Mathematics - 2017-12-07 (page 4 of 4)

Leaky Nun

9:09 PM

hi everyone

Theo

any help with this?

0

Q: Creating a DFA that only accepts number of a's that are multiples of 3 and no substring aba

I wanted to better understand dfa. I wanted to build upon a previous question: Creating a DFA that only accepts number of a's that are multiples of 3 But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...

finite-automata

Leaky Nun

@Theo so would baaa be accepted?

Astyx

@MaryStar hint : write $\lambda = x+iy, (x,y)\in \Bbb R^2$

Mary Star

Ahh ok!! @TobiasKildetoft

@Astyx I see. And at the property that I showed do we have to substitute z=a+bi, or is it correct as I did it?

Theo

@LeakyNun yes

Leaky Nun

9:14 PM

@Theo so it's actually b*(aaa)*b* that you're trying to match?

Theo

@LeakyNun so would aaa or bbb (since 0 is divisible by 3)

@LeakyNun yes

Leaky Nun

I think you would need five states, with three accept states

can you do that yourself?

or six states

Theo

@LeakyNun its possible with 5 states? hmm okay I'll try, i thought i would need 7-8.

Leaky Nun

that was a rough estimate, it doesn't matter

Theo

okay i

i'll try and send it here

Eric Silva

9:17 PM

@Daminark lol

@Balarka I had a long talk with Farb and it looks like I'll probably end up doing a reading course with him or his wife on gromov stuff

Daminark

Damn, you and Benson are best mates, it seems

Balarka Sen

ooh neat

Astyx

@MaryStar I don't think so

brot

How can I show that $n\ln(1+(\frac{x}{n})^\alpha)$ diverges for $\alpha \in (0,1)$ and $x>0$?

Eric Silva

Benson is a bro

Literally one of my favorite profs

His math philosophy lines up so close to mine

Astyx

9:26 PM

@brot By using an asymptotical equivalent of $\ln(1+x)$ as $x\to 0$

Theo

@LeakyNun I believe this works?

Leaky Nun

accept states?

Theo

@LeakyNun a = 0 and b = 1, it was faster for me to add numbers

@LeakyNun q0 and q8, and q5 is a dead state

@LeakyNun so, with {0, 1}, we want a DFA that only accepts number of 0's that are multiples of 3 and no substring 010

Leaky Nun

oh, my regex was wrong

1*(0(11+)?0(11+)?0(11+)?)*1*

is this right?

Theo

@LeakyNun we could also have 1*(000)*1* can we not?

Leaky Nun

9:34 PM

@Theo that wouldn't accept 01100

brot

@Astyx And then l'Hopital, right?

Theo

@LeakyNun oh right, yes then your regex is correct

Astyx

L'Hopital ?

How ?

Theo

@LeakyNun wait my dfa is incorrect, there is no 1 going from q6

Daminark

@Eric yeah I get the feel that you're something of a Benson/Neves person, also to a lesser degree Schlag

brot

9:39 PM

If I write $\frac{1}{n} \ln(1+(xn)^\alpha)$ and consider $n \to 0$ I can use l'Hopital?

Eric Silva

Neves and I are actually weirdly alike as people

Theo

@LeakyNun q7 is incorrect too since we get 010, i'm getting stuck on q6 and q7

Leaky Nun

@Theo I'd probably just read my regex and implement the DFA state by state

Theo

@LeakyNun okay will try

Daminark

9:51 PM

@Eric I don't have much of a picture of Neves as a person, aside from being a total comedian

Eric Silva

he's chill and disorganized and we have similar music tastes

and speak the same language obviously

Daminark

Can relate to the disorganization bit

I mean I dunno if I've seen any profs who are organized, aside from like, Schlag

Eric Silva

also have similar math philosophies i think

Danny is pretty organized in my experience

Daminark

Ah, well I haven't interacted with him much but I could buy that

Eric Silva

his lectures and notes are super structured

Daminark

9:54 PM

Oh he uploads notes?

Eric Silva

no

but like his lecture notes he uses in class

Daminark

Ah

Eric Silva

i sit in front of the table usually so I've seen them laid out and it's extremely well put together

Daminark

Well that's at least two people then (I at least think Schlag is?)

Eric Silva

his office is sometimes a mess but i dont think it's because he's disorganized he just has tea everywhere probably cause he likes tea a lot

Daminark

9:58 PM

I see. Oh by the way, after a quarter of the guy's class, would you say you recommend it if he's teaching next year?

Eric Silva

absolutely

user193319

0

Q: Pointwise Limit of a Sequence of Measurable Functions

Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...

real-analysis complex-analysis measure-theory

Eric Silva

@Daminark are you gonna take the undergrad course in the spring

mick

Why is this not accepted ??

0

Q: Factoring non-ufd methods ??

I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd. I can come up with arguments for that , but I also have arguments in the opposite direction. For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...

ring-theory norm computational-complexity factoring unique-factorization-domains

Also I wonder about this

0

Q: Algorithm to count irreducibles in a non-UFD?

Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? ) I wonder about counting the irreducible elements bounded by a lower...

ring-theory algorithms factoring recursion

Daminark

@EricSilva probably not, between this reading course and just reading through some book, prob Aguilar, I'll say that I'm good

I may audit that but I don't want to use up a slot in the spring

Eric Silva

10:16 PM

That's fair

Do you know who's teaching it

Daminark

Nagpal

And I heard he's pretty good so that's something

Narcissus jewel

The sidebar always confuses and amuses me.

Eric Silva

Oh yeah Rohit is great I TAed for him

Daminark

10:36 PM

So I've heard. I'm possibly auditing his commutative algebra

MatheinBoulomenos

commutative algebra is great, you have to hear it

Daminark

Yeah I definitely will at some point, it'll be a question of whether being in commutative algebra and basic ring theory simultaneously is a bad idea

MatheinBoulomenos

hmm

Daminark

@Balarka so it seems the plan is to give something closer to a sketch of the quasifibration part of Dold-Thom and focus instead on the resulting homology bit

Theo

How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are
0's, or all the symbols in even positions within w are 0's, or both.

I want to construct a nfa from this, but I'm struggling with the regex part

Leaky Nun

10:44 PM

@Theo so it matches 010001010 and 101110101?

Balarka Sen

@Daminark Ah OK

Theo

@LeakyNun the first one yes, not the second one since it has 111

Leaky Nun

oops

I meant, 1010001010

Theo

@LeakyNun yes

@LeakyNun that matches

Leaky Nun

but then you never know whether you're in the first pattern or the second pattern

Balarka Sen

10:47 PM

@Daminark Whatever you said to me so far proves that $\pi_n \text{SP}$ is a homology theory in the Eilenberg-Maclane sense though, no?

Well modulo Mayer-Vietoris

That's how you show $\pi_n \text{SP} \cong H_n$, right?

after noting they agree at a pt

Daminark

Well, so the theorem itself I mentioned should be the key to verifying the axioms, yeah

Theo

@LeakyNun which is why im struggling with it

Daminark

But you'll need that quotient bit to get the exact sequence

Semiclassical

Just got back from a math talk: math.umn.edu/seminar/…

Balarka Sen

SP(A) --> SP(X) --> SP(X/A) thing, yeah

@Semiclassical oh schnoop dawg

Semiclassical

10:49 PM

the technical details were beyond me, but I followed it well enough

So that was neat

Daminark

Sick! Hope it was fun

Semiclassical

Yeah

Though attending talks at 3:30 pm are always a bit of a risky endeavor

Daminark

Why so?

Kevin Driscoll

soooo sleeeeepy

Semiclassical

It’s about the time of day where it’s easy to fall asleep :p

Daminark

10:52 PM

Ah

Semiclassical

Ding

Theo

@LeakyNun 1*(01*01*)* + 1*01*(01*01*)* + 0*

@LeakyNun would this do it?

Leaky Nun

how do you even +

Semiclassical

They also mebtioned quadratic differentials which is something I’d wanted to understand a while back

(I didn’t succeed, alas. But somehow it connects the kind of integrals I was interested in with Teichmuller theory?)

Balarka Sen

yeah you can think of the Teichmuller space as the space of quadratic differentials

It's a proof-worthy statement

nbro

11:00 PM

:D

What's up, guys?!

Theo

@LeakyNun yeah sorry, i think the one i gave is incorrect

Ted Shifrin

Demonark: I'm the very model of disorganization :P

Theo

@LeakyNun do you have any approaches for the regex?

Semiclassical

@BalarkaSen yeah, they had statements like that

Leaky Nun

@Theo no idea

it's still three things union together

Semiclassical

11:03 PM

Something like quadratic differentials are the cotangent bundle for the moduli space of genus-g curves

Balarka Sen

Sounds about right

MatheinBoulomenos

What does the half-plane modulo the action of $SL_2(\Bbb Z)$ like? does it have some nice description?

Semiclassical

A precise statement found via google: “The moduli space of holomorphic quadratic differentials is isomorphic to the cotangent bundle over the moduli space of complex structures.“

Theo

@LeakyNun no worries! thanks for the help though! :)

Balarka Sen

@MatheinBoulomenos It's well uh like a cone

Semiclassical

11:05 PM

So there :p

Balarka Sen

with the base having two non-manifold points

Daminark

@TedShifrin nice

Balarka Sen

(which are the "orbifold points" of order 2 and 3)

and the top of the cone going to infinity

MatheinBoulomenos

Wait why is this not a manifold? is the action of $SL_2(\Bbb Z)$ not properly discontinuous?

Balarka Sen

It fixes points on the boundary of the upper half plane

ABcDexter

11:15 PM

hi

does anyone know baout conditionalvariance?

MatheinBoulomenos

I meant the upper half plane without the real line

ABcDexter

*about

Ted Shifrin

I once did, @AbcD

ABcDexter

ohh, i had few questions

Balarka Sen

@MatheinBoulomenos well, sure, then it is properly discontinuous. But that's not the moduli space of the torus...

Ted Shifrin

11:17 PM

Depends on what your questions are.

MatheinBoulomenos

You don't want to include the real line. The points on the real line don't give you a lattice

Ted Shifrin

@Balarka: But you're looking at $1$ and $\tau\in\Bbb H$ to get the lattice, and $\tau\notin\Bbb R$.

Balarka Sen

Oh fair enough

Blah

So you get three cusps

I get it

Ted Shifrin

But the fundamental domain will have 3 points "at $\infty$"

Balarka Sen

and the two cusps at the bottom have that Z/3 and Z/2-symmetry

Ted Shifrin

11:20 PM

Hmm ... Maybe one of 'em isn't at infinity. You get the intersection of the vertical ray and the semicircle at the top of the semicircle, I recall.

Balarka Sen

Oh

Right

It's on the middle of the semicircle

Eric Silva

sup chat

Leaky Nun

"o" <-- this isn't a circle, not because it's imperfect: even if I manage to arrange the atoms such that it's perfectly continuous and perfectly circular, it still isn't a circle

MatheinBoulomenos

sup

Leaky Nun

a circle is the abstract concept in the Platonic universe

it's the equivalence class of all those things that look like a circle

@MatheinBoulomenos hi

what's the Galois group of the general quintic in F_p?

MatheinBoulomenos

11:26 PM

if you have an irreducible quintic it's cyclic of order 5

Leaky Nun

no, not that

MatheinBoulomenos

the general quintic over any field has Galois group $S_5$, if I remember the definition of a general quintic correctly

Leaky Nun

"the general quintic", a well-defined thing

oops, I forgot the definition

MatheinBoulomenos

The definition I know of general polynomials involves the extension $F(x_1, \dots, x_n)$ over the field of symmetric rational functions over $F$ in $x_1, \dots x_n$

this will always have Galois group $S_n$, regardless of the field $F$

the general polynomial of degree $n$ is then just $(T-x_1) \dots (T-x_n)$

which has symmetric polynomials as coefficients

because the symmetric rational functions are those rational functions fixed by the action of $S_n$ permuting the variables, this has Galois group $S_n$

By a theorem of Artin

Ted Shifrin

DogAteMy!

Leaky Nun

11:34 PM

@MatheinBoulomenos thanks

MatheinBoulomenos

the theorem of Artin I meant is if $G$ is a finite group of field automorphisms acting on a field $F$, then $F/F^G$ is Galois with Galois group $G$

that's an important step in the modern proofs of Galois theory

and also interesting on its own