Mathematics - 2018-02-03 (page 1 of 2)

Balarka Sen

00:02

@0celo7 Watching melon's roast of Logan right now

He put up a new Cringing With video

Ted Shifrin

@Balarka @EricSilva (for later): This seems a cool question to me.

Semiclassical

@0celo7 if only

physics.umn.edu/events/calendar/spa.HSTC/thisweek/index/…

That particular colloquium is hit or miss

Ted Shifrin

Was this the colloquium with the nuclear war discussion a few days ago?

0celo7

00:17

@BalarkaSen it's a repost

Semiclassical

No

0celo7

"school of architecture"

Semiclassical

That one was for the general Physics and Astronomy colloquium

Ted Shifrin

Oh, right.

Semiclassical

This one was the History of Science/Technology/Medicine colloquium

Ted Shifrin

00:19

Ohhh ...

Semiclassical

Sometimes they’re good, but sometimes they’re just dreadful

Ted Shifrin

Well, I applaud you for going to colloquia. I was astounded in my days at Berkeley by how few of the grad students went even semi-regularly. Same for UGA students (and faculty, for that matter).

Semiclassical

Yeah

0celo7

I'm amazed I'm the only undergrad who goes to anything

PVAL-inactive

We have a gerrymandering workshop tomorrow at least partially run by an algebraic topology professor (who incidentally was a student of JPM). Maybe he'll retire and become a politician but I don't think he lied on his thesis so he's got work to do.

0celo7

00:20

@PVAL-inactive wot

Semiclassical

I suspect I’m one of the few physics students who is on the mailing list for the math seminars

Ted Shifrin

I didn't follow that sentence, PVAL.

Well, Semiclassic, given that you hang out here permanently ...

PVAL-inactive

There's a politically active algebraic topologist here who was a student of JPM.

Ted Shifrin

"I don't think he lied on his thesis"?

PVAL-inactive

There's a student of JPM who is now a politician who lied on his thesis.

or seemingly so.

0celo7

00:22

Another student?

Ted Shifrin

Wait, who's JPM?

0celo7

@TedShifrin May, presumably.

PVAL-inactive

j peter may

Ted Shifrin

Oh, Peter May.

I don't know the J.

"Lied on his thesis"?

I'm glad some people are finally getting unapathetic about the international scandal that is the US.

PVAL-inactive

heck what's that guys name

en.wikipedia.org/wiki/Daniel_Biss This guy.

Maybe the really trivial error he had was a legitimate error.

Ted Shifrin

00:25

Oh, right. You mean lied in his thesis ... You've confuzled me.

PVAL, you have any cool ideas on this? It seems like it should be super concrete.

PVAL-inactive

I don't know shouldn't Chris know how to do whatever he wants to.

Ted Shifrin

LOL ... you know him?

PVAL-inactive

He's pretty talented.

I don't know him personally.

Ted Shifrin

That was an old question. I didn't really try it very hard. But now I'm more intrigued.

I think I want to complexify the real Gauss map to start with.

PVAL-inactive

So for S^2 this thing is independent of the embedding by a simple homotopy argument.

Why is this thing independent of the embedding for any other surface?

It seems like for S^2 you could maybe find an embedding where this thing is actually O(-1) with the usual subbundle definition of O(-1) inside S^2 \times C^2.

but I don't know how to recognize this thing otherwise.

assuming he computed c_1 correctly.

Ted Shifrin

00:38

I'm not used to thinking about Pauli matrices. :(

But, yeah, this mixing of real and complex bugs me. I want to work in the holomorphic category. :)

PVAL-inactive

I'm sort of having trouble picturing what a convenient embedding to use is.

Ted Shifrin

Hi again, @Antonios.

Antonios-Alexandros Robotis

00:58

oh hi @TedShifrin

when was that message sent

Ted Shifrin

Oh, when your icon descended. Probably 20 minutes ago.

Eric Silva

@PVAL-inactive the couple of people I know who actually know biss are of the opinion that he just made mistakes

Daminark

Hey there everyone!

Antonios-Alexandros Robotis

I wonder why that happened.

Ted Shifrin

Back from Chinatown, Eric?

Antonios-Alexandros Robotis

01:01

I don't think I logged in...

strange

Eric Silva

@TedShifrin yup

Ted Shifrin

Was it good, Eric? :)

Eric Silva

Veryy

I love dim sum

Ted Shifrin

Cool — now you don't have to hate me any more.

I don't usually do dim sum for dinner, though :)

Although that's a good idea :D

Eric Silva

We have a midnight dim sum thing in my circle of friends

We just go get it at ungodly hours

Ted Shifrin

01:04

Oooh, I should visit you :)

Eric Silva

Usually to procrastinate

Ted Shifrin

Yes, you're good at that :)

Eric Silva

Lol yes visit and we'll eat like royalty

Ted Shifrin

Hmm, well, maybe when I visit Michigan this summer (if you're in Chicago then).

Eric Silva

I will be at least some time here but I might spend time going elsewhere

Antonios-Alexandros Robotis

01:13

its a cold evening in NYC

Daminark

F

Balarka Sen

C

Daminark

Heh, that was a good one

Balarka Sen

@TedShifrin I looked at the question. I have no intuition for those matrices, admittedly

Ted Shifrin

Just the usual basis for $\mathfrak{su}(2)$, @Balarka.

Balarka Sen

01:14

I see

Daminark

Actually it reminds me, earlier we were waiting for tea time and someone asked how much longer it'd be

Response was t-6 minutes

Balarka Sen

Instant clap worthy

Daminark

Witnessing that was the high point of my existence tbh, it's downhill from here

Balarka Sen

Here's a thing we have been thinking about a little: math.stackexchange.com/questions/2284443/…

DeVito posted a new answer a few hours ago

Ted Shifrin

I like Jason. We've overlapped a number of times, and I met him a few years ago.

Leaky Nun

01:17

how many dimensions do we need if we want to embed $SO(n)$?

Balarka Sen

He's pretty cool

Leaky Nun

@TedShifrin

I feel like having asked this before

Ted Shifrin

I have no idea @Leaky. Obviously it fits in $\Bbb R^{n^2}$, but I don't know the minimal dimension.

(Other than Whitney, of course.)

Leaky Nun

$\Bbb R^{n(n-1)}$ as we previously established that its dimension is the same as $\bigwedge^2 \Bbb R^n$ @TedShifrin

Balarka Sen

Whitney doesn't give the minimal dimension here, too

Ah no wait probably

Ted Shifrin

01:19

Right, you can do one better with Whitney.

Daminark

Okay so I see that this person doesn't want to use AT but would the idea be to use that you can't write Z = G^2 for some group G?

Balarka Sen

@TedShifrin Right, right

Ted Shifrin

@Leaky: The Lie algebra is $\mathfrak{o}(n)$ which is isomorphic to $\bigwedge^2\Bbb R^n$. You lost a 2, of course.

Leaky Nun

you need to multiply the dimension by two to embed it

according to some theorem

Balarka Sen

@Daminark Z is a free group

Ted Shifrin

01:20

That's Whitney embedding, Leaky.

Balarka Sen

Hm, maybe not relevant

Leaky Nun

ah

Ted Shifrin

@Balarka: Perhaps free of minimum rank?

Leaky Nun

does (the number of dimension needed to embed a Lie algebra) have anything to do with (the dimension of the Lie group)?

Balarka Sen

Something of that sort. Ah, yes, G needs to be abelian

So that would do

Ted Shifrin

01:22

A Lie algebra is a vector space, Leaky.

Leaky Nun

oops

Ted Shifrin

It is the tangent space of the Lie group at the identity.

Balarka Sen

dimension of Lie algebra = dimension of Lie group

Leaky Nun

I swapped the two things apparently

does (the number of dimension needed to embed a Lie group) have anything to do with (the dimension of the Lie algebra)?

Ted Shifrin

No.

Leaky Nun

01:22

nice

Daminark

So yeah in that case if you could write the sphere as the square of a space, its homotopy groups would be the squares of those of the underlying space, but for Z you can't do that so rip

Leaky Nun

tangent space is a local property right

Ted Shifrin

Other than the fact that, as Balarka and I both said, the Lie group's dimension is the Lie algebra's dimension by definition of dimension of manifold.

Balarka Sen

@Daminark Correct

Well, $\pi_2$ actually but yeah

Leaky Nun

(but every point in SO(n) is the same I think)

Daminark

01:23

And actually that generalizes to all spheres

Leaky Nun

where by same I mean there is an auto-homeomorphism that maps any point to any point

Daminark

That's neat

Balarka Sen

Sure

Ted Shifrin

A diffeomorphism, Leaky, but this is unnecessary

All you're using is connectedness.

Leaky Nun

[0,1] is connected

Ted Shifrin

01:24

Huh?

If you take a connected locally Euclidean space, it has the same dimension everywhere.

Leaky Nun

alright, although every point is the same, tangent space is still a local property so we still can't infer the number of dimensions needed

Ted Shifrin

Embedding dimension of a manifold is very hard.

That's in part what characteristic classes help tell you.

Leaky Nun

is there a canonical isomorphism from the tangent space at the identity to the tangent space at any point in SO(n)?

Balarka Sen

@LeakyNun Any manifold has a self-diffeomorphism that takes any point to any given point.

Ted Shifrin

Yes, of course, using group multiplication, Leaky.

@Balarka, but far from canonical. Again, better say connected! Or it's wrong.

Balarka Sen

01:26

Right, for Lie groups it's a special identification

Thanks, connected

Also without boundary while we're at it :P

Leaky Nun

do things like cohomology help?

Ted Shifrin

My language is clear. Manifold is without boundary. I have to say with boundary if I want to allow boundary.

Balarka Sen

Sure @Ted

Ted Shifrin

Not cohomology by itself — that's all intrinsic.

Leaky Nun

1 min ago, by Ted Shifrin

That's in part what characteristic classes help tell you.

what do we know about that?

Balarka Sen

01:28

If $M$ is a smooth manifold of dimension $n$ that embeds in $\Bbb R^{n+k}$, taking the "normal bundle of $M$ in $\Bbb R^{n+k}$" gives a vector bundle $n$ of rank $k$ such that $TM \oplus n$ is the trivial bundle of rank $n+k$

\oplus

Ted Shifrin

To embed in certain codimension requires vanishing of certain Stiefel-Whitney and Pontryagin classes.

Balarka Sen

That's called being "stably trivial"

It's what certain elements of the cohomology group detects

It gives obstructions to which dimensions of Euclidean spaces you cannot embed your manifold in (but does not tell you much about which dimensions you can embed in, if it passes obstruction)

Ted Shifrin

For example, $\Bbb RP^n$ cannot be embedded in codimension $2$ unless there are certain formulas for $n$.

Leaky Nun

2 mins ago, by Balarka Sen

If $M$ is a smooth manifold of dimension $n$ that embeds in $\Bbb R^{n+k}$, taking the "normal bundle of $M$ in $\Bbb R^{n+k}$" gives a vector bundle $n$ of rank $k$ such that $TM \oplus n$ is the trivial bundle of rank $n+k$

$n$ is a very unfortunate name

Balarka Sen

Sorry :P

Make that $\mathbf{n}$ please

Daminark says I should use the thonk emoji to denote my normal bundle

Ted Shifrin

01:32

I use $N$, but topologists typically use $\nu$. But $n$? thwack

Balarka Sen

lmao

Leaky Nun

@BalarkaSen do that in your papers

Balarka Sen

I have already done it

(Vacuously, I have no papers)

Leaky Nun

hmm, a logic reference from Balarka

Daminark

^^ I vacuously practice what I preach

Ted Shifrin

01:35

Oy ... visualizing Demonark as a preacher.

Balarka Sen

Daminark

I'd rather not visualize that. Mostly because I'd rather not visualize period

But still

Leaky Nun

> I'd rather not visualize period

not sure who would rather visualize that

Ted Shifrin

Well, Demonark, you'd be one-dimensional, and even you can visualize that.

Balarka Sen

gotta love those 5th grade jokes

Oh @Daminark you just got roasted

Daminark

01:37

@TedShifrin you overestimate my abilities

Balarka Sen

@Daminark Do you know the loss meme?

Ted Shifrin

That's probably true, Demonark.

Daminark

Nope

Balarka Sen

Look it up in knowyourmeme

Im not going to say it

Daminark

Oh that

Balarka Sen

01:39

now check this out

Daminark

Well then

Bob

01:56

I recently posted this question, if anybody cares to look at it:

0

Q: Problem with Two Geometric Random Variables

Please consider the following problem and my attempt to solve it. Let $X$ and $Y$ be independent random variables having geometric densities with parameters $p_1$ and $p_2$ respectively. Find $P(X >= Y)$. Answer: \begin{eqnarray*} P( X >= Y ) &=& P( X = 0)P(Y <= 0) + P(x = 1)P(Y <= 1) \\ &+& P...

Narcissus jewel

@BalarkaSen loss

Adeek

@Narcissusjewel do you know little bit about what is the meaning of cycle class map ?

I was thinking if you get an element in the homology and you want to somehow define a cycle

orbit-stabilizer

04:37

1 + 3 is 4 minus 1 is 3

quick maffs

Xander Henderson

I... what?!

Daminark

04:51

Xander is confused

Secret

06:13

arxiv.org/pdf/1511.09230.pdf

A Type Theory for Probabilistic and Bayesian Reasoning

16

Q: How do we express measurable spaces using type theory?

A measurable space $(X,\mathcal X)$ consists of a set $X$ equipped with a $\sigma$-algebra of subsets $\mathcal X$. I would like to write computer programs involving measurable spaces, but to the best of my knowledge, no language (including Haskell) has the ability to deal with completely unstruc...

pr.probability measure-theory computer-science type-theory

Shago

08:28

Hey guys if we throw 4 dice what is the probability of getting one 6 ?

Anonymous

Which branch of math deals with solving recursion equations? And are there any popular books on that subject? I might need one because the CS books are doing crazy hand waving whenever it comes to recursive analysis of algorithms

Secret

Recurrence relation

In mathematics, a recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given: each further term of the sequence or array is defined as a function of the preceding terms. The term difference equation sometimes (and for the purposes of this article) refers to a specific type of recurrence relation. However, "difference equation" is frequently used to refer to any recurrence relation. == Examples == === Logistic map === An example of a recurrence relation is the logistic map: ...

so probably finite differences and numerical analysis in general

Anonymous

@Secret Ah, it seems Dover has a book on it: Batchelder, Paul M. (1967). An introduction to linear difference equations. Dover Publications

Anonymous

I should check it out the pdf first maybe

jublikon

08:51

@Shobhit Sorry I forgot to answer two days ago. I was very tired and went to gym and after that I forgot about my task.... :/

Secret

09:10

The $\omega_1$ discussion 0: Predicative uncountable well orderings

Oct 7 '17 at 6:25, 3 hours 7 minutes total – 133 messages, 4 users, 0 stars

Bookmarked 23 secs ago by Secret

The $\omega_1$ discussion 1: Physical meaning of a mathematical object with cardinality $\omega_1$

Nov 29 '17 at 15:49, 1 hour 33 minutes total – 76 messages, 2 users, 0 stars

Bookmarked 26 mins ago by Secret

The $\omega_1$ discussion 2: Necessity of Borel algebra in defining probability theory and stochastic processes

4 hours ago, 3 hours 14 minutes total – 77 messages, 3 users, 2 stars

Bookmarked 30 mins ago by Secret

The Great Duck

@Secret teach me how you did that?

Secret

click arrow next to "room" tab -> create a bookmark

after that, you can paste the links containing the bookmarked conversations

The Great Duck

ah

Bennett

Just wasted time on a question I didn't read properly

0

A: About an equivalent of $\displaystyle \int_{0}^{+\infty}\frac{\text{d}x}{\left(\alpha^2+x^2\right)^n}$ [ Checking a result ]

If $\displaystyle \mathscr{H}(\sqrt{\alpha}) =\int_{0}^{\infty}\frac{\text{d}x}{\left(\alpha+x^2\right)} = \frac{\pi}{2\sqrt{\alpha}}$ then $\displaystyle \mathscr{H}^{(n)}(\sqrt{\alpha}) = \int_{0}^{+\infty}\frac{(-1)^nn! }{\left(\alpha+x^2\right)^{n+1}}\,{\text{d}x} = \frac{(-1)^n(2n)!}{2^{2n+...

:( Shouldn't do maths on friday night!

jublikon

09:50

Hola :)

can someone help me veryfying if my proof is correct? That would be VERY nice !

I would like to show the convergence of $\sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}}$.

$\begin{align}
\sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} &= \sum_{k=0}^{\infty}{\frac{(2i)^{k+1}}{5^{k+1}}} = \sum_{k=1}^{\infty}{\left( \frac{ 2i}{5} \right)^{k+1}} \\
\end{align}$

Let $a_k = \left( \frac{ 2i}{5} \right)^{k+1}$ and thus a monotonously decreasing sequence.
And using the Leibniz criterion the sequence $\sum_{k=1}^{\infty}(-1)^ka_k$ converges. Thus $ \sum_{k=1}^{\infty}{\left( \frac{ 2i}{5} \right)^{k+1}}$

Secret

[Random]

0

[0,1,2,3,...,n]

[0,1,2,3,...

ω=[0,1,2,3,...]

Bennett

10:07

@jublikon what's $i$?

jublikon

$i \in \mathbb{C}$

@Bennett

Secret

[0,1,2,3,...,ω]

The Great Duck

@Secret can you review edits?

Secret

I am not sure, my rep probably not high enough. Let me ehck

The Great Duck

Ur rep low

Secret

10:14

yeah, I am way too lazy to answer questions on main

The Great Duck

I thought you had 100000+ rep

i have only 170 rep

@Secret i wish i had 2k rep so i could review

Secret

Restart:

⠀

$\underbrace{[]}_{\text{null}}$

$\underbrace{[0]}_{0}$

$\underbrace{[0,1,2,...n]}_{n}$

$\underbrace{[0,1,2,...}_{\text{Huge}}$

$\underbrace{[0,1,2,...]}_{ω}$

$\underbrace{[0,1,2,...]}_{m}$

The Great Duck

@Secret id actually have 2000 rep but i knew people wouldnt want me in the review queue so I intentionally made overrated bounties to drop my rep down to 0.

I.E. I lowered my rep intentionally to remove my priveledges

@MatheinBoulomenos hello

Secret

10:30

$\underbrace{[0,1,2,...]}_{0}$(check)

MatheinBoulomenos

@TheGreatDuck hello

Secret

$\lim_{m \to 0}\underbrace{[0,1,2,...]}_{m}$

The Great Duck

Huh?

Secret

I am testing something infinity related

The Great Duck

Ok

Secret

10:31

Short version: Trying to formalise finite time blowup as a foundational mathematical object

The Great Duck

Lol

u cant

Thats hypercomputation related

humans cannot do hypercomputation or understand it

Dattier

Hi,

$f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

The Great Duck

I think i may have to fake flag myself soon lest i die from lack of sleep

ugh

@Dattier i dont know anymore

Dattier

$\frac{2(a_1+...+a_n)!}{(a_1!)\times ...(a_k !)(k!)} \in \mathbb N$ with $a_i \in \mathbb N^*$

The Great Duck

Im going to bed

skullpatrol

10:42

cya

Dattier

Is it true that : $\frac{2(a_1+...+a_n)!}{(a_1!)\times ...(a_n !)(n!)} \in \mathbb N$ with $a_i \in \mathbb N^*$ ?

@TheGreatDuck good night

Secret

10:54

Is there really any way to algorithmetically answer functional equation questions:

MatheinBoulomenos

@Dattier $n=3$ $a_1=a_2=3$ $a_3=1$, so $a_1+a_2+a_3=7$, we get $\frac{2 \cdot 7!}{3!3!1!3!}=\frac{140}{3}$

Secret

I need to revise what happens when one integrate a convex function

Dattier

@MatheinBoulomenos : bravo

Now, $f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

Secret

$g=f''$

$g(tx_1+(1-t)x_2) \leq t g(x_1) + (1-t) g(x_2)$

Dattier

@Secret : do you talk with me ?

Secret

11:03

not yet, cause the message is incomplete

$f''(tx_1+(1-t)x_2) \leq t f''(x_1) + (1-t) f''(x_2)$

$\int_{[0,1]^2} f''(tx_1+(1-t)x_2)dx \leq t \int_{[0,1]^2}f''(x_1)dx + (1-t) \int_{[0,1]^2}f''(x_2)dx$

Shobhit

@jublikon its alright.

Mats Granvik

11:17

@AkivaWeinberger I have a number I want to show you.

https://oeis.org/A214533

$$\sum _{n=1}^{\infty } \left(-\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}-\frac{1}{\sqrt{n+3}}+\frac{1}{\sqrt{n+4}}+\frac{2}{\sqrt{n+5}}+\frac{1}{\sqrt{n+0}}\right)$$

=-2.54913

Secret

Ok I have no idea how to integrate the above

Mats Granvik

The sum has the same numerators as here: math.stackexchange.com/questions/883233/a-certain-harmonic-sum

$\frac{1}{2}+\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{3}}
=-2.54913$

Mats Granvik

11:39

$$\begin{array}{ll} 14.13565056860325 & \text{=Im[ZetaZero[18]]/2/2.5491277293} \\ 14.13472514173469 & \text{=Im[ZetaZero[1]]} \\ 21.02064364000642 & \text{=Im[ZetaZero[33]]/2/2.5491277293} \\ 21.02203963877155 & \text{=Im[ZetaZero[2]]} \\ 25.01182706734213 & \text{=Im[ZetaZero[42]]/2/2.5491277293} \\ 25.01085758014568 & \text{=Im[ZetaZero[3]]} \end{array}$$

$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1‌}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$

Dattier

12:31

$\text{Determinate the }n\in \mathbb N, P_1,...,P_n\in\mathbb Z[x],\text{deg}(P_i)\geq 2 \\
\text{with }\forall a\in\mathbb Z,\exists k\in \mathbb N,\exists (\alpha_i)_{i=1...k}\in \mathbb [1,n]\cap\mathbb N, P_{\alpha_1}(P_{\alpha_2}(...P_{\alpha_k}(0)))=a $

Dattier

12:45

Is it true that : $2^{k-a} | k!$ with $k=\sum\limits_{i=0}^n a_i2^i$ in base $2$ and $a=a_0+...+a_n$ ?

Secret

$$\frac{2^k}{2^{a_0}2^{a_1}2^{a_2}\cdots 2^{a_n}}$$

$k! = (a^{(n)}_i2^i)(a^{(n-1)}_j2^j)\cdots 2 \cdot 1$

$(a^{(n-1)})^+ = a^{(n)}$

0,1,10,11,100,101,110,111,1000,...

$a = a_0+\cdots + a_n \implies$

$$\begin{pmatrix}a^{(0)}\\a^{(1)}\\\vdots \\ a^{(n)}\end{pmatrix} = \begin{pmatrix}a_0^{(0)}\\a_0^{(1)}\\\vdots \\ a_0^{(n)}\end{pmatrix} + \cdots + \begin{pmatrix}a_n^{(0)}\\a_n^{(1)}\\\vdots \\ a_n^{(n)}\end{pmatrix}$$

Tuki

13:02

hmm what is going on ?

Secret

$k! =\text{sum} \left( \begin{pmatrix}1 \\ 0 \\ 0 \\ \vdots \\ 0\end{pmatrix}\otimes \begin{pmatrix}0 \\ 1 \\ 0 \\ \vdots \\ 0\end{pmatrix} \otimes \begin{pmatrix}0 \\ 0 \\ 1 \\ \vdots \\ 0\end{pmatrix} \otimes \cdots \otimes \begin{pmatrix}a_0 \\ a_1 \\ a_2\\ \vdots \\ a_n\end{pmatrix}\right)$

Silent

How to show that $2^x<1$ if and only if $x<0$?

Secret

Show that $2^x$ is strictly increasing, hence you can justify that $2^x$ is order preserving thus you can apply $2^x$ both sides without need to flip the inequality sign?

$$\frac{2^{a_0}2^{a_12}2^{a_22^2}\cdots 2^{a_n2^n}}{2^{a_0}2^{a_1}2^{a_2}\cdots 2^{a_n}}$$

Silent

Is it true that if $a>1$ then for $0<x<y$, $a^x<a^y$ and if $0<a<1$, then $a^x>a^y$?

user777

Hi! I have a book that says the path has 3 vertices. Does he mean that we have triangle like a-b-c-a or a-b-c-d?

this is in graph theory

I mean he says P3 (does he mean the number of edges is 3 or number of vertices is 3 but we have 2 edges only)

Silent

13:19

@Secret Got it!

thanks

Secret

graphclasses.org/smallgraphs.html#nodes3

user777: If that is all he said, then I don't think a path with 3 vertice is necessary be a closed path

user777

@Secret Thank you for this resource. So, I understand that it could be triangle or could be line of 3 vertices!

Secret

I am not sure what terminology your book use (you might want to check), but if they are talking about a vertex with 3 edges connected to it, the terminology (I think) should be "3-vertex connected"

An example of that will be a graph:

Here, the central vertex is 3-vertex connected, while all 3 possible paths involving 2 edges has 3 vertices each

user777

I see! but I don't think they say "3-vertex connected", I will cope paste the statement.

they say the following:

Observe that a graph is a cluster graph if and only if it does not contain a path on three vertices (P3) as an induced subgraph. As long as there exists a P3 as an induced subgraph, branch by choosing one of its vertices to delete

this is hint to solve a problem called "Cluster Vertex Delation"

This problem says that If we have a G that is not Cluster (Cluster means that we have several connected component of complete graph), then we need to design an algorithm to look for some forbidden induced subgraph in order to get a cluster graph

he says that: look for path on three vertices. Now, I'm confused about what exactly path on three vertices means? in Diestel's textbook if we have Path with 7 vertices, then the length of path is 6 edges. Does he mean the same definition of Diestel's or different!

Secret

13:37

A path is a sequence of edges, thus it cannot have branching structures like the graph I showed above

user777

I think the first path with 3 vertices means that a-b-c (but not a-b-c-d) so even the link you give it to me it controdict with Diestel's textbook

Secret

Thus this agree with the definition of a cluster graph, since as soon a path passing through 3 vertices is found in it, there are subgraphs that are not complete

user777

yes! so it must be "a-b-c"!

Secret

yup

user777

Thank you @Secret

Alessandro Jacopson

14:51

Hi chat

Hi @AlessandroCodenotti I heard about a plan of an army of Alessandros

Alessandro Codenotti

Ah, we meet at last! Together we shall conquer the chat

Alessandro Jacopson

:-)

Alessandro Codenotti

Where are you from?

Alessandro Jacopson

Imola

and you?

Alessandro Codenotti

Brescia, but I'm studying in Trento

Alessandro Jacopson

14:55

Bruno Kessler?

Alessandro Codenotti

Nope, university of Trento

Kasmir Khaan

My worst fear came true, double Alessandro , double trouble

3

I Heard rumors about this, never thought id see it in my own Eyes ._.

Alessandro Jacopson

@KasmirKhaan :-)

Kasmir Khaan

:)

Alessandro Jacopson

@AlessandroCodenotti PhD?

Alessandro Codenotti

15:06

Oh, no, just a third year undergrad!

What about you?

Alessandro Jacopson

A programmer... almost my questions here at math.stackexchange.com are in some way (sometime very loose way) related to my job

Alex K Chen

Hello @LeakyNun

Hello all

Hello @SimplyBeautifulArt

Simply Beautiful Art

Heyo

Anything I missed?

Alex K Chen

Remember when I an you and ETA were toddlers, we used to play games in a chat room ?

Simply Beautiful Art

Mhm

$$0!+1!=2!\\\Gamma(1)+\Gamma(2)=\Gamma(3)$$

Alex K Chen

15:18

Is the sequence $\{ \nu_p(2^{p-1}-1) \}_{\text{p prime}}$ bounded above by some large integer ?

Leaky Nun

@SimplyBeautifulArt ...

Simply Beautiful Art

@AlexKChen nu?

Hey Leaky

Alex K Chen

$\nu_p(x)$ Exponent of $p$ in the prime decomposition of $x$. For example $\nu_2(24) = 3$

Leaky Nun

@AlexKChen are you sure it isn't just 1?

Alex K Chen

@LeakyNun There was a prime in the range 1900-2000 (I forgot the exact value) which gave 2

Alessandro Codenotti

15:22

@AlessandroJacopson Cool, what kind of programming do you do?

Leaky Nun

oh

Alex K Chen

most prolly 1951

Leaky Nun

[[1093, 2]]

Alex K Chen

yeah

Leaky Nun

[[1093, 2], [3511, 2]]

Alex K Chen

15:23

So is the sequence bounded ? Is it ever more than 4 ?

Leaky Nun

no other found below 30000.

bye

Alessandro Codenotti

Ok, I think I'm missing something here. All the variables range over rationals: I have $j<rs$ and I want to show that $j=ab$ with $a<r$ and $b<s$. I did so by calling $q=rs-j$ and picking $k$ such that $s-k>0$ and $rk-q<0$ and considering $a=r-\varepsilon$ and $b=s-k$, with $\varepsilon=\frac{rk-q}{k-s}$. I'm sure there's a simpler approach, any idea?

Alessandro Jacopson

15:39

@AlessandroCodenotti machine vision

Del Monte

16:16

I have a programming doubt.. Can I ask it here?

Secret

17:32

[Random]

⠀

$\underbrace{[]}_{0}$

$\underbrace{[0]}_{0}$

$\underbrace{[0,1,2,...,n]}_{n}$

Shobhit

what are you doing?

Secret

testing sequence stuff, since chat is quiet atm

$\underbrace{[0,1,2,...}_{\text{Huge}}$

$\underbrace{[0,1,2,...]}_{\omega}$

$\underbrace{[0,1,2,...]}_{m}$

$\underbrace{[0,1,2,...,n]}_{0}$ (check)

$\underbrace{[0,1,2,...,\omega]}_{\omega+1}$

$\underbrace{[0,1,2,...,\lambda]}_{\lambda +1}$

$\underbrace{[0,1,2,...]}_{\aleph_0}$

$\underbrace{[0,1,2,...,n]}_{\omega}$

$\underbrace{[0,1,2,...,n]}_{\aleph_0}$

The $\omega_1$ discussion 0: Predicative uncountable well orderings

The $\omega_1$ discussion 1: Physical meaning of a mathematical object with cardinality $\omega_1$

The $\omega_1$ discussion 2: Necessity of Borel algebra in defining probability theory and stochastic processes

Transcript for

$Mathematics$ Mathematics