}^\infty {4 c_{n-2} x^{n+r-1}}$$

$$\implies r(r-2)c_0 x^{r-1} + (r^2-1)c_1x^r +\sum_{n=2}^\infty {[(n+r)(n+r-2)c_n + {4}c_{n-2}]x^{n+r-1}}=0$$

In this case, I have two indicial equations, $$r(r-2)=0$$ and $$r^2-1=0$$, giving $$r = 2, 0, 1, -1$$.

At this point, I would have expected to only have two values for $$r$$, but now I have four values.

How should I proceed in terms of using the values of $$r$$ to obtain a series solution? Is it correct to simply set $$c_1=0$$ and just use the indicial equation $$r(r-2)=0$$, and using the larger root $$r=2$$? If yes, then is there a general rule for which indicial equation to use when faced with more than two values of $$r$$ in the case of a second order differential equation?

Note: Maple gave one solution to be $$y_1(x) = x^2\{1 – \frac{1}{2}x^2 + \frac{1}{12}x^4+O(x^6)\}$$

## Prints information about a restaurant (provided by user input) using a .txt file

My code works, but not the way I want it to. If a user provides a string, even if the string doesn’t match a name in the array, it will print information about all 5 restaurants. If I put a break statement, it will only print the information for the first one (as I would expect it to).

Code:

#include <stdio.h> #include <malloc.h>  typedef struct restaurants{     char name[50];     char type[50];     float avg;     int rating; }restaurant;  int readRestaurantsFromFile(restaurant **restaurants, char fileName[]) {     int size, i;     restaurant *arr;     FILE *fp;     fopen_s(&fp, fileName, "r");     if (fp) {         fscanf_s(fp, "%d", &size);         arr = (restaurant *)malloc(sizeof(restaurant) * size);         for (i = 0; i < size; ++i) {             fscanf_s(fp, "%s", arr[i].name, sizeof(arr[i].name));             fscanf_s(fp, "%s", arr[i].type, sizeof(arr[i].type));             fscanf_s(fp, "%f", &arr[i].avg);             fscanf_s(fp, "%d", &arr[i].rating);         }         *restaurants = arr;         fclose(fp);     }     else {         printf("Can not open %s to read...\n", fileName);     }     return size; }  void printDetails(restaurant *arr, int size, char res_name) {     for (int i = 0; i < size; ++i) {         if (res_name = arr[i].name) {             printf("Food: %s\n", arr[i].type);             printf("Avg. Meal: $%.2f\n", arr[i].avg); printf("Rating: %d\n\n", arr[i].rating); } } } int main() { restaurant *arr; int size = readRestaurantsFromFile(&arr, "C:\Users\Owner\source\repos\FedericoReadLab6Problem2\FedericoReadLab6Problem2\reviews.txt"); for (int i = 0; i < size; ++i) { char buffer[50]; printf("Enter a restaurant: "); char res_name = fgets(buffer, sizeof buffer, stdin); printDetails(arr, size, res_name); } system("pause"); }  Sample output (just for the first iteration): Enter a restaurant: Chili's Food: American Avg. Meal:$  10.95 Rating: 2  Food: American Avg. Meal: $4.50 Rating: 1 Food: Breakfast Avg. Meal:$  9.50 Rating: 1  Food: AmericanizedItalian Avg. Meal: $11.00 Rating: 1 Food: American Avg. Meal:$  6.75 Rating: 2 

## [ Politics ] Open Question : What is your opinion about the Notre Dame fire? Is it just a coincidence it occurred on Easter week? Is there a coverup underfoot?

It is known that Muslims destroy the holy sites of others. https://www.youtube.com/watch?v=VbnPT3jeWL4

## A problem about the field of rational functions over finite field

Let $$p$$ be a prime number, and let $$F_{p}$$ be the finite field with $$p$$ elements. Let $$F=F_{p}(t)$$ be the field of rational functions over $$F_{p}$$ . Consider all subfields of $$F$$ such that $$F/C$$ is an finite Galois extension. 1.Show that among such subfields, there is a smallest one $$C_{0}$$ , i.e.$$C_{0}$$ is contained in any other $$C$$. 2.What is the degree of $$F/C_{0}$$?

## Is 2(2^(p) − 1) a divisor of n? How about 2^2(2^(p − 1))? Finish the proof.

A positive integer is called whole if it equals the sum of its positive divisors.

Example: 6 is whole because the divisors of 6are 1,2 and 3 and 6=1+2+3.

Show that 28 is whole.

We want to show that the number n = 2^(p−1)(2^(p) − 1) is a whole number when 2^(p) − 1 is prime.

What are the divisors of 2^(p−1)? (it might help to try various values: what are the divisors of 2^3 or 2^4…?)

What is their sum? Hint: 1+2+2^2 +…2^k =2^(k+1) −1(geometric series)

Is 2(2p − 1) a divisor of n? How about 22(2p − 1)? Finish the proof.