Mathematics

5:00 AM

@EdwardEvans I had a question about programs in Germany. I was just looking at TUM's website with regards to Masters in Statistics. Are the two intake periods for new students the Winter and Summer semesters in Germany? They don't do fall in Germany?

Edward Evans

Usually just winter and summer semester

In my university the winter semester starts in October and summer semester in April :)

dc3rd

Ok. Something to be aware of then........wait what?.....lol

Edward Evans

whut

dc3rd

That's kind of how I envisioned it.

Edward Evans

TUM might be different because it's in Bavaria but afaik it's the same everywhere, best to just look it up on their website

dc3rd

5:02 AM

I'm looking it up again. I had it open

Edward Evans

Okie dokie

Leonhard Euler

I bethink not yond the univ'rse is a fractal

dc3rd

SO this means that You would start in October of 2022?

@EdwardEvans

StudySmarterNotHarder

@LeonhardEuler it appears at small scales like it's not, but it's so grandiose and huge that we can't make out its structure from 4D spacetime

Edward Evans

@dc3rd if you apply now for the winter semester you'll start in October 2021. The winter semester will usually be called smth like WiSe 21/22.

Or whenever their semester dates are

Leonhard Euler

5:07 AM

@StudySmarterNotHarder yea

Edward Evans

On their site it says winter semester 21/22 starts on 01.10.2021

dc3rd

Ok....that's what I wanted clarified. I'm not thinking about it until next year, but I just wanted to get my bearings straight

StudySmarterNotHarder

@LeonhardEuler what if you took $\lim_{n \to \infty}\sum_{i = 1}^{\pi(n)} \dfrac{p_{i+1} - p_i}{n^2}$

Edward Evans

Okay nice. Then winter semester starts around the end of September/start of October

dc3rd

Got it and then if I were to apply for summer I would start summer of 2022. Now I got an idea of how to plan out my application for the following year.

Thanks for the clarification.

Edward Evans

5:10 AM

No worries, make sure you get all of your application documents PERFECT. The Germans are ridiculous bureaucrats.

dc3rd

I'll definitely have more questions closer to time. Just getting the ducks lined up from now.

Leonhard Euler

@StudySmarterNotHarder I'm not on a computer right now

Edward Evans

No worries, lemme know

Leonhard Euler

Lemme switch it on

StudySmarterNotHarder

@LeonhardEuler I think the sum just goes to $(p_{\pi(n)} - p_1)/n^2$, so it's probably a divergent series

But if there are are only $c_k$ consectutive prime pairs differing by $k$, then there is finite summation for (all of) those pairs, then oops a giant leap into $c_k = \infty$.

After you group terms

What about taking all differences $p_i - p_j, i,j \leq \pi(n)$

Then what denominator $n^k$ would you need to make the series converge instead

Define $k \gt 1$ to be the least real number such that the series converges

Actually you'd have to square the gaps $p_i - p_j$ other wise they all would sum to zero

Alessandro Codenotti

7:15 AM

@StudySmarterNotHarder Departure Songs - We Lost The Sea

StudySmarterNotHarder

@AlessandroCodenotti thx

Andrew Micallef

Neat, although $75 seams like a lot, I hope I didn't somehow manipulate you into purchasing another book that won't get read :P
(I'm just reading through the responses from @leslietownes and @TedShifrin re my question earlier, I appreciate the pointers!)
For the moment I feel like my maths is the weakest link in my understanding, hence asking here not physics :P

StudySmarterNotHarder

Nice

Andrew Micallef

also, my copy of griffiths came without the front and back covers so I have no idea what the integrals are in the tables!

StudySmarterNotHarder

Yeah, it's fun to ask never-before asked questions or at least not posted yet

and since you purchased it, it would be ethical

Um.. yeah, I removed that

Andrew Micallef

7:25 AM

bad boy

StudySmarterNotHarder

don't want to get banned

Andrew Micallef

better

:P

StudySmarterNotHarder

There's nothing like a hardback book copy, it seems that flipping through pages though primitive is more efficient than adobe reader

I wanted to get the purple 2nd edition but I clicked the link on amazon and the book turns blue

I thought purple was an unusual math book color

The third edition has some chapters deleted and some chapters added

Andrew Micallef

@leslietownes I realise now on reflection that my strategy may not be particularly profitable in simplyfying the question, but I think my goal is to get to an understanding by following as many blind paths as possible, not efficiently getting the right answer

StudySmarterNotHarder

That's how you learn

And possibly one day stumble upon a fruitful path that no one has taken

Andrew Micallef

7:28 AM

I've got a digital edition of I think the latest one, borrowing my partners library card (so dunno how ethical that is all things considered) but the html version has no integral tables :(

Yeah, maybe

Makes me think of the knowledge mines in Greg Egans novel "Diaspora"

I'm a huge fan of hardback textbooks too, but they are a bit too a) heavy and b) conspicous for everyday consumption

StudySmarterNotHarder

If $0 \leq f(n) \lt 1$ does $\sum_{n = 1}^{\infty} f(n)^{n}$ converge?

@AndrewMicallef they're nice for sitting on the couch with

Andrew Micallef

Gotta go agian now, have ddinner to cook

thanks again though :p

StudySmarterNotHarder

great. Enjoy your dinner!

Studying math is all about taking good breaks

robjohn

@StudySmarterNotHarder Not necessarily. If $f(n)=\left(1-\frac1n\right)$, then the terms tend to $1/e$, not to $0$.

StudySmarterNotHarder

@robjohn that is okay

with me

:)

I let the numbers do what they want and just observe

robjohn

7:42 AM

Just answering the question. The answer was "no".

StudySmarterNotHarder

@robjohn does $$\sum_{n = 1}^{\infty} \dfrac{ p_{n+1} - p_n}{n^s}$$ converge for all $\text{Re}(s) \gt 1$ or something like that?

robjohn

$p_n$ is the $n^\text{th}$ prime?

StudySmarterNotHarder

Yes

I think we could even include $(p_{n+1} - p_n)^s$ into the power

robjohn

@StudySmarterNotHarder I believe so.

Though prime gaps are tricky.

StudySmarterNotHarder

Dirichlet series

In mathematics, a Dirichlet series is any series of the form ∑ n = 1 ∞ a n n s , {\displaystyle \sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}},} where s is complex, and a n...

According to that subsection of the article we can safely take $\text{Re}(s) \gt 2$ since $p_{n+1} - p_n = O(n)$

and it will converge absolutely there

robjohn

7:50 AM

@StudySmarterNotHarder sure, for $\mathrm{Re}(s)\gt2$

StudySmarterNotHarder

It's real part

so $s$ can be in the complex numbers

robjohn

In Dirichlet series, imaginary parts usually aid convergence.

$\sum\limits_{n=1}^\infty\frac1{n^{1+i}}$ converges conditionally.

StudySmarterNotHarder

So collect terms and write: $$\text{the sum above} = \sum_{ k = 1}^{\infty} k\sum_{n \in \Bbb{N} \\ p_{n+1} - p_n = k} \dfrac{1}{n^s}$$

I honestly don't know where I'm going with this :)

Just exploring, new to analysis

robjohn

8:06 AM

primes are not an easy place to start, but if you can get a feeling for them, it might be good

porridgemathematics

good day, I am seeking verification of the following :

Let $X$ be a CW complex and $Y \subset X$ a subcomplex such that all cells of $X \setminus Y$ have dimension at least $n$, where $n \geq 2$. Then $X^{n-1} \subset Y$ and $X^{n-1} = Y^{n-1}$.

frankly i also don't see why $n \geq 2 $ is necessary

username

8:36 AM

Suppose I were to start a mathematics related project, some sort of rating of professional mathematicians in departments around the world. Suppose rhat I was starting to program it on github, built a proof of concept, and then felt it was time to ask for some help and gauge the level of interest in such a thing.
It would be a very controversial thing to do, as it could be read to mean this mathematician is better than that mathematician, which is a very big no no,
By nature, it would not ask said mathematicians' opinion, or their department's submission, like funding agencies. It would ju…

Anyhow, suppose I did it anyway. Would this chat room be the right place to recruit volunteers to help me make it better?

Prithu biswas

9:03 AM

I seached for 'what is a tensor' in the internet and found this interesting answer from William Oliver in Quora (it is the first answer). Can someone take a look at it?
https://www.quora.com/What-is-a-tensor

I found it interesting because this answer goes "kind of" contrary to almost all 'introduction to tensors' type videos in the internet.

Astyx

9:43 AM

@ShaVuklia @TedShifrin Sorry I vanished yesterday, I saw Ted got involved and had some business to attend to. Is everything settled?

porridgemathematics

ooo I have a question if you have some time

Astyx

idk CW complexes sorry

porridgemathematics

ok np

StudySmarterNotHarder

10:09 AM

Say you have $f(s) =\sum_{n = 1}^{\infty} \dfrac{\liminf_{k \geq n} (p_{k+1} - p_k)}{p_k^s}$

Can we somehow bring this liminf outside the summation?

Btw, by Zhang's result, $f(s) = C + Z(P(s) - \sum_{k=1}^{Z - 1} \dfrac{1}{p_k^s})$ for some $Z \lt $ 70 million where $P(s)$ is prime zeta

I mean it should be $f(s) = \sum\limits_{n=1}^{\infty} \dfrac{\inf_{k \geq n}(p_{k+1} - p_k)}{p_n^s}$ i.e.

The twin prime conjecture is that $f(s) = 2P(s)$ for all $s$ I guess

$\text{Re}(s) \gt 1$

robjohn

@StudySmarterNotHarder assuming the twin prime conjecture, your liminf would be $2$.

StudySmarterNotHarder

Yes that is correct

Instead it's $Z$ by Zhang

so far

robjohn

@StudySmarterNotHarder doh!

porridgemathematics

10:36 AM

the suspension of a space minus one of the tips of the cones is contractible, right?

70 million prime conjecture

Andrew Micallef

10:57 AM

Okay, after following this path through I can conclude that this is exactly one of those cases. By integrating with respect to $x$ I wind up injecting a second derivative with respect to $t$, which in this particular case is much less good to work with.

But this dead end has given me an idea for where the profitable path may lie. which will have to wait till my lunchbreak to explore.

satan 29

hi'

Andrew Micallef

hi

@TedShifrin Just as I am about to go to bed I realise I am denser than a lead brick. The freaking problem was of the form $\int{f(x,y) \cdot g'(x,y)}dx$...I can use integration by parts to solve it!

satan 29

11:12 AM

some test examples were:

(a,b): (f/inf)

10,10 inf

1,10 f

6,9 inf

7,3 f

i tried to come up with a solution by purely looking at these examples, and it worked lmao

the ans is inf if gcd(a,b) >1, and f if gcd(a,b) =1

now, I tried to come up with a rationale for this

after messing around for a bit, it is clear that, for a no. to be coloured white, it must be of the form

$$ax+by$$

now, if gcd(a,b) is not 1, we can come up with infinite numbers such that they are nonzero mod( (gcd(a,b)))

which means there will be an inf no. of (non white), i.e black squares.

However, I am not sure if we can simply use this argument to conclude that if gcd(a,b) is 1, then we are guaranteed to have finite solutions?

Rover

Hey Hi Guys Can anyone please tell what really is Maths? What significance it has of it's own and is it just about numbers and no visualisation of what that equation is really signifying ?

Ashish Ahuja

11:40 AM

@satan29 codeforces.com/blog/entry/71080

satan 29

@AshishAhuja oh I see

Ashish Ahuja

@Rover maths does have a lot of visualization, I personally like 3b1b videos; they show plenty of visualization, and I really liked the linear algebra series since it showed the visual side which usually isn't introduced in high school.

satan 29

@AshishAhuja wait

oh no nvm

Prithu biswas

12:00 PM

(commenting again because it got behind a lot of comments) I seached for 'what is a tensor' in the internet and found this interesting answer from William Oliver in Quora (it is the first answer). Can someone take a look at it?
https://www.quora.com/What-is-a-tensor
I found it interesting because this answer goes "kind of" contrary to 'almost' all 'introduction to tensors' type videos in the internet.

geocalc33

12:18 PM

Hello

Rover

@AshishAhuja Thanks the series are cool ! Focuses on visualisation!

Srijan M.T

1:00 PM

@AshishAhuja Amazing bro.

qwerty uiop

1:26 PM

hi

hello

Sha Vuklia

@Astyx No worries! I still have to figure sth out, but for now I'll leave it as it is

Alessandro Codenotti

@EdwardEvans can you ping me when you're around if you want to help me sort out an ANT problem?

geocalc33

1:41 PM

$X_2=\big\{(x,y)\mapsto\big(ax,\frac{y}{a^2}\big)\big\}.$ Does this transformation have a name?

copper.hat

2:01 PM

jimmy?

Srijan M.T

When finding trigonometric ratios for 90+theta. Why don’t we make the diagram like this ? I see we are not getting an angle of 90 degree. So , can’t we say there is no values of trigonometric ratio possible for 90+theta.[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/Lm9O9.jpg

https://www.math-only-math.com/trigonometrical-ratios-of-90-degree-plus-theta.html. So , on this site I checked the proof for this. Below is the passage needed to understand this proof. What I didn’t get is how is sin theta = FE/OE

Rover

2:17 PM

The two methods of same integration and I think both are correct then also I am getting different answers !!

How's that possible?

@robjohn @copper.hat

Lucas Henrique

2:31 PM

either they agree everywhere or you made a mistake.

or ZFC is inconsistent but I do doubt that

:p

robjohn

@Rover They are the same answer.

$\frac{\frac{2(x-1)}3}{2\left(\left(\frac{x-1}3\right)^2+1\right)}\cdot\frac{9/2}{9/2}=\frac{3(x-1)}{(x-1)^2+9}$

leslie townes

one of the most accursed lessons of integral calculus is how many different ways there can be of writing the same thing.

rational functions, expressions with trig functions in them, on and on. it's maddening.

Lucas Henrique

and that can be quite interesting. like proving that $\frac{\pi^2}{6} = \sum 1/n^2$

robjohn

The most commonly worrisome "different" answers are $\arcsin(x)+C$ and $-\arccos(x)+C$

Lucas Henrique

ok, I probably need help. my book says that the derivative $f^(k)$ of a function $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$ can be inductively defined as a $k$-linear form

robjohn

2:40 PM

Is $f^{(k)}$ the $k^\text{th}$ derivative?

or the coordinate $k$?

Lucas Henrique

for $k = 2$, I did the math and it's ok. how about the general case? how do you apply the canonical isomorphism $\mathrm{Hom}(\mathbb{R}^m, \mathrm{Hom}^k(\mathbb{R}^m, \mathbb{R}^n)) \cong \mathrm{Hom}^{k+1}(\mathbb{R}^m, \mathbb{R}^n)$?

robjohn

As soon as you wrote Hom, my eyes glazed over.

Lucas Henrique

lol

@robjohn: it's the $k$-th derivative

ok, so $f^{(k)}(x) = \sum\limits_{i=1}^m \underbrace{\frac{\partial f^{(k-1)}}{\partial x_i}(x)}_{k-1 \text{ linear}} \overbrace{dx^i}^\text{linear}$

robjohn

That looks like the chain rule for one derivative.

Oh, I see. You are doing that one derivative at a time

Lucas Henrique

my book talks about derivatives in any finite-dimensional normed spaces

the norm doesn't matter since they're all equivalent

Thorgott

2:55 PM

should just be $\sum_{i_1<\dots<i_k}\frac{\partial^kf}{\partial_1^{i_1}\cdot\dots\cdot\partial_k^{i_k}}dx^{i_1}\wedge\dots\wedge dx^{i_k}$, methinks

Lucas Henrique

so the "actual" definition of $f^{(k)}$ is $(f^{(k-1)})'$ which belongs to the functions from U to Hom(R^m, Hom(...))

however, the author says that you can inductively define $f^{(k)}$ as a function from $U$ to the $k$-linear forms

my question is: how?

(this definition of derivative makes sense since $\mathrm{Hom}(E, E')$ is also a normed space when $E, E'$ are normed spaces)

Lucas Henrique

3:08 PM

ok, so if you define $df$ as the (unique) linear transformation s.t. (blablabla...), then you can define recursively $f^{(k)}(x_1,\ldots,x_k) := d(f^{(k-1)})(x_k)(x_1,\ldots,x_{k-1})$

copper.hat

even at the simplest level (Boolean functions) its a big deal in (digital) circuit design to check if two functions (finite automata really) are the same.

Edward Evans

3:30 PM

@Alessandro hey I just woke up lol, ask the Q and I shall think about it over a guilt-coffee

Leonhard Euler

3:49 PM

whom'st'd've'dist'd'n't'st'd've'll's'd've're'n't'y'all'll'ven't is the father of topology?

Alessandro Codenotti

4:12 PM

@Edward Still here? I was at a seminar

Edward Evans

Yeah I'm here, I can't think about it in real time though, I'm still trying to work through probability lol

Alessandro Codenotti

No problem, I need some setup anyway, let me write it down

I have $K$ a number field with its ring of integers $O_K$ and for $P\in\mathrm{Spec}(O_K)$ I defined the valuation $v_P\colon K\to\Bbb Z\cup\{\infty\}$ in the usual way. Now I fix $Q\subseteq\mathrm{Spec}(O_K)$ cofinite and $\Sigma=\mathrm{Spec}(O_K)\setminus Q$. I consider the ring $A_Q=\{x\in K\mid v_P(x)\geq 0\,\forall P\in A\}$ and I want to show that its group of units $U_\Sigma$ is finitely generated and that $U_\Sigma/O^\times_K\cong \Bbb Z^{|\Sigma|}$

I showed that $x\in U_\Sigma$ iff $xO_K$ factors as a product of stuff in $\Sigma$

So now the natural map $\varphi\colon U_\Sigma\to\Bbb Z^{|\Sigma|}$ is to take $x\in U_\Sigma$ and send it to $(v_{P_1}(x),\ldots,v_{P_{|\Sigma|}}(x))$, where the $P_i$'s enumerate $\Sigma$

This is a group hom with kernel $O^\times_K$, so I get an iso $U_\Sigma/O^\times_K\to\varphi(U_\Sigma)$

Unfortunately $\varphi$ doesn't look surjective at all to me in general

But I still want to argue that it's image is a maximal rank subgroup of $\Bbb Z^{|\Sigma|}$ which I don't see how to do

@AlessandroCodenotti woops in the definition of $A_Q$ I meant $\forall P\in Q$ rather than $\forall P\in A$.

Transcript for

$Mathematics$ Mathematics