Mathematics - 2019-03-26

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12:07 AM

Hi, my name is 🍌😹

Hi 😜

I'm interested in category theory diagrams

I'm writing this software tool (see my user profile for link)

for category theorists to diagram chase with using a database

I'm not a spam user, I'm really a regular user who's working on this cool tool (open source) for the community.

I would like some help on it if there are any Python coders in here?

I do python coding, but not much category theory. What is the problem?

Oh, the problem is me. I'm coding all alone

and I like social coding

I would like some people to work on it with

So I need those people who know python and think that what they see when they google "commutative diagrams" is beautiful

Looking at the project you might want to consider using something like pipenv for the environment.

pipenv.readthedocs.io/en/latest

global installation paths are not usually recommended.

12:12 AM

Nothanks, got to keep it simple ya know

what's that for anyway?

Oh

Well, I don't suspect those that will work on it will have any issue

likely a guru Python person with many envs will know exactly what to do in that case

it would simplify things further. lol.

?

but ok. It's not super important since you have only a couple dependencies i guess.

How ?

Dev Dependencies: PyQt5

so basically, I do pip install pipenv once for a lot of different projects.

as in I have pipenv installed.

12:13 AM

End User Deps: MikTeX-like package, Neo4j Browser

ohh... You have dependencies outside of using pip too...

Yeah, this software cannot be too easy to install, my installer is very professional looking, currently not tied into that code, but directs the user how to search for their MikTeX and or install it and does a test LaTeX rendering

interesting. you may want to look in the codereview chat room...

it's very snazzy, so what's left there is to do the same thing but for Neo4j graph database software

(the installer I mean)

Some body like Zeta (on codereview) might be able to help a lot... I'm not sure if he does a lot of category theory, but he does a lot of Haskell (not that I'm trying to conflate the two)... so he would probably be one of the better bets for asking for revision of code.

he is usually on the 2nd monitor chat room. There are a lot of people on those chat rooms that help each other with projects.

i'm not sure how many of them are adept at category theory though... still, this chat tends to emphasize a lot of small problems and occasionally goes off tangent.

you're project is probably too large for an actual question on codereview, but there is a lot of github activity in the chat rooms. gl.

12:20 AM

thx

12:59 AM

@AkivaWeinberger so we want $f$ to be positive at the maxima of $|T_n|$... and also negative

2 hours later…

Buddhini Angelika

3:09 AM

If a graph has two edge disjoint 1-factors, does the union of the two 1-factors always give a 2-factor for the graph? (2 regular spanning subgraph)

P.S. If it is a simple graph.

3:22 AM

@TobiasKildetoft nothing is sheltered and nice in academia when it comes to proper (= permanent + independent) positions. Then it turns into the really ugly shark pool.

the rationals are not evenly distributed

4:01 AM

That's obvious

I think a more interesting question is: Can there exists a dense set in the euclediean metric such that every point is roughly the same distance away from each other

2 hours later…

5:37 AM

@AlessandroCodenotti have you come here on two feet? :P

Alessandro Codenotti

Nope, I'm still asleep

stai venuto qui su due piedi

Akiva Weinberger

5 hours ago, by Leaky Nun

@AkivaWeinberger so we want $f$ to be positive at the maxima of $|T_n|$... and also negative

If $f$ were negative at all of $T_n$'s maxima and positive at its minima then $|T_n+\epsilon f|$ would have a smaller supremum than $|T_n|$

so positive at maxima and minima

Akiva Weinberger

?

Positive at the maxima would increase the supremum

5:45 AM

which is what we need

Akiva Weinberger

If at some of $T_n$'s extrema $f$ had the same sign as it, and at others $f$ had the opposite sign, then $|T_n+\epsilon f|$ would have strictly larger supremum

Doesn't need to be positive at the maxima and negative at the minima

There are loads of $f$s that don't do that

So now what you need to prove is,

all $f$s of degree less than $n$ satisfy this

(other than $f=0$)

what

Akiva Weinberger

That is, none have opposite sign as $T_n$ at all its maxima

This will prove the theorem

@LeakyNun I didn't like this because it's not general enough

I mean, a degree n polynomial can only have so many extrema

Akiva Weinberger

Well we're looking at its behavior at $T_n$'s extrema

not at $f$'s own extrema

5:50 AM

oh right

Akiva Weinberger

user image

$T_5$^

if f has opposite sign at all Tn's extrema

then f would have zeroes between the extrema

Akiva Weinberger

Er, up to a constant factor

which, uh, can happen?

oh, less than n

so it can't happen, qed

Akiva Weinberger

Yes!

And why do we want $\deg f<n$

5:52 AM

because it's always monic

Akiva Weinberger

Yeah

the polynomial that we optimize

Akiva Weinberger

So $2^{1-n}T_n$ minimizes it

uniquely

and the smallest supremum is $2^{1-n}$

nice job to have figured that out all by yourself...

Akiva Weinberger

Oh and we didn't even need the fact that the roots are in $[-1,1]$

As a corollary, $T_n$ is the only degree $n$ polynomial with all its maxima and minima level

(again up to a constant factor)

5:54 AM

and the optimized value is $2^{1-n}$

Akiva Weinberger

Oh and replace "maxima" with "local maxima" everywhere

Sorry

@LeakyNun Mhm

@LeakyNun Except for $n=0$

in which case the only monic polynomial is $1$ and that has supremum $1$

is there $f \in C^\infty[-1,1]$ with $\frac{1}{n! 2^n} |f^{(n)}|_\infty \to \infty$?

$\exp(-1/x^2)$ seems like a good candidate

$\sum e^{-\sqrt{2^k}} \cos(2^k x)$ is smooth and nowhere analytic

just how crazy are Fourier series

Akiva Weinberger

6:47 AM

Fabius function

In mathematics, the Fabius function is an example of an infinitely differentiable function that is nowhere analytic, found by Jaap Fabius (1966). It was also written down as the Fourier transform of f ^ ( z ) = ∏ m = 1 ∞ ( cos ⁡ ...

Defined as the probability that $\sum_{n=1}^\infty2^{-n}\zeta_n$ will be less than $x$, where the $\zeta_n$ are chosen randomly and independently from the unit interval

I don't know how I would compute that

It's defined on $[0,1]$

It says that $f'(x)=2f(2x)$ for $x\in[0,\frac12]$

I don't know why

This is also smooth and nowhere analytic

(again, no idea why)

3 hours later…

Akiva Weinberger

10:19 AM

Revisiting Do Carmo

There's this theorem that says it needs to be dimension 3 or higher

but I don't see where in the proof this is used

@Akiva for why the fabius function is not analytic, here is a calculation
https://math.stackexchange.com/questions/2087361/how-to-show-that-the-fabius-function-is-nowhere-analytic

@AkivaWeinberger are you familiar with the theory behind Fourier series?

anyway here's a food for thought

for $f : S^1 \to \Bbb C$ square-integrable, let $c_n := \displaystyle \int_{S^1} f(\theta) \exp(-i n \theta) (\mathrm d\theta/2\pi)$, and $f^\ast := \displaystyle \sum_{n \in \Bbb Z} c_n \exp(in\theta)$. It is known that $f^\ast = f$ almost surely.

(a) is $-^\ast$ idempotent? i.e. is it true that $f^{\ast \ast} = f^\ast$?

Akiva Weinberger

10:36 AM

@s.harp Oh that's clever

Still makes me wonder why that differential equation holds

(b) Can we "express" $f^\ast$ in terms of $f$? i.e. if $f$ is piecewise continuous then $f^\ast(\theta) = \frac12(f(\theta-) + f(\theta+))$, but what if $f$ is just square-integrable?

Akiva Weinberger

I thought $f^*$ is just $f$ except at discontinuities where it's $\frac12(f(x^+)+f(x^-))$

(by which I mean the left- and right-hand limits of $f$ at $x$)

Oh

This is what I get for writing before reading

Sorry

@AkivaWeinberger You need to use the definition of $F$ as the cumulative function of the random variables. $C^\infty$ was a simple step, but I don't have access to the paper right now so I don't recall it.

Akiva Weinberger

So the derivative is the distribution function, right?

yesh

but you need that it is itself the cumulative distribution to see that it is monotone and so $F(x)=0\iff x=0$

Akiva Weinberger

10:42 AM

@LeakyNun Is there a name for $\lim\limits_{\epsilon\to0}\frac1{2\epsilon}\int_{x-\epsilon}^{x+\epsilon}f(t)dt‌$

'cause that seems like a reasonable guess

aha yes that's exactly it

there's something wrong with me

Akiva Weinberger

I seem to remember that if $f$ and $g$ are square-integrable then their product is integrable

that's what I get by not recalling proofs

that is true

Akiva Weinberger

so that integral should always be defined

Dunno if the limit is always defined

yes, welcome to Fourier analysis

so does this seem idempotent to you?

Akiva Weinberger

10:45 AM

I sure hope so

that would involve interchanging limit with integral and also fubini, I guess

and also changing between 2D cartesian and 2D polar

ultimately I can see it working out

thanks

Akiva Weinberger

Polar?

you know, the same trick to evaluate $\int_\Bbb R \exp(-x^2) \ \mathrm dx$

ok so $\displaystyle f^\ast(0) = \sum_{n \in \Bbb Z} \int_{\theta \in S^1} f(\theta) \exp(-in\theta) (\mathrm d\theta/\tau) = \lim_{N \to \infty} \int_{\theta \in S^1} f(\theta) \frac{\exp(i(N+1)\theta)-\exp(-iN\theta)}{\exp(i\theta)-1} (\mathrm d\theta/\tau)$

now cut the last integral at $\pm \varepsilon$

$= \displaystyle \lim_{N \to \infty} \left( \int_{\varepsilon}^{-\varepsilon} + \int_{-\varepsilon}^{\varepsilon} \right) f(\theta) \frac{\exp(i(N+1)\theta)-\exp(-iN\theta)}{\exp(i\theta)-1} (\mathrm d\theta/\tau)$

the first integral tends to zero by Riemann-Lebesgue lemma (which is not hard to prove)

so we have $\forall \varepsilon > 0, f^\ast(0) = \displaystyle \lim_{N \to \infty} \int_{-\varepsilon}^{\varepsilon} f(\theta) \frac{\exp(i(N+1)\theta)-\exp(-iN\theta)}{\exp(i\theta)-1} (\mathrm d\theta/\tau)$

Akiva Weinberger

11:00 AM

OK question

Let $\mu$ be $1$ on a nonmeasurable set and $-1$ elsewhere

Is $\mu f$ square-integrable?

If $f$ is

but $\mu$ is not measurable...

Akiva Weinberger

This only applies to measurable functions?

@AkivaWeinberger yes, because the square of $\mu f$ is the same as the square of $f$

Akiva Weinberger

Is it in the $L^2$ space though

'cause $\mu f+f$ is not square integrable

and I'm pretty sure we want the sum of two square-integrable functions to be square-integrable

> In mathematics, a square-integrable function, also called a quadratically integrable function, is a real- or complex-valued measurable function for which the integral of the square of the absolute value is finite.

thanks lol I never knew that

Akiva Weinberger

11:04 AM

Ah OK

So why is the sum of two square-integrable things square-integrable

Equivalently why is their product integrable

probably some Cauchy-Schwarz thing

if $a_n \to L$ and $b_n \to M$ then $a_n b_n \to LM$

estimate $f$ and $g$ by simple functions

reduce to a problem about limits of real numbers

Akiva Weinberger

Ah OK cool

Lebesgue integral is very reductionistic

reduce to the case $f \ge 0$ a.s.

then reduce to the case $0 \le f \le N$ a.s.

then reduce to simple functions

I hope

Akiva Weinberger

11:20 AM

Oh I just noticed something

$2(T_n)^2-1=T_{2n}$

Proof (a) - write $T_n$ as $\cos(n\cos^{-1}(x))$ and use the double angle formula

nice

Akiva Weinberger

Proof (b) - the expression on the LHS has degree $2n$ and has its maxima at $1$ and its minima at $-1$

nice

Akiva Weinberger

and we proved earlier that that describes $T_{2n}$ is uniquely

so we know where the maxima of |T_n| are!

Akiva Weinberger

11:25 AM

$\cos(2\pi k/n)$ for $0\le k\le n/2$

For the minima it's the same except $k$ is half-integers in that range instead of integers

(Er, maxima and minima of $T_n$, not its absolute value)

(But same thing really)

11:57 AM

@MatheinBoulomenos hi

MatheinBoulomenos

@user123 hi

i wrote today the proof today we talked about $Q_{2^n} \to S_k$

for $k=2^n$ , i want to build an embedding into $S_k$ , so i want $\phi(x) = \sigma$ and $\phi(y) = \tau$ s.t $\sigma ^{2^{n-1}} = 1$ and $\tau^2 = \sigma^{2^{n-2}}$

so to find $\sigma$ is easy

$\sigma = (12\dots 2^{n-1})$ can work

can you help me find $\tau $ ?

MatheinBoulomenos

you don't need to think so concretely about $Q_{2^n}$. Do you know the proof of Cayley's theorem? you can build an embedding $G \to S_{|G|}$ for any finite group $G$

Ahhhh.

so that make it simple lol

would've been nice to write it explicitly though

MatheinBoulomenos

you can enumerate the elements of $Q_{2^n}$ and then write the embedding from Cayley's theorem explicitly

12:03 PM

yeah i will refresh my memory of the proof of this theorem (and maybe some properties of $S_n$ too )

thanks.

2 hours later…

1:53 PM

Are these the simplicial homology groups of the klein bottle or just the homology group (is there any difference?): topospaces.subwiki.org/wiki/Homology_of_Klein_bottle

Alessandro Codenotti

What is "just the homology group"?

I think what you're getting at is that, for instance, you can consider both singular homology and simplicial homology on a given space

@AlessandroCodenotti Don't know.

and these don't look the same at first glance

Yes, Hatcher talks about singular and simplicial homology.

1:57 PM

but the homology groups you get turn out to be identical up to isomorphism

Alessandro Codenotti

All homology theories (satisfying the dimension axiom so no extraordinary theories) agree on CW-complexes, so in particular singular and cellular homology are the same for CW-complexes

Oh, I see. So the link does contain the simplicial homology groups of the klein bottle.

There's a Q&A question on the main site on singular vs. simplicial homology:

34

Q: Difference between simplicial and singular homology?

I am having some difficulties understanding the difference between simplicial and singular homology. I am aware of the fact that they are isomorphic, i.e. the homology groups are in fact the same (and maybe this doesnt't help my intuition), but I am having trouble seeing where in the setup they d...

algebraic-topology homology-cohomology simplicial-stuff

the comments seem worthwhile

the bottom line seems to be: the construction of the homology group depends on what theory you use (e.g. singular vs. simplicial) but the homology group itself doesn't

which is handy

Are the course notes of my university professors necessary to revise for the competitions or is it better and more than sufficient to work on books?

What competitions, do you mean exams?

Usually it is a great advantage to consult the notes, as they tell you exactly what has been done. A book will teach you the field, but not necessarily help you understand the style that the prof. (who creates the exam) creates questions.

2:12 PM

aren't exams competitions?

no

they are examinations, as the name suggests

2:42 PM

@AkivaWeinberger having thought about it a little, I think the best way to approach the geometry problem is to argue that the relevant condition (centroid is on the incircle) is preserved by similarity transformations

hence you're free to rescale the sides, and therefore the (semi)perimeter as well

so one may (for instance) choose $s=(a+b+c)/2=1$ without loss of generality

that makes a lot of the formulas simpler, e.g. the inradius is identical to the area

competitions=contest that made not by your professor by Institution or other professors (after your graduation

Is this known?

$$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left(\sum\limits_{n=1}^{n=k} \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}+\sum\limits_{r=1}^{q-1} \frac{B_{2 r} k^{-2 r-s+1} \left(\prod _{i=0}^{2 r-2} (i+s)\right)}{(2 r)!}\right)$$

appears to be true whenever $\Re(s)>-(2q-1)$ where: $q=1,2,3,4,5,...$

In particular is the condition $\Re(s)>-(2q-1)$ known?

That looks an awful lot like an Euler-Maclaurin result

see for instance the expansion on the bottom of page four here: math.ucdavis.edu/~tracy/courses/math205A/…

I'm not sure it's identical but it like it should be related

3:05 PM

Does it say anything about strips of convergence?

The Euler Maclaurin formula it is yes.

It is asking how many terms of the Euler Maclaurin formula do we need in order to compute the Riemann zeta function in the complex plane?

$q$ is the upper summation index in the sum with the Bernoulli numbers.

This appears to answer it in the positive:
"By repeating the above argument we see that we have analytically continued
the Riemann zeta-function to the right-half plane σ > 1 − k, for all k = 1, 2, 3, . . .."

anyone wants to play the Noetherian ring game?

how does one win with Z[ε] where ε^2=0?

3:25 PM

Picking p^2 seems to work

I think you lose if you pick p^2

because I would pick pε afterward and give you (Z/p^2Z)[ε]/(pε)

that's the same as Z/p^2Z i believe

and I can pick p

(or ε)

I don't think (Z/p^2Z)[ε]/(pε) = Z/p^2Z

arxiv.org/pdf/1205.2884.pdf says that I always win with Z[X]/(X^2) = Z[ε]

problem solved

3:41 PM

whoops

Alessandro Codenotti

@LeakyNun Thinking about it as a scheme in light of Remark 5.2 makes it clear, I like this interpretation

I see that we're all nerds

Does it tell too much about me that I think this game would be kind of fun? — Olivier Apr 6 '12 at 8:06

When finding critical points for an optimization problem $\frac{df(x)}{dg(x)} = 0$, can you just set the differential of $f(x)$ to $0$ as the algebra suggest and ignore the "denominator"?

4:12 PM

Does $S_4$ contain any elements of order greater than $4$?

Write an element of $S_4$ in terms of its cycle decomposition. See what that implies.

2 hours later…

Akiva Weinberger

6:15 PM

@Perturbative Hint: find an element of $S_5$ of order six

3 hours later…

8:57 PM

@LeakyNun while looking at comments on a Q&A question I came across this statement I thought was amusing (in regards to a paper they were working on):

“But my proof is not complete yet. Actually it is just a bunch of conjectures which I could not refute so far. ”

lol

MatheinBoulomenos

9:08 PM

@LeakyNun fun fact: the math department board games night in Bonn has a list of games to vote on, and the Noetherian ring game is on that list

Alessandro Codenotti

9:27 PM

@MatheinBoulomenos I confirm

10:20 PM

🍌😹

Yes I'm here

@AkivaWeinberger nice avatar

Akiva Weinberger

Hi

Thanks

I'm 🍌😹

Are you into category theory?

Akiva Weinberger

user image

Unfortunately I don't know much of it

I know like the definition of a category but not much else

Well, there's not much to know for starters, but for advanced stuff there's a crap load

Akiva Weinberger

It's on my list of things I want to learn eventually

10:25 PM

Do you know python?

Akiva Weinberger

Yeah

Or at least I used to

O__O

Akiva Weinberger

I haven't touched it in years

Want to help me on my project?

Thta's what she said

Akiva Weinberger

It's pretty late where I am

I think I'm gonna go to bed soon

Sorry

10:26 PM

No worries

See my profile for link later if interested

It's about basic category theory and learning category theory and can lead to advanced stuff if the community builds up

It's a piece of software

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