Mathematics - 2018-07-19

Daminark

00:10

Hey everyone!

Ted Shifrin

hi Demonark

Daminark

How's it going?

Stat lover

Hello everyone .. can anyone help me to figure out the difference between non -degenerate limit and non -degenerate random variable ? thank you

geocalc33

anyone know graph theory

Jeff

05:21

Anyone know why we need the function notation $(f+g)(x)$ versus $f(x)+g(x)$? That is, what's the difference in the two notations?

bolbteppa

The map $f+g : A \to B$ which sends $ x \mapsto (f+g)(x)$ is such that $(f+g)(x)$ is defined as $(f+g)(x) = f(x) + g(x)$

Jeff

05:49

Can you explain that in plain language? I still don't quite understand the difference.
Is it something like $f+g$ is the name of the new function, whose variable is $x$, and the new function is created by the operation of adding two functions $f(x)+g(x)$?

Eulb

06:33

my name's jeff lol

he's saying they're the same

Balarka Sen

d e d m e m e

Eulb

no u

Balarka Sen

06:47

U got me
Right now cant think of a response
Maybe I lose
On the other hand
My word of condolence
Gotcha
Another word is needed...
Yolo

Eulb

what is this sorcery

Sebastiano

07:00

@TedShifrin Thank you very much for your help. Thank you very much.

Leaky Nun

07:45

@BalarkaSen Never mind I don’t get it

Or maybe I’m too dumb

Unicorns might be related though

Balarka Sen

Only people of high IQ will understand what I said

Leaky Nun

:)

Balarka Sen

read vertically

Leaky Nun

read vertically

Balarka Sen

*red

did u try first letters

Leaky Nun

07:48

read my response vertically :)

Balarka Sen

OH shit

Fuckin

@LeakyNun give me some of your IQ points senpai

Leaky Nun

lol

Alessandro Codenotti

08:07

We're gathered here today to mourn the loss of Balarka, savagely rekt in chat

Balarka Sen

08:34

@Alessandro I will have my revenge

Leaky Nun

"No You Won't"

►

Balarka Sen

only soy boys watch sherlock

(gottem, by the way)

my vengeful revenge has been accomplished

Leaky Nun

lol

philmcole

10:27

Hi, how do you know that the set defined by $x(y^2-x)=0$ has vertical lines at $x=0$?

For $x \neq 0$ I can divide by $x$ and see that $y=\pm \sqrt{x}$. But how do I see what it does at $x=0$?

Leaky Nun

@philmcole if $x=0$ then $x(y^2-x) = 0$

philmcole

Ah and $y$ can be whatever it wants. I see, thanks.

Rudi_Birnbaum

10:52

hi leakynun and all!

hi @anon! Saw your also interested in the "Mochizuki case".

Jeff

12:04

@Eulb Hi Jeff. I had gone to bed. So what's the use or advantage of the different notations? Surely it's not for the sole purpose of confusing students?

Silent

12:35

@BalarkaSen, if differentiable function $f$ is $\Bbb R\to\Bbb R$, then we can talk about whether derivative continuous or not. But can we do something similar with total derivatives? I am not sure because, now derivative at each point itself is a function.

Alex K Chen

13:05

4

Q: Learning math historically

What is meant by learning math historically (not learning math history only, but learning math with a historical development perspective)? I've seen some sources say that to learn a math topic X, you need to look at the historical development of the topic X and go over the famous questions by yo...

user193319

13:24

Claim: If $X$ is a topological space and $(A,B)$ a separation, then $X$ intersects both sets in the intersection. Proof: Suppose that $X \cap A = \emptyset$. Then $X = X-(X \cap A) = (X-X) \cup (X-A) = X-A$ and therefore $A \subseteq A \cup B = X = X-A$, which is absurd. Hence, $X \cap A \neq \emptyset$. The same work shows that $X \cap B \neq \emptyset$...How does that sound?

Leaky Nun

what is a separation?

rschwieb

@LeakyNun There is now a FAQ page at the site, and I've adjusted the (+) notation we discussed

Leaky Nun

wtf @rschwieb

rschwieb

beg pardon?

user193319

@LeakyNun $A$ and $B$ are open, disjoint sets such that $X = A \cup B$.

rschwieb

13:26

Welcome To the Forum i hope?

Leaky Nun

I hadn't even started typing when you tagged me, and you posted the message immediately after I posted mine :o

@user193319 just that?

user193319

Yes, I believe that's the whole definition.

Leaky Nun

can A and B be empty?

> Do all the rings here have identity?
Yes: the reason is that a great deal of the logic connecting properties here uses the assumption that the ring has identity.

no, the right reason is that the other rings don't exist

user193319

Whoops, forgot that! They cannot be empty.

Leaky Nun

@user193319 and A and B are subsets of X?

user193319

13:28

Yup!

Leaky Nun

so $X \cap A = A$?

user193319

Hmm...you're right...Perhaps the result should be rephrased in terms of subspaces of $X$, say $Y \subseteq$, so that $A$ and $B$ are not necessarily subsets of $Y$. I think the same proof works in that case, right?

Leaky Nun

no

take Y=B

user193319

Something can't be right then...If $Y=B$, I don't see how $(A,B)$ could be an actual separation of $Y$.

Leaky Nun

please rephrase your question precisely

rschwieb

13:34

@LeakyNun heh

@LeakyNun You might change your mind when you see Smoctunowicz's simple nil ring

Leaky Nun

@rschwieb redet Sie Deutsch?

@rschwieb where?

rschwieb

I can't speak german, no. I could only guess at meanings

I had to request a copy through my library

Leaky Nun

hmm

rschwieb

@LeakyNun But maybe your institution has access: tandfonline.com/doi/abs/10.1081/AGB-120006478

oo, I'll have to get my hands on this: tandfonline.com/doi/full/10.1080/… I had not heard about it

Actually, I'm no longer certain the second paper refers to the same example as the first

It seems that people who come up with amazing examples usually come up with more than one (e.g. Bergman)

Yup: I made a mistake: the second paper is about a different example, although I should get that one too :)

rschwieb

14:47

Anyone have a good feeling for how cohomology works?

Really broad ideas

Balarka Sen

@rschwieb What kind of cohomology do you have in mind

rschwieb

@BalarkaSen I don't know anything really about homology or cohomology of any sort. Just a few abortive attempts to read it back in the day

I had a lot of basic homological algebra but never developed a good feeling for using it

Balarka Sen

I mostly think about singular cohomology of topological spaces. The intuition I have in mind is it extracts out the cochain complex $C^\bullet(X)$ from a given topological space $X$ where elements of $C^k(X)$ ("$k$-cochains on $X$") are functions on the "$k$-skeleton of $X$"

If $X$ is a simplicial complex, then this "$k$-skeleton of $X$" is nothing but the union of $k$-simplices of $X$.

Think of piecewise linear functions on $X$ if you like

rschwieb

15:11

So "homology" is a topic, "chain complexes" each have "homologies", and "cochain complexes" each have "cohomologies"

and I can think about homology/cohomology of a complex as a linear approximation of sorts?

Balarka Sen

It's never clear to me how algebraists think of homology of a chain complex. I will inevitably think of cobordisms when I think about it

Leaky Nun

> It's never clear to me how algebraists think of homology of a chain complex.

we think of them in terms of the axioms :)

rschwieb

It's tough not to have any handhold in a new discipline

Leaky Nun

I like to use the long exact sequence

rschwieb

@BalarkaSen Recently I learned how an algebraic result about group rings had some reflection in group cohomology, and I wished I understood more. mathoverflow.net/q/297043/19965

Balarka Sen

15:28

I suspect there are many ways to understand group cohomology

I only know of one or two

rschwieb

Can group co/homology be described purely in terms of co/homology on the modules of group rings? or is it different?

Balarka Sen

It can be described as cohomology of a resolution of the group ring, yes (or equivalently Ext or Tor over a group ring)

Jake S

16:15

Excuse me, I come for a reference request. I finished my first linear algebra course and I would like to brush up on linear maps and projections because I feel I didn't really understand them

Does anyone have a book to recommend?

rschwieb

@JakeS I would recommend Kaplansky's Linear algebra and geometry: a second course

Jake S

Thank you! I'll look into it.

loch

I think one intuition for many stuff in derived functor cohomology is to measure how far away a left/right exact functor is from being exact

e.g. often in AG people you want to understand global sections, and cohomology inevitably comes up because the global section functor is not right exact

same goes with Tor, Ext etc.

rschwieb

It's both inexpensive and a pretty good book.

@loch I understand the face value of that, but can you give me an intuition for what "far from exact" and "close to exact" means for functors?

"close to exact" mostly means a close reflection of properties on both ends of the functor, right?

Daminark

So, for reference, apparently number theory often cares about group cohomology with a focus on H^2, for the sake of group extensions

rschwieb

16:28

@loch Btw, I think I saw you register at the DaRT site :) any plans to submit a ring suggestion? Your help, especially with commutative algebra would be appreciated.

i'm at a little bit of a disadvantage because I don't know a lot of AG. I don't know how to wield the weapons of homological algebra... I mainly engage problems with fisticuffs.

Not that I don't appreciate examples given in that vein... I'll probably eventually catch onto them!

Mike Miller

@rschwieb We are dual; I can't do anything down to earth with rings. :)

Silent

@Daminark, will you please look at this:

4 hours ago, by Silent

@BalarkaSen, if differentiable function $f$ is $\Bbb R\to\Bbb R$, then we can talk about whether derivative continuous or not. But can we do something similar with total derivatives? I am not sure because, now derivative at each point itself is a function.

Daminark

16:47

Okay so

Let's say you have a smooth function $\mathbb{R}^n \to \mathbb{R}^m$

Then the derivative of the function at a point is a linear map $\mathbb{R}^n \to \mathbb{R}^m$

So the map that takes in a point $p$ and spits out $Df(p)$ is a function from $\mathbb{R}^n$ to $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m) \cong \mathbb{R}^{nm}$

That's the analog of thinking about $f'$ as a function $\mathbb{R}\to\mathbb{R}$. So this does allow you to talk about a function's derivative as itself being a continuous/differentiable function

Ted Shifrin

@JakeS: I'll also add some self-horn-tooting and suggest my Linear Algebra book with M. Adams. (I'm assuming you can find it in a college library, rather than buying it.) We do lots more than most books with geometric stuff, and there are several discussions of linear maps and projections.

howdy, Demonark

Daminark

How's everything going Ted?

Ted Shifrin

@Silent: To add to Demonark's reply, you can also then think about differentiating the function $g(p) = Df(p)$ as a map $g\colon \Bbb R^n\to \mathscr L(\Bbb R^n,\Bbb R^m)$. This will be the second derivative, and the deep result is that when you view $Dg(p)$ as a linear map $\Bbb R^n\to \mathscr L(\Bbb R^n,\Bbb R^m)$, you can associate this to a bilinear map $\Bbb R^n\times\Bbb R^n\to\Bbb R^m$, and, as such, it is symmetric.

pretty well, Demonark, and how's your summer research going?

Fargle

17:10

Heya chat.

Ted Shifrin

heya @Fargle

Fargle

Ooof, actually, make that "heya chat in about ten minutes".

Gotta run

Ted Shifrin

Figures.

Fargle

:|

Daminark

Hey @Fargle! In 10 minutes

@Ted it's going quite well. Right now I'm reading some algebraic number theory, after which I'll get to the main paper I'm reading, on Kronecker-Weber for $\mathbb{Q}(i)$

Ted Shifrin

17:24

Cool stuff. Good thing you have Mathein to chat with about these things :P

Did I miss anything interesting in my week-long absence?

Daminark

Haha, this is definitely true

Hmm, I'm not on as much, and when I am on I usually just tab in for a second to see what's up, maybe say hi to people, then tab out. So with that noted, I haven't seen much go on

Ted Shifrin

Well, I just can't count on you ... Sigh. :D

I assume no luck with the police.

Daminark

Nope, and they did say that in all likelihood at this point the guy doesn't even have the laptop anymore. So I got my dad's old chromebook that I'll be using for the time being

Ted Shifrin

Ah, that's good. Lesson learned.

(On the latter, not the former.)

Daminark

Yeah

Anything interesting going on on your side?

(I thought "that that" was bad enough and now we have "on on")

Ted Shifrin

17:37

Nah, puzzling over an answer on main at the moment. (Can't figure out how to get two "over"s.)

Semiclassical

Which question?

Erik M

Hey all

Alessandro Codenotti

@Daminark What kind of ANT?

Daminark

Neukirch chapter 1

Hey Semi, Erik, and Alessandro!

loch

I think what "far from exact" and "close to exact" means for functors kind of depends on context --- for example something similar to what @Daminark - but for modules, is that if you have two $A$-modules $M,N$, then $Ext^1_A(M,N)$ is in bijection with short exact sequences of the form
$$0\rightarrow N \rightarrow P \rightarrow M \rightarrow 0$$
up to isomorphism.

So here how far away $Hom_A(-,N)$ is away from being right exact is the same thing as saying how many ways can you have an exact sequence of the above form (if Ext is trivial, then $P$ is isomorphic to $M\oplus N$.

Alessandro Codenotti

17:40

@Daminark Cool, I read a little bit of it

Erik M

Question: Is there a word for a 1-dimensional slice? I have a solid with a line going through it. The word slice (to me) conjures up a 2-dimensional (planar) section, is there an analogous word for slice for the 1-dimensional case?

Semiclassical

interval?

or just 'segment'

Ted Shifrin

@Semiclassic: this one.

hi again, demonic @Alessandro

hi @loch

Alessandro Codenotti

Hi @Ted

Erik M

@Semiclassical thanks!

loch

17:42

hi @TedShifrin

Daminark

Nice! But yeah I haven't made much progress, Mathein told me some stuff on DC but in terms of working through things carefully and all I basically just made it through a section and a half. Hopefully I can get through a few more, and ideally reach ramification and the like soon

Semiclassical

@TedShifrin thx

Ted Shifrin

@Semiclassic: i've never thought about scaling the polar variable $\theta$. If anything, I'm inclined to say we should scale with $1/2$ rather than $2$ here. I'm in the middle of writing my fourth paragraph or so. :)

Semiclassical

neat

I'd more have expected some kind of offset in $\theta$, since (at first glance) that looks like some ellipse with rotated axes

Ted Shifrin

It's very much not an ellipse, but, yes, an offset is what we get.

Semiclassical

17:46

oh, ^4

yeah, fair

Daminark

One thing I will say, I think Neukirch tends to multiply vectors on the left which threw me off

He wrote a matrix for something and it was the transpose of mine

Fargle

I'm back!

Semiclassical

@Daminark so his vectors are row vectors by default?

Daminark

So it seems. I've only seen one instance of this

So it's possible this was a typo or something

Alessandro Codenotti

@Daminark Imho Neukirch goes really fast in the first few sections if you're not a bit familiar with the material already

Semiclassical

17:47

The fact that certain authors default to row vs. column vectors probably has some historical roots

Alessandro Codenotti

@Daminark My abstract algebra professor does as well, maybe it's an algebraists's thing

Semiclassical

could be worse. could be talking about (reverse) polish notation

Ted Shifrin

Now that @Fargle is back, it's clearly time for Ted to leave. :P

Fargle

:(

Daminark

Rip

Ted Shifrin

17:50

@Alessandro, Demonark: Actually, computer science folks (doing computer graphics) do row vectors and right multiplication. I've never quite known why.

Aw @Fargle. You wanna talk to me? :)

Alessandro Codenotti

@Semiclassical I've encountered logic textbooks using $\to\phi\psi$ rather than $\phi\to\psi$ because clearly using prefix notation is waaaaaay less of a pain than having to write brackets

Fargle

lol, well, maybe not in particular. I just like going around and being offended by things.

Daminark

So I programmed in Racket for a little bit and you used Polish notation, but you still used parentheses (the joke, or maybe not even joke, is that racket is supposed to say "bracket")

Ted Shifrin

@Semiclassic: I don't think the fourth power is relevant. I said in my answer that perhaps the difference in degrees of homogeneity of the two sides should lead to an $\alpha$, but I don't believe it now.

Daminark

So it was (+ 2 3) for 2+3

Ted Shifrin

17:51

Well, @Fargle, geez, I should have plenty of offense in the bank by now! :P

Alessandro Codenotti

@Daminark parenthesis should be superfluous in rpn

rschwieb

@loch hmmmmmm

Ted Shifrin

@Fargle: Whenever you wanna chat analysis or geometry, I'm at your disposal. Oh, or algebraic topology if the pros aren't here.

Semiclassical

I guess my line of attack would be to rescale the axes and then rotate by 45 degrees

Fargle

Does algebraic geometry count?

Semiclassical

17:52

and see what it looks like in those coordinates

Ted Shifrin

You're up to algebraic geometry now? How so?

Daminark

One thing which I wonder actually, what tends to make people want to do topology over geometry and vice versa?

loch

@rschwieb I guess what I'm trying to say is that I think the failure of exactness of functors manifests itself in different ways depending on context

rschwieb

Yeah, functors can have many different shapes

loch

@Daminark topology is too hard :(

Fargle

17:53

@TedShifrin By way of Reid's UAG, which I'm reading alongside Atiyah-Macdonald and some other stuff

Ted Shifrin

Demonark: I was never particularly drawn to topology ... I liked the interplay of complex variables and differential geometry.

rschwieb

@loch How many useful, practical, distinct functors do you have to keep in mind?

Alessandro Codenotti

Interesting question @Daminark, no idea, but I know for sure I prefer topology

Ted Shifrin

Oh, @Fargle, well, as you know, algebra isn't my strength, but I know some of the stuff in Reid.

Fargle

@Daminark I don't know which, or whether, I prefer.

Ted Shifrin

17:54

@Fargle: You still haven't learned much geometry :P

Do you have one more year of school?

Fargle

I haven't learned much topology either.

Alessandro Codenotti

@Fargle If you like Reid's UAG he also has a commutative algebra book which is less terse than AM

Fargle

One more class, one more semester.

Ted Shifrin

Oh, not much.

loch

@rschwieb Hmm off the top of my mind in AG the derived functors that usually show up are sheaf cohomology H^i, Ext , Tor, higher direct images etc.

Fargle

17:54

@AlessandroCodenotti Yeah, I've seen UCA--I didn't take to it last time, but I can take another look.

Ted Shifrin

@Alessandro: Way less terse is Eisenbud.

But he has so many examples of things ...

I wish I had held on to my copy.

rschwieb

@loch Does scalar extension fit in as a special case of one of those?

taht and localization

I think if I could look at those through the lens, I might learn something

Fargle

I'm not sure why I didn't take to it. Maybe I accidentally forgot the woffle was a woffle and got intimidated?

loch

yeah - after all scalar extension is just taking tensor products

I guess so is localization (although localization is exact - so in that sense derived functors don't really show up)

Ted Shifrin

@Semiclassic: If you have any improvements or ideas on that post, I'll be glad to amend.

Alessandro Codenotti

17:59

@TedShifrin Yeah but Eisenbud is a huge book, while AM and Reid's are small and short, despite having a lot of stuff

Semiclassical

I'll see what comes to mind

the fact that the left-hand side is a function of $x+y$ is what suggests "rotate by 45 degrees"

Alessandro Codenotti

(notably Reid omits tensor products though)

Ted Shifrin

@Semiclassic: Yes, of course. I made that point eventually :P

I can't see why the difference in exponents is helped by scaling $\theta$.

Semiclassical

yeah

Ted Shifrin

@Alessandro: Mostly, students complained about my texts that I didn't do enough examples (even though I thought I did plenty) ...

Semiclassical

18:01

it's just not that natural for the kind of problem being done

Daminark

Yours meaning, the ones you wrote?

Ted Shifrin

I suppose an appropriate scaling will turn $\pi/4$ into $\pi$ ... so maybe that simplifies things?

Yeah, Demonark.

Semiclassical

yeah

Alessandro Codenotti

I like books with the important examples (rather than no examples at all), but I also like having to think about (counter)examples myself sometimes

Semiclassical

If I substitute $x=a(u-v)/\sqrt{2}$, $y=b(v+u)/\sqrt{2}$ (I can't remember if that's the canonical rotation by 45 degrees, but close enough)

Alessandro Codenotti

18:04

I really like it when an author introduces important examples (and sometimes even important theorems!) as guided exercises and lets the reader work through them (Of course that's risky because people might just skip the exercises :/ )

Ted Shifrin

@Semiclassic: Sure, that's the rotation.

Semiclassical

then $(x/a+y/b)^4=4xy$ becomes $4u^4=2ab(u^2-v^2)$

Ted Shifrin

@Alessandro: Jeanne Clelland's new book on geometry done with moving frames is exactly that. @EricSilva loves it. I was a reviewer for it and made various suggestions, but that style is better for beginning grad students than for typical (US) undergrads.

Semiclassical

which has reflection symmetry along the two axes

Ted Shifrin

@Semiclassic: Which turns into the trig in my answer. :P

Semiclassical

18:06

which in turn means that it's enough to look at the first quadrant, corresponding to an angular range 0<theta<pi/2

that's where I could see maaaybe an advantage to rescaling the angle

Ted Shifrin

So the double-angle is natural on the RHS of what you wrote, but doesn't help when you put it on the left as well :P

Alessandro Codenotti

@TedShifrin Interesting! I was thinking about Shoenfield's logic book, he put results as important as Łoś's Theorem as a guided exercise!

Ted Shifrin

well, and of course there's Halmos's Hilbert Space Problem Book.

Semiclassical

there's a name for this figure as well, which I'm forgetting

Ted Shifrin

I don't know it, Semiclassic.

Semiclassical

18:09

it comes out looking like a figure eight, at least when a,b>0

Ted Shifrin

Yeah, flatter than most :P

Daminark

Somehow I never found myself using books that much, I will say Schlag's psets second quarter of analysis were often of that sort

Alessandro Codenotti

@TedShifrin never heard of it, but I like Halmos's style, I'll see if I can find a copy!

Daminark

Like, I reference so many of his pset problems, more than anyone else's

Ted Shifrin

As you've heard me say, Demonark, problem sets/exercises are what makes a course/textbook great. I mean, the text/lectures are significant, too, but ...

Silent

18:15

@TedShifrin I am sorry, but which symmetry are you talking about?

Ted Shifrin

Oh, that bilinear form is symmetric (in the two vectors in the product in the domain).

That's a fancy way of saying mixed partial derivatives are equal for a reasonable function.

But if $g$ is differentiable, it gives you that symmetry.

Hmm, maybe I need continuous differentiable.

Yeah, I need $Dg$ to be continuous.

Silent

Thank you very much, @ted

@Daminark Thank you

Daminark

No problem!

Ted Shifrin

Sure thing, @Silent.

quallenjäger

Hi chat

Daminark

18:24

Yo

philmcole

Hello chat

If I have an open set and exclude finitely many points, is it then still open?

Semiclassical

@TedShifrin en.wikipedia.org/wiki/Lemniscate_of_Gerono, maybe

huh, check out that "another representation"

which in turn points to the connection with Lissajous figures

Daminark

@philmcole in what context? Metric spaces? $\mathbb{R}^n$? General topological spaces?

Semiclassical

so that's sorta neat

philmcole

@Daminark Yeah metric spaces. I know that finite sets are closed. Does that help me?

Alessandro Codenotti

18:34

How's the complement of that finite set?

philmcole

open

Alessandro Codenotti

If $A$ is your open set, can you write $A\setminus \{x_1,\dotsc,x_n}$ in terms of $A$ and the complement of the $x_i$?

philmcole

Ah yes I can write it as $A \cap \{x_i\}^C$

And both are open so the intersection is open and hence the relative complement is open.

Thanks

Alessandro Codenotti

(Note that the same works for $A\setminus C$ with $A$ open and $C$ closed, regardless of how many points $C$ has)

philmcole

right

thanks

Mike Miller

18:42

(The path I wanted to lead you down was the same proof written differently: whatever the distance is from x in U to some n points, there is a minimum distance, say $d$. If you pick $\epsilon$ small enough so that $\epsilon < d$ and $B(x, \epsilon) \subset U$, then...)

Alessandro Codenotti

I was trying to avoid using the metric since it works in general (I mean it doesn't for finite sets of points but for closed sets)

EnjoysMath

Instead of saying $A \text{ true}$ all the time is there a simple notation on the proposition $A$?

Alessandro Codenotti

$T\vDash A$ if $A$ is true in the theory $T$, $\mathfrak A\vDash A$ if $A$ is true in the structure $\mathfrak A$ but I don't know if that's what you're looking for

EnjoysMath

more like $\underline{A}$

Should I learn Coq, Isabelle, Haskell, or something else if I had to choose one?

I already know D, Python, Assembler, some category theory

I'm going with Haskell for now

since there is a HoTT library for it and it's a full-fledge language where as Coq seems more like an embedded DSL

Daminark

Probably depends on what your aim is

EnjoysMath

18:54

My aim is to integrate visual diagrams into the workflow of the app

That you can also manipulate and have fed back into the "Coq editor"

Coq-like

Possibly full expressiveness with the diagrams or at least of accademic interest

Idk exactly my end goal

I just know I like playing with blocks and arrows -_-

My goal is a tool that makes a mathematician's or students life easier

It would be nice to use Haskell itself instead of me having to get into parsing

for the app's DSL

I hate parsing. It took a month or more to debug my EDID parser for HDMI (work stuff)

Semiclassical

@TedShifrin in retrospect, I was misreading the question: I assumed that it was something like $x=a r\cos\alpha \theta,$ $y=br\sin\beta \theta$

which would make a lot more sense to me, frankly, since with the form given you can just absorb $\beta$ into $r$.

and then would lead to the Lissajous shenanigans I mentioned earlier

Fargle

@EnjoysMath I'm a fan of Haskell, at least from an aesthetic standpoint.

Semiclassical

19:10

(though the lemniscate only shows up after the rotation, so...hmm. dunno)

Abcd

19:44

$$I (k)= \displaystyle\int_\infty^0\dfrac{k}{x^2 +k^2}\ln x ~dx$$

How to integrate this one?

Semiclassical

I guess what I'd note is that, for $\ln x$, $x=1$ is always a special point

which suggests (to me) splitting the integration range into (infty,1] and [1,0)

and then considering how ln(x) behaves on both regions

Abcd

i dont think its beneficial to split it

Semiclassical

as an alternative (which contains some of the same ideas) one can make the u-sub x=e^x

Abcd

doesnt help

Semiclassical

okay then.

Abcd

20:00

not integrable

integrand given by integral calculator has something like $L_i$ and other imaginary numbers

not of high school level.

Semiclassical

The result I get in Mathematica, assuming $k>0$, is $I(k)=\frac{\pi}{2}\log k$

Abcd

oh

@Semiclassical yeah k>0

Semiclassical

Which is an interesting answer...

Abcd

ya

Semiclassical

One approach would be to show that $k\frac{dI}{dk}=\frac{\pi}{2}$ and $I(1)=0$

Not sure how realistic an approach that is tho

Abcd

20:03

@Semiclassical are you sure there's no minus sign?

Semiclassical

ah, yeah, there is

forgot that I swapped the integration limits

The fact that evidently $I(k)+I(1/k)=0$ is interesting as well

Abcd

@Semiclassical I got this:

1

A: Evaluating the integral $I (k)= \int_\infty^0\dfrac{1}{x^2 +k^2}\ln x ~ dx$

You can let $x= k t$ to find $$ I(k) \equiv\int \limits_0^\infty \frac{-k \ln(x)}{k^2+x^2} \, \mathrm{d} x = \int \limits_0^\infty \frac{-\ln(t)}{1+t^2} \, \mathrm{d} t + \int \limits_0^\infty \frac{-\ln(k)}{1+t^2} \, \mathrm{d} t \, .$$ Your substitution $t \to \frac{1}{t}$ shows that the first ...

x=kt, wow

Semiclassical

whistle

I mean, I had thought to do that substitution, but it hadn't seemed sufficiently productive

so I didn't bother writing it out

Abcd

@Semiclassical how to think this way :(

I could have never thought x=kt

:(

Semiclassical

well, note that the scale of the rational function k/(x^2+k^2) is set by x=k

that's the only point which seems at all special for that function

Abcd

20:16

@Semiclassical what do you mean by "scale of the rational function"

Semiclassical

I'm being vague, yeah

I guess I mean that the only thing I can really compare x with in that function is k

e.g. if I had k=100, then I'd naturally be most interested in the case when x=100

anyways, once you have it in your head that x and k are more or less on the same footing in that denominator, that to me does suggest the substitution x=kt

since in that case one has k/(x^2+k^2) = 1/k*1/(t^2+1)

so the k-dependence has been factored out of that rational function

this comes at the cost of dx = k dt and log(x) = log(k)+log(t)

What I hadn't accounted for was being able to show that the log(t) term vanished

Dattier

21:30

Hi,

Let $(u_n)_n \in \mathbb Q^{\mathbb N}$, with $\forall \alpha>0, |u_{n+1}-u_{n}|=o(\alpha^n)$.
Is it true that the limit of $(u_n)_n$ is a transcendental real ?

anon

21:47

$u_n=0$ or $u_n=1/n!$ fit the bill, both with limit $0$

geocalc33

22:02

anyone know graph theory

Dattier

22:12

@anon thanks

Dattier

22:39

Let $\alpha \in\mathbb C, \exists R\in\mathbb Q[x], R(\alpha)=0$. Is it true that $\exists Q\in\{0,1\}[x],\exists P\in\mathbb Q[x], \exists a\in\mathbb C, Q(a)=0,\alpha=P(a)$ ?

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$Mathematics$ Mathematics