Mathematics - 2017-06-04 (page 3 of 4)

SteamyRoot

12:03

@Astyx Well, according to general relativity, they're the same (locally).

Astyx

It's weird really

SteamyRoot

So I'm guessing that if you wanted to find a difference, you'd have to work either on quantum scale, or work with black holes.

Either way, I don't know if there's any serious experiments going on.

Astyx

I read an article some time ago where they hoped to find a difference

SteamyRoot

12:25

Finding a difference would certainly be very interesting!

But, at the same time, it's probably wishful thinking

(for now)

Astyx

If you take $$g:\begin{cases}x\mapsto (x+1)^2\tab \text{ when }x\in]-\infty, -1]\\x\mapsto 0 \tab\text{ when } x\in [-1, 1]\\ x\mapsto (x-1)^2\tab \text{ when } x\in[1, +\infty[\end{cases}$$, $f:t\mapsto(t, g(t))$ and $A$ as the $x$ axis, then $f^{-1}(A)$ is not a manifold (it's $[-1, 1]$) and $f$ is not transversal to $A$ right ?

Ugh, chatjax is not working ?

SteamyRoot

you started with double $$

Astyx

Oh thanks

ignore those $\tab$s

SteamyRoot

I think you can use &'s in cases to align... not sure though, I tend to use array

Astyx

Okay I give up

SteamyRoot

12:30

$$g:\begin{cases}x\mapsto (x+1)^2 & \text{ when }x\in]-\infty, -1]\\x\mapsto 0 &\text{ when } x\in [-1, 1]\\ x\mapsto (x-1)^2& \text{ when } x\in[1, +\infty[\end{cases}$$

Astyx

Thanks

SteamyRoot

hmmm, well, $f^{-1}(A) = [-1,1]$ isn't a manifold but it is a manifold with boundary, I guess.

Astyx

yes, I'm trying to grasp the concept of transversability at the moment

I guess the issue comes when dealing with surfaces shrinking into curves

SteamyRoot

So, you mean to ask whether $f$ (the "stretched" parabola) and the $x$-axis are transversal as submanifolds of $\mathbb{R}^2$ ?

Astyx

I'm probably not making much sense

Then answer is no unless I'm mistaken

SteamyRoot

12:37

Yeah, they indeed aren't.

Astyx

I'm trying to find an example where they are not transversal and they do not form a manifold

But I guess I found an answer

Locally you want them to look like the whole ambient space, otherwise their intersection might not have the same dimension everywhere, right ?

SteamyRoot

Not sure if I'm following here

Secret

Dissecting Hypercubes with Pascal's Triangle | Infinite Series

►

I have trouble figuring out what a cross between a tetrahedron and octahedron is like

and it seems that 4D shape is not documented anywhere

Astyx

@SteamyRoot If you don't have transversality, you might end up with something like this when taking the intersection (or the reciprocal image, depending on the context), and this cannot be a manifold. But when you impose transversability, you fix the dimension so you don't get this kind of problem, and therefore you always get a manifold

Terrible drawing by the way, sorry, for that

Balarka Sen

12:53

What are we talking about?

Astyx

Why transversality allows you to create manifolds

I'm trying to improve my intuition on transversality mainly

Balarka Sen

Oh, you wanted an example where things are not transverse, and intersect in a non-manifold?

Astyx

For instance

But I think I got it

Balarka Sen

OK

I think G&P has good pictures of this phenomenon.

Astyx

They do, I was trying to understand the general mechanisms instead of examples

Secret

12:57

o wait, found it:

Rectified 5-cell

In four-dimensional geometry, the rectified 5-cell is a uniform 4-polytope composed of 5 regular tetrahedral and 5 regular octahedral cells. Each edge has one tetrahedron and two octahedra. Each vertex has two tetrahedra and three octahedra. In total it has 30 triangle faces, 30 edges, and 10 vertices. Each vertex is surrounded by 3 octahedra and 2 tetrahedra; the vertex figure is a triangular prism. The vertex figure of the rectified 5-cell is a uniform triangular prism, formed by three octahedra around the sides, and two tetrahedra on the opposite ends. == Structure == Together with the simplex...

Balarka Sen

@Astyx Yeah, your idea is correct. If $M, N$ are transverse submanifolds of $\Bbb R^n$, then for all $p \in M \cap N$, $T_pM$ and $T_pN$ are transverse subspaces of $T_p \Bbb R^n$, and $T_p (M \cap N) = T_pM \cap T_pN$, so the dimension of the "tangent space" of $M \cap N$ is locally constant.

Astyx

$T_p$ meaning the tangent plane at $p$ right ?

Balarka Sen

Which is the intuitive reason why $M \cap N$ is a submanifold.

Yes, @Astyx.

Astyx

Thanks a lot

Balarka Sen

I'm making a heuristic argument here though.

Astyx

13:03

If it's locally constant, it has to be constant

Balarka Sen

Right.

Secret

Ok I give up, it becomes way too cluttered to find out where I am

Balarka Sen

@Astyx If you don't require transversality, intersection can be arbitrarily bad. It is a fact that for any closed subset $C$ of $\Bbb R$, there is a curve $X$ in $\Bbb R^2$ such that $X \cap \Bbb R = C$ (here $\Bbb R$ is the x-axis of $\Bbb R^2$).

Eg C can be a Cantor set :P

Astyx

Gotta love Cantor sets

Sard's theorem states that if you have a smooth map of manifolds $f:X\to Y$ then the set of critical values of $f$ has measure zero. But what if you take $f$ to be constant ? Isn't the set of critical values $X$ ?

Simply Beautiful Art

@Secret O.O

Balarka Sen

13:11

@Astyx Value lies on the codomain. Point lies on the domain.

If $f$ sends everything to $y_0 \in Y$, critical values of $f$ is just a single element $y_0$ in $Y$

Astyx

Ugh, I'm being very silly

Balarka Sen

Nah, terminology takes time to get used to.

Astyx

G&P even stated that points not in the image are regular by vacuity

Balarka Sen

Yup.

Astyx

Right, yeah it makes much more sense now

I probably should have finished the chapter before asking, G&P answered my question just after

@Hippa Est-ce que tu as le numéro de JeSuis ? Histoire qu'on organise la journée avec lui

Hippalectryon

13:26

@Astyx Oui. Je vais te donner mon numéro, comme ça tu peux m'envoyer un messaage pour que je récupère le tien, puis je t'envoie celui de JeSuis

Astyx

Il me l'a déjà donné, mais je veux bien le tien

C'est bon

J'ai envoyé un message

Hippalectryon

@Astyx reçu

Astyx

Parfait

@Balarka Did you know Calculus before reading G&P ? I never really worked with Hessians

Balarka Sen

@Astyx Yeah I did Ted's course before GP

Helped me a lot

Astyx

Huh, I'll look at that sometimes then

How long did it take you ?

Balarka Sen

13:37

I think you only need to look at Hessians if you want to learn Morse theory, which you can skip freely

Ted and Mike would tell you to not learn that section if you're doing diff top

@Astyx Multivariable Calculus?

Astyx

Yes

Balarka Sen

I spent two years on it, just because I sucked :P I imagine you could learn all of it in a few weeks

Astyx

I guess I'll just read through it and see what catches my interrest

I already know some, just not Hessians etc

Balarka Sen

yeah

You should pick up the stuff whenever required, I think. You know enough multicalc that doing a full course could potentially be a waste of time

Astyx

I agree. I guess I'll also adapt to what possibilities I might have in my school next year too

Assuming I pass the entrance exams

The main result of G&P on Morse functions means they are dense in the set of smooth functions, correct ?

Balarka Sen

13:47

yup

actually stronger than that; it says you can perturb any smooth function to a Morse function by a linear perturbation

Astyx

Right

Balarka Sen

self-ad + quick summary of the proof

Astyx

That's basically the proof in the book

I almost finished chapter 1

Balarka Sen

right

I think one should always emphasize more that Hessian is nothing but derivative of the gradient.

Astyx

That's for real valued functions right ?

SBM

14:01

hmm

Astyx

Otherwise you'd get a much bigger matrix

Balarka Sen

@Astyx Hessian is defined as the matrix of mixed partials of order 2 for real valued functions, so.

Astyx

Fair enough

Balarka Sen

Potentially you could define it for functions $f : \Bbb R^n \to \Bbb R^m$ in general: then $Df$ is a function $\Bbb R^n \to L(\Bbb R^n, \Bbb R^m)$; the latter being the space of linear maps. Define $Hf$ to be exactly the derivative of that.

$L(\Bbb R^n, \Bbb R^m)$ can be identified with $\Bbb R^{nm}$.

Astyx

Yup

Fargle

14:32

Heya chat!

SBM

Hey @Fargle

Fargle

What's everyone up to this fine day?

Astyx

Hi Fargle. Manifolds for me, what about you ?

Fargle

@Astyx Probably more group theory.

SBM

Manifold?

Fargle

14:42

@SBM A manifold is roughly a surface that locally looks like Euclidean space everywhere. For example, the surface of a sphere is locally flat.

SBM

oh

Balarka Sen

@AkivaWeinberger Remember you were telling me about the observation at one point that you found it weird how $z^n$ has antiderivative $z^{1+n}/(1+n)$ for all $n$ but $n = -1$?

Here's a pretty interesting physical-chemical consequence of that: say $A \to P$ be a chemical reaction (reactant -> product). Rate of the reaction is $r = -d[A]/dt$, where $[A]$ is the concentration of reactant at time $t$ (the minus indicates the rate of dissipation of reactant).

Furthermore, one knows $r \propto [A]^x$ where $x$ is a number known as the "order" of the reactant, which agrees with the stoichiometric coefficient (eg $n$ in $nA \to A_n$) of $A$ for "really simple" reactions, but is an experimental constant in general. Let $k$ be the proportionality constant, so write $r = d[A]/dt = k[A]^x$. If you start with initial concentration $[A]_0$ of $A$ at time $t = 0$, then you can solve the differential equation with that initial condition.

In particular you can find the "half-life", which is a time $t := t_{1/2}$ at which point the concentration $[A]_{t = t_{1/2}} = [A]_0/2$ is exactly half the initial concentration. Really simple computation shows only order 1 (x = 1) reaction has $t_{1/2}$ independent of the initial concentration $[A]_0$

Astyx

Hi Mike

Balarka Sen

So in particular, as a consequence of your observation, radioactive decay (an example of order 1 reactions) have half-life independent of initial active mass (analogue of concentration for non-liquid reactant). Or rather, using that fact you can identify radioactive decay as order 1 reactions. Cool, eh?

Fargle

@Balarka Cool indeed. I didn't know why half-life was independent of sample size, or whether that held true for other kinds of decay.

Balarka Sen

14:49

Yeah, apparently it is so for order 1 reactions. Really strange.

Astyx

Chapter 1 is finished, I'll let every thing cool down before continuing

Balarka Sen

It's a consequence of the decay being exponential, @Fargle, as you can see if you solve $d[A]/dt = -k[A]$. (EDIT: Damn, I messed up a minus sign above. Fix that mentally)

Fargle

Right.

Balarka Sen

Nice, @Astyx. Don't forget to do some exercises.

Astyx

I figured I'd do them once I reread the whole chapter (still before moving on to chapter 2)

Fargle

14:52

@BalarkaSen I'll just pretend $k < 0$.

Balarka Sen

:P

Astyx

I always found cool how you derive the rate from the elementry reactions

Balarka Sen

Hmm, I am currently wondering if I know examples of unimolecular reactions which are not elementary.

Astyx

I have one in my notes somwhere

Akiva Weinberger

That is cool

Astyx

14:55

Wait, what do you mean by "unimolecular reaction" ?

$$N_2O_3 \to 2NO_2 + {1\over 2}O_2$$

Akiva Weinberger

The strange thing isn't that it's not of the form $\frac{z^{n+1}}{n+1}+C$; $~\ln z+C$ already is of that form, sort of. $\displaystyle\lim_{n\to-1}\frac{z^{n+1}-1}{n+1}=\ln z$. The strange thing is that, over $\Bbb C$, $~z^n$ has an antiderivative at all except for $n=-1$, where it has no global antiderivative.

Balarka Sen

@Astyx Only one reactant molecule is involved. Thanks, that's exactly what I wanted!

Astyx

The elementary reactions being $$\begin{align}N_2O_3 &\to NO_2 + NO_3\\NO_2 + NO_3 &\to NO + NO_2 + O_2\\NO + N_2O_3 &\to 3NO_2\end{align}$$

Balarka Sen

Yeah. Very cool, man.

@AkivaWeinberger I agree. Also, fun observation on it being already of that form.

Akiva Weinberger

(The reason the above limit fails for $\Bbb C$, of course, is that $z^{n+1}$ isn't globally defined for noninteger $n$.)

Astyx

15:14

Absorbing elements are weird

That's why you should stick to groups and forget about fields

And don't even get me started on rings

Dodsy

15:33

Oh I hate chemistry.

Semiclassical

I don't remember the parts of chemistry which aren't physics so well

Naming conventions, for instance, I remember little of

Balarka Sen

as in organic stuff?

I hate that

Dodsy

I hate balancing equations.

and answering those little questions that are basically just there to save the world.

like "calculate when the geothermal system would pay for itself"

or "write a 500 word essay on the benefits of a hybrid car"

"write an essay about fossil fuels and create an action plan to reduce your carbon footprint in your home"

Zophikel

Does anyone have the book: Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals

Dodsy

15:52

oh that's not the full book.

Semiclassical

I don't mind balancing equations. It appeals to the puzzle solver in me.

Abdullah UYU

is there a way to show latex ex. in mobile

Zophikel

@Dodsy you have the book

Dodsy

@Zophikel anyways I found it online, I won't post it here. I thought that would've been okay because it was right off the cambridge website but it turns out it was a preview. You can definitely find it online, no problem. But I don't condone "stealing" books and don't really want to implicate myself. if you're determined enough, you'll find the book online. That's all I will say.

And silence fell over the room.

@Semiclassical I don't even know how to do it.

Zophikel

all right @Dodsy

Dodsy

16:05

@Semiclassical s2.quickmeme.com/img/bc/…

Akiva Weinberger

Balancing equations is basically linear algebra

Dodsy

lmfao, I just got my chemistry unit back

I received 100%

I just handed it in yesterday too.

Akiva Weinberger

They should teach people how to convert them to matrices should they find it easier

Dodsy

That'd be cool

BAYMAX

Hi

Dodsy

16:11

Hey baywatch

Semiclassical

Well, the other way to view it is as vector algebra

BAYMAX

I was thinking about how to solve this ODE $(1-x^2)y'' -2xy' + 6y = 0,y(1) = 2 $?

Ha ha :) @Dodsy

Semiclassical

Namely, how can I take an integer combination of one set of basis vectors and get an integer combination of other basis vectors

BAYMAX

Like I was thinking something is there in that ODE

Semiclassical

That looks Bessel-ish?

BAYMAX

16:13

like I can see that there derivative of $x^2$ as $2x$ there

oh

Dodsy

Is that how it works with equations like this?

You find the anti derivative of the coefficients?

BAYMAX

No

Dodsy

I see.

BAYMAX

Like just an observation

Besselish

mathworld.wolfram.com/BesselDifferentialEquation.html

Dodsy

Y(1) = 2

Semiclassical

16:15

Not really. It's more that certain equations show up so much that their solutions pick up specific names

Dodsy

So some function $f(x) = 2$

or $f(1) = 2$

Semiclassical

Eh, it's nothing that simple

BAYMAX

@Semiclassical In Bessel the coefficient of $y$ is $(x^2 - n^2)$

Dodsy

I see.

Semiclassical

Ah, so n=1 might work

Abdullah UYU

16:17

Let $\vec{a}$ and $\vec{b}$ are vectors on $\mathbb{R}^2$. For the angle bisector vector of $\vec{a}$ and $\vec{b}$, do we have to choose the small angle between them? does it's defition state it, or can we take which one we want?

BAYMAX

and here it is just a constant

Semiclassical

I don't know off the top of my head if n=1 will be Bessels ODE directly

Abdullah UYU

Dodsy

If it's a bisector

then it's right in the middle of the angle that a and b make.

Semiclassical

If it bisects the small angle, it also bisects the large angle

So it shouldn't matter

@BAYMAX hmm, I'd missed that 1-x^2 was the coefficient of y'' not y

Dodsy

16:23

@Semiclassical so then $x^2 - n^2 = 6$

BAYMAX

Yes

Dodsy

Is that how you solve these bessel functions?

Semiclassical

@Dodsy Which really means that it doesn't work. You'd need the coefficients to agree for all $x$

Dodsy

Ah, I see.

Abdullah UYU

so, how do we know it's length?

Dodsy

16:25

Of the bisector?

Well first I'd choose the small angle.

Abdullah UYU

yeah

Semiclassical

So while it's reminiscent of Bessel, the solutions aren't Bessel functiond

So probably Bessel's ode is the wrong bet. But I think it's still be worth looking at a table of known second-order ODEs. Might get lucky.

Dodsy

and find a third vector that connects the two vectors

Semiclassical

Ahah: mathworld.wolfram.com/ChebyshevDifferentialEquation.html

BAYMAX

actually it is Legendre Differential Equation :)

mathworld.wolfram.com/LegendreDifferentialEquation.html

Semiclassical

16:29

Nuts

BAYMAX

Like it would be interesting if we gather all set of popularly known second order differential equations :)

Semiclassical

Well, if it works it works

BAYMAX

yes

Semiclassical

Well, Chebyshev and Legendre polynomials are big orthogonal polynomials on the interval [-1,1]

Abdullah UYU

actually my problem is that does it's length matter? we can find many vectors that bisects the angle. so they are all angle bisector vector?

Semiclassical

16:31

Though with different weight functions

BAYMAX

yes

Like $\int_{-1}^{1}P_{n}P_{m} dx = 0$

if $m \neq n$

Semiclassical

Right.

BAYMAX

and = $\frac{2}{2n + 1}$ if $m = n$

Dodsy

@AbdullahUYU I was under the impression that a bisector bisects an angle equally, however, i think I know what you mean. If we mark the vector as being (1,4) or (2,8) it's the same vector. Does the length matter? No, not really.

Semiclassical

They both should fall into the Jacobi family of orthogonal polynomials

BAYMAX

16:34

Jacobi family of orthogonal polynomials ?

Semiclassical

Yeah. As such they satisfy ODEs of the form of Jacobi's ODE: mathworld.wolfram.com/JacobiDifferentialEquation.html

BAYMAX

ha ha

class of classical

:)

Semiclassical

Yes, and you're learning about it from Semiclassical :>

BAYMAX

yes

Dodsy

I'm sure you'd usually make an equation for the line which bisects the angle, where any input into t still bisects the angle. And you could find infinitely many points on the bisector line.

BAYMAX

16:36

but it was a sense of humour due to a wiki article line

Dodsy

So, I'd say that the length does not matter.

Abdullah UYU

yeah, i got @Dodsy

BAYMAX

Jacobi polynomials

In mathematics, Jacobi polynomials (occasionally called hypergeometric polynomials) P(α, β)n(x) are a class of classical orthogonal polynomials. They are orthogonal with respect to the weight (1 − x)α(1 + x)β on the interval [−1, 1]. The Gegenbauer polynomials, and thus also the Legendre, Zernike and Chebyshev polynomials, are special cases of the Jacobi polynomials. The Jacobi polynomials were introduced by Carl Gustav Jacob Jacobi. == Definitions == === Via the hypergeometric function === The Jacobi polynomials are defined via the hypergeometric function as follows: ...

first line :)

Semiclassical

I figured, heh

There's also the Gegenbaur polynomials

Dodsy

But if the question said "what is the length of the bisector from the angle made by A and B to vector AB" that's a different question entirely.

BAYMAX

16:37

Hey is there any source which lists all kinds of these special differential equations

@Semiclassical

Semiclassical

There's probably a list of common Sturm-Liouville ODEs somewhere,yeah

BAYMAX

THese are Sturm Liouville ODES?

Semiclassical

Yeah.

Sturm-Liouville theory

BAYMAX

thanks

how you did that for the link?

like simply a l ine

except a block para as above in my case

Semiclassical

Ah. [ link name ] ( html )

except without the spaces.

BAYMAX

16:43

Chebyshevs DE

done

Akiva Weinberger

[Chebyshevs DE](http://mathworld.wolfram.com/ChebyshevDifferentialEquation.html)

Semiclassical

Yup. It's handy.

BAYMAX

ya

Semiclassical

Not sure why that didn't work for you, @akiva

Akiva Weinberger

'Cause I did it on purpose

Semiclassical

16:44

ah :P

BAYMAX

thanks

Is there anywhere on chrome I can save the links

like nott bookmark

Semiclassical

Bookmark is the only one I know off hand

Dodsy

I love bookmark manager

I have a little folder with my last name on it

BAYMAX

any other interactive browser

Dodsy

that includes everything in the world!

Astyx

16:49

Hello Ted

Even though you're probably not here

BAYMAX

hololulu @TedShifrin

Dodsy

Where'd he go

Ted Shifrin

Salut, @Astyx et tout le monde.

Balarka Sen

Hi @Ted

Astyx

C'est bien l'Italie ?

Ou c'est plus tard ?

Dodsy

16:50

Bonjour Ted Shifrin, Comme sa va?

Ted Shifrin

Balarka, you're supposed to be gone for weeks!

Balarka Sen

I am. Somewhat.

BAYMAX

Hi = Hololulu for BAYMAX :)

Ted Shifrin

Hi, Nate. Moving on to Paris early in the morning.

Dodsy

He won't even talk to me, Ted.

And that's great!

Ted Shifrin

16:51

Berlin so far, Astyx.

Dodsy

Germany would be great.

Astyx

Oh, so Italy is after France

Balarka Sen

did you get banned earlier today

Dodsy

Do you speak any german?

@BalarkaSen Who, me?

Last night, yes.

Balarka Sen

huh

Dodsy

16:52

Because of ducky.

Balarka Sen

ez to guess

Ted Shifrin

Yes, Nate, but disappointingly.

Astyx

@Dodsy This is getting annoying

Ted Shifrin

Is the flagging continuing unflaggingly?

Dodsy

@Astyx Oh, I didn't mean to annoy, friend.

Astyx

16:53

I'm not sure one can flag flags

@Dodsy Sure, I'm not blaming you

Ted Shifrin

No.

Astyx, you organizing pour moi? :)

Dodsy

@TedShifrin My parents go to Germany a lot to visit paper plants, I heard they're very accommodating to people who cannot speak German.

Astyx

@TedShifrin We are slowly but surely coming to an agreement :p

Ted Shifrin

Yes, but I want to speak decent German. The folks working the bar in

Astyx

Is your trip purely touristic ? Or are you there to attend some conference ?

Ted Shifrin

16:56

my hotelhave been very nice :)

I am retired, dammit. I came to see the city and to meet Danu.

Dodsy

haha it sounds like you're enjoying yourself!

Nothing better than going to european bars and relaxing, Ted.

I googled "Barbin" trying to figure out what you're talking about.

Astyx

You can be retired and still enjoy conferences :p

Ted Shifrin

Nah.

Astyx

Oh. Okay ..

Where will you be going in Italy ?

Ted Shifrin

Genova, Cinque Terre, Como, Trieste

Astyx

16:59

Cool, I've never been to any of these cities

Ted Shifrin

@Astyx, you're lucky I tolerate you all ... or unlucky.

Astyx

You're lucky we tolerate your tolerance

Dodsy

haha :o)

Well, I should go do homework

I want to finish it tonight so i can start studying functions next week.

Ted Shifrin

Well, less tolerance is easily arranged.

Dodsy

Have a good trip Ted.

Enjoy the german brew.

Astyx

17:01

I'll have to leave too

Ted Shifrin

It is disappointing.

Astyx

See ya on saturday (I hope we'll have arranged something by then)

Ted Shifrin

We'd better, Astyx:) I'll treat you guys to lunch.

Astyx

So it was Hippa, JeSuis and LeGrandDoo right ?

Ted Shifrin

Someone's missing.

Astyx

17:03

I'll check LeGrandDodo knows what's going on

There you and me too, I'm not sure anyone else is missing

Anyway, bye !

Ted Shifrin

Bye!

Fargle

Awww, just missed him.

Oh wait, no, I can't read. Hi @Ted.

Krijn

17:25

Heya @Ted/chat

Ted Shifrin

Actually, @Fargle, you can read — I was gone. Hi @Krijn.

Fargle

@TedShifrin Ah. How is it so far?

Ted Shifrin

Doing fine. On to Paris on the morrow (early).

Krijn

Where are you at the moment?

Somewhere in Europe then, I guess

Ted Shifrin

Yup, München.

Krijn

17:35

Oh, I was there two weeks ago

Fargle

Jealous! I've always wanted to see Muenchen--my grandmother is from Bavaria.

Krijn

Pretty close really

Ted Shifrin

Did you get a reply/retort, @Fargle?

Fargle

@TedShifrin I did!

Ted Shifrin

See?

Fargle

17:36

Oh, hush. I'm 22, this is the prime time of my life to be overly worried about interpersonal interaction.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Well, fine, then. Was he mean?

Hi @Alessandro.

Fargle

Not at all. He even said he'd be glad to look at the things I've been working on research-wise to lend a second hand.

Ted Shifrin

See? Nicer than me :)

Hippalectryon

17:52

@TedShifrin o/

Ted Shifrin

Salut @Hippa

Faraad Armwood

hello

Ted Shifrin

Howdy

Faraad Armwood

how are you

Ted Shifrin

Recovering from the 9 hour time change — dinnertime in Europe ;)

Transcript for

$Mathematics$ Mathematics