Mathematics - 2016-08-31 (page 1 of 2)

The whole point is that if $M$ is noncompact, $g$ might get closer and closer to $0$ so that trick no longer works. That's why you do partition of unity to do the same trick but varying $\epsilon$ (by doing it on compact subsets of $M$ inside charts, say) and gluing those $\epsilon$'s togather.

This is the idea for all partition of unity proofs: do on small pieces, then glue.

0celo7

@BalarkaSen Actually, one can prove the more powerful: Let $F:M\to \Bbb R$ be cont. and $\delta:M\to\Bbb R$ be positive and cont. Then there is a smooth function $G:M\to\Bbb R$ such that $|G-F|<\delta$.

Balarka Sen

@0celo7 Agreed. That's why complex manifolds are harder, I have been told.

No partition of unity there.

0celo7

12:19 AM

Are they analytic?

Dumb question.

Balarka Sen

Holomorphic functions are analytic, yes.

0celo7

THAT'S WHY I SAID DUMB

Balarka Sen

:P

0celo7

So, given that theorem, pick $F=\delta=\frac{1}{2}g$

Then you have a smooth $F$ such that $|F-\frac{1}{2}g|<\frac{1}{2}g$

To my surprise, this doesn't seem to be in Hirsch

Balarka Sen

Probably because all of that is not really hard.

0celo7

12:22 AM

But the result was known in '61

He should at least state it

Maybe it's an exercise

I'm analyzing the Nomizu-Ozeki paper for my advisor and he doesn't like when I reference Lee

If it's in Hirsch, I'm good...

@BalarkaSen long shot, but do you know what the symbol $d(x,X)$ on the bottom of page 55 of Hirsch means?

Balarka Sen

@0celo7 I don't have Hirsch with me but it usually means the infimum of $d(x, y)$, $y$ varying over $X$.

0celo7

@BalarkaSen Agreed but I was asking what $d(-,-)$ means.

Balarka Sen

Eh. The metric?

0celo7

What metric?

Balarka Sen

Dunno. I'd have to look at Hirsch.

0celo7

12:37 AM

it's a manifold, not a Riemannian manifold

and he never talks about Riemannian distances

it's not in the list of symbols

oh, he seems to use it in some proofs

@Huy Jesus christ go to bed

Balarka Sen

I should sleep too

0celo7

I should finish this proof

I should buy my prof Lee for his birthday

So he can get some good math

Forever Mozart

12:54 AM

poopoo

0celo7

eat da poopoo

Forever Mozart

dont do that

i think

0celo7

@ForeverMozart Nah. Builds the immune system.

Forever Mozart

i am quasi-famous now

but there are still many problems in my area that I cannot solve

0celo7

you're famous?

Forever Mozart

1:05 AM

not quite

its hard to get people interested in my research area

but I think it is fascinating

0celo7

what is it

Forever Mozart

weird topological spaces

but they can be metric

or even in the plane

Balarka Sen

@ForeverMozart did you publish your paper

@Soham Hey

Forever Mozart

I sent it off

no word yet

Balarka Sen

aha

Forever Mozart

1:13 AM

i have no idea when I will hear back

0celo7

@ForeverMozart lol

my topology prof said there are people like you

your dream is to make textbook exercises

Balarka Sen

@ForeverMozart you should do shape theory

Forever Mozart

no i dont care about books

Balarka Sen

the goal is the same: to study weird topological spaces. but a lot more tools - useful if you want to get more people interested

also good research possibilities

0celo7

> Shape theory is a branch of topology, which provides a more global view of the topological spaces than homotopy theory.

...

that's pretty global

Forever Mozart

1:18 AM

what is an example

of a shape

Balarka Sen

It tries to generalize algebraic topology to weird topological spaces

0celo7

this is a circle

Balarka Sen

@ForeverMozart I don't know much about it. You should look at Mardesic's works.

The basic example I think is the closed topologist's sine curve.

All of it's standard homotopy groups are 0 - yet it's not homotopy equivalent to the circle

Forever Mozart

can you apply it to non-compact spaces?

Balarka Sen

I think one proves that by showing the first shape group (analog of fundamental group) is nonzero or something

@ForeverMozart I'd bet.

0celo7

1:22 AM

the hell is a shape group

Balarka Sen

you should be able to do this for very disconnected spaces too - that's Cech theory

homotopy theory for non-path connected spaces. good stuff.

Forever Mozart

where do I learn that?

Balarka Sen

dunno; google is your friend.

Balarka Sen

1:42 AM

Um.

You'd think people would lower their tone down a few notches, given who's answering the question?

Forever Mozart

lol that is overkill and circular

0celo7

2:22 AM

> Your password may not contain more than 3 consecutive characters from a singles class (Upper, Lower, Number, Special). Please enter a different password.

smashes keyboard

Cbjork

that sounds absolutely awful. It seems to make the password easier for a brute force attack and harder to remember.

0celo7

why easier for brute force?

I'm not remembering this

It's scribbled on this paper with some geometry

Cbjork

You're more likely to try to use a shorter password, also there are fewer combinations of length $n$ than just random. for example, qqqqqqqqqq is a 10 character string not included in the set of passwords required.

0celo7

min of 14 characters

Cbjork

allowed, not required

0celo7

2:31 AM

also they make you use upper, lower, numbers, and special

How does brute force work on websites anyway

they lock you out after 3 tries

Cbjork

I'm just saying that putting these requirements makes the list of possible passwords much smaller.

0celo7

yeah I guess

Cbjork

It's probably a good idea to use upper, lower, numbers, and specials anyway

0celo7

my go-to password is geometryBest

Cbjork

Well I hope you don't use it on anything important, because these chat logs are public, I'm pretty sure

0celo7

2:34 AM

that's not actually my password

user360471

Hello

user61230

3:29 AM

DID YOU KNOW?

user61230

If you take any two coprime integers $a$ and $b$, their product is an integer that has a square.

0celo7

do you mean a square root

Secret

4:44 AM

@Emrakul $gcd(a,b)=\frac{a\cdot b}{lcm(a,b)}$, coprime means $1=\frac{a\cdot b}{lcm(a,b)}$ but there's nothing in terms of $c^2$ or $a^2$ or $b^2$ here?

arctic tern

@Secret every integer has a square (the square of x is x^2, which exists)

#humor

Secret

O i see, I mistaken as the integer obtained above is a perfect square of another integer

Agawa001

9:21 AM

good @daniel you saved the situation

Vrouvrou

10:49 AM

hello, i want to study the continuity of \begin{align*}
\varphi: & E \longrightarrow \mathbb{R}\\
& f\mapsto \varphi(f)=f(0)
\end{align*}

in $(E,||.||_{\infty})$

Tobias Kildetoft

What is $E$?

Vrouvrou

a normed space

(E,||.||_{\infty})

Tobias Kildetoft

then $f(0)$ does not make sense

Vrouvrou

sorry

$E=C([0,1],\R)$

Can i say that

Agawa001

@TobiasKildetoft hmm identicon falsification ?

Tobias Kildetoft

10:51 AM

@Agawa001 ???

Agawa001

just see how do u look from the icon-lineup

Tobias Kildetoft

@Agawa001 there is apparently some problem with gravatar (I have different icons in the lineup and next to where I write this)

Vrouvrou

can i say $|f(0)|\leq \sup_{[0,1]} |f(x)|=||f||_{\infty}$ then there exists $c=1$ such that $|\varphi(f)|\leq c||f||_{\infty}$ ?

@TobiasKildetoft ?

Agawa001

yes appearently , but members as balarka arent affected

(nepotism)

or he does pay a regular fee maybe

Vrouvrou

can someone help me ?

Balarka Sen

11:00 AM

@Agawa001 lol

@Tobias I fixed it by getting a snapshot of the "right" avatar and saving it as gravatar instead of the randomly generated identicon.

Maybe you'd want to try that out. Mike did the same.

mick

This was put on hold :/ . Its a good question ( 2 upvotes , yet on hold ?! )

2

Q: Find $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $

I wonder about a closed form for $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $ I tried solving the related equation $ f(x) ^3 = 1 + (x+y) f(x+1) $ for various fixed values integer $y$ , but I failed. It appears $$ ? = \sqrt[3] {1 + \sqrt[3] 5} $$ But I am not able...

:/ meh

Tobias Kildetoft

11:19 AM

@BalarkaSen Nah, either it corrects itself or I end up with a new icon. Either way is fine by me.

Balarka Sen

11:39 AM

@TobiasKildetoft Fair enough.

What're you upto?

Tobias Kildetoft

@BalarkaSen Preparing a grant application as well as a job application

Balarka Sen

I see.

Tobias Kildetoft

What about you?

(ohh, and my coauthor just dropped by with the print copies of our Advances paper)

Balarka Sen

Reading a lemma I need to prove the Poincare-Hopf theorem. Almost done.

Tobias Kildetoft

nice

0celo7

11:55 AM

@BalarkaSen Suppose $f:E\to F$ (Banach spaces) is continuous along every continuous curve through $x_0\in E$. Is $f$ continuous at $x_0$?

Balarka Sen

Yes: if it was discontinuous at $x_0$, there is a limit of points approaching $x_0$, say $\{a_n\}$ such that $f(a_n)$ does not approach $f(x_0)$. Look at the piecewise path joining $a_n$.

0celo7

That's the idea.

But making "piecewise path" precise takes a bit of work

Balarka Sen

Nah.

Your spaces are all vector spaces.

0celo7

So?

Balarka Sen

So you think.

I am not going to work it out for you.

It makes sense to join $a_k$ and $a_{k+1}$ by a line is the point.

user 1618033

12:01 PM

@JasperLoy Indeed, I was pretty busy. Then I spent a part of the night on very important research.

Huy

@BalarkaSen: there's an even more elementary solution to the geometric problem. :P

I can't believe none of us saw it

Balarka Sen

hm?

user 1618033

@TobiasKildetoft related to a previous message of yours, imagine this situation: you have at a large table the most important mathematical figures (from the mathematics you're interested too) from all history, all alive, and waiting for you to join them. Would you dare to bother them with what you produced so far in mathematics? Would they need to see your work?

No need to answer me.

Tobias Kildetoft

@user1618033 For most of them, no, they would not care or even have any idea why the sort of questions I look at could ever be called interesting.

user 1618033

@TobiasKildetoft OK

Tobias Kildetoft

12:06 PM

On the other hand, Lusztig would probably belong on that list (even though he is still alive) and he has cited some of my work and will certainly care about several of the things I have done

Balarka Sen

Oh, Lusztig has cited you? Then you're semi-famous, @Tobias.

Tobias Kildetoft

@BalarkaSen Heh, yeah. Now I am just waiting for the money to come rolling in (not sure what is keeping it).

0celo7

@BalarkaSen I don't need you to work it out for me, I solved it.

Balarka Sen

@0celo7 OK, good to hear.

0celo7

But proving that thing is continuous without the gluing lemma is not fun

Balarka Sen

12:09 PM

Why not use the gluing lemma?

0celo7

And proving it is continuous at $x_0$ requires some convexity magic

@BalarkaSen because we haven't covered it in class?

Balarka Sen

Oh well.

Yes, maybe you'd take off a point because my curve doesn't pass through $x_0$. But one can fix that.

Tobias Kildetoft

@0celo7 Then prove it by writing first a proof of the gluing lemma, then follow with a proof using the gluing lemma

Balarka Sen

@TobiasKildetoft Nice.

Money as in, you're getting some prize or a grant?

Tobias Kildetoft

@BalarkaSen that was meant as a joke (being famous should equate to getting money, right?)

Balarka Sen

12:11 PM

Ah.

Yes, indeed.

user 1618033

Thinking seriously to get out of any kind of mathematical community, I've seen enough, and continue mathematics absolutely alone (away from any nonproductive opinion). It would be enough to publish once in a while some papers, books (if the case).

0celo7

@TobiasKildetoft No, I just said "due to the gluing lemma"

I doubt the prof will object to it

he might say he doesn't know what it is though...

I don't think it's a standard name

Balarka Sen

I think it doesn't have a standard name

Or rather, it's so standard it doesn't have a name

0celo7

Gluing lemma is what Lee calls it

Munkres calls it the pasting lemma?

Balarka Sen

I thought Munkres also calls it the gluing lemma. Maybe I misremembered, you're probably right.

0celo7

12:16 PM

It's my topology prof who calls it the pasting lemma

Balarka Sen

Yes, I have seen both being used.

0celo7

gluing lemma is not in the index of Munkres

yup, pasting lemma

Someone sent me an email with the wrong name attached, but it's definitely directed at me

hmm

Agawa001

12:35 PM

@user1618033 have you ever planned to wake up early in daily basis to practise some yoga in middle of the forest, recall some juvenile moments, process some mathematical calcs with perception alone, try to forget that you are in forest or even in romania or earth itself, just focusing on the abstract, then see what wonderful results would come into you

user 1618033

@Agawa001 never tried practising yoga. Do you practise yoga?

Agawa001

@user1618033 i learn with internet lol

bad student i am

user 1618033

@Agawa001 hehe, well, it's a beginning.

Agawa001

@user1618033 it's wonderfuuuul, once you get to that stage of 'forgetting the surrounding' you can consider that you made halfway of the session

user 1618033

(It might help, as some claim)

Agawa001

12:39 PM

it is like a "just-in-mind picnic"

user 1618033

@Agawa001 But this takes time, and I need time to invest in math, especially these days when I have a lot of stuff to do.

Agawa001

@user1618033 well i do this because sometimes my mind refuses to turn on

user 1618033

@Agawa001 And you feel it works?

Agawa001

@user1618033 at some point yes

user 1618033

I see. Interesting.

Despite my belief, I actually believe in almost nothing that cannot be proved in a rational manner. I need to convince myself that Yoga works, and then I don't believe in those energy acumulation points called in Yoga.

Agawa001

12:42 PM

@user1618033 this is better than self-shocking one's brain

brain stimulation

user 1618033

@Agawa001 :-)))

Agawa001

@user1618033 its good for constant thinking, and uninterrupted concentration

user 1618033

@Agawa001 I reached this stage by self-education, then being able of constant thinking and uninterputed concetration for many hours in a row

But if there is room for more, it's great.

Agawa001

also for that phenomenon when ur thinking switch to something else as another topic somehow connected to the main one unwillingly

@user1618033 gooood, that is an achievement

user 1618033

It seems I have a problem with concentration when typing in English. :-)

Agawa001

12:49 PM

@user1618033 well it is different for me cuz of the irresistible noise i have here

user 1618033

@Agawa001 you live in the city?

Agawa001

@user1618033 no! but there is unwanted noise

user 1618033

@Agawa001 I see.

Gigili

hallo

Why does the following equality hold where $W_k=\text{diag}(w^{(k)}_i=\lvert x^{(k)}_i+\sigma\rvert^{p-2})$?
$$\sum_iw^{(k)}_ix^2_i=x^TW_kx$$

user 1618033

1:45 PM

I'm honest, I want to revolutionize my mathematical area, to make the life of people studing this area much much better than ever before. I have results, of course, and there is still an incredible amount of work to do.

Danu

→ 135 messages moved to Trash

Time to cut it out, guys.

This discussion was completely unconstructive, and it's bad for the general atmosphere of the chat room.

Agawa001

i ll get lost to pepare some coffee @chriss : my beforelast note , if you want to take depart, do it with ur own dignity, by

Danu

→ 8 messages moved to Trash

user 1618033

@Agawa001 OK

@Agawa001 I'm at risk of becoming pretty unproductive today. Back to my work to finish some proofs.

BBL

Pentapolis

2:02 PM

Is there any one interested in Fractional calculus here?

0celo7

I've always thought that was a party trick

Jose Manuel Ferreira Benavides

I do not know what that is, Pentapolis. My knowledge goes as far as Integral calculus' basics.

user 1618033

At the beginning of this week I sent by email to some students I train the following integral

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x$$

They have a week to come up with a very simple solution to it that use no complex numbers.

So far I received mail that it seems impossible, of course it is not, but it might be just a bIt more difficult.

Please let me know, if the case, how you teach in a class your student to very easily calculate such an integral.

Thanks.

I'm around working on my research stuff.

BBL

i change my mind.

Please calculate it without pen and paper. How?

BBL

Balarka Sen

2:26 PM

Another drama?

user 1618033

@BalarkaSen sorry if my integral is a drama to you. Just skip it as usual.

Balarka Sen

They aren't.

But apparently I missed a big fight.

user 1618033

@BalarkaSen This is only a place of peace and love. Don't think negatively anymore. ;)

BBL

0celo7

::steps over corpse::

whoa...

user 1618033

I think it would be positively nice to also give you the closed-form

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

So lovely, isn't it!

@Agawa001 give it a try when you're back :-)

OK, back to my research, I feel I spend too much time on this chat. BBL

Semiclassical

2:39 PM

I forget what $G$ is

ah, catalan's constant

0celo7

never heard of it

Semiclassical

I've heard of it, but only that. reference: en.wikipedia.org/wiki/Catalan%27s_constant

Mithlesh Upadhyay

Should be deleted? math.stackexchange.com/questions/1900288/…

Semiclassical

looks to be both not general enough and circular at the same time

i.e. it applies only to numbers which are products of prime powers

@user1618033 i think a good hint for that would be which integral representation of Catalan's constant they should be looking for

Jose Manuel Ferreira Benavides

tries to solve with u sustitution, then by parts. Gets stuck. Finds out this involves infinite series. Flips table.

Please restrict your problem professor. A warning of sorts @user1618033

Semiclassical

2:47 PM

e.g. $G=\int_0^1 \arctan{t}\,\frac{dt}{t}$

Semiclassical

3:07 PM

@user1618033 Playing around with Mathematica, I find that the substitution $x=\frac{e^u-1}{e^u+1}$ gives the equivalent form $$\frac{1}{2} \int_0^\infty \frac{u\,\text{gd }u}{e^u+1}\, du$$ where $\text{gd }u$ is the so-called Gudermannian function.

which is cute if likely useless

Agawa001

3:54 PM

@user1618033 i didnt recognise u fom ur new look, ur identicon is misbehaving buy a new one

Semiclassical

there's been a bunch of that going around

Agawa001

yes appearently , it is better to adopt a fixed picture

ok what we 'v got here

i could solve this without that annoying arctan

log(1+x)/(x+1) , variable substitution e=log(x+1)

for log(1-x)/(1+x) , same

amWhy

@MithleshUpadhyay Thanks for providing the link. Yes, that accepted answer with currently score at -4 should be deleted. I've put my vote (to delete) in the mix. But, as long as it remains the accepted answer, we'll need 27 more votes to delete

Jose Manuel Ferreira Benavides

4:13 PM

d = x_B y_C - x_C y_B

Guys, please, can i get an explanation about the scalar d in the second answer of this? It is not explained in the answer. (Maybe someone can add a edit).

4

A: Determining if an arbitrary point lies inside a triangle defined by three points?

Method To test any ppoint $P=(x,y)$, first move the origin at one vertex, like $A$ such that $$ B \rightarrow B - A $$ $$ C \rightarrow C - A $$ $$ P \rightarrow P - A $$ Then I calculate the scalar $ d = x_B y_C - x_C y_B $ and the three Barycentric weights $$ \begin{matrix} w_A = \frac{ ...

user 1618033

4:26 PM

Back.

@Semiclassical OK

@Agawa001 To me it seems all is unchanged.

user 1618033

4:39 PM

I wonder if robjohn ever cared to calculate this integral $$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

user 1618033

4:59 PM

Maybe I should say once in a while what I do in mathematics, that is making artwork, turning my mathematics into artwork (to avoid possible misunderstandings).

The integral above shoud be included in some textbooks at least (together with a shining proof).

user227867

5:20 PM

@user1618033 Today I went out for a haircut and walk, lol.

user 1618033

@JasperLoy You made yourself beautiful? :D

@JasperLoy Walking is GOOD!

user227867

@user1618033 Yes, even though I am already very beautiful, lol.

user 1618033

@JasperLoy hehe, sure. But to add some more I meant.

user227867

@user1618033 Was it a Monica who came yesterday to your house?

user 1618033

@JasperLoy lol, not really ::D

@JasperLoy How is it going? Rest of the things. Some date then? (after haircut and walk)

user227867

5:26 PM

@user1618033 So so. I hope to sort out more thoughts in September. No dates. =) There is no Laura here.

user 1618033

@JasperLoy Too bad then. :-)

user227867

@user1618033 When the time is right, Laura will appear, and so will Monica, lol.

user 1618033

@JasperLoy I'm not concerned about such things these days (anyway I find it very easy to solve :-)), but about finishing more important objectives from my research.

user227867

@user1618033 I am trying to ignore that user but her stupid comments keep appearing everywhere on the site, hehe.

user 1618033

@JasperLoy I also received a lot of strange comments today, and I think it comes from an old users that uses a different account. But I won't talk more about it. I have some ideas about that who.

user227867

5:32 PM

@MithleshUpadhyay It seems that the answerer cannot delete the answer if it is accepted.

user227867

@user1618033 You mean ideas about your strange comments or my user?

user 1618033

@JasperLoy hehe, my comments might be sometimes strange, it depends on what one understands, no matter what I say. ;)

user227867

@user1618033 Sorry, what I mean is you think you have ideas about who I am referring to?

user 1618033

@JasperLoy No. I was only referring to comments that were addressed to me.

user227867

@user1618033 I see. Anyway, no more 'negative energy'! I am flagging that user's stupid comments. Hope she gets suspended from the site forever...

user227867

5:38 PM

Everywhere I go, I see her attacking other users using sarcasm.

user 1618033

@JasperLoy Forever? It means a lot. :-)

user227867

@user1618033 Well, this user does not contribute to the site at all, even though she has a huge rep. The points mean nothing.

user 1618033

@JasperLoy I don't contribute to the site either these days. :-) I come in chat and fight with some users. :D

user227867

@user1618033 I don't want to mention this user, but she keeps abusing me and other people through comments and emails.

user 1618033

@JasperLoy There is so much negativity in this chat sometimes, it almost seems unreal.

user227867

5:42 PM

This nonsense has not stopped for years and years and years, despite people being nice to her again and again and again.

user 1618033

@JasperLoy Well, don't pay much attention to it. Focuse yourself on positive stuff. ;)

user227867

@user1618033 Yes, let's talk about other things then. =)

user 1618033

@JasperLoy OK. I'm leaving now for a jogging session, but we can talk later.

@JasperLoy Very long time hard work said a word to me lastly. I need to make many changes, and follow rules like in a army.

@JasperLoy Dressed and ready to go.

Talk later.

BBL

user61230

6:17 PM

DID YOU KNOW?

user61230

For all numbers $a$ and $b$ coprime with the exception of a shared factor of 7, the Collatz sequence contains only integers.

Balarka Sen

um

0celo7

what

Balarka Sen

dunno what collatz sequence of a pair of integers mean.

Agawa001

google+ has changed its style

user61230

6:21 PM

I have started to confuse the math chat. Soon, they shall be under my control.

user61230

Uh. Wrong room.

Balarka Sen

low-quality troll tbh

user61230

Not really trying to troll, honestly. If it's seen that way, then I'll stop.

Balarka Sen

Nah was joking.

I think the crackpot mathematics mathgen comes up with is better.

user61230

I'm glad my math is better than an algorithm designed to generate crap ;)

Balarka Sen

6:29 PM

Hah.

Prototank

Folks, I am having a crisis. Is there a way to deduce from $s^2=e$, $r^n=e$, $srsr=e$, that $sr^j\neq e$?? I am running in circles here on my paper

For instance, I come up with $sr^j=e \Rightarrow r^j=s$, which I hope isn't true...but it doesn't seem to be leading to any contradictions.

I can also determine that $r^{2j}=e$, which forces $n|2j$, but I still see no problem.

Prototank

6:44 PM

WAIT

I think I got it

go on with your lives per usual

user 1618033

7:05 PM

Let me change my username

@Agawa001 I changed my username

Let's come back to the integral above

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

How I do math art? Well, it's easy and complicated at the same time. The easy part is to explain the steps, that is working extremely hard for years (mainly), and the hard art, of course, is to put that into practice. What's next? After that hard work I meantioned we do the math art. How? We look at this integral, we notice something cool that comes from those hellish research days and we are done.

Agawa001

@Idomathart nice, a beginning of new cycle ?

I do math art

@Agawa001 After the discussion with Juan I considered that it better to simplify my life in this chat, and to better explain what I do, that is mathematical art (all this by using a proper username).

A case: the students come to you and beg you, implore you to teach them how to calculate the integral above in a very nice way. What do you tell them? Telling it's not an important question?

Agawa001

@Idomathart i doubt you are that vulnerable to online trolling

I do math art

@Agawa001 There is a different thing here. Having a misconception on my mathematics is a really bad thing. One miss so much beauty that cannot be compared to anything (I admit I might have exaggerated a bit).

Agawa001

@Idomathart i dont think he was here for seeing you with a different suit

but i like this new name keep it

I do math art

7:19 PM

@Agawa001 Yeah, it represents me fully.

Trying now to finish some proof.

BBL

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$Mathematics$ Mathematics