Mathematics - 2020-10-08

user193319

01:43

0

Q: Closed/Exact Forms and Complex Functions

If $f : D \to \Bbb{C}$ is a function, where $D$ is simply connected, what does it mean to say that $f(z)dz$ is either closed or exact? I know what it means for $Pdx + Qdy$ to be exact or closed when everything is real, but what happens when you move over to $\Bbb{C}$?

feynhat

03:03

@BalarkaSen Insert two-spidermen-pointing-at-each-other meme.

Also, that 'joke' is terrible.

Sayan Chattopadhyay

03:17

Is the action of Z_n defined on S^3 given by $\psi_k : Z_n \times S^3 \to S^3$ , $\psi_k(a, (z_1,z_2)) = (az_1, a^kz_2)$ where $k$ ranges from 1 to n-1 free?

Where by $Z_n$, I mean the nth roots of unity.

feynhat

(0, 1) is a fixed point, no?

Sayan Chattopadhyay

Ah yes and this won't be a fixed point when n is prime

feynhat

Yeah. I guess, more generally, when k and n are coprimes.

Sayan Chattopadhyay

Hmm yeah. What is the orbit space of this action?

Say for n prime

feynhat

03:34

idk man... i guess some quotient of lens space

geocalc33

04:04

I don't understand the Mellin transform that well. Once you take the Mellin transform of a function f(x) what information do you get from it?

geocalc33

04:50

0

Q: Mellin transform yields Bessel function

Consider the Mellin transform on bounded support $(0,1).$ I computed the following: $$\mathscr M[f;s]=F(s)=\int_0^1 x^{s-1}e^{\frac{1}{\log(x)}}~dx=\frac{2K_1(2\sqrt{s})}{\sqrt{s}} $$ Where $K_1$ is the modified Bessel function of the second kind. Is that the correct result? I'm not sure I got i...

solution-verification transformation bessel-functions integral-transforms mellin-transform

geocalc33

05:05

How do you construct a metric in which consecutive primes have gaps of 1 unit?

Akiva Weinberger

What do you mean by metric?

You mean you want $d(p_n,p_{n+1})=1$?

You could use the discrete metric, in which the distance between any pair of numbers is 1, I guess

geocalc33

Yeah that ^^

that's a good point about the discrete metric...

Ted Shifrin

Howdy, DogAteMy.

geocalc33

I thought about the primes as degenerate natural numbers but apparently that's mathematically incorrect

so I stopped

AMDG

05:47

You know, I've heard of dogs eating homework, but haven't you heard of homework eating dogs? I wonder why that is...

This is just one of many of my deep contemplations.

geocalc33

I have asked 283 questions in my life

32 percent of them are 24 percent meaningful

AMDG

@geocalc33 Is this going to be on the test?

geocalc33

10 percent of this material will be on 29 percent of the next 56 percent of the last 33 percent of the chapters

AMDG

ya lost me at chapters

geocalc33

reminds me of an actuarial exam question

AMDG

05:59

Ah, yes, exams. Takes me back to when I was younger and had exams... in high school... about six years ago... now I'm old and 20. Time flies.

You know, with all this finding circular functions stuff, I just feel like I've been going in circles...

Also, the implicit equation of a circle is rather square for something so round.

AMDG

06:18

Is reciprocal sum the correct term for something like $\frac{1}{x+a}$ for real values of x and a?

I'm thinking I would like to find various simple but non-trivial identities for reciprocals of linear expressions given by $\frac{1}{a_{1}x_{1} + ... + a_{n}x_{n}}$ so that I can then find something simple enough to integrate with $\ln(x)$ that is real-valued and is an identity of $\arctan(x)$.

I might have to add or modify some axioms to make it work, but I mean... it's not like that hasn't happened before cough $\frac{i}{2} \ln(\frac{i+x}{i-x})$ cough

jeea

07:34

Hello is there a general method for doing the following: suppose that I have 3 linearly independant vectors in 5 dimensional space, is there a quick way to find remaining 2 independant vectors

feynhat

@Astyx mfw

3

AMDG

08:10

I've set some limitations to help me design this system of mine: 1. Everything is a norm. ($|x|$). 2. Everything is a wave. 3. What's a circular function? 4. All operations isomorphic (but not strictly commutative or associative). 5. What's Calculus of Infinitesimals?

athos

May I ask a quick question, for 3 Brownian motion, if the correlation between $W_1$ and $W_2$ is $\rho_{12}$, the correlation between $W_1$ and $W_3$ is $\rho_{13}$, what is the correlation between $W_2$ and $W_3$ ($\rho_{23}$)?

AMDG

Now how in the world do I get a function like odd(x) or even(x) that are equal to x & 1?

Balarka Sen

@feynhat HAHAH

@MikeMiller Ah shucks. Thanks though. We can try to read, yeah. I found this, seems readable.

athos

is $\rho_{23}$ determined by $\rho_{12}$ and $\rho_{13}$ or all three correlations are independent?

Balarka Sen

08:28

@athos You can get inequalities between $\rho_{ij}$ even just for three random variables of standard deviation 1 let's say, no need to go to Brownian motions. The variance-covariance matrix of $(X, Y, Z)$ is positive-definitive

Taking the determinant gives you an inequality

$\rho_{23}$ won't be determined by $\rho_{12}$ and $\rho_{13}$ but you can write an inequality.

Edward Evans

wagwan

feynhat

08:47

@BalarkaSen Which wacom do you have?

Balarka Sen

09:50

@feynhat One

once you get it install the driver and use Bamboo paper to write on

feynhat

eeh... I don't use windows so it might not be that easy for me.

Balarka Sen

Oh I see. I'm sure there must be some equivalent

feynhat

Yeah. There is.

Astyx

There's some software you need to install if you're on mac

athos

@BalarkaSen thank you

feynhat

09:53

nah... I don't use mac either.

Balarka Sen

feynhat only uses TempleOS

feynhat

built by the smartest programmer in the galaxy (RIP)

Astyx

I don't have an OS, I live in harmony with the computer

Balarka Sen

The smartest programmer who has ever lived

►

feynhat

what a legend

Balarka Sen

09:56

godsent

at least thats what he said

porridgemathematics

hello, does anyone know why the lower bound provided in the last paragraph of the image in this post math.stackexchange.com/questions/3856343/… holds?

im not entirely convinced it is true, (although it is not necessary as far as I can see since the cantor space is compact and the target space of this embedding is hausdorff, so just showing it is a continuous injection should be enough)

uh that is , why the lower bound in (15.38) holds

Astyx

10:42

By definition of $r_n$ I think

porridgemathematics

but since those points are inside the same ball, it may have minimum radius r_n, then couldn't those points be smaller than a minimum radius r_n apart?

they seem to be implying that the shortest distance between the two disjoint (n+1) balls inside the n ball is >= r_n, but i dont see why this needs to be true

maybe im missing something obvious

Astyx

Let me think for one sec

Alessandro Codenotti

11:21

@porridgemathematics I don't have the time to check the details in your questions, but there are easier ways to prove that the Cantor space embeds into uncountable Polish spaces

porridgemathematics

@AlessandroCodenotti oh, i've already proved that, its just a particular detail in the given proof that I don't understand, the detail itself is not needed to prove that the embedding exists

Alessandro Codenotti

(also you're right that getting a continuous injection is enough by compactness)

porridgemathematics

the details are just to construct a sequence of nbhoods, $U_{\sigma}$, for each$ \sigma$ a finite (or empty) sequence in ${0,1}^{\infinity}$, such that for each $\sigma$ we have$ \overline{U_{\sigma0}} \subset U_{\sigma} $and the same for $\overline{U_{\sigma1}}$ and the two 'split' sub neighbourhoods are disjoint , with diameter going to 0

any construction that does the above will provide an embedding

Alessandro Codenotti

That's sometimes called a Cantor scheme on $X$ by the way, your book doesn't seem to mention this

Mike Miller

@AlessandroCodenotti Never heard of this

porridgemathematics

11:26

but the proof then , rather than stop at showing that this provides a continuous injection (and thus by what you just said about compact -> hausdorff injections being homeomorphisms being enough), tries to show the continuity of the inverse directly @AlessandroCodenotti

its this bit that is what confuses me

ah yes, i've seen that in 'descriptive set theory' some springer verlag book

Alessandro Codenotti

@MikeMiller Lusin schemes (indexed over $\Bbb N^{<\Bbb N}$) is more common as a term

@porridgemathematics Classical Descriptive Set Theory by Kechris I guess

porridgemathematics

yup, that's the one

and yes this proof makes no mention of this general scheme (in fact this is a proof contained in a note where the entire note is dedicated to proving kuratowksi's theorem on borel isomorphism of standard borel spaces)

this fact (that every uncountable polish space contains a homeomorph of cantor space) is the last piece of the puzzle in the note

Astyx

It's worth noting $2r_{n+1}\le r_n$ I think

porridgemathematics

ah yes i tried to use that somehow

but i didn't get anywhere with it

Alessandro Codenotti

@porridgemathematics Ah right, because you want to use the Cantor-Schroeder-Bernstein theorem for standard Borel spaces

porridgemathematics

11:32

exactly @AlessandroCodenotti

Alessandro Codenotti

Yeah I remember now, I looked at this proof some time ago

Astyx

Oh but that can't work

porridgemathematics

@Astyx yeah because at best that will give an upper bound I think? we want a lower bound anyway

Astyx

Yeah

geocalc33

12:15

hi

Astyx

12:32

@porridgemathematics I think the lower bound is just wrong. What's true is that $\rho(y_{\sigma}, y_{\tilde\sigma})>0$ because they are in two disjoint subsets, and that's enough to conclude that $\sigma\mapsto y_\sigma$ is one-to-one

Mike Miller

@SayanChattopadhyay Unless I misunderstand certainly yes. Suppose $a \neq 1$. If $az_1 = z_1$ then $(a-1)z_1 = 0$, so that $z_1 = 0$. Because $a$ is an $n$th root of unity and $1 \leq k \leq n-1$ we also have $a^k \neq 1$.

So from the second term we get $(a^k - 1) z_2 = 0$ so that $z_2 = 0$.

So that the only fixed point of this action on $\Bbb C^2$ is (0,0)

When $a = e^{2\pi i/n}$ this quotient is literally $L(n, k)$.

When $a = e^{2\pi i m/n}$ then let $mq = 1 \pmod n$ for some $q$. Taking everything to the $q$th power there is a diffeomorphism taking your action to $\psi_{qk}$, so that your quotient is the lens space $L(n, qk) = L(n, m^{-1} k)$

Oh nevermind.

I misunderstood.

porridgemathematics

@Astyx sure, the lower bound isn't to conclude injectivity though, but continuity of the inverse

Not A Zoomed Image

12:51

What is the name of the theorem "$ ab =0$ if only if $a=0$ or $b=0$"? Who invented this?

porridgemathematics

this is not a theorem but a property of integral domains, which are rings that have no non-trivial zero divisors

that doesn't hold in all algebraic structures, for instance 3*4 = 0 mod 12, but neither 3 nor 4 are 0 mod 12

Mike Miller

I suspect the Greeks would just tell you to draw a rectangle

Astyx

Your question is bugging me so much lol

Mike Miller

By any sensible notion of area, the area of a rectangle (nondegenerate, so positive side lengths) is positive

porridgemathematics

@Astyx some more thoughts on this, I can see that in a 'nice' metric space , two disjoint closed balls will have a positive smallest distance , i.e. $\inf \{d(x,y): x \in B_1, y \in B_2 \}$ between them, where $B_1$ and $B_2$ are closed balls, so if polish spaces have this property for disjoint closed balls, then this should be sufficient to prove the continuity of the inverse by a similar means

just choose $y_{\sigma}$ and $y_{\tilde{\sigma}}$ to be less than the smallest positive distance between all pairs of closed balls up to some level n, and then $\sigma$ and $\tilde{\sigma}$ would need to agree on the first $n$ digits

Astyx

13:04

The balls are open here

porridgemathematics

yes but you can modify them so that they are small enough s.t. the closed versions of them are disjoint

by that i don't mean the closure but the ball inclusive of its natural boundary

the thing is i don't know how to prove that if two closed balls are disjoint, that they are a positive distance apart

i can prove it in spaces with heine borel

Astyx

"The completeness of $X$ ensures that the intersection $\bigcap_{n>1} B_{(\sigma_1,\dots,\sigma_n)}$ is non-tempy and contains exactly one point $y_\sigma$." Isn't that true only if you take closed balls by the way ?

porridgemathematics

right, or if you use the modification I propose in my post

Astyx

If I take $B_i = (1-1/n,1)$ it doesn't work

Oh yeah I hadn't read that

porridgemathematics

i.e. define $i(n) = \inf \{ m > n : \overline{B(x_m,2^{-m})} \subset B(x_n,2^{-n}) \setminus B(x_n,2^{-m}) \}$

so now you have the closure of the ball as a subset of the annulus

and similarly for $k(n)$

then you have that $\overline{B_{\sigma0}} \subset B_{\sigma}$ and the same for $B_{\sigma1}$

so that intersection is really the intersection of the closures of those balls, which is nonempty and a singleton

Astyx

13:11

yup

Anway this proof is way too elliptical on this step for me

porridgemathematics

as in the definitions of $i(n)$ and $k(n)$?

you could just see them as a way to get that binary tree structure

they aren't used for anything more than that

Astyx

No, as in "eq 15.38 lets us conclude that $r_n < \rho (y_1,y_2)$"

porridgemathematics

ah of course

Astyx

The rest I understood

porridgemathematics

yeah I totally agree

definitely kinda handwavey there

considering a lot of explanation went into the previous stuff, and then suddenly no explanation at all

Astyx

13:14

yup

If this is true (and I'm not even sure it is) then I suspect it has something to do with the order of the $x_k$

holahola

Hi !

I have a quick question

what would be the divergence of the norm of a vector?

(assuming it has a quick answer)

Astyx

What does that mean ?

holahola

e.g. $\vec{\nabla} \cdot \frac{1}{\|\vec{x}\|}$

am I explaining myself?

Astyx

Well what you call the norm of a vector is a function $E\to\Bbb R$ where $E$ is your vector space

So its divergence is the divergence applied to this particular function

holahola

$\vec{\nabla} \cdot \frac{\vec{y}}{\|\vec{x}\|}$ that's better I guess... where $\vec{y}$ doesn't deppend on $\vec{\nabla}$

but... shouldn't it be 0 or something?

Astyx

13:21

Divergence

In vector calculus, divergence is a vector operator that operates on a vector field, producing a scalar field giving the quantity of the vector field's source at each point. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. As an example, consider air as it is heated or cooled. The velocity of the air at each point defines a vector field. While air is heated in a region, it expands in all directions, and thus the velocity field points outward from that region. The divergence of the velocity field...

Why would it be zero ?

holahola

I see... it has no sense asking that haha. I'm confused about the dependence being only in the modulus and not a vector.

Do you know if there is any identity related to grad(div(f)) ?

my question is related to: math.stackexchange.com/questions/3856562/…

Astyx

I don't think so

holahola

ok... I'll keep thinking on it

thanks for your help Astyx!

geocalc33

13:37

0

Q: Conjugating summands with diffeomorphism and looking at analytic continuation

Consider a function $f_n(x)=x^n$ and the summatory function $\sum_{n=0}^\infty f_n(x).$ Now note that $f_n(x)$ is diffeomorphic to $g_n(x)=e^{(\log(x))^n}.$ The explicit diffeomorphism is given by: $\psi:\Bbb R^2\to \Bbb R^2$ with $\psi(x,y)=(e^x,e^y).$ Now place the summation back into the mix. ...

sequences-and-series complex-analysis analysis analytic-continuation diffeomorphism

any nice examples?

AMDG

14:03

120 messages :O

Good morning

AMDG

14:23

So... is there an arithmetic way to describe using modulus and things like floor and ceil to somehow get me even and odd number generator waves? I just need something that will graph in desmos so that I can actually work with existing systems (and port to and from). These two (square) wave functions are fundamental to the whole thing because I can make any number and function from just modifying and/or combining even and odd.

I suppose I could just make my own graphing utility, but... I mean, this would be convenient for obvious reasons as making my own at this time would be time consuming.

It's very simple: given a real value of x, let x be +1 if it matches the chosen polarity, 0 otherwise, so in this case, the even function would be one for all even real values of x, and odd would be one for all odd real values of x.

Then $even(x) + odd(x) = 1$

It has the ever so convenient property that $(f(x))even(x) + (f(x))odd(x) = f(x)$ .

That's pretty nice if you ask me.

Pythagorean identity: $\cos^2(x)even(x) + \cos^2(x)odd(x) + \sin^{2}(x)even(x) + \sin^{2}odd(x) = 1$

feynhat

15:55

@MikeMiller Why is $a^k \ne 1$? Take $-1$ for example, it is 4th root of 1, but certainly $(-1)^2 = 1$.

Ted Shifrin

@feynhat Maybe I'm missing your point (and I only quickly looked back to see what this is about), but $a\ne 1$, so the action is not trivial anyhow.

feynhat

Sayan's question was whether the action of $\Bbb Z_n$ on $S^3$ given by $a \cdot (z_1, z_2) = (az_1, a^kz_2)$ is free ($a\in \Bbb Z_n$ and $(z_1, z_2)\in S^3$).

This is not true. For example take $n=4$ and $k=2$, then, $(0, 1)$ is fixed by $-1$.

(obviously, here Z_4 is {1, -1, i, -i})

Sayan Chattopadhyay

16:37

Yeah, what Mike does is only true if you take your $n$ to be a prime number. Then you have $a^k \neq 1$

AMDG

I still don't get the logic for why 1 is "no longer considered" prime.

I can also put as many zeroes as I want in front of something in a sum... wasn't even prime to begin with but I mean you know... similar logic here...

What is with math and all of these exceptions?

If math is so powerful, why does it have exceptions? Why are there things that are undefined or indeterminate? And why is research mathematics apparently guess-and-check?

I guess I can summarize all those questions as "Why isn't mathematics more intuitive?"

At least, formal mathematics I find to be not so intuitive compared to just counting quantities, and definitely not as simple as it could be.

feynhat

@SayanChattopadhyay Btw, there is similarly an action of $S^1$ on $S^3$ given by $a\cdot (z_1, z_2) = (a^{k_1}z_1, a^{k_2}z_2)$ called the weighted Hopf action. Putting $k_1=1$, $k_2 = 1$ you recover your usual Hopf action. The resulting quotient space is denote by $\mathbb{WP}(k_1, k_2)$, the weighted projective space.

The space $\mathbb{WP}(1, k)$ for any $k>1$, has a natural orbifold structure, with underlying space $S^2$ and one singularity of order $k$. Thurston calls this a teardrop.

Sayan Chattopadhyay

Oh neat! My initial guess was that the orbit space (in my case) would have something to do with the Hopf action.

feynhat

$\mathbb{WP}(k_1, k_2)$ for both $k_1$, $k_2$ greater than $1$ is called a football (because apparently that's what 'footballs' look like in US).

Sayan Chattopadhyay

16:54

Is this mentioned in Thurston's Topology of 3-Manifolds book? This teardrop thing?

feynhat

Yes. Chapter 13, I think.

He doesn't mention weighted projective spaces though.

I read those in context of Lie groupoids. See 'Orbifolds and Stringy Topology' by Adem, Ruan and a third author whose name I forgot.

Sayan Chattopadhyay

Ah I see. I will have a look at these, thanks!

feynhat

Oh man... ever since I returned from taking GRE, I have been experiencing The Symptoms™.

I am actually scared now.

Sayan Chattopadhyay

Damn, they can be psychosomatic. I have extreme health anxiety and OCD, so I keep on getting "The Symptoms" of a whole basket of shit. But it is hard to differentiate between a true one and a psychosomatic one.

Mike Miller

@feynhat Yeah that's why I said "I misunderstood" after.

I've made worse mistakes.

feynhat

17:11

@SayanChattopadhyay Yeah. That might be true, considering I hadn't moved out of my place in past 6 months, I was already a bit anxious. Anyway, I am visiting a hospital tomorrow. (for what is worth, I can still taste and smell stuff. I read that loss of sense of taste and smell was observed in almost all patients.)

Sayan Chattopadhyay

17:33

@feynhat Yeah it doesn't hurt to visit a doctor, but I guess one has to be cautious now. Also, just some advice, DO NOT google symptoms, they wreck your mind up.

feynhat

Too late bro. I already went past the first page of results and have atleast two dozens tabs open...

Sayan Chattopadhyay

Damn, yeah its never good to do that. I had once spend entire months googling rabies symptoms and thinking I have rabies. It is not good at all.

feynhat

Yeah. I shouldn't have. It only dawned on me this evening that I could have caught the 'rona. I panicked and started looking shit up.

Anyway, you heard anything thing from Mohali yet? When are they letting you guys back in?

Sayan Chattopadhyay

There were some rumors that they would start getting people in small batches into the campus starting Oct 15. Don't have anything conclusive yet. Anything over in Bhubaneshwar?

feynhat

Well they gave a one week window in Sept for final year kids to return. Almost 2/3 of my batch has gone back.

It was voluntary ofcourse. So I didn't go. My advisor was fine with it.

One of the Deans told us that they will have the end-sem exams in Feb'21, and their plan is to get everyone back before that. (lmao. Seems unlikely this will happen)

Sayan Chattopadhyay

17:48

Does it though? Hasn't there been something about that colleges can open from Oct 15?

anakhro

According to google I have been pregnant for the last 3 years of my life. :((((((((((

feynhat

hmm... I haven't been following news lately. Maybe. But still.

Sayan Chattopadhyay

Ofcourse they cannot enforce this everywhere, I guess the state has the final decision. But to me it seems likely that we might be back before December

feynhat

@anakhro lmfaooo

Okay man. I gotta crash. G'night Sayan, anakhro (you should consult a gynecologist not google).

Sayan Chattopadhyay

G'night man, take care!

anakhro

18:06

I don't think a gynecologist would take on a male client.

rschwieb

20:00

Remember the joker with the "disproof" of Bell's theorem? Turns out he's still up to it. Lately he published what purports to be a counterexample to Hurwitz's theorem. Ugh!

Tobias Kildetoft

@rschwieb And that is even (or at least was previously) a decent journal. Not top tier, but at least reputable.

orientablesurface

0

Q: measure theory problem and epsilon

Let $(X,S,\mu)$ be a measure space. Let $\epsilon>0$. Show that there exists $E\in S$ such that $\mu(E)<\infty$ and $\int f \leq \int_{E}f d\mu +\epsilon$. Where $f\in M(X,S,\mu)$ is non-negative. My idea consisted of doing: Let $\epsilon>0$. As $\int f d\mu-\epsilon$ is not an upper bound, there...

integration measure-theory geometric-measure-theory

rschwieb

@TobiasKildetoft i know :'(

Tobias Kildetoft

I have two papers in it. Certainly my weakest papers, but at least they are correct.

geocalc33

20:16

0

Q: A strange process involving a diffeomorphism, a summation and an analytic continuation. Simpler example?

Consider a map $\psi:\Bbb R^2\to \Bbb R^2_+$ with $\psi(x,y)=(e^x,e^y).$ Looking at $\varphi(x)=e^{\frac{1}{\log(x)}}$ we see that it's related to $h(x)=\frac{1}{x}$ via $u=\log(x)$ and $v=\log(\varphi),$ yielding $uv=1.$ I'll try to describe a process: $(1)$ Change the coordinate system of the p...

complex-analysis sequence-of-function analytic-continuation diffeomorphism

actually that's a bad question /:

Ted Shifrin

22:45

@TobiasKildetoft So it shouldn't be named Miscommunications in Algebra?

2

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$Mathematics$ Mathematics