Mathematics - 2016-01-31 (page 1 of 2)

user276387

00:56

How do you do on latex the differential equations notation when they use dots above the x's to indicate the order of the DE. Like $x^{..}$ but where the dots are aligned on top of the $x$?

usukidoll

anyone here

Semiclassical

@user276387 that's usually known as 'fluxion notation' (going back to Newton) and it's done as $\dot{x}$ for one derivative and $\ddot{x}$ for two

not sure if $\dddot{x}$ works---oh, yep---but past that fluxion notation isn't as convenient as $x^{(n)}$

usukidoll

anyone know if it's legal to split the inequality for
$x < \frac{x+y}{2} < y$ ?

cuz if it's not, there goes my proof :(

Semiclassical

what does 'split the inequality' mean?

usukidoll

like in two parts

user276387

01:08

@Semiclassical Thank you! I looked everywhere for that!

usukidoll

$ x < \frac{x+y}{2}$
$\frac{x+y}{2} <y$

Semiclassical

well, yes. $1<2<3$ means both $1<2$ and $2<3$.

usukidoll

my original question is
If $ x < y $, prove that $x < \frac{x+y}{2} < y$
And by Axiom 8
$ x < y \rightarrow x+z < y+z$

Ted Shifrin

@usukidoll: The only way to prove (or interpret) such a thing is as two separate statements

Semiclassical

exactly

usukidoll

01:09

OH wait dang no scratch that. I don't think I can split them up T_T

Ted Shifrin

Yes you can

L33ter

hi @TedShifrin

Ted Shifrin

hi Karim

L33ter

I am enjoying algebraic topology btw

usukidoll

ohhhh yeah I could

L33ter

01:10

very nice

usukidoll

:P

L33ter

but hard though haha

usukidoll

It's been forever since I've done that though, so I was wondering if it was still legal

Ted Shifrin

no comment, Karim

L33ter

btw @TedShifrin

youtube.com/channel/UCErLELnXehsJ7ycW4OJgfQQ

I found this channel very nice

I wish there was other channels like this all over youtube that would be so nice

usukidoll

01:13

ah! I think I know where this is at

starting at $ x < y$ and by Axiom 8 which is $x < y \rightarrow x+z < y+z$
so there must be at least 2 unique z's to create these end results
$x < \frac{x+y}{2}$ and $\frac{x+y}{2} <y$

so if I start at x<y for the first case I need z = x
if I start at x<y for the second case I need z = y

therefore

first case
x+z <y+z (since z = x)
x+x < y+x
2x<y+x
and then dividing 2 by both sides I should get
$x < \frac{x+y}{2}$

second case
x+z < y+z (since z = y)
x+y <y+y
x+y < 2y
dividing 2 by both sides I should get $\frac{x+y}{2} <y$

Ted Shifrin

I don't know who he is, Karim.

L33ter

he has some nice set of videos on advanced math

which is nice

usukidoll

therefore
$x < \frac{x+y}{2} < y$

@Semiclassical I only saw your comment after I typed I couldn't split them up, when actually it is possible with that example you gave me.

Semiclassical

gotcha

Ted Shifrin

So I see ... Presumably he knows what he's talking about. He appears to be a graduate student at George Washington U in Washington DC.

L33ter

01:17

is it good uni ?

Ted Shifrin

Not known for its mathematics ...

Semiclassical

more for law, I think?

Ted Shifrin

yeah, maybe social science

Semiclassical

i think the following is just a reordering of the proof given above, but you can also do the following

$x< y\implies x-y<0<y-x$ by Axiom 8 with $z=-x,-y$

Ted Shifrin

Karim: For what it's worth (probably very little), I've heard of zero of the faculty members in that department.

Semiclassical

01:20

then adding $x+y$ to each term preserves the inequalities, and dividing by two gives the desired results

Ted Shifrin

Clever, @Semiclassic.

Semiclassical

eh, only difference is that i subtracted rather than adding $x,y$

so it's not a reordering, but it's really no different

Ted Shifrin

well, you made it into one simultaneous proof.

L33ter

oh

Semiclassical

eh, i guess. but you could also say the first part of the other proof as $x<y\implies x+x<x+y<y+y$

so mine is actually longer :p

Mike Miller

01:24

@Ted: Were you here for the involution question? I forget.

Ted Shifrin

Yes, @MikeM ... I made the irrelevant submanifold comment.

Mike Miller

Ah, right.

Semiclassical

going back to what i was talking about during that conversation

@ted do you know much about Clifford algebras re: K-theory?

Ted Shifrin

I know nothing.

Semiclassical

figured as much, but thought i'd ask.

Ted Shifrin

01:25

Generally it's cool to figure I know nothing :)

Semiclassical

hah. i think the problem for me is that the physics usage of K-theory is not the same as the obvious math one (i.e. topological K-theory)

L33ter

there is many seminar going about those stuff here @Semiclassical

but it way over my head

Semiclassical

yeah

L33ter

I want to get into it over the summer though

Semiclassical

what i'd like to be able to appreciate is the following

there's a table which you see bandied about in discussions of topological condensed matter systems, a so-called 'periodic table' for topological insulators and superconductors

Mike Miller

01:29

@Semiclassical: I think it is, because tgere's a physics student who's been asking me about topological K-theory for a talk he's giving about the classification of topological insulators.

Semiclassical

oh, i don't mean they're not related.

Mike Miller

But I mean I'm pretty sure his talk is to actually give the classification and he's been asking me about bundles and not about Clifford algebras.

Semiclassical

the table lays out, in various dimensions, the kinds of topological invariants (usually Z- or Z/2Z-valued) which go with the 10 'classes' of such systems

Ted Shifrin

Good night, all.

Semiclassical

night, @ted

and that seems to be about the fact that one can classify 3 real Clifford algebras and 7 complex ones

arxiv.org/pdf/0901.2686v2.pdf

is probably the most obvious reference

though i really just work with $d=1$ :P

L33ter

01:42

very nice graphics

to imagine the CW complex of how we build the sphere

by identifying the boundary of $D^2$ to single point $x_0$ which in the picture the top point of the sphere

Akiva Weinberger

@L33ter Now do one for the 3-sphere

L33ter

any n sphere can be done by same intuition same way

Akiva Weinberger

But give me a picture @L33ter

L33ter

lol

I can't give a picture in 4d

this is from wikipedia btw

Akiva Weinberger

Yeah I know

L33ter

01:53

I guess the n-sphere can be realized as a CW complex structure by attaching a n-cell to $e^0$

where $e^n = \{ x \in R^n : ||x|| \leq 1 \}$

by attaching what I mean is we identify the boundary of $D^n$ to a single point

so $S^n = S^{n - 1} \cup D^n / x \sim \phi(x)$

where here $x \in S^{n - 1}$ and $\phi(x) : S^{n - 1} \rightarrow e_0$

Akiva Weinberger

02:16

A nice quote:

"Pigeonhole principle: When you have $n$ pigeons and $n+1$ holes, there is at least one pigeon with two holes in it."

6

Also: "$\log(\log(x))$ is bounded above by $6$."

Random Variable

02:44

@Semiclassical Are you aware of an Abel-Plana-like formula involving cosh? I derived a formula that seems to be not quite correct.

Semiclassical

I forget what the Abel-Plana formula is

Random Variable

@Semiclassical en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula

Semiclassical

ah. can't say i know any

Random Variable

02:56

@Semiclassical I derived it using the Boole summation formula, which is like the alternating version of the Euler-Maclaurin summation formula. But I made a bunch of assumptions. That's probably why it's not quite correct.

Semiclassical

mmkay

L33ter

03:24

hey @Semiclassical

here

L33ter

05:30

Hey @MikeMiller available ?

drawing picture in lower dimension as a way to understand something is effective

Balarka Sen

05:59

@L33ter I am here.

L33ter

good :D I am learning a lot

just want quick question

Balarka Sen

Ok?

I have to leave, so be quick, please.

L33ter

Hatcher defn of real projective space is just the all vectors that go through the orgin topologized as the final topology obtained from R with the relation x ~ cx where x is a vector passing through the orgin.

Balarka Sen

Yes.

$c$ is a nonzero real.

L33ter

The reason we can restrict to only vectors of norm 1 is that we only go in two opposite direct and so if we have two vectors in same directions we don't really get any new information

so that is why we can restrict to vectors of norm 1

right ?

yeah exactly c is non-zero real

Balarka Sen

06:03

@L33ter Eh. Well, OK. But more rigorously, $v \sim c \cdot v$ means the equivalence class of $[v]$ consists of two distinct representatives: one of norm 1 and one of norm -1.

But the intuition is right.

L33ter

oh oke good

I see ok

the reason we can now restrict to only the hemisphere with anti-podal points on the equator identified is because of also any vector that pass through the orgin will also to have to pass through the equator

Balarka Sen

That's not true...

There are lots of lines through $(0, 0, 0)$ in $\Bbb R^3$ which doesn't pass through the equator of the unit sphere.

L33ter

hm

why are we allowed to restrict then to the upper hemisphere or lower hemisphere then ?

Balarka Sen

We aren't. Any line through the origin must either pass through both the (open) hemispheres or the equator.

Anyway, I have to go now.

Mithlesh Upadhayay

07:22

Hello ! everyone, would you like to check this integral question. Thank in advance.

http://math.stackexchange.com/questions/1579690/the-value-of-double-integral-int-01-int-0-frac1x-fracx1y2-dx

Forever Mozart

07:41

m

Huy

n

Forever Mozart

I think I have an interesting result now

it is fairly intuitive, but it hasn't been done before

i will show my advisor next week

Martin Sleziak

It seems that American politicians enjoy learning mathematics. For example, we have Sarah Palin and Ben Carson among users.

Huy

thanks Obama

Forever Mozart

the Palin profile is funny

"a penchant for going rogue" lol

is there an Obama?

Huy

07:47

@ForeverMozart: you know some basic topology, right??

Forever Mozart

yes

Huy

I have a question

Forever Mozart

@Huy ok

L33ter

@ForeverMozart I want to ask a topology question

specifically in algebraic topology

Huy

say that two transverse simple closed curves $\alpha, \beta$ in a surface $S$ form a bigon if there is an embedded disk in $S$ (the "bigon") whose boundary is the union of an arc of $\alpha$ and one of $\beta$ intersecting in exactly two points

L33ter

07:48

Huy

I have the following Lemma: if transverse simple closed curves $\alpha, \beta$ on $S$ do not form any bigons, then in the universal cover, any pair of lifts intersect in at most one point

L33ter

here I don't understand the statement thus as a set $X^n = X^{n - 1} \cup e_{\alpha}^n$, where here the union is taken to be disjoint.

Martin Sleziak

@ForeverMozart It seems that there used to be, but deleted his account: math.stackexchange.com/questions/377211/…

Probably this is a better link: normalhuman.github.io/oldusernames/…

Huy

the proof for $\chi(S) \leq 0$ starts as follows: note that $S \cong \mathbb{R}^2$ in this case. let $p: \tilde{S} \to S$ be the covering map. suppose $\tilde{\alpha}, \tilde{\beta}$ intersect in at least two points. then, there is an embedded disk $D_0$ in $\tilde{S}$ bounded by one subarc of each of the two lifts.

by compactness and transversality, the intersection $(p^{-1}(\alpha) \cup p^{-1}(\beta)) \cap D_0$ is a finite graph, if we think of the intersection points as vertices.

---

compactness of what is used here? I don't see why this intersection should contain only finitely many …

L33ter

@BalarkaSen maybe you can explain this to me

when you come back

Forever Mozart

07:52

@L33ter you dont understand disjoint union?

L33ter

no no

what I don't understand is the following

Forever Mozart

@Huy what is meant by universal cover?

L33ter

here the final set is just $X^{n - 1} \cup D^n / \sim$

Huy

@ForeverMozart: the simply connected cover

L33ter

i.e $X^n = X^{n - 1} \cup D^n / \sim$ why is that equal to

$X^{n - 1} \cup e^n$?

Forever Mozart

07:57

@L33ter I think you are taking closed discs and bending them so that the boundary goes to a point, then attaching that point to $X^{n-1}$

so as a set its the same as $X^{n-1}$ with some open discs

(the boundaries of the closed discs are identified with points of $X^{n-1}$, leaving open discs)

@Huy I'm thinking about it. Algebraic topology is my weakest subset of topology.

Huy

no rush, I'm thinking about it too. I just want to finish this chapter of a geometric topology book and then I'll go and do some real algebraic topology so I can proceed

Forever Mozart

@L33ter does that make sense?

L33ter

1 moment reading

ohh

I see

ok good I understand

just one more question

do you know about the projective plane @ForeverMozart ?

Forever Mozart

I think

L33ter

so RP^n = S^n / (v ~ -v), where v and -v are antipodal points identified

why are we allowed to restrict ourself to hemisphere with antipodal points identified

Forever Mozart

08:12

I dont understand the question

Huy

because of the identification?

L33ter

when they mention this is equivalent to saying that RP^n is the quotient space of hemisphere with antipodal points identified

why is are they equivalent ?

Huy

because the other hemisphere is identified with the upper?

so you only need to worry about the equator

L33ter

ohh I see

oh ok I understand

thank you

Huy

note that (for n=2) if you draw the equator as a square, you get the fundamental square of the real projective plane and see that it contains the mobius strip and hence isn't orientable (in case you didn't know yet)

L33ter

08:18

cool

hey huy can you give me some geometric intuition of the following

Then I will very satisfied for today

Huy

sorry, but I can't satisfy you for today

L33ter

:S

I have been trying to develop intuition of why we get it like this

I understand the torus issue

like I understand the geometry of identification of torus,projective plane,klein bottle

but trying to understand the geometrical intuition or have it for my mind for genus g orientable surface

Mike Miller

@Huy CompCtness of the disc $D_0$

Huy

hey Mike, you're up early (or still

I assumed, but how is that used?

Mike Miller

Poker was long.

Huy

08:27

hope you won

Mike Miller

@Huy: That thing is a closed subspace so compact.

Huy

so you got more than 1 dollar and 33 cents in your bank account now

Mike Miller

If you believe the vertices are discrete we then see that they're a finite set.

Huy

hm

Mike Miller

The discreteness should come from the transverse intersextion.

Huy

08:29

yes, that I believe

Mike Miller

Well, a compact discrete set

Huy

ah

um

so the compact set we're looking at is actually the set of intersections

Mike Miller

Yea

Huy

that makes more sense

ok thanks

Mike Miller

We just know it's a compaxt set because it's a closed subset of $D_0)

Huy

08:31

yes

are you drunk?

Forever Mozart

lol

L33ter

lol

probably his c doesn't work

so, he is using x instead

Mike Miller

No.

Huy

@MikeMiller: wait a second. the set we're looking at is $(p^{-1}(\alpha) \cup p^{-1}(\beta)) \cap D_0$, not $(\dots \cap \dots)$, so it's not the intersection points of the curves but the union intersected with the disk. why should this be discrete?

Mike Miller

@Huy: The discrete thing is the intersection points of the curves. What's left is then a finite number of points with arcs between them. Aka, a finite graph.

Huy

08:40

ah, ok

Vrouvrou

hello, i have a small question please : here http://math.stackexchange.com/questions/1038406/convergent-squence-in-topology
from where we get the idea to use the set A ?

Balarka Sen

@L33ter What is le question?

Vrouvrou

from where we get the idea to use the set A ?

Balarka Sen

If you haven't figured it out already that is.

L33ter

yeah I figured it out first my question was why is $X^n = X^{n - 1} \cup D^n / \sim = X^{n - 1} \cup e^n$ as a set

Vrouvrou

08:44

have you an idea @Huy ?

L33ter

but I figured it out with help of mozart

Balarka Sen

Okie.

The thing is attaching maps leaves interior of attached disks alone. That's all.

L33ter

yeah

Balarka Sen

By the way, they are not equal as sets. Merely in bijective correspondence.

L33ter

I see

hey @BalarkaSen maybe you can help me with some geometric intuition of the following picture

Balarka Sen

08:46

@ForeverMozart Herro.

L33ter

Balarka Sen

@L33ter What is unclear?

L33ter

I can't see it

why does it generate those shapes

I see it from the CW complex if we started with a point and keep attaching n-cell

but I don't see why the picture above produce those shapes

Balarka Sen

Look at the top octagon. Cut off from the diagonal going from the small dot to the opposite vertex. Identify. You'll get 2 torii with big punctures. Then paste along punctures.

You'll get the genus 2 surface.

You should try it out with a piece of paper.

Forever Mozart

@Vrouvrou suppose a sequence is not eventually constant. let $a$ be any point. pick an increasing subsequence of points in the sequence different from $a$. the complement of this subsequence is open containing $a$. Thus the sequence does not converge to $a$. since $a$ is arbitrary, the sequence does not converge at all.

Balarka Sen

08:48

Well, very elastic piece of paper, I think.

@Huy What's up.

Huy

@BalarkaSen: I wanna finish this section before I go to Hatcher

L33ter

I will try it with a paper

I am gonna try it with clay

I think that is better

Balarka Sen

@Huy Sure. But note that I can't help you much for the next - uh - 10 days because of my exams.

Then of course there's Mike.

:)

Huy

sure, maybe Mike will be around

when's your last exam?

Balarka Sen

9th.

Forever Mozart

08:52

@BalarkaSen every point of the interval $(0,1)$ is a cut point

and each point divides it into two connected components

Balarka Sen

@ForeverMozart Is that a recent breakthrough?

:P

Forever Mozart

lol no

but this was open question:

is there a metric space so that every point divides it into 3 connected components

Balarka Sen

um um um

Forever Mozart

it was open until i solved it ;)

Huy

wow

that must be like

Balarka Sen

08:53

Wow, really?

Huy

$$\huge{\text{THE MOST AMAZING RESULT YOU GOT THIS YEAR!!!!!}}$$

3

Forever Mozart

lol

Balarka Sen

@ForeverMozart What was your result?

Vrouvrou

@ForeverMozart so we do by contradiction

Balarka Sen

Did you come up with an example or did you prove there's none?

Forever Mozart

08:54

there is an example

Balarka Sen

Hm.

Actually, let me ponder a bit. I know lots of topological space of that form but I am not sure if any of them are Hausdorff.

Forever Mozart

yeah think about it maybe you will come up with something different

Balarka Sen

@ForeverMozart Are you going to publish this?

Forever Mozart

I don't know, I have to show it to my advisor first

Balarka Sen

Ah, OK.

Who proposed this problem first?

Forever Mozart

08:57

previously there were some hausdorff examples

Balarka Sen

i.e., which paper?

Forever Mozart

but none of them were metric

Balarka Sen

ah, so not second countable.

Forever Mozart

i dont think so, cause they were regular

the hardest thing was proving the triangle inequality for my metric

Balarka Sen

Interesting.

Well, congratulations, @ForeverMozart. It's exciting.

Forever Mozart

09:00

thanks, hopefully i dont find any errors

i think for each $n$ you can get each point to divide the space into $n$ components

Balarka Sen

cool

Forever Mozart

but i havent proved that

Mike Miller

how about for $\kappa$ components

Forever Mozart

hmm

maybe but I'm not sure

Huy

Ted is so popular on mathed.SE: "My favorite textbook for an undergraduate course in Abstract Algebra, Ted Shifrin's Abstract Algebra: A Geometric Approach, uses a rings-first approach."

3

and yet another "geometric approach"

Forever Mozart

09:04

I used Hungerford's book

almost have it memorized cover to cover

Balarka Sen

Ted's books are great.

Forever Mozart

@Vrouvrou yes or contrapositive

Balarka Sen

Joyce is a good mathematician. Have heard of him.

iirc, he's Silverman's student

Forever Mozart

I think it is contrapositive

i feel like getting some donuts

L33ter

alrighty guys I am going to sleep

cya all tomorrow

Forever Mozart

09:10

l88ter

@Vrouvrou does it make sense?

we pick a sequence which is not eventually constant, and show it does not converge to any point

Vrouvrou

@ForeverMozart why an increasing subsequence , A is the set where x_n do not equal to a no ?

why we work with a subsequence ?

Forever Mozart

because then the complement of the subsequence is an open set containing $a$, but it misses infinitely many points of the sequence... so that the sequence does not converge to $a$

we need to work with a subsequence possibly, so that we avoid $a$. ($a$ might be a term of the sequence)

try to think about it visually

Vrouvrou

but the limit of a sequence is not necesserly a term of the sequence

like (1\n) and 0

@ForeverMozart

Forever Mozart

yes. $a$ is just any point of the space.

but it MAY be a term of the sequence

but we can choose a subsequence that avoids it because the sequence is not eventually equal to $a$

Vrouvrou

@ForeverMozart when we suppose that $x_n$ is not constant at a by definition $\forall n\in \mathbb{N}, \exists n_0\geq n, x_n\neq a$

so we can directly set $A=\{x_n\in E, x_n\neq a\}$

Forever Mozart

09:26

i lost the link to your question

what is $E$?

Vrouvrou

$E$ is an uncountable set

Forever Mozart

ok so $E$ is the entire space?

Vrouvrou

yes

Forever Mozart

look...
Let $(x_n)$ be a sequence in $E$ that is not eventually constant. Let $a\in E$. There is a subsequence $(x_{n_i})$ of $(x_n)$ such that $x_{n_i}\neq a$ for each $i$. Let $U=E\setminus \{x_{n_i}:i\in\omega\}$. Then $U$ is open and contains $a$. Also, $U$ misses infinitely many terms of $(x_n)$, so that $(x_n)$ does not converge to $a$.

@Vrouvrou

Vrouvrou

we must use a subsequence ?

Forever Mozart

09:33

because what if $a$ is a term of the sequence?

Then $E\setminus \{x_n:n\in\omega\}$ would not contain $a$

Vrouvrou

we take it from the definition of not constant $\forall n\in \mathbb{N}, \exists n_0\geq n, x_n\neq a$

Forever Mozart

so there would be no contradiction

$x_{n_0}\neq a$

you mean

Vrouvrou

i don't understand

we take the set $A=\{x_n\in E, x_n\neq a\}$ then $E\setminus A$ is open containig $a$

then $x_n$ do not converge to a

Forever Mozart

not eventually constant means for each $a$ and each $n$ there exists $n_0\geq n$ such that $x_{n_0}\neq a$.

@Vrouvrou that works, but you need to show $A$ has infinitely many terms of the sequence... that's why I used the subsequence

Vrouvrou

ok i understand

Forever Mozart

09:40

you are french?

Vrouvrou

mmm Algerian

Forever Mozart

you know this? youtube.com/watch?v=Q6omsDyFNlk

Vrouvrou

yes

Forever Mozart

i just saw that today

i have lots of french movies

Godard

jean luc godard

he is the best director

Vrouvrou

yes

I'm an artist

11:11

@Huy nice starred message. But there is an interesting thing though: when I say the most amazing result this year I omit to say that I get amazing results every day which counts a lot. :-)

@Huy and one more thing: at the present moment I have far more results, numerically speaking, than Ramanujan had.

What is the meaning of that? Who knows?

BBL

Balarka Sen

11:37

It means you are about to die.

Tobias Kildetoft

11:52

@BalarkaSen It means you have actually counted the two things.

I'm an artist

12:21

@BalarkaSen you're right, very hard work can be dangerous. Hope to survive and produce very good work for a long while.

Balarka Sen

You misinterpret me.

But true nonetheless.

I'm an artist

@BalarkaSen I'm sure you have only noble thoughts to me. ;)

Balarka Sen

It was a mere joking reference to Ramanujan's early death, nothing more.

I'm an artist

@BalarkaSen I know, and I was pretending I didn't get your point. ;)

No more work. Trying to watch a movie.

BBL

Huy

12:52

@I'manartist I'm sure you understand the concept of quality vs. quantity - not that I'm saying anything about the quality of your work. I leave that to experts in the field.

@BalarkaSen: are you studying for your exams atm ?

iwriteonbananas

13:12

@BalarkaSen Oh hi

Partly Putrid Pile of Pus

@I'manartist I find this extremely unlikely. Where can I see expositions of your work?

I'm an artist

@PartlyPutridPileofPus Do you think I should have an exposition with my work?

Partly Putrid Pile of Pus

13:28

If your results are comparable in quality, to those of Ramanujan, of course.

I'm an artist

@Huy By experts I understand those that can finish these integrals in no time, say

17

A: Two integrals involving logarithm and polylogarithm function

$$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\ \\ {\...

@Huy and far harder integrals. But wait I wouldn't call these ones hard.

They are more of a joke.

@Huy and at least any other questions answered by Cleo with much ease. Most of the people that show amazing faces on main to the Cleo's answers is because they are not trained enough.

It's OK, but I just wanted you to know this little point.

Partly Putrid Pile of Pus

So your field of research is calculus?

I'm an artist

I've tried to specialize myself in the calculation of integrals, series and limits.

Partly Putrid Pile of Pus

I see, I am surprised to hear someone could find an advisor for such a specialization.

I'm an artist

@PartlyPutridPileofPus I'm not affiliated to any institution, I didn't follow a math uni, I'm self-educated, and I'm not advisor for anyone. I teach (tutor) some kids, but some easy math.

Partly Putrid Pile of Pus

13:38

I think I am misunderstanding. You are doing research in the field of calculation of integrals, series and limits, and you are not working with anyone else on this? While your results are as copious as those of Ramanujan?

I'm an artist

@PartlyPutridPileofPus Yes, I do everything alone. I was only saying that numerically I have more results than him.

Balarka Sen

@Huy Somewhat. But if you have questions, feel free to ask.

Hi @iwriteonbananas.

Idle001

14:03

helpability decreased quite remarkable level inhere

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