12:44 AM
@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

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 ...

1:19 AM
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.

2:25 AM
4

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...

How to do this? Any ideas?
I don't understand the answer there at all.

2:42 AM
@geocalc33
hey

0

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...

3:06 AM
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.

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

3:21 AM
I don't understand how surjectivity of the map is related to path connectedness.

3:36 AM

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

I said why.

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

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

how did you obtain that sentence?

3:47 AM
You should wash your mouth out with soap for the previous question.

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.

4:02 AM
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.

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

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

@Koro Because of your hypothesis that $X_a$ is disjoint from $A$.
Definition of path component, perhaps?

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?

4:22 AM
So there is a path from $x_a$ to $\alpha$.

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.

What is a $1$-chain?

elements of $C_1(X)$

Grrr

i.e., linear combination of 1 singular maps $\Delta^1\to X$.
or one could just therefore say paths.

4:32 AM
Precisely, at long last

1-chains are precisely paths in X.

Or formal linear combinations thereof, yes
So we’re done.

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

What does that actually mean?
How do you subtract points in $X$?

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 $\Delta^0\to X$.

4:43 AM
Fine … and so $\gamma$ is a path where? From where to where?
You need to understand what all these symbols actually mean.

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

Now reread what I said 20 minutes ago.

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

You just said it was. So, huh?

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...

4:51 AM
You said $\gamma(0)=x_a$ and $\gamma(1)=a$.

general topology is better

Hell no, general topology is boring as all hell.

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

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

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

5:03 AM
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.

I had lot of expectations from Algebraic topology, it turned out to be such a disappointment.
2

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?

At least I enjoyed AT problems in quals.

5:22 AM
quals?

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.

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

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

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

5:48 AM
I finally found it =D

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

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?

anyone??

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

@PNDas On a nbd

pndas: what one potato says.

6:01 AM
Infinite differentiability implies analyticity?

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.

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.

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.

I'll just complete one book first. Then go to internet
I'm following Rudin now.

6:16 AM
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.

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.

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.

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 $f(z)=\sum c_n (z-a)^n$.

6:32 AM

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 $U$?

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.

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

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

0

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...

6:35 AM
@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?

@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 $a=b$?

yes

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.

1 hour later…
7:47 AM
okay i see now thank you !!
is Rudin implicitly using reals with standard metric to evaluate the limit of the diam$E_N$?

7:59 AM
Yes (as it's always the case for a sequence of real numbers unless otherwise specified)

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?

5 hours later…
12:52 PM
@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.

1:04 PM
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 $f(z)=\sum d_n (z-b)^n$?

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

@PNDas Yes with possibly smaller ROC

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

Yes by uniqueness

1:38 PM
@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.

2:10 PM
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?

f is inclusion here?
f is inclusion here?
$H_0(X)= Z${ path connected components of X}
$H_0(X)= Z${ path connected components of X}

2:54 PM
@PrithuBiswas Yes, of course they are assuming $f$ is integrable on $[ca,cb]$.

3:10 PM
@TedShifrin Oh ok. Thank you.

3:57 PM
you have to consider $H_0(A)$ too
$f$ is a map from the basis of $H_0(A)$ to the basis of $H_0(X)$

4:08 PM
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?

@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

4:49 PM
I want an example of a complex function which is continuous everywhere also it's analytic everywhere except a point.

Have you read up on removable singularities?

Hmm, but it has to be continuous.
Lemme see the definitions again,

@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.

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

@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

4:59 PM
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.

@robjohn Sorry but I don't understand. Can you please tell me if there exists such functions or not?
@schn It's just $\sum_{j=1}^2\sum_{k=1}^2$.

@PNDas makes sense, thanks

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.

5:34 PM
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

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?

5:50 PM
@Schn I'm not that knowledgeable but does a surjection between infinite sets make sense?

@Obliv hmm, probably not

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

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

this is probably an elementary matter, but...
suppose i have some unknown discrete distribution which i repeatedly sample from

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.

5:55 PM
what's the bayesian way to incorporate each new sample? i'm guessing i should be looking up prior vs posterior distribution

@semiclassical what do you mean by incorporate?

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

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

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

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

6:04 PM
and want to reflect what you learn about the distribution sample-by-sample, rather than taking many samples and judging after (frequentist)

just been given the distribution

yeah, it's not something i've dealt with much myself
i've heard the phrase but never had to actually use it

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

6:48 PM
@Obliv Of course.

2 hours later…
8:38 PM
$X=S^1 \times (0,1)$ and $X$ accumulates to $p=(0,0)$ and $q=(1,0).$ where $S^1$ is $|z|=1.$ Can $X$ be a complex manifold?

That makes no sense. What does $X$ accumulates mean?
Any open subset of $\Bbb C$ is a $1$-dimensional complex manifold, trivially.

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

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

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

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.

8:53 PM
Oh wow, I was thinking since $S^1$ is a circle in the complex plane that it would be a subset of $\Bbb C$

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

Oh okay I gotcha now

9:12 PM
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...

Why unsuccessful?

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 $x\neq 0$??

9:32 PM
@schn Precisely.

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

What is the most basic kind of series?

Geometric?

Yes. Now think.

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

9:40 PM
12?

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

Schn, think. I’m saying no more.

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!

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

oh my god... infinite?

9:48 PM
What is the actual question you’re trying to do?

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

That’s right. It isn’t.

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

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

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.

9:51 PM
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 $1$.

true...

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

phbt

10:07 PM
You can take over, leslie. You like cos = 12. Give it to Munchkin.

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

And $3+4$.

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

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

Fancy schmancy, Morera.

10:18 PM
Yeah, all you really need is Cauchy's Integral Formula.

I would do the Laurent series, yeah.
I do love Morera, though.

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.....

where else would one start?

Dedekind cuts.

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

10:23 PM
suicidal that

this october, we start over from the ground up

Better to dig down, WW III and all.

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

10:41 PM
@leslietownes wouldn't that be base 1?

depends on how you pile 'em

11:12 PM
1

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...

a typical mick post :)
hope that sounds good and not bad lol