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12:17 AM
could you possibly take a look at my question? it seems it has received very few attentions and probably is out of interest of community...
0
Q: calculate geometry properties using boundary coordinates

Enthusiastic EngineerCalculation of geometry properties of various shapes are useful specifically in engineering purposes. This is very easily formulated when the shapes of the geometry are rectangular, circle, etc. However, when the shape of becomes a combination of these and becomes more complicated, such calculati...

 
 
3 hours later…
2:56 AM
I thought I read somewhere that you can define ln(x) with exp(x). Is this true and what is the identity (besides the obvious e^y = x)?
 
For $x>0$, $\ln(x)$ is the unique real number satisfying $e^{\ln(x)}=x$
 
Yeah, but what is that inverse? Can you not define logarithms using exponential functions?
A closed form not using any logarithms
from what I can see based on things such as the infinite series of tanh(x) etc., we don't really need ln(x) to compute ln(x)... we can use exp(x) somehow. It also means we can compute arctan without i using exp(x) for all real values of x--somehow. We'd have to simply adjust the coefficient, not that I'm saying that would be easy :)
Some unknown function f(x) for all real values of x as exp(f(x)*x) maps to arctan(x), therefore.
 
3:29 AM
"infinite power tower"
That sounds so good and so funny at the same time
 
Hello all! I'm trying to prove that if $(B,+,\cdot)$ is a Boolean Algebra, then for all $x,y\in B$, $x+y=x+y'$ then $x=1$
But I can't prove it
I started with $x+y=x+y'$, then $x+y+y=x+y'+y$, so $x+y=1$, then?? Also, starting from LHS, then $x=x+xy=(x+x)(x+y)=x(x+y')=x+xy'$, then??
 
 
5 hours later…
8:17 AM
If two matrices have equal traces and determinants will they necessarily be similar?
ok so if they also have same characteristic polynomial then?
 
9:20 AM
There is a website where you can enter an (approximate) real number, and it returns (algebraic?) expressions equal or close to that number. – I can't remember the name of that site, can someone help me?
 
9:58 AM
https://math.stackexchange.com/a/3840927/811225
How was the expansion written?
 
10:19 AM
is dirichlet theorem hard to prove?
 
10:40 AM
what should I learn before understanding dirichlet theorem's proof ?
 
10:57 AM
why don't you first prove if an+b is odd for a and b is positive integer coprime and n is just positive integer
alright I have a hypothesis that tells math community doesn't really give a sh
 
@jeea No.
 
Sup @feynhat
 
Yo. Tell me something about free group with countably infinite generators.
 
Uhoh
What do you want to know
 
What do its quotients look like?
 
11:05 AM
Yikes dude. You get all countably generated groups.
 
That sucks.
What does a puncture in the surface with infinite genus deform to? I mean surely it is some product of commutators, right? But can it be an infinite product of commutators?
 
@feynhat That's a good question. Remember that the surface of infinite genus (let's say, one-ended infinite genus surface, for the sake of simplicity) deformation retacts onto a wedge of circles.
If you add a puncture to the first copy of $T^2$ in $T^2 \# T^2 \# \cdots$, what happens to it under this 'deformation retraction from infinity'?
It won't be a product of commutators, product of commutators aren't nullhomotopic in the infinite genus surface anymore.
 
Hmm... I can't see what happens to this puncture under the def ret.
 
11:20 AM
Here's a general question. Suppose $M$ is a noncompact manifold which deformation retracts to a subcomplex $X \subset M$ (this is usually called the core of $M$). What is the homotopy type of $M \setminus p$ where $p$ is some point on $M$?
 
@BalarkaSen Does the def ret take the puncture to two meridians in the first copy of T^2?
 
What do you mean by "the puncture is taken to ____"?
 
I mean the image of the loop around the puncture under final stage of def ret.
 
But the image of the deformation retaction of $\Sigma_\infty$ and $\Sigma_\infty \setminus \{p\}$ are not comparable, are they? Their core subcomplexes are different!
If it's a wedge of $2\infty$-many circles for the first case, it's a wedge of $2\infty + 1$-many circles for the second case. Do you see that?
 
Oh yeah. That makes sense. Where exactly is this extra circle being introduced? A meridian of the first T^2 right?
 
11:33 AM
Yup, exactly. Let me draw some picture.
So on the right I have a picture of an apeirogon. It's the standard picture of a $4\infty$-gon in hyperbolic plane, which is drawn so that the vertices lie on a horocycle in $\Bbb H^2$, which is a circle in $\Bbb H^2 \cup \partial \Bbb H^2$ tangential to $\partial \Bbb H^2$ at exactly a point (the point drawn in dirt green)
The horocycle is drawn in red, the apeirogon is drawn in blue. You can quotient the apeirogon by hyperbolic isometries to get $\Sigma_\infty$, paste the edges consecutively like you'd do for a finite $4n$-gon, except for infinitely many quadruplets of edges.
I have marked the domains which quotient to handles on $\Sigma_\infty$ downstairs, so that $\Sigma_\infty = T^2_1 \# T^2_2 \# T^2_3 \# \cdots$
The deformation retraction of $\Sigma_\infty$ to the wedge of $2\infty$-circles is given, at the level of the apeirogon (i.e. before quotienting) by pushing inwards from the dirt green point (the point at infinity), the def ret is shown by the dirt green arrows.
Now if you have a puncture at the orange point, you deformation retract to wedge of $2\infty$-circles, wedge a lollipop. This is because of what happens to deformation retracts if you add a puncture: see the image on the left. The top image is the deformation retraction of the "glass of water" $D^2 \times I$ to $D^2 \times \{0\} \cup \partial D^2 \times I$, by throwing rays from a point above the "glass".
If you have a puncture in the interior of the glass though, you get deformation retracted to a bubble around the puncture, with a stick connecting the bubble to the base of the glass.
 
11:52 AM
Damn.
I was drawing a literal T^2 # T^2 #...
I didn't know this could be realized as a quotient of a $4\infty$gon.
 
Yeah, it's a fun picture. Of course, you can also see it from your picture, with more concentration.
 
what does it means by order of magnitude of prime? Does it mean magnitude of prime number?
size of prime number is $\pi(x)$ but it says the order of magnitude of $pi(x)$ is x/log(x)
so what does it mean?
 
What if it is infinite on both ends? @BalarkaSen
Does it still def ret to wedge of $2\infty$ circles? I suppose, it does.
 
@feynhat Yup. You might find it interesting to draw the corresponding apeirogon
 
12:14 PM
In general if $M$ is a noncompact smooth manifold you can always find a deformation retraction to a codimension $1$ embedded subcomplex. Choose a triangulation (blue), and take a simplex, it's barycenter, and draw a smooth embedded path starting at the barycenter and escaping to infinity (red)
Now "insert a finger" along the red path by drawing a tubular neighborhood of it (orange) and deleting it from $M$. There's a deformation retraction from $M$ onto $M$ with orange tube deleted, and the deformation retraction looks like a finger move.
Finally project outward from the boundary of the orange tube to deformation retract that simplex to its boundary (top right, deformation retraction shown in green).
One has to do this systematically to all simplices to find a def ret of $M$ to a subcomplex. To do this you need to find a red path which connects all the barycenters of all the simplices, and this can be done. Then you insert a finger along this path "from infinity" and do the above procedure.
 
Curious, is there any particular reason why using the function of a circle creates such a good approximation of the circular functions? Even putting them into the circular functions, you get approximations of the hyperbolic functions.
Maybe I should just make an AI to solve this for me...
 
1:04 PM
@Stupidquestioninc it means $\pi(x)$ grows like $x/\log x$
 
@EdwardEvans their limit is 1 as x reaches infinity
 
the limit of their ratio approaches 1, yes
 
ok I understand I had problem with notation so I was confused
 
What was the notation? $\mathcal{O}(x/\log x)$ ?
 
George Floyd
Black Lives matter
 
1:07 PM
@EdwardEvans No book gave a wrong notation. It should actually be big O notation.
 
I see
 
Yeah O notation
 
what notation was given in the book?
 
Edward is right
I think it was theta
 
There is a big theta notation too
 
1:08 PM
oh yeah
 
I mean big theta is equivalent to omega and big O
CLRS has very good explanations
 
you can see (vi)
 
yeah, thanks @stupid
what is main topic here?
 
Hardy writes rubbish
 
1:09 PM
Main topic is people complaining to each other about hating math
@Balarka pretty broad question man
 
@BalarkaSen Damn this is beautiful.
 
Hi @ACuriousMind
 
29 messages deleted
@BalarkaSen Hi :P
 
...and today marks the day when a bunch of mathematicians defeated China.
 
1:16 PM
@ACuriousMind Awkwaaard
Why can't you come to this chat at normal times
 
If only wars were math duels... imagine...
Hey!! You, country! I have conflict with you!! Let us duel at noon!* Later: Awww, where did you get a quintic formula!? Here, take my land!
 
@BalarkaSen I'm not sure I know what "normal times" are anymore ;P
 
Truth, me neither.
 
but maybe I should drop by at times when it's not a flag summoning me...
 
Indeed, feel free.
@EdwardEvans Let me introduce you to a metalhead. He listens to... JINJER and ARCH ENEMY
 
1:20 PM
is it @Alessandro
 
ACuriousMind
 
it me
 
o
hahaha
waddup @ACuriousMind
 
not much, slow weekend, but that's fine
 
Nothing wrong with slowing down
I love watching a classical performance and a movement ends and literally everyone coughs loudly and clears their throats
weird grammar in that sentence
 
1:23 PM
So, why's my taste in music noteworthy? Have I stepped into a den of classical music lovers?
 
@BalarkaSen (I don't know anything about the hyperbolic plane) Can I still embed this apeirogon on a horocycle in the Poincare disk? I don't think it should be possible...
 
@ACuriousMind @Balarka and I regularly share DSBM artists so your taste in music is fine
 
lolol
@feynhat Yeah that's not possible anymore, the horocycle has only one end.
 
Because on that apeirogon, one of the end T^2 is always like reachable.
@BalarkaSen Yeah. That's what I wanted to say.
 
Yup
I don't actually know the right way to see it in $\Bbb H^2$ immediately. It should be a bunch of hyperbolic quadrilaterals truncated at two antipodal vertices and pastes side by side, both sides shrinking at infinity
But I would be interested if you could see this as part of a tessellation of $\Bbb H^2$ and make $F_{2\infty}$ act on by hyperbolic isometries
 
1:29 PM
I don't know what many of those words mean.
But as far as the def ret goes, I think it should again be $2\infty$ circles + lollipop.
 
That's true
But what I'm saying is the fundamental domain is gonna be an infinite Batman logo
 
bro that's not a batman logo
 
lol aight
 
It is if Batman has been the victim of some horrible experiments
 
lol
it's like a stretched astroid
 
1:37 PM
The Batman of Zur-En-Arrh is a fictional character in the DC Comics universe. In this story, the character is the alien Tlano from the planet Zur-En-Arrh. == Publication history == Batman of Zur-En-Arrh first appeared in Batman #113 (February 1958), in a story titled "Batman—The Superman of Planet-X!". It was written by France Herron and drawn by Dick Sprang. In the story, Tlano, the Batman from Zur-En-Arrh, brings Earth's Batman to his planet to help him battle giant robots piloted by an unidentified alien race. While on the planet, Earth's Batman discovers he has "Superman-like" powers through...
 
@Thorgott That's $\infty$-batman logoid.
 
lol fair
 
??? Let $\Bbb Z \to F_{2\infty + 1}$ be the inclusion induced map $\pi_1(S^1) \to \pi_1(\Sigma_\infty - p)$ (where the $S^1$ is a loop around $p$) and $\Bbb Z \to F_{2\infty}$ is trivial. Then, the pushout $F_{2\infty + 1} \ast_{\Bbb Z} F_{2\infty}$ is again $F_{2\infty}$?
 
That's equivalent to saying if you forget a petal in a bouquet of infinitely many flowers it's still a bouquet of infinitely many flowers
$\infty - 1 = \infty$
 
let's compute the symbolic dynamics in the fundamental region of the $\infty-$ batman logoid. It'll be fun they said
 
1:43 PM
@BalarkaSen Bye.
 
Lol
 
1:57 PM
@BalarkaSen quick math
 
If in my 4x4 matrix I get second row as zero can I say vector $[0\,1\, 0\, 0]^T$ is part of null space? I know this can be said if my second column would have been zero, but what if the second row is zero, thanks!
 
try some examples
 
No its not, I tried a simple example and it seems if any second column elements are non zero then $[0\, 1\, 0\, 0]^T$ is not part of nul space... am I correct
 
indeed
now convince yourself that $(0100)^T$ being part of the null space is in fact equivalent to the second column being zero
 
@Thorgott Yeah now I understood after looking at example. thanks !!
 
 
1 hour later…
3:19 PM
Can an ideal of a ring $R$ be a free $R$-module of rank greater or equal $2$?
Necessarily, $R$ would have to be non-commutative
 
@Thorgott If $V$ is a countably infinite dimensional vector space, $R = \text{End}(V)$ has the property that $R \cong R^2$ as left $R$-modules.
Rank is an awkward notion in noncommutative rings. Look up invariant basis number property, and Leavitt path algebras.
 
3:35 PM
@MartinR Answering my own questions: It is the Inverse Symbolic Calculator.
 
right, I shouldn't have said rank, I just mean free on $2$ or more generators
but of course, for rings with $R\cong R^2$, $R$ is an ideal of itself and free on $2$ generators, so that takes care of it
 
 
1 hour later…
4:45 PM
Let $u(z)$ be a harmonic function on a domain $D \subseteq \Bbb{C}$ containing the disc $\{z \in \Bbb{C} : |z-z_0| < \rho \}$. Using the polar form of the Cauchy Riemann equations, I want to show that $0 = r \int_{0}^{2 \pi } \frac{\partial u}{\partial r} (z_0 + re^{i \theta}) d \theta$.
Here is my work: One of the CR equations in polar form says $r \frac{\partial u}{\partial r} = \frac{\partial v}{\partial \theta}$, so $r \frac{\partial u}{\partial r}(z_0 + re^{i \theta}) = \frac{\partial v}{\partial \theta}(z_0 + re^{i \theta})$. Hence, integrating from $0$ to $2 \pi$ with respect to $\theta$, we get
$$r \int_{0}^{2 \pi} \frac{\partial u}{\partial r} (z_0 + r e^{i \theta}) d \theta = \int_{0}^{2 \pi} \frac{\partial v}{\partial \theta} (z_0 + re^{i \theta}) d \theta = v(z_0 + re^{2 \pi i}) - v(z_0 + re^{0i}) = 0$$
And why does $$\int_{0}^{2 \pi} \frac{\partial v}{\partial \theta} (z_0 + re^{i \theta}) d \theta = v(z_0 + re^{2 \pi i}) - v(z_0 + re^{0i})$$?
 
4:58 PM
isn't that just
the fundamental theorem of calculus
naively
 
algebraist solves difficult analysis problem, subsequently earns fields medal
 
using FACTS and LOGIC
also I'm flattered that you use the word algebraist, but please refer to me in future as "master student out of his depth"
 
FacTs and logiC
 
black hole forms
 
professional corner-crier solves difficult analysis problem..
 
5:03 PM
hahaha
 
@EdwardEvans I too might consider calling you an algebraist
 
So then why does the problem I am working say that I need to use the polar CR equations?
 
You already used them didn't you?
Arithmetician is the correct terminology
 
Oh, wait! I am being a dummy...I thought $\frac{\partial u}{\partial \theta}$ was in the first integral...
 
@EdwardEvans You do arithmetic geometry?
 
5:06 PM
Now I see why the CR equations are needed.
 
@Tobias nah I like algebraic number theory and spend most of my time not understanding papers about it
I guess that falls under arithmetic
 
why are you calling that arithmetic
 
that does sound close to arithmetic geometry (potentially)
 
arithmetic is like addition and multiplication smh
 
@Tobias I haven't even taken "bAsIc" algebraic geometry yet so I'm not sure if I know if I'll enjoy it, but elliptic curves are cool so
arithmetic is the way of saying number theory without letting people know that you like number theory
 
5:08 PM
Then just get started. It is not like Hartshorne is hard to read or anything :)
 
Idk if that's irony or not, I got a copy though and I'll start reading it before the semester starts
 
It was somewhat ironic
I have read parts of it, and I found it very hard going
It assumes a decent backgrouns in what the main ideas of AG are
It is not that you can't understand it without that, but it is hard to figure out why things are done the way they are without it
 
lol, actually I had a lecturer during my bachelor who recommended some introductor AG books that were geared toward arithmetic geometry
but I can't remember the author
 
I think Ravi Vakil's notes might be a better startign point these days
 
Eisenbud perhaps
 
5:11 PM
I have Vakil's notes too, the skim I did of the first few pages was entertaining
nah it wasn't Eisenbud
 
Vakil's notes are the best imho
 
maybe I should read something as well
 
I think the authors were Ulrich Görtz and Günter Harder
 
Goertz-Wedhorn maybe?
 
yeah that's the one
 
5:14 PM
scary name
 
Goertz is a professor somewhere in NRW iirc
 
That a person could be a master baker and simultaneously an algebraic geometer is astounding
Düsseldorf I think
Na Essen
Modulräume abelscher Varietäten in positiver Charakteristik, Shimura-Varietäten
Affine Weyl-Gruppen und Iwahori-Hecke-Algebren
Seems pretty arithmetic to me
 
Well, except the last parts
Which are distinctly algebraic in nature
 
(only knows of the Hecke algebra in a number theoretic context)
 
Yeah, these are slightly different
There is some connection, but I never really understood it
Actually not just slightly different, Iwahori-Hecke algebras are non-commutative
 
5:23 PM
OH SHIT algebra fest
bunch of NERDS
Fuck outta here Imma post pictures
 
Flagged
 
yeah there's another Hecke algebra in this other book which is defined as a space of locally constant compactly supported functions on a unimodular group together with some convolution
rofl
 
To heck with your hecke algebra dude
 
there's probably a connection there too
weeeey
 
hecking algebra
 
5:26 PM
Iwahori-Hecke algebras arise the the endomorphism algebra of the module induced from the trivial module of a Borel subgroup (as far as I recall)
Mostly, they get defined in terms of generators and relations
 
bleh
 
Let me post a picture to trigger y'all algebraists
Oh shit wait wrong picture
 
some kind of
spectral shit
 
Shhhh
You were not supposed to see that
 
hahaha
 
5:30 PM
I have blackmail material now
 
Hi all.
 
Good
 
Hey Anakhro
 
lol when ur gf swiping thru ur phone and sees some algebra
 
hahah
 
5:32 PM
Hey that's cool, apparently the Hecke algebra is kind of the extension of the Reps <-> group algebras equivalence but for smooth reps of locally profinite groups
 
smooth?
 
p-adic smoothness lmao
 
oof
 
yeah it just means open stabilisers
idk
 
where is the picture
 
5:33 PM
What's the motivation for calling it smooth? Does it yield similar results to smoothness in the usual sense?
 
i literally do not yet know, I'm on read 0, if you know what I mean
 
I know exactly what you mean.
 
The Iwahori-Hecke algebra if the decaterification of projective functors on the principal block of category $\mathcal{O}$.
 
i'm out
 
decaterification
 
5:34 PM
Dekaterifikation is the German word for "hangover cure"
 
@geocalc33 I do not have the image, but here is an alternative image: i.imgur.com/Yt0wnJ0.png
 
something inside me just died
 
@TobiasKildetoft ...
 
@anakhro haha good one
 
I seem to have forgotten a syllable. Ahh well.
 
5:38 PM
oh decategorification? now it makes perfect sense
 
Actually, I should probably have specified that it needed to be the graded version of that category. But eh, details
 
6:05 PM
@Thorgott good
 
7:04 PM
Is Nash Equilibrium a "solution" of Prisoner's Dilemma?
Can we say that rational prisoners will confess?
 
7:55 PM
I am trying to show a matrix B is similar to A by using diagonalisation, but my problem is that eigenvectors of A, B had complex elements. So I first set eigenvalue matrix $D = T A T^{-1}$ and $D = U B U^{-1}$ from here I took $A = T^{-1}U B U^{-1}T$ and took $P = T^{-1}U$ so that $A = P B P^{-1}$ but now the last equation is not holding correct
Does this mean that complex eigenvectors mean $A$ and $B$ cannot be similar??
Sorry ignore this it was calculation blunder
 
 
3 hours later…
10:38 PM
@Archer Yes, since defection is a dominant strategy.
 
10:57 PM
Using function counting as an example, would the loosest sense of the term "permutation" apply to every case in which the codomain is distinguishable, and "combination" to every case in which the codomain is indistinguishable, regardless of whether or not the domain is distinguishable, and whether or not the functions are restricted to have left-uniqueness and/or right-totality, or are these definitions too loose?
 
I am regularly frustrated by the fact that I can't write an inline summation symbol $\sum\limits_{i=1}^n$ without messing up the spacing of the lines above and below (at least not without writing it incorrectly as $\sum_{i=1}^n$).
Also, /smash is a bad idea
 

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