For example, how I do go about solving $\int \frac{x^2 arctan(x^3)}{1+x^6}$ ?

Make the substitution $u=x^3$

Oh and above it's $x^2+3x+6$ of course

@Cristopher Well, that's what I did.. but after nothing seemed to happen

@Apoorv What did you get? After the substitution the integral becomes $\int \frac{\arctan(u)}{3(u^2+1)}\,du$

Then you make another substitution, $u=\tan(v)$

ohh.. I didn't see the second substitution

I guess, in these type of problems you just have to hack your way with substitutions until you get it

I guess you could say that

@iwriteonbananas did you figure out what the map is?

Thank you very much :)

Sure

One more thing, how do you figure out how many terms should be there in a partial fraction decomposition?

it's actually pretty natural. it sends $\sigma : \Delta^n \to X$ to $C_0\sigma - C_1\sigma : \Delta^{n+1} \to X$ where $C_i\sigma$ are the cone maps of $\sigma$, i.e., the upper cone and the lower cone of the suspension map $S\sigma$

this is precisely what you get by exploiting the snake map, btdubs

hi @PaulPlummer.

@Apoorv The number of fractions is the same as the number of factors

At some place, I saw even a quadratic denominator had three terms..

What to do when the factors are same?

Do we take different constants like A and B or just twice constant like 2A?

Hello @BalarkaSen

A and B, I think

@PaulPlummer So, what kind of math have you been thinking about?

Nothing to much, been doing a few problems from Ch0 Hatcher, review some ring theory, and I looked a bit at PL Morse theory by Bestvina @BalarkaSen Although at the moment I am actually looking for a place to live... I hate it, feels like such a waste of time

I agree that it's a waste of time.

Any problem from Ch0 you found that you like?

PL Morse theory sounds interesting. I've been intrigued by classic Morse theory after reading bits from here and bits from there, but I don't think I know -gulp- enough multivariable calculus to understand all of that.

hi @Ted

Hi @Balarka and all

Well I have not gotten to the ones you have talked about but, none that I thought were really fun, but problem 5 was a good review of some stuff from point-set and definitions intrudocued ($X$ def retracts to a point then every neigh $U$ around that point contains a neigh $V$ st $V \to U$ is null)

@TedShifrin I'm still (almost) trying to come up with homotopy theory as a consequence of covering space theory. :P

I don't believe it.

Though I'm stuck at the point where I need a good notion of fibers for galois theory of fields. Pondering about it.

@TedShifrin PS : by homotopy theory, I mean just the notion of loops and path-homotopies.

The motivation for this is to see loops in the context of galois extensions

It does not really look like you need much of the classical theory to get started with PL, it defines something called affine polytope complexes and you do the Morse stuff there

@PaulPlummer: What does Bestvina use it for?

Sounds like a fun project @BalarkaSen

Don't call that homotopy theory. That's a defined term.

@PaulPlummer ah, I see.

@PaulPlummer it is. I believe it'd be pretty interesting if I can come up with something (although I don't believe I will)

@TedShifrin yeah, I realized that after saying it. sorry.

@MikeMiller It is just an overview of the theory, but I think the original motivation was to study finiteness properties of groups, by studying morse functions on $K(G,1)$ spaces

(whatever that is)

Any particular $G$?

Goodnight @MikeM

Morning.

I think some of the main applications were to groups like $F_n \rtimes \mathbb Z$

Ok, thanks.

Quick probability question if anyone can help: I have a probability space with measure $P$ and some r.v. $X$. I condition it by some event $E$ with $P(E) > 0$ and so get a conditioned measure $P(\cdot|E)$. Is the distribution of the conditioned random variable given by $\mu_{(X|E)}(F) = P(X \in F | E)$?

@TedShifrin hey ted!

@MikeMiller for examples, as a simple, probably unneccessary use of this Morse theory he shows that the kernel of $F_2 \to \mathbb Z$, generators map to $1$ is free of infinite rank

Sounds pretty cool.

But an analogous function $F_2 \times F_2 \to \mathbb Z$ has a kernel that is finitely generated but is infinitely presented

The paper seems quite accessible, and I also saw some examples in "The Geometry of the Word Problem for Finitely Generated Groups" too, although I probably won't be putting to much time into this stuff till I study fundemental group and $K(G,1)$ stuff, and what it "really" means

Yah you might want to take a look at it, you would probably have a better idea of what is going on in it. There is some stuff on Tits spherical building in the paper, showing that it has homotopy type of wedges of spheres (not really sure what that is useful for though) @BalarkaSen

ooooooooooo, yeaaaaahhhhhhhhhhh!!!! WHAT AN AMAZING RESULT I JUST GOT!!!!

@PaulPlummer I'd love to. I will have a look when I get time.

For now, I have to think about the thing I want to think about :)

Is anyone here familiar with ordinary least squares for simple linear regression?

I am working on such a cool feature

I need to add colour to the statebars but!

@PaulP: The first thing you said can be proved with a simple picture; the second thing is super interesting.

@AlecTeal what is this latest creation?

@StanShunpike it goes with that book thing I mentioned to you yesterday

It'll look a lot better when the blocks have colour

@Hippalectryon

?

O_o where do you get those ideas xD

@Hippalectryon They naturally come to mine. :-)

hi @Stan

@Hippalectryon I'm at the very very very beginning of my mathematical activity. This so far means almost nothing.

Well I can't say unless I actually try to do it :-)

@Hippalectryon I was saying that in general. I have a lot of ideas but I need time to develop them and create stuff.

ok

@Hippalectryon If you ask me, I would teach series in Flajolet and Salvy even in high school. I'm confident I can make all those kids fully understand the easy methods I propose say. Actually I can do that for the worst integrals and series ever.

@MikeMiller It is interesting. The "outline of strategy" the is given in the paper is 1) find a "nice" $K(G,1)$, 2) find a nice Morse function $f:K(G,1) \to \mathbb R$, 3) compute descending links and hope that finitely many fail to be $(k-1)$-connected. Maybe this gives you a better idea of when it can be used.

@Chris'ssistheartist They wouldn't have time for the rest of the program though

One thing at a time

Sure, in this case our $K(G,1)$ is $(S^1 \vee S^1)^2$, which you'll give a tilted embedding in $\mathbb R^3$, whose height function will be the Morse function. I don't know what a descending link is. Where's the article/notes you got the example from? I'd like to take a look.

@Hippalectryon The art of teaching.

@Chris'ssistheartist That doesn't change a lot. You can only do so much in a fixed time

here is the paper @MikeMiller

Thanks, I'll take a look later.

I originally came across this paper looking for geometric techniques that might be useful for this question. I actually have a feeling this kind of attack will end up being useless for that group.

the $F_2 \times F_2 \to \mathbb Z$ is example 6.2

Yes, I'm much happier now, @Stan. I think you did say it before, ages ago :P

hi @Huy

@MikeM: Did you have your meeting today with Danny yet?

@Stan: Now consider the appropriate portion of the derivative matrix of $DF$.

Yes, wasn't much more to say.

oh :( @MikeM

hey @Ted

@MikeM: Scarily, in just about exactly 3 weeks, I'll be in SD.

You'll roast.

@TedShifrin right, so I have $\mathbf{y} = \begin{bmatrix} \lambda \\ \mathbf{x} \end{bmatrix}$ and this makes sense because it yields the appropriate values for the BH matrix.

I will? It's been over 100º here and something like 95% relative humidity.

That is, when I take the derivative wrt to y

@TedShifrin: You'll surely know this: I want to compute the volume of the following solid using sums. When I just cut along lines similar to the ones in the sketch, I end up with an expression including $\sum_{k=1}^n \sqrt{k}$ and I don't know any way to evaluate it by hand. I'm sure there must be some geometric trick to simplify this? i.imgur.com/ZxNvVnl.png

Righto, @Stan.

OK, you'll do great. Get the hell out of there.

@TedShifrin that is abhorrent

The weather

Yeah, it's been nicer (and rainy right now) the last few days ...

I will not miss summer in the south.

So then because the BH is invertible, the IFT applies and guarantees the existence of $\psi$

@Huy: You're not supposed to do it numerically before you get to the actual integration?

You need to evaluate the sum, or the limit with a $1/n$ in front of it?

@TedShifrin: It's an old exercise from when I was at high school, it was before we learned that an integral is just an "anti-derivative".

Oh, actually $\sqrt{k/n}$, probably, too.

Yeah, I will get such an expression.

And the $\sqrt{k}$ really bothers me.

@TedShifrin which allows us to then have $(\lambda,\mathbf{x})$ vary smoothly as we vary $c$ because $\psi$ is continuous

$\psi$ is in fact $C^1$, @Stan.

Yeah

Well, @Huy, when I used to assign such problems like $$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sqrt{\frac kn},$$ the students were supposed to recognize this as an upper sum for an integral and do the integral.

I've been thinking about the relationship between braid groups and mapping class groups. In the process I've implicitly assumed the following claim: if $G$ is a simply connected group, $H$ a subgroup, and $H^\circ$ the connected component of the identity in $H$, then $\pi_1(G/H)\cong H/H^\circ$ via $[\gamma]\mapsto\gamma(1)H^\circ$. Just a sanity check, does this seem reasonable?

@anon, you're now out of my league (as usual).

oh, well, maybe not with the $\pi_1$, just with the braid group. :P

@TedShifrin: Yes, I know what you mean. But no, this is not knowing how to actually integrate, just cutting along something into $n$ parts and then summing up.

Unless you do the integral with Fermat's geometric partition, rather than the uniform partition, I don't know a way to do $\sqrt x$ ... Other than to recognize (I've forgotten the person's name) that you can relate $\int_0^a f^{-1}(x)\,dx$ to $\int f(x)\,dx$.

Wait.

@Huy: So I'm suggesting you can relate that volume to what's outside of it in the obvious rectangular box.

When you say $G$ is simply connected, are you assuming $G$ is connected?

Yes.

OK, then yes.

@TedShifrin: I thought about that too but when I try to compute the outside volume, I seem to end up with a $\sqrt{k}$ too.

No, you should slice the other way. Do the area in the plane first, then modify.

You can check that by hand. If you want an absurdly high-powered reason, pass to the homotopy long exact sequence of the fibration $H \to G \to G/H$.

What exactly do you mean?

@Huy: I'm thinking of the way you see geometrically how to relate the area under $\sqrt x$ to the area under $x^2$. I'm assuming (but haven't tried drawing it out) that you can modify that for this tent volume problem.

(Note that $H/H^\circ = \pi_0(H)$.)

well, I feel safe in finishing writing my question then :)

can you ping me when it's up?

yeah

Who's the star lecturer today/tomorrow, @MikeM?

oscar randall-williams is the starting speaker for both

wow, don't know of him

homotopy theorist

that explains it :P

it was about some results he got with boris botvinnik and johannes ebert on the space of PSC metrics on a high-dimensional manifold

Ah, he's worked with the brother of my (former) colleague Noah Giansiracusa.

giansiracusa isn't at UGA anymore? where'd he go?

He just got here. I'm gone :P

ah.

(the result, by the way, is that the space of PSC metrics, provided it's nonempty, is complicated)

PSC? positive sectional curvature?

scalar

oh, duh ... so homotopy theorists have some Riemannian geometry in 'em these days :)

I guess Gromov-Lawson started thinking about these sorts of things ages ago, too.

if $M$ is a spin manifold of dimension $d \geq 6$, there is a map $\pi_k(\mathcal R^+(M)) \to \pi_{k+d+1}(\mathbb Z \times BO)$ whose image is nontrivial whenever the latter group is nontrivial

Yeah, Gromov-Lawson definitely were doing stuff in that flavor, as I recall. But I've thrown out all those papers.

I wonder if @Huy fell asleep.

I will do that soon.

to my understanding Boris is an index theorist and Johannes does lot of geometry, so put together they can write great papers on using index theory to study the homotopy theory of Riemannian geometry

Huh? Who are those?

@MikeMiller aight, I'm getting tired of adding thoughts and it's a long enough post anyway, so I uploaded the question: math.stackexchange.com/questions/1347497/…

@TedShifrin the other authors...

oh, last names?

i wrote those down above!

@ configuration spaces : that thing even has a name?

didn't even know.

I'm confused. Neither Randel-Williams nor Giansiracusa is a Boris.

c'mon

ohhhh .. sorry, I totally missed that

spanks self

@anon: I would be surprised if this was true outside of the realm of surfaces. Maybe it's true for surfaces. I'll think about it when I get a chance

I've only thought about it for surfaces, tbh

@TedShifrin: Yes, I'm on my way to falling asleep.

@TedShifrin: I guess I'll try tomorrow if you think that works.

I basically got no sleep last night, @Huy (because of a medical procedure this morning), so I sympathize.

@anon do I have a reason to believe that would hold for a big, interesting class of spaces?

(speaks someone who is unfamiliar with the mapping class group)

I know what I said works for area. So it has to work for the volume, since the height is constant.

@TedShifrin: I hope everything's okay.

@TedShifrin: Just to make sure: You want to slize horizontally instead of vertically?

I'll know more in a while, @Huy, but seems ok so far, thanks.

@BalarkaSen well, the vague pictures all feel convincing when I put them on paper. and I used the pictures in my head to craft the argument that I wrote out. presumably the argument can be repaired to be applicable to some class of spaces. the whole reason I was doing the argument in the first place was to convince myself that it was true for ${\cal S}=\overline{\Bbb D^2}$ (which we know to be true, assuming this $\Bbb C$ has the same braid groups as $\overline{\Bbb D^2}$, which looks clear).

Draw the graph of $y=x^2$ on $[0,a]$, say. Then you can split the rectangle $[0,a]\times [0,a^2]$ so that part is the area under $x^2$ and part is the area under $\sqrt x$. That's what I'm talking about.

@Gridley: "small" = "same"

Yes sorry. Can't edit on the mobile site.

This sounds like Ramsey theory. You need @Kaj.

so, the minimum number of chromatic triangles in a 2-coloring of the complete graph Kn

@anon: fyi, you're usually going to want to delete small balls instead of points to get compactness, at least if you ever want to increase dimension. for surfaces there shouldn't be a difference but in general there will be different behaviour for noncompact and compact things

oooh, @MikeM has turned British.

what? balls v disks?

It's in from a high school text book (albeit in the miscellaneous problem section). Graph theory isn't covered (and I don't know any other than the basics one can deduce from first principles)

no, behaviour

@MikeMiller I was wondering if compactness was relevant. I couldn't figure out why it should be, although obviously it's the biggest difference between C and D^2 (and we've got compact surfaces classified...).

Ok, I'll try again in 6 hours I guess.

ok, @Huy. Schlaf gut.

Have a nice evening!

Do the Canadians have that "u" around? Maybe he has become Canadian instead @TedShifrin

ok, @Paul :)

I don't know how mapping class groups of $\Sigma_g$ minus finitely many points look like.

AFAIK, $\pi_1$ of symmetric product of $\Sigma_g$ coincides with $H_1$.

What's $\Sigma_g$?

anyway, I am not willing to think about this.

@anon surface of genus g

of the $g$th symmetric product...

@BalarkaSen well, basically it would look like indian rugburns and dehn twists

it occurs to me indian burn is not common terminology

probably especially not for Indians

shurgs

but yeah. it's a rubber sheet, so just twist things around until all of the marked points and boundaries are back where they belong, with the marked points possibly permuted

@TedShifrin :)

shrugs too

i have no idea what you're talking about, @anon

-_-

@BalarkaSen can you visualize how Mod(D^2 - {x,y}) is B_2?

nope.

i guess it has something to do with permuting the removed points.

the nontrivial generating braid in B_2 that swaps the two strings corresponds to putting your two fingers on x and y inside D^2 and then twisting until x and y have swapped, but with the rubber sheet attached to the unmoving boundary of D^2.

(btw removed points are basically the same as marked points)

ok.

@anon and that's the key difference; removed points behave like marked points in the nice case of a surface

do they not in higher dimensions?

probably not

weird

For $\pi_1$ you'd have to remove codim 2 submanifolds, not points. :)

i'll let you carry on with these. i'm not the person you could discuss mapping class groups with

you're asking whether diffeos of $S^n \times [0,\infty)$ are isotopic to ones that extend to the ball, and then that isotopies should also extend

@BalarkaSen you're the one that started the conversation.

@TedShifrin well, we're not taking pi1 of n-punctured spaces, we're taking mcg's of them and then comparing with pi1s of n-configuration space

ok, @anon.

@MikeMiller yes, but i am chickening out of it.

As usual, @Balarka, it's hours past your bedtime.

ok, back to wreck open my brains on the correct analogy of fibers for field extensions

@TedShifrin yes, that's an indication that i've finally fully recovered from my illness.

What did you recover from?

@anon: You get $n$-transitivity by using that $\text{Aut}_c(S)$ - automorphisms of compact support - acts transitively on any boundary-free surface $S$.

then induct

oh, actually $\pi_1$ of symmetric product of any simplicial complex is isom to $H_1$

Is that really right? Symmetric products are singular in dimensions $>1$.

I don't know. I just wrote down what this answer said.

I think I have seen a branched covering argument for this.

That said surfaces. Which is complex dimension 1.

You wrote simplicial complex.

Look at Kallel's answer.

Yeah, I just went back and looked. Hmm.

Why are we talking about symmetric products?

anon wants an equivalence between $\pi_1$ of symmetric products of things and mapping class groups.

No he doesn't. A configuration space is not a symmetric product.

oh?

it's the space of n-subsets of X

see the definition of F_n(X) and SF_n(X) in my question

@Balarka: You have to remove all the diagonals.

ok, fair. distinct coordinates.

bah.

also see the answer I linked to on visualizing braid groups to understand it

bookmarked it, will have a look in the morning.

sounds like something altogether out of my cup of tea, though, so probably wouldn't be able to contribute anything to it

hey chat

hi @Semiclassic

Hi pal :-)

You're right @anon it is considered "bad form," but I gave you a star anyways ;)

Except by the "elite" like Zhen.

ok, I think I have figured out what I want. gonna sleep on it.

Okay, @skillpatrol you must look at this

Is that not brilliant?

@AlecTeal :O

You see what it is right?

Not really

Elaborate please @AlecTeal

Remember that book thing I wrote, This is a progress through a book

Yes

hello, if i have J''(u)=id-A'(u) to get J\in C^2, it is sufficient to have that A' is continuous right ?

Blue is unread, green read once, yellow = 2, ---> red (max read, so say you read one page 8 times, it's on a gradient with red being 8)

Ok

How does that help @AlecTeal?

realized I forgot to meet someone I was supposed to tutor at my office today, but I checked my text messages and he cancelled beforehand. bullet dodged.

@skillpatrol I have to choose what books to take home for study at some point. The more "linear" the progression through it is, the least useful it is as a reference book

Interesting approach...

For example:

VERY sawtooth, surprisingly linear

But!

Exhibits a lot of jumping about

Do you read with a pencil and paper @AlecTeal?

Yes?

Do you do calculations, draw sketches, and take notes?

Yes

And I write [a-b] in the top right where a and b are page numbers and add them to this thing later

@Chris'ssistheartist: you might like this answer. It uses the integral we chatted about a while ago: $$\int_0^1\left\{\frac1x\right\}\,\mathrm{d}x$$

@robjohn Nice. :-)

You should give "this thing" a name and copyright it @AlecTeal :)

....I really need someone more than just me to use it first :P

@Rememberme doesn't seem to get it!

Give it time, it will catch on.

How can somebody use this to improve their reading skills @AlecTeal?

It's more so you can jump around freely within a book

I see.

Not waste time rereading stuff you think understand.

When does term end for you @MaryStar (sigh)

What I am asking is not part of a subject... I want to solve this for personal interest. @AlecTeal

Are you a student? @AlecTeal

I'm thinking of becoming a murderer. @MaryStar

-_-

Hello @skillpatrol

How are you?

Hi @MaryStar fine thanks. How are you doing?

Fine.

here @MaryStar