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12:07 AM
Hi, my name is šŸŒšŸ˜¹
 
Hi šŸ˜œ
 
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.
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…
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.
 
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
 
Nope, I'm still asleep
 
stai venuto qui su due piedi
 
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
 
?
Positive at the maxima would increase the supremum
 
5:45 AM
which is what we need
 
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
 
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
 
Well we're looking at its behavior at $T_n$'s extrema
not at $f$'s own extrema
 
5:50 AM
oh right
 
if f has opposite sign at all Tn's extrema
then f would have zeroes between the extrema
 
Er, up to a constant factor
 
which, uh, can happen?
oh, less than n
so it can't happen, qed
 
Yes!
And why do we want $\deg f<n$
 
5:52 AM
because it's always monic
 
the polynomial that we optimize
 
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...
 
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 $2^{1-n}$
 
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
just how crazy are Fourier series
 
6:47 AM
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 ) = āˆ m = 1 āˆž ( cos ā” ...
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…
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 $f^{\ast \ast} = f^\ast$?
 
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?
 
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.
 
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 $F(x)=0\iff x=0$
 
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
 
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
 
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?
 
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
 
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 $\forall \varepsilon > 0, f^\ast(0) = \displaystyle \lim_{N \to \infty} \int_{-\varepsilon}^{\varepsilon} f(\theta) \frac{\exp(i(N+1)\theta)-\exp(-iN\theta)}{\exp(i\theta)-1} (\mathrm d\theta/\tau)$
 
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...
 
This only applies to measurable functions?
 
@AkivaWeinberger yes, because the square of $\mu f$ is the same as the square of $f$
 
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
 
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
 
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
 
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
 
Proof (b) - the expression on the LHS has degree $2n$ and has its maxima at $1$ and its minima at $-1$
 
nice
 
and we proved earlier that that describes $T_{2n}$ is uniquely
 
so we know where the maxima of |T_n| are!
 
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
 
@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 $ ?
 
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
 
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
 
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
 
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?

user48168I 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...

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 Ļƒ > 1 āˆ’ k, for all k = 1, 2, 3, . . .."
 
anyone wants to play the Noetherian ring game?
how does one win with Z[Īµ] where Īµ^2=0?
 
3:25 PM
Picking p^2 seems to work
 
I think you lose if you pick p^2
because I would pick pĪµ afterward and give you (Z/p^2Z)[Īµ]/(pĪµ)
 
that's the same as Z/p^2Z i believe
and I can pick p
(or Īµ)
 
I don't think (Z/p^2Z)[Īµ]/(pĪµ) = Z/p^2Z
arxiv.org/pdf/1205.2884.pdf says that I always win with Z[X]/(X^2) = Z[Īµ]
problem solved
 
3:41 PM
whoops
 
@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…
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):
ā€œBut my proof is not complete yet. Actually it is just a bunch of conjectures which I could not refute so far. ā€
 
lol
 
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
 
9:27 PM
@MatheinBoulomenos I confirm
 
10:20 PM
šŸŒšŸ˜¹
Yes I'm here
@AkivaWeinberger nice avatar
 
Hi
Thanks
 
I'm šŸŒšŸ˜¹
Are you into category theory?
 
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
 
It's on my list of things I want to learn eventually
 
10:25 PM
Do you know python?
 
Yeah
Or at least I used to
 
O__O
 
I haven't touched it in years
 
Want to help me on my project?
Thta's what she said
 
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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