Mathematics

Ilya_Gazman

12:05 AM

hello

Can someone help me with a modular problem?

Ilya_Gazman

1:18 AM

hmm....

Avantgarde

Looks like no one's around

Ilya_Gazman

There is me ;)

Avantgarde

My comment was a response to your hmm :P

but yeah

Ilya_Gazman

@Avantgarde So how is yours modular skills?

Avantgarde

@Ilya_Gazman I don't exactly know what you mean. I'm not a math student

Ilya_Gazman

1:29 AM

Ok, I will wait for some one then ;)

user193319

0

Q: Finding a Bounded Measurable Set and Sequence of Simple Functions

Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m(E) < \infty$. Show that for each $\epsilon > 0$, show that there is a measurable set $F$ contained in $E$ and a sequence $\{\varphi_n \}$ of simple functions on $E$ such that $f$ is bounded on $F$ and $m(E-F) < \epsilon$...

Captain Bohemian

moduli space?

Avantgarde

Ok good luck. Maybe post a question

Ilya_Gazman

0

Q: How to simplify $2^{a\left(2^{bc}-1\right)}-1\mod{N}$

How to simplify $$2^{a\left(2^{bc}-1\right)}-1\mod{N}$$ Where a,b,c are big integers with relatively small primes and N is a big integer that is hard to factor(it does not have small primes). I have been trying doing the following: $$2^{a\left(2^{bc}-1\right)}-1\mod{N}=$$ $$2^{a^{\left(2^{bc}-...

number-theory elementary-number-theory prime-numbers modular-arithmetic prime-factorization

Did that too ;)

Secret

2:02 AM

(cont.) Ok, what I said yesterday does not work since a set $S$ with cardinality $\kappa$ only has a $\mathcal{I}(S)$ of cardinality $2\kappa+1$, which is not enough to reach uncountability.

Hmm...

$a+1 := a \cup \{a\}$

$a+(b+1) = (a + b) + 1 := (a+b) \cup \{a+b\}$

$ab := a \times b = (a,b)$

$a^n :=\prod_{k=1}^n a=(a,a,a,...,a)$

$\lambda = \sup (n < \omega : \lambda [n]), |\lambda| = \aleph_0$

So succession will take us from 0 to any naturals, and from any limit to successors

Multiplication will take us from $\omega$ all the way to $\omega$ towers

actually no

Just succession and supremum of countable increasing sequences of ordinals (fundamental sequences) alone will reach any computable ordinal

Multiplication and exponentiation are ultimately shorthands for a lot of successions and supremums, so as Veblen functions and other recursive increasing functions on the ordinals

So eventually all the computable ordinals will be produced

gian

2:45 AM

Whenever you get the chance, please check this @BalarkaSen. $H_2(P, P-p) \cong \Bbb{Z} \oplus \Bbb{Z}$. $H_1(P, P-p) \cong 0$. $H_0(P, P-p) \cong \Bbb{Z}.$

$P$ is the pinched torus and $p$ is the pinch point.

Manolis Lyviakis

IF any1 is up for some topology

2

Q: Prove $\mathbb{R}$ is connected

Prove that $\mathbb{R}$ is connected. PLease i have found other ways to prove it but i want to make this way work. Proof: 1) Strategy : If i show that a arbitrary interval is connected then i can take the colection of intervals around zero that make up $\mathbb{R}$ and have a common point So...

real-analysis general-topology analysis connectedness

im really stuck trying to use the specific arguments

i can draw the possibillities but i cant formally prove it

Caddyshack

3:30 AM

Does anyone know what the notation $C_{3v}: E, C_3, C_3^2, 3\sigma_v$ means w.r.t point groups?

It's supposed to describe trigonal symmetry of sorts.

It might even be the dihedral group $D_2$ for all I know, but I can't decipher the notation.

Kevin Driscoll

4:13 AM

@Caddyshack $C_{3 v}$ is isomorphis to $D_{3}$

The $E, C_3, C_3^2$ are the identity and rotations, isomorphic to $C_3$ group. Each of the $\sigma_v$ is a reflection. There are 3 axes you can reflect about

(Some people call it $D_6$ dunno which you use

Caddyshack

@KevinDriscoll Makes sense, thanks. With $D_3$ I know how to do the operations - with this notation, however, it doesn't seem clear.

Kevin Driscoll

What about it seems unclear?

Caddyshack

Actually the notation is effectively the same. What initially threw me off is the fact that $3\sigma_v$ is three elements.

I had also seen it written as $C_{3v}: E, 2C_3^2, 3\sigma_v$, which confused me even more.

Caddyshack

4:39 AM

@KevinDriscoll I'm reading that the nontrivial subgroups of $C_{3v}$ are $H_1 = \left\{E, \sigma_h\right\}$ and $H_2 = \left\{E, C_3, C_3^2\right\}$. But $\sigma_h$ is not in the group, right? It should be $\sigma_v$?

Kevin Driscoll

@Caddyshack I think its just notation. They proably mean $h$ to index the 3 $\sigma$ elements. So each of the sets $\{E, \sigma_1 \}$ $\{E, \sigma_2 \}$ and $\{E, \sigma_3 \}$ is a subgroup

or maybe you want to call them $\sigma_{v 1}$ etc....

Caddyshack

@KevinDriscoll Thanks, again. Makes sense.

Secret

8:11 AM

in Mathworks (Not the main chat!), 4 mins ago, by Secret

We now learnt from an ordinal arithmetic perspective, $\omega_1^{CK}$ is in fact the first ordinal inaccessible by succession, cartesian product and supremum from any computable ordinals. This is why we cannot find a fundamental sequence for it

Maneesh Narayanan

Really, apologize for the wrong link.

5

Q: which of the following is/are true for the entire function $f$?

Let , $f$ be an entire function. Let, $g(z)=\overline{f(\bar z)}$. Let, $D=\{z:Im(z)=0\}\cup\{z:Im(z)=a\}$ for some $a>0$. Then which are correct ? (A) If $f(z)\in \mathbb R$ for all $z\in \mathbb R$ then $f=g$. (B) If $f(z)\in \mathbb R$ for all $z\in D$ then $f(z+ia)=f(z-ia)$ for all $z\in \m...

complex-analysis

The answer is given by the timon in (B). I have doubts.
f(x−ia)=g(x−ia)=f(x+ia)
How it is coming
please help me.

Balarka Sen

9:11 AM

@gian Quite right. You can compute like this: say $U$ is a cone neighborhood of $p$, then $P - U$ is a subset of $P - p$ is a subset of $P$. Excise that little piece of shit. You get $H_2(P, P - p) \cong H_2(U, U - p)$. Now, you can prove homology $(U, U - p)$ is isomorphic to the homology of $(U, \partial U)$ - this is because $U - p$ deformation retracts to $\partial U$

(Hint: write down the two long exact sequences, and set up isomorphism $H_*(U) \to H_*(U)$, $H_*(\partial U) \to H_*(U - p)$, use the five lemma to conclude)

But $H_*(U, \partial U)$ is isomorphic to $H_*(U/\partial U)$, and that quotient space is just $S^2 \vee S^2$

Daminark

Hey everyone!

Balarka Sen

This is the algebraic nonsense. Geometrically, you do this: Bro you're just looking at singular simplices in $P$ containing $p$ in the interior. It is clear that there are just two such simplices upto homology inside $P - p$

One in "one side" of the cone, the other in "the other side" of the cone

So it's $\Bbb Z^2$ :P

I'll let you decide if algebra is efficient or geometry is efficient

@Daminark Heya

Daminark

Lol speaking of geometry

After literally 6 weeks I only just understood the dihedral group

Alessandro Codenotti

What do you mean?

Daminark

But I get it now so that's something

Alessandro Codenotti

9:15 AM

Hi chat

Daminark

Like my prof was deriving all the properties of D_n and I was like "Uh... I guess I'll take your word for it"

But I just couldn't see why, for example, $srs = r^{-1}$

Balarka Sen

so you mean you prefer blackboxing the presentation of $D_n$ instead of thinking of it as the isometry group of the polygon?

or rather, you did

Daminark

I never realized that all the dihedral groups were the isometry groups of the square. Math is truly a magical subject

Alessandro Codenotti

How did you define $D_n$?

Daminark

Dammit Balarka with your time traveling

Balarka Sen

9:18 AM

not the square, the $2n$-gon. sorry

i'm in a loose mode

Daminark

Kek the 2n gon. But I defined it just as the symmetries and then immediately used the presentation

So I did black box the proof of the presentation

Balarka Sen

gotcha

Alessandro Codenotti

Ah, I see, you can construct it as a subgroup of $S_n$ in a way that I find more intuitive to work with than "symmetries of a polygon"

Balarka Sen

I don't actually think about the 2n gon most of the time

Daminark

But about a week and half, maybe 2 weeks ago

Discrete math was doing graph theory and one of the problems I was grading was to find how many automorphisms there are of the path, the cycle graph, and the complete graph

Now, the automorphism group of the cycle graph is $D_n$

And the argument for why there were 2n automorphisms made total sense to me

You have to preserve adjacency of vertices, so you can send any vertex to any other vertex by rotations (turns out I'm okay with rotations in 2d), and at that point, take one of the vertices that was adjacent to it, it has 2 places it can go

Either to the left or to the right. That was the first time I was comfortably convinced that $|D_n| = 2n$

And tonight I realized why $srs = r^{-1}$ after a friend was explaining it to me for like, literally half an hour

@Alessandro what do you do?

Alessandro Codenotti

9:28 AM

Consider the monoid of "affine" functions on $\Bbb Z/(n)$, that is $\{f_{a,b}:\Bbb Z/(n)\to\Bbb Z/(n)\quad a,b\in \Bbb Z/(n)\}$ with $f_{a,b}(x)=ax+b$, then $D_n$ is the group of units of that monoid, that is $f_{a,b}$ with $a=\pm 1$. If you write $\Bbb Z/(n)$ along the vertices of an $n$-gon you recover the geometric interpretation, $f_{1,b}$ are rotations, $f_{-1,b}$ reflections

Daminark

I see

Alessandro Codenotti

The product then is just composition of functions

By taking $\Bbb Z$ instead of $\Bbb Z/(n)$ you get $D_\infty$

Daminark

I can take that

Balarka Sen

I quite like that

Daminark

If I ever teach group theory (which I probably will) I'll try to explain it like so

Balarka Sen

9:39 AM

@Daminark Here's another shitpost for you

Daminark

Oh lord

Alessandro Codenotti

That's how my professor introduced the dihedral groups (we also saw the presentation and stuff later, but we did all of the geometrical interpretation and conjugacy classes things with the group written as a group of functions)

Daminark

Sky cowboy is making a comeback

Balarka Sen

Truly

Daminark

Like for a while on r/Youtubehaiku every single video on "hot" had that in some form

@Alessandro our prof just toyed with polygons and I'm like "Ya cool kthxm8"

Daminark

10:02 AM

Alright everybody imma take a geometric approach to sleeping so see you!

Alessandro Codenotti

Good night and differentiable dreams?

Daminark

Combinatorial dreams kthx

Honestly it seems like once an algebra problem becomes combinatorics or just putting theorems together I handle it so much better than I do anything else

Like I don't know if later number theory/combinatorics looks at all like this stuff but honestly that might just be the subject I should go for, I can reason about numbers

So yeah smooth dreams are all nice nd shit but like discrete is where it's at

Secret

[To be checked]

in Mathworks (Not the main chat!), 2 mins ago, by Secret

$$\omega_1 = \sup (m,n < \omega : S^m[n])$$

In words: Grab any computable countable ordinal already constructed, let $\omega_1^{CK}$ be the first ordinal not constructible from the countable union of any computable ordinal already constructed (inserted by some axiom I added), make a fundamental sequence using $\omega_1^{CK}$ as the base with cartesian products, succession and supremums. Iterate the above process countably many times. After that, one get the recursively inaccessible ordinals.

Iterate the above process countably many times with these ordinals as a starting base

Since $\omega_1$ is an admissible, eventually this will get you to it

In order to do the above, two axioms were inserted to create this particular model of ZF:

Existence of smallest supremum inaccessible ordinals

> For every set of ordinals, there exists a smallest ordinal that is not a successor ordinal, and inaccessible from any supremum of ordinals smaller than it

Axiom of smallest inaccessible ordinals

> For every ordinal with cardinality $\kappa$, there exists a smallest ordinal inaccessible from $\kappa$ many iterations of the following: Iterated construction of fundamental sequence and application of the axiom of Existence of smallest supremum inaccessible ordinals

Secret

10:45 AM

Therefore:

Let E,Reg,U,R,I,P,C be the axioms: Extensionality, Regularity, Union, Replacement, Infinity, Powerset, Choice.
\begin{align}
\Bbb{On}^{\text{ZF-E}} & = \text{I have no idea}\\
\Bbb{On}^{\text{ZF-Reg}} & = 0\\
\Bbb{On}^{\text{ZF-U}} & = \{0,1,2\}\\
\Bbb{On}^{\text{ZF-R-I-P}} & = \Bbb{On} \cap [0,\omega)\\
\Bbb{On}^{\text{ZF-I-P}} & = \Bbb{On} \cap [0,\omega 2 )\\
\Bbb{On}^{\text{ZF-P-2nd order definitions}} & = \Bbb{On}^{\text{Predicative}} = \Bbb{On} \cap [0,\Gamma_0 )\\
\Bbb{On}^{\text{ZF-P}} & = \Bbb{On}^{\text{Predicative}} = \Bbb{On} \cap [0,\omega_1^{CK} )\\

Waiting

11:17 AM

@Hippalectryon hey! How are you doing? :-)

@Secret how is it going?

Secret

Currently deep in set theory stuff

Waiting

Secret

My logic background is not very strong, which is why a lot of the stuff I said here sounds like rambles

It also does not help that I am actually procrastinating from writing the final data analysis code for my PhD, which is why I am too guilty to be able to spend time reading the logic textbook called forallx

Above all, I am currently deeply obssessed in trying to define $\omega_1$, the first uncountable ordinal explicitly

so in conclusion, not a very productive month

Hippalectryon

12:30 PM

@Waiting Hi :-) I have a PC now that's always on so I might appear online even when I'm not in front of the computer

Waiting

@Hippalectryon haha, that's slightly deceiving

:-)

Hippalectryon

It's been a long time since I've been here, anything new ? @Waiting

Waiting

@Hippalectryon The same here. I didn't enter the room for a long while. Well, except the fact these days are my best days ever, not that much. I'm just one step away from finalizing my project.

Hippalectryon

Great :D

Waiting

@Hippalectryon I made a couple of fascinating discoveries these days. I'm still overwhelmed by the beauty of the results.

@Hippalectryon You?

Hippalectryon

12:36 PM

Not much about maths, I've been doing more physics and CS lately

Waiting

I see.

Hippalectryon

Any news about the book project ? Last time we talked if I recall it hyad been called off ?

Waiting

@Hippalectryon Tell you more privately when there are news. ;)

user193319

0

Q: Finding a Bounded Measurable Set and Sequence of Simple Functions

Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m(E) < \infty$. Show that for each $\epsilon > 0$, show that there is a measurable set $F$ contained in $E$ and a sequence $\{\varphi_n \}$ of simple functions on $E$ such that $f$ is bounded on $F$ and $m(E-F) < \epsilon$...

real-analysis measure-theory

Hippalectryon

@Waiting great :-)

Waiting

12:42 PM

@Hippalectryon I wondered a couple of time if I would going to get stuck at some point, not being able to advance anymore. It's so nice things continously come to me.

@Hippalectryon It's as if mathematics comes to me naturally, without doing a particular effort, like you navigate on the sea, you're pushed in all directions by the wind.

I love what I do, that's clear, and every new discovery is a moment of great joy!

Hippalectryon

Still only on integrals/sums/..., or have you started other fields ?

Waiting

@Hippalectryon I'm looking a bit over fractals lately, but not doing research, more learning.

Hippalectryon

Ah yeah I've studied those in the past :-)

Waiting

@Hippalectryon Nice. If you have some links with free-to-read materials, let me know.

Hippalectryon

Ah that was some time ago, and I don't think I've stored any special papers on that subject :(

Waiting

12:55 PM

Ah, it's OK.

@Hippalectryon I also have a result that I think would help me to clarify once and for all the irrationality of Catalan's constant.

Hippalectryon

@Waiting That would be pretty awesome

Waiting

I need more time and energy investment in it, but all that after finishing my first project (just one step away).

user193319

Is it true that if two different metrics $d$ and $\rho$ on the same set induce the same topology, then they are equivalent, in these that $ad \le \rho \le bd$ for positive constants $a$ and $b$?

Waiting

@Hippalectryon I'm out with some tutoring. Talk later. ;)

user685252

cya

Hippalectryon

1:12 PM

See ya

user685252

How about a math bot @Hippalectryon?

Hippalectryon

@user685252 I'm currently fixing chemobot and despite its name, chemobot can be adapted to pretty much any chatroom

But I'm not sure what use it would have here

user685252

or we could just use the chembot

Hippalectryon

What would we use it for in this room ?

user685252

Wikipedia articles

Hippalectryon

1:16 PM

Other than that ?

user685252

dunno, that's a start :-)

There was a network wide bot that brought up the articles.

and definitions

Hippalectryon

I mean, having a bot for just one command is not really worth it

imo

user685252

true

(just a thought :)

Maneesh Narayanan

1:53 PM

math.stackexchange.com/questions/1103634/… Using Parseval's inequality. I can show that it is equality.Right? How could he say that (c) is true and (d) is false? He pointed out the counterexample also, I don't understand. Please help me.

user685252

2:22 PM

From "The Mathematician's Shiva," by Stuart Rojstaczer. While mathematicians can be expected to be difficult and unsuitable for human interaction, PDE people, as they are sometimes known for short, are certainly among the most normal of the lot. Topologists ...are the worst."

Tweeted by NGhoussoub on November 25, 2017 at 5:23 AM

Balarka Sen

@0celo7 Hmmmmmmm

I think not.

user685252

he won't get notified of that message

Gabriel Romon

2:48 PM

has anyone in here taken the GRE ?

Alessandro Codenotti

Suppose $X_0,X_1,\cdots$ are completely metrizable spaces (even Polish actually) and that $d_n$ is an equivalent metric on $X_n$ bounded by $1$. We define $\hat{d}$ on $\prod X_i$ by setting $\hat{d}(f,g)=\sum\limits_{n=0}^\infty \frac1{2^{i+1}}d_n(f(n),g(n))$ where $f,g\in\prod X_i$ and $f(n)=\pi_n(f)$. I can see that with the topology induced by $\hat{d}$ the product is a complete space, but I'm not sure why this topology is the same as the product one.

Ender Look

Does $a_n=a_1^{\log^{n-1}r}$ means $a_{n}=a_{1}^{(\log{r})^{n-1}}$?

Transcript for

$Mathematics$ Mathematics