Mathematics - 2023-03-26

D Left Adjoint to U

00:44

@shintuku I know how to relate arithmetics modulo $n$ and modulo $m \gt 2(n-1)$. With a polynomial!

What you get is a polynomial $f \in \Bbb{Z}[X]$ such that $f(x + y) = f(x) +_n f(y) \pmod m$ for all $x, y \in \{0, \dots, n-1\}$.

So it's as if the structure $(\Bbb{Z}/n ,+)$ sits as a polynomial image inside of the abelian group $(\Bbb{Z}/m, +)$. For divisorially unrelated $m,n$ that is usually unheard of to relate them!

0

Q: Trick to have a polynomial relationship of $(\Bbb{Z}/n, +)$ with $(\Bbb{Z}/m, +)$ for $2 \leq n$ and $m \gt 2(n-1)$?

Define: $$ g(z) = (z(z - 1)(z-2) \cdots (z-n + 1))^{\varphi(m)-1} \equiv \begin{cases} 0 \text{ if } 0\leq z \leq n \\ 1 \text{ otherwise}\end{cases} \pmod m $$ We can use $g(z)$ to define polynomially instead of piece-wise the following map: $$ f(z) := \begin{cases} z \text{ if } g(z) = 0 \\ z ...

one potato two potato

01:19

Let $G$ be a group and $X$ be a $G$-set. If $F$ is a forgetful functor from the
category of $G$-set to the category of set, then we can define a bijection $(\eta^g)_X = \varphi_X(g):F(X)\to F(X)$ where $\varphi_X:G\to S_X$ is a permutation representation of $X$. If $\alpha:F\to F$ is a natural isomorphism, then I'm trying to show $\alpha = \eta^g$ for some $g$.

So I need to find some suitable $g$. If $X = G$, considering $G$ as a $G$-set with left-multiplication, then $\alpha_G:F(G)\to F(G)$ so $\alpha_G(1) = g$. Now I'm hoping $\eta^g = \alpha$.

Is this correct? I can't proceed from here. Naturality of $\alpha$ can be used to show that $\alpha_X:F(X)\to F(X)$ is a $G$-equivariant map I think but don't know what to do next.

geocalc33

@DLeftAdjointtoU what's up

Koro

02:25

4

Q: $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$

I've been doing some more exercises in Hatcher, in particular the following: Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$. "$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to...

algebraic-topology

How to do this? Any ideas?

I don't understand the answer there at all.

D Left Adjoint to U

02:42

@geocalc33

hey

Koro

0

Q: $H_0(X,A)=0$ iff $A$ intersects every path component of $X$.

We have a long exact sequence $...\to H_0(A)\to H_0(X)\to H_0(X,A)\to 0$ If $H_0(X,A)=0$, then by exactness at $H_0(X)$: Image $(H_0(A)\to H_0(X))=H_0(X)=$ ker $(H_0(X)\to H_0(X,A))$ so the map $i_*:H_0(A)\to H_0(X)$ is surjective. Let $X_a$ be a path component of $X$. Suppose on the contrary tha...

algebraic-topology homology-cohomology exact-sequence

leslie townes

03:06

koro, given the relation between your question and the earlier one, it might make sense to link to the previous question in your post (maybe with commentary about what you don't understand about the earlier answer, or how your question is different). if only to avoid having someone respond by referring you there or marking it as a duplicate.

one potato two potato

I remember it's one of the exercise problem in hatcher

Koro

03:21

I don't understand how surjectivity of the map is related to path connectedness.

Ted Shifrin

03:36

I answered it. I commented, but it’s an answer.

Koro

Thanks for the comment.

I don't understand why $x_a$ is not the image of any element in $H_0(A)$.

Ted Shifrin

I said why.

Koro

yes, but how do I get that sentence from equations?

Ted Shifrin

Can it be homologous to an element of $A$?

Koro

how did you obtain that sentence?

Ted Shifrin

03:47

You should wash your mouth out with soap for the previous question.

Koro

If $[x_a]=[\alpha]$ for some $\alpha \in A$, then $x_a-\alpha $ is in Im ($C_1(X)\to C_0(X))$

I don't understand why this is not possible.

one potato two potato

04:02

Btw I think I solved my problem using the fact that $(G,1)$ represents $F$ with Yoneda lemma. It seems nobody cares so I won't put it in detail.

Koro

:'(

somebody might care. you may post it with your answer on mse.

one potato two potato

Well don't need to I think the proof is correct

It's totally fine that nobody cares about whatever question I ask in chat. Free questions free answers

Ted Shifrin

@Koro Because of your hypothesis that $X_a$ is disjoint from $A$.

Definition of path component, perhaps?

Koro

I don't understand how disjointness of $X_a$ from A prevents $x_a-\alpha$ is in Im ( ).

$x_a\ne \alpha$ fine. But their difference could be a boundary?

Ted Shifrin

04:22

So there is a path from $x_a$ to $\alpha$.

Koro

that will give me contradiction by path components definition.

But why is there a path from $x_a$ to $\alpha$?

This is the part I don't understand.

Ted Shifrin

What is a $1$-chain?

Koro

elements of $C_1(X)$

Ted Shifrin

Grrr

Koro

i.e., linear combination of 1 singular maps $\Delta^1\to X$.

or one could just therefore say paths.

Ted Shifrin

04:32

Precisely, at long last

Koro

1-chains are precisely paths in X.

Ted Shifrin

Or formal linear combinations thereof, yes

So we’re done.

Koro

But don't we only have: $x_a-\alpha =\gamma (1)-\gamma(0)$ for some path $\gamma$?

Ted Shifrin

What does that actually mean?

How do you subtract points in $X$?

Koro

that's formal sum. This operation is happening in free Abelian group generated by maps $\Delta^0\to X$.

$C_0(X)=$ free Abelian group gen'd by maps $\Delta^0\to X$.

Ted Shifrin

04:43

Fine … and so $\gamma$ is a path where? From where to where?

You need to understand what all these symbols actually mean.

Koro

$\gamma $ is a path in X starting at $\gamma (0)$ and ending at $\gamma (1)$ in X.

yeah sure.

Ted Shifrin

Now reread what I said 20 minutes ago.

Koro

But $\gamma$ may not be a path from $x_a$ to $\alpha$, right?

Ted Shifrin

You just said it was. So, huh?

Koro

no, I said that it is a path in X. It does not have to start at $x_a$ and end at $\alpha$. If there were such a path, then I understand how we get the contradiction.

:(

what a boring subject this is...

Ted Shifrin

04:51

You said $\gamma(0)=x_a$ and $\gamma(1)=a$.

Koro

general topology is better

Ted Shifrin

Hell no, general topology is boring as all hell.

Koro

No, I did not say that $\gamma(0)=x_a$ and $\gamma(1)=\alpha$.

Ted Shifrin

It sure looks like you did. You’re what’s boring.

Koro

I only said $x_a-\alpha$ is in Im ($C_1(X)\to C_0(X))$.

one potato two potato

05:03

My impression of general topology is a formal setup to make a mathematical statement clear and rigorous, not to actually prove something.

Or a formal setup to avoid some pathological examples.

Koro

I had lot of expectations from Algebraic topology, it turned out to be such a disappointment.

2

Prithu Biswas

If A is a bounded subset of ℝ and c ∈ ℝ, And B = {x : ∃y ∈ A (x = c.y)},
Then if c > 0, then sup(B) = c.sup(A) and inf(B) = c.inf(A)
And if c < 0, then sup(B) = c.inf(A) and inf(B) = c.sup(A)
And if c = 0, then sup(B) = inf(B) = 0
Is this correct?

one potato two potato

At least I enjoyed AT problems in quals.

Koro

05:22

quals?

Prithu Biswas

I remember seeing a animated topology series by a channel by the name Bourbaki[?] which showed stuff like simplex, open balls etc. But I can't find it anywhere.

one potato two potato

@Koro qualifying exam. Maybe you should take it to be a PhD candidate.

Koro

I see.

I will not do Ph.D. I have decided.

one potato two potato

But anyway, those problems consist of selected good problems in general.

Prithu Biswas

05:48

I finally found it =D

You Could Have Invented Homology, Part 1: Topology | Boarbarktree

►

one potato two potato

I don't believe youtube shits except for lecture and seminar videos.

PNDas

Hi I'm just confused. People say complex differentiable implies infinite differentiability.

Is this true? Do they mean complex differentiable at a point? or a neighbourhood?

Does infinite differentiability implies analyticity?

eman yalpsid

anyone??

https://math.stackexchange.com/questions/4666478/prove-that-a-simple-connected-graph-having-exactly-two-vertices-that-are-not-cut

one potato two potato

@PNDas On a nbd

leslie townes

pndas: what one potato says.

PNDas

06:01

Infinite differentiability implies analyticity?

leslie townes

books vary in how they define 'analyticity' (often in terms of series representations but not always). also in how they define 'differentiable' (often just the existence of the derivative, but sometimes, on an open set and not a point, and sometimes also at least initially required to be continuous on that open set)

but putting aside the static around which definitions you pick, yes. if a function is infinitely complex differentiable on an open set it will be analytic on that set.

and as a lot of textbooks approach it, they usually prove something from a weaker hypothesis, e.g. that f be twice differentiable or at least C^1.

they often assume enough derivatives at the outset so that the ingredients of the cauchy riemann equations will exist without a lot of leg work.

PNDas

Hmm, I saw an MSE answer. Where the answerer said infinite differentiability doesn't imply analyticity. math.stackexchange.com/questions/2135331/…

May be he's talking about real case.

leslie townes

yeah, who knows. all of this stuff, "differentiable," "holomorphic," "analytic" - is not really a good fit for internet discussion, unless it's the kind of discussion where everybody states what definitions they are using at the outset.

you can't explain how or whether "holomorphic" is any different from "analytic" without that context, for example. and if people are just meeting this stuff by working off of wikipedia and other aggregations of various sources, i'm not sure that any of these differences would be clear to them.

PNDas

I'll just complete one book first. Then go to internet

I'm following Rudin now.

leslie townes

06:16

if you're dealing with both real and complex stuff at the same time, you might use different definitions than you would if you were only doing intro complex analysis.

PNDas

He didn't make any distinction. For him differentiable in an open set is the definition of Holomorphic or analytic in that set. He didn't name the functions representable by power series.

leslie townes

we ran into this a little bit a while ago, someone was asking a question relating to harmonic functions. and some books deduce facts about harmonic functions from analogous facts about analytic functions, and other books do it the other way around.

PNDas

I'm a stupid. I have 4 complex analysis books. Ahlfors, Sarason, Kumaresan, Rudin.

I have also studied complex analysis from Churchill in BSc, from Conway in MSc

Suppose $f$ has a power series expansion in an open set $U$ means that for every $a\in U$, there exists an open ball centered at $a$ s.t. in that ball $f(z)=\sum c_n (z-a)^n$.

Franklin

Can anyone please help me with this: math.stackexchange.com/questions/4666542/… ?

Silly Goose

06:32

PNDas

Now suppose there exists a point $a_0\in U$ such that $f(z)=\sum c_n (z-a_0)^n$ for every $z\in U$ then does that mean $f$ has a power series expansion in $U$?

Silly Goose

I do not really understand 1) the last line of the proof. Isn't this saying that $\epsilon = 0$? I assume that we are trying to prove equality by showing geq and leq.

PNDas

@SillyGoose If $a\leq b+c$ for every $c>0$ then $a\leq b$.

Silly Goose

2) Why do we need to do this geq leq proof when it seems like the argument starting at "Fix $\epsilon > 0$" is sufficient? Since the final inequality tells us that $diam\bar{E} - diamE \leq 2\epsilon$ where $\epsilon$ is arbitrary positive real. which implies that the LHS of the inequality must be equal to 0

Franklin

0

Q: Let $\alpha$ be a cut and $r\in\Bbb Q^+.$ Then $\exists p\in \alpha$ and $q\in\alpha ^c$ such that $q$ is not the least element of $\alpha^c$ and

Let $\alpha$ be a cut and $r\in\Bbb Q^+.$ Then $\exists p\in \alpha$ and $q\in\alpha ^c$ such that $q$ is not the least element of $\alpha^c$ and $r=q-p.$ (The definition of cut is attached at the end of this post for reference) The solution presented is as follows: Let $s\in \alpha$ be any elem...

real-analysis proof-explanation

Silly Goose

06:35

@PNDas How does this not prove that $a = b$?

oh wait hm okay i think my problem is that the presence of $\leq$ does not allow for us to say $a = b$. but if it were $<$, then it would?

PNDas

@SillyGoose In the second line of the proof, we already have $\text{diam}E\leq\text{diam}\overline{E}$. Next as I said, we get the other way inequality. So they are same.

@SillyGoose But no this doesn't prove that a=b.

@SillyGoose are you talking about $a<b+c$ for all $c>0$ implies $a=b$?

Silly Goose

yes

PNDas

1<2+c for all c>0 but 1 is not 2

We can only say that $a\leq b$.

The proof is also easy. Let $a>b$ then take $c=a-b$. Get $a<a$ which is a contradiction.

Silly Goose

07:47

okay i see now thank you !!

is Rudin implicitly using reals with standard metric to evaluate the limit of the diam$E_N$?

Alessandro Codenotti

07:59

Yes (as it's always the case for a sequence of real numbers unless otherwise specified)

Prithu Biswas

My book tells me to prove that:

$\int_{ca}^{cb}f(t).dt = c\int_{a}^{b}f(ct).dt$

But aren't there supposed to be some assumptions. Like, what is integrable etc?

Thorgott

12:52

@Koro It's not that complicated. If you have a map of sets $f\colon S\rightarrow T$, it induces a map of free abelian groups $\bigoplus_{s\in S}\mathbb{Z}e_s\rightarrow\bigoplus_{t\in T}\mathbb{Z}e_t$ uniquely defined by mapping $e_s\mapsto e_{f(s)}$. The image of this map is then precisely the free abelian subgroup of $\bigoplus_{t\in T}\mathbb{Z}e_t$ consisting of those summands where $t\in f(S)$. From this, deduce that the cokernel of this map is a free abelian group with basis $T\setminus f(S)$. In particular, the map between the free abelian groups is surjective iff $f$ is surjective. …

PNDas

13:04

Suppose we $f(z)=\sum c_n(z-a)^n$ in $B(a,r)$. Then does this mean for any $b\in B(a,r)$, we have another power series $f(z)=\sum d_n (z-b)^n$?

Ajay

Can someone check my answer to this post: math.stackexchange.com/questions/4666702/…

one potato two potato

@PNDas Yes with possibly smaller ROC

PNDas

@onepotatotwopotato $n!d_n=f^{(n)}(b)$? It's just Taylor expansion. Right?

one potato two potato

Yes by uniqueness

Koro

13:38

@Thorgott I know that the map $H_0(A)\to H_0(X)$ is surjective (assuming $H_0(X,A)=0$). But I don't see how to conclude A intersects all path components of X from here.

Thorgott

14:10

You've said as much before, but you're not carefully reading what I'm saying. What are the bases of $H_0(A)$ and $H_0(X)$ as free abelian groups? What is the map $f$ from my message above in this context?

Koro

f is inclusion here?

$H_0(X)= Z${ path connected components of X}

Ted Shifrin

14:54

@PrithuBiswas Yes, of course they are assuming $f$ is integrable on $[ca,cb]$.

Prithu Biswas

15:10

@TedShifrin Oh ok. Thank you.

Thorgott

15:57

you have to consider $H_0(A)$ too

$f$ is a map from the basis of $H_0(A)$ to the basis of $H_0(X)$

Koro

16:08

I don't know why messages are repeating.

@Thorgott: I'm afraid I don't understand what you mean.

is it possible to bounty a closed question?

robjohn

@Koro You'd just be tossing rep. If it is closed, no one can answer. I guess if there were an answer, you could add a bounty and, after it's possible, give the bounty to that answer.

> A bounty can be started on any question 48 hours after the question was asked, provided the question isn't closed, locked, or deleted.

So, no, you can't put a bounty on a closed question.

You can read more on bounties here

PNDas

16:49

I want an example of a complex function which is continuous everywhere also it's analytic everywhere except a point.

robjohn

Have you read up on removable singularities?

PNDas

Hmm, but it has to be continuous.

Lemme see the definitions again,

Thorgott

@Koro I described a specific class of morphisms between free abelian groups with fixed bases in my first messages. I'm saying the map $H_0(A)\rightarrow H_0(X)$ can be described as such. Be specific about which parts are not clear to you.

PNDas

@robjohn But then it won't be continuous everywhere.

robjohn

@PNDas you are misreading my intent. Look also at the Cauchy Integral Formula

You need to use a keyhole contour, but you end up proving stuff about removable singularities

schn

16:59

An extract from my lecture notes. I do not understand the sigma notation. Why does it extend to 2?

The last parentheses belongs to a parentheses not shown in the screenshot.

PNDas

@robjohn Sorry but I don't understand. Can you please tell me if there exists such functions or not?

@schn It's just $\sum_{j=1}^2\sum_{k=1}^2$.

schn

@PNDas makes sense, thanks

PNDas

Okay, I got it. From theorem 10.13 of rudin, my conditions imply that integration is zero over all triangles and then by Morrera's theorem $f$ is holomorphic.

Obliv

17:34

what is the general definition of implicit differentiation?

I'm looking for something abstract not "equations of the form (insert example)"

nvm I think if I understand general application of the differential operator i don't even need this

schn

I'm reading through the article about countable sets on Wikipedia. They say a set is countable if there exists an injective function from the set into the natural numbers. However, doesn't one require a bijective function for it to be called countable?

Obliv

17:50

@Schn I'm not that knowledgeable but does a surjection between infinite sets make sense?

schn

@Obliv hmm, probably not

PNDas

@schn They are taking countable and almost countable in one go.

Obliv

Disregard @Schn I don't know what I'm talking about :P

Semiclassical

this is probably an elementary matter, but...

suppose i have some unknown discrete distribution which i repeatedly sample from

PNDas

If it's bijective we get countably infinite and if it's not then it may be bijective with a finite set so you get finite set.

Semiclassical

17:55

what's the bayesian way to incorporate each new sample? i'm guessing i should be looking up prior vs posterior distribution

Obliv

@semiclassical what do you mean by incorporate?

Semiclassical

"bayesian updating" seems to be what i'm looking for

Obliv

i've only dealt with random samples that dont affect the distribution

Semiclassical

the typical context for it is when you're trying to determine an unknown distribution by sampling from it

Obliv

oh i haven't got to that part yet in my stats class

Semiclassical

18:04

and want to reflect what you learn about the distribution sample-by-sample, rather than taking many samples and judging after (frequentist)

Obliv

just been given the distribution

Semiclassical

yeah, it's not something i've dealt with much myself

i've heard the phrase but never had to actually use it

Obliv

interesting though, because if the samples do affect the distribution, you may learn about the affect of your sampling on the distribution but if it's not reversable you can't learn about the original distribution

Ted Shifrin

18:48

@Obliv Of course.

geocalc33

20:38

$X=S^1 \times (0,1)$ and $X$ accumulates to $p=(0,0)$ and $q=(1,0).$ where $S^1$ is $|z|=1.$ Can $X$ be a complex manifold?

Ted Shifrin

That makes no sense. What does $X$ accumulates mean?

Any open subset of $\Bbb C$ is a $1$-dimensional complex manifold, trivially.

geocalc33

I was using the term accumulate in the sense of real smooth surfaces i.e. when a surface accumulates to a point it's in someway converging from all directions to that point

but I can use accumulate in the complex setting as well I thought

Ted Shifrin

I’ve only heard of accumulation point as a limit point, never this language.

geocalc33

I guess I could say the cylinder $ S^1 \times (0,1) $ 'collapses' to those points at either end..not really sure of the exact terminology

Ted Shifrin

But you need to define the subset of $\Bbb C$. The cartesian product of $S^1$ with $(0,1)$ is not a subset of $\Bbb C$ at all.

geocalc33

20:53

Oh wow, I was thinking since $S^1$ is a circle in the complex plane that it would be a subset of $\Bbb C$

Ted Shifrin

No, as you defined it, it’s a subset of $\Bbb C\times (0,1)$.

geocalc33

Oh okay I gotcha now

schn

21:12

I'm trying to determine for which $x$ the series $\sum_{k=0}^\infty (1+x^2)^{-k}$ converges and what it converges to. I have been unsuccessful so far with the ratio test...

Ted Shifrin

Why unsuccessful?

schn

It leads me to $\frac{(1+x^2)^{-(k+1)}}{(1+x^2)^{-k}}=\frac{1}{1+x^2}$. Now, the ratio test says the series converges if $\frac{1}{1+x^2}< 1$, or equivalently, $0<x^2$. So I guess it is saying $x\neq 0$??

Node.JS

I am wondering can someone please help with this question math.stackexchange.com/questions/4666799/…

Ted Shifrin

21:32

@schn Precisely.

schn

OK. Now, how do I calculate what it converges to?

Ted Shifrin

What is the most basic kind of series?

schn

Geometric?

Ted Shifrin

Yes. Now think.

parvin

Hello. How can I find the x in Cos(x)=12 with an online calculator?!

Ted Shifrin

21:40

12?

schn

In a geometric series we have a positive exponent though...

Ted Shifrin

Schn, think. I’m saying no more.

parvin

Cos(x)= y then how am I supposed to reverse it at find x?!

I mean google is not helping me at all! I did look up. and the last time I was studying these was like 10 years ago. long time for remembering anything!

Ted Shifrin

The 12 is impossible. You do the inverse function (arccos) and that gives you one of infinitely many possible answers.

parvin

oh my god... infinite?

Ted Shifrin

21:48

What is the actual question you’re trying to do?

parvin

the thing is,
for example
cos (12.86) = 0.95719
but arccos (0.95719) is not 12.86 !

Ted Shifrin

That’s right. It isn’t.

parvin

T_T
How is one supposed to solve these stuff then...

Ted Shifrin

I asked you what question you’re trying to solve.

parvin

I'm trying to find a way to find x in Cos(x) for a number of coordinates.

for instance 3.5 = Cos(a)
32.5 = Sin(a)

and some others.

Ted Shifrin

21:51

You cannot do that. No answer without crazy complex numbers.

If you review trig at all, you will recall that sin and cos must be between $-1$ and $1$.

parvin

true...

Ted Shifrin

So if you start with $y$ between $-1$ and $1$, then $\arccos(y)$ gives you the unique $\theta$ between $0$ and $\pi$ with $\cos\theta = y$.

leslie townes

phbt

Ted Shifrin

22:07

You can take over, leslie. You like cos = 12. Give it to Munchkin.

leslie townes

she doesn't learn complex numbers until next week. we're still working on the axioms for a complete ordered field.

Ted Shifrin

And $3+4$.

leslie townes

she might actually be able to do that. she can do some of those.

robjohn

@PNDas How do you show that the integral over a triangle containing the exceptional point is $0$?

Ted Shifrin

Fancy schmancy, Morera.

robjohn

22:18

Yeah, all you really need is Cauchy's Integral Formula.

Ted Shifrin

I would do the Laurent series, yeah.

I do love Morera, though.

D.C. the III

Math and its need for scratch paper and a pen by your side......love it and hate it...

@leslietownes surely you start her with the ZF axioms.....

robjohn

where else would one start?

Ted Shifrin

Dedekind cuts.

leslie townes

nah. there's definitely a time for naive set theory. before age 5 is it.

D.C. the III

22:23

suicidal that

leslie townes

this october, we start over from the ground up

Ted Shifrin

Better to dig down, WW III and all.

leslie townes

haha, base 0. piling stones outside of our cave.

robjohn

22:41

@leslietownes wouldn't that be base 1?

leslie townes

depends on how you pile 'em

mick

23:12

1

Q: Special numerical method for $\sqrt 2$ with rational functions

Numerical methods for approximating Pythagoras' constant $t =\sqrt 2$ by fractions. (This is an idea from my mentor while he was barely $13$ yo, as a response to a challenge). We all know Newton's method for finding $t$. It converges quadratically meaning like $o( C x^{2^n} )$ where $C$ is a cons...

sequences-and-series numerical-methods radicals rational-functions

a typical mick post :)

hope that sounds good and not bad lol

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