yeah I am gonna write about all of the connection it made

lol fermat is so funny saying stuff like it won't fit the margin

he probably didn't figured it out

That is good, that is the real important part anyways. Haha, yes

Guys, can someone please explain voting pattern of user "pizza" aka "thisismuchhealthier"? Thousands downvotes a month?? I am not criticizing him, just want to understand the behavior.

@VividD Sure I can

@VividD Basically the idea is if you vote on something but it is then subsequently deleted on the same day you get the vote back

So every day (around 3UTC) closed questions satsifying a certain criteria are deleted from the site (see here ), and once a week questions satsifying a certain criterion get deleted.

I think I understand what degrees of freedom are, however there are a few definitions. Can anyone explain this one to me: Degrees of freedom is the number of dimensions of the domain of a random vector.

For example if at 0UTC I start voting on some closed questions (typically to get them to satisfy the criterion), say 20 questions, maybe only 10 satisfy the criterion by 3UTC, so I "recover" 10 votes and at the end of the day I could end up with 50 votes. Also just getting questions deleted in general give you any votes back. @VividD

hmm, so this guy wants to be accepted in Guinness Book of Records for downvoting?

Today pizza actually became the number one voter on the site @VividD

math.stackexchange.com/users?tab=Voters&filter=all @VividD

Yesterday I got 80votes, its not too difficult.

Thanks @DiscipleofBarney

@VividD I think pizza is just helping the site the way pizza knows how

@VividD This also explains the idea meta.stackexchange.com/a/239868

@DiscipleofBarney I am fairly sure you misunderstood Vivid's question

He wanted to know why someone would vote like that, not how they got so many :P

You think VIvid wants the psychology

>I am not criticizing him, just want to understand the behavior.

I admit I must have skipped that part

yes, psychology

Funnily enough, one of his old usernames was Behaviour :P

I am not sure, after all I am not pizza, but like I said I think pizza just wants to help the site the way he knows and thinks there is too much trash on the site

Okay guys I am gonna go to sleep finished 3 questions out of 9 in my assignment today for number theory tomorrow 6 :D

then that is it ! xD

Sounds good

ok

@VividD Well I would say he makes up 40% of downvotes by himself right now

So really people don't downvote enough

People don't vote enough

in general

It's probably closer to 10% but still

good night guys I will see you tomorrow

@KarimMansour Goodnight, best of luck

Good luck @KarimMansour

All the high voters are largely upvoters, so I wouldn't be surprised if pizza was 40% of downvotes

good night from me too guys

@Vivid

That graph quickly before you go

Upvotes are blue

10 times the downvotes is red

Wow

Pizza starting is on the right

Ten times and it still doesn't match

It's aged

So I will try to find the query

incredible

we need 10 pizzas

I'm graphing now

Haha, I wouldn't mind 10 pizzas

at the end only couple of hundred questions should remain at this whole site, in lets say 2017 hah

good night

and this room should be on the home page of the site hah

Yup....

I think that is sort of Reopen? Undelete? Close? Delete? although it is not too active, and it is "inlcusive" for other voting, and opening

Cumulative top, daily entries bot

Not times 10 for comparison for bottom(since it is hard to see the magnitude difference)

I wonder if me starting can be seen? :D

In the times ten graph, surely

@pizza: Thanks. I'm fairly computer illiterate; how do I get this to run automatically? I use Chrome, if that helps.

Well I will get back to work, cya later

@Incurrence Which question?

@MikeMiller My initial idea for this problem was to lift loops in $X$ based at $x_0 \in A$ to $\widetilde{X}$, and by simple connectedness homotope the lifted path to some path in $p^{-1}(A)$, and finally pushing down that homotoped path to a loop in $A$ based at $x_0$. This, I think, constructs a well-defined map $\pi_1(X) \to \pi_1(A)$ which is probably the right inverse of the map $i_\#$.

What do you think?

@Balarka: You want any two ways of doing this to be connected by a homotopy downstairs. Why can you connect any two such constructions by a homotopy upstairs that lives in $p^{-1}(A)$?

upstairs

@MikeMiller You mean that there may be two homotopic loops in $A$ downstairs that aren't homotopic upstairs in $p^{-1}(A)$?

Other way around. You're homotoping your path to one that lives in $p^{-1}(A)$. For two different choices of which path you homotope to, it seems plausible that they may not map to homotopic loops in $A$.

I guess more to the point - loops that are homotopic in $X$ needn't be homotopic in $A$.

yeah, i thought about that. what if i choose the the path to which i homotope to for every generator of $\pi_1(X)$?

I guess you don't care if it's a group homomorphism. That makes me feel gross, though.

No, the question just wants $i_\#$ to be surjective.

Right.

It's gross nevertheless.

:P

But I am not entirely sure if it's loop(+)hole-free. (+ no pun intended)

Ugh.

lol

OK. You have an inclusion $A \subset X$, and $p^{-1}(A)$ is connected. You want to show that $\pi_1(A) \to \pi_1(X)$ is surjective. Pick a loop in $X$; lift, as you said; this gives a path in $\tilde X$ whose endpoints are both in $p^{-1}(A)$. Because $p^{-1}(A)$ is connected, these two points are indeed path-connected, and because we're simply connected, we can homotope our path to a path that lives in $p^{-1}(A)$. Pushing downwards, we just represented a loop in $X$ by a loop in $A$.

That's just what we wanted.

It's exactly what you're doing, without bothering to construct a right inverse explicitly; of course a surjective map of sets has a right inverse.

But how can you prove something is surjective without constructing the right inverse? I don't understand your objection.

We literally just proved it was surjective.

ah, dumb me.

yeah.

:P

I haven't really come upon a situation where the right inverse was helpful unless it was actually a group homomorphism in a situation where those show up naturally.

well, i guess i should post this as an answer, then.

@MikeMiller yeah. i get you.

Did the one that was there not answer it? I didn't read.

I guess it answers it, but not sure if it uses my argument. Let me have a look.

hmm. off topic, but this statement of the question looks oddly familiar to me. i have seen it somewhere, but it's definitely not in Hatcher.

Dunno. It's a cute fact.

yeah. let me google it, see where it came from.

ah, i got lucky and hit the right thing. there it is.

Smale's result is a very nice fact.

Hey @Balarka how can i geometrically say that integration is just the inverse of differentiation

you can't. fundamental theorem of calculus is not entirely a geometric fact. that's why it's nontrivial.

So its just on pen and paper not visually true

i don't know what you mean by that.

thanks anyways...........

it's true. "visually true" doesn't make sense to me, as visualization is just a representation, not a method. you can prove fundamental theorem by drawing, but it'd essentially be equivalent to doing the calculations.

@BalarkaSen: But drawings help understanding. Surely you prefer the picture that the fundamental group's product is associative to the formula?

Yes, but it's still a representation. Not a rigorous method.

You take the picture and turn it into rigor. But nonetheless, the picture is where one should start.

Fair, but that's not relevant anymore :)

@MikeMiller actually, that's not enough. we need to show that the loop on X we just made outta the loop in A is actually the preimage of the loop in A we started with, right?

I can't understand what you mean by that.

"representing" the loop in X by a loop in A isn't enough.

Yes it is...

but that's essentially the same as constructing the right inverse...

That $\pi_1(A) \to \pi_1(X)$ is surjective is equivalent to saying that every loop in $X$ is homotopic to one in $A$.

but we just constructed a map $\pi_1(X, x_0)$ to $\pi_1(A, x_0)$, right?

If you want to say you constructed a map, sure, whatever. It's the same thing.

okay.

btw, your loop space thing $\pi_n(\Omega X) \cong \pi_{n+1}(X)$ was bothering me. this seems to be true, but i dunno how one plans to go about proving this.

can't have anything to do with suspending stuff, can it?

It's precisely the fact that $\Sigma S^n = S^{n+1}$, where I'm taking the reduced suspension here. Prove that there is a natural bijection $[X, \Omega Y] = [\Sigma X, Y]$.

You could also do this with the fibration $X \to PX \to \Omega X$, where $PX$ is the space of based paths.

ooh.

I prefer the former.

@Balarka Mike answered it in the most part

I was under the impression that

The green line has a homotopic equivalence class with any other line that goes around both singularities

yes, not true.

and the red has a homotopic equivalence class to any that goes around just one of the singularities

The green line isn't meant to connect twice like that

So there isn't some equivalence class thing?

Just "homotopy class".

Homotopy equivalence is something different.

Oh ok

So there is no such thing? My lecturer definitely led me to believe there was such a thing, but I couldn't remember the names

homotopy equivalence is an equivalence relation of spaces.

you just mean (path?) homotopy. those are the correct equivalence relation of paths.

The reason I asked, was originally a curiosity about a path like this:

@BalarkaSen Okay yes, path homotopy

That blue line just goes around and joins up

Is this acceptable? Can paths move through the singularities?

depends on what you mean by singularities.

Is this equivalent path homotopically with any other path going through both of those wholes

@BalarkaSen I guess handles

the blue line certainly is a path in the genus two surface.

@Incurrence consider the path the goes inside one singularity, comes back from another singularity and then goes inside the first singularity and then again comes back from the other.

Ahhhhh

This is looking like mobius strips a little now

not in the least. it's just the blue line looped twice.

xD

Alright, thanks. Do I have heaps of study in Algebra to do before I start any of this?

not really.

but it's good to know some algebra.

the point-set topology requirement is very small too, but as usual, it's good to know some topology.

So I should do munkres topology, then hatcher(which together I understand will take me years)

Greetings

Greetings

@robjohn I have amazing news

@r9m look at the following statement

you have found something @Chris'ssis

I'm done with this one in 2 lines by simple manipulations $$\sum_{n\ge 1}\left(\frac{H_n}{n}\right)^3$$

This is hard to accept, yeah I can guess that, but this is the way the reality looks like sometimes! :-)

(Well, some clever manipulations, indeed)

what does $H_n$ represent here@Chris'ssis

@Rememberme harmonic number

Oh

@Incurrence Hi

That's why I say I prefer to consider myself an artist, I only like the solutions done in spirit of the art. Who cares about only solving problems? I don't.

$$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3 =\frac{31}{5040}\pi^6 -\frac{5}{2}\zeta^2(3) =\frac{93}{16}\zeta(6) -\frac{5}{2}\zeta^2(3)$$

When writing an article, of course, it takes more, maybe 2 pages since you explain a lot.

@Incurrence right

BBL

@BalarkaSen But Munkres deals primarily with point-set I believed?

Oh nvm

Part II switches over

yes, but it talks about algebraic topology in it's second part.

Why are 31 people in the chat room tonight(tonight for me)

they forgot to leave when they were done

:P

Over twice as many as usual

@MikeMiller So if you don't mind me asking, is Ted paranoid in thinking you are avoiding him?

You have been black listed @Incurrence

@DiscipleofBarney By?

Mike

@DiscipleofBarney Mike?

@DiscipleofBarney Probably not, but I won't ask about it again

@DiscipleofBarney Can you help me with trivial crap?

@DiscipleofBarney For a few min?

Give it a go, who knows, I might be able too

You definitely can haha

Looking at past exams right now, and I have the answers, and I asked this yesterday but I am still sucking

$\sin z =6$ I want in terms $x+iy$

So I set $\frac{e^{iz} - e^{-iz}}{2i}=6$

$e^{iz} - e^{-iz} = 12i$

Oh wait

I never considered changing the RHS to exp form...

Wow

lol

Let me go play around again lmao

Okay

Okay so I have $e^{iz} -e^{-iz} = 12e^{i\frac\pi 2}$

$6+i \cdot 0$ :D

Is that a hint to do with matching left and right real and complex components^?

I was just writing $6$ in terms of $x+iy$

oh lmao

Okay, more really embarrassing questions.......

Well really, I just don't get how the hell they did this

That just comes of nowhere

The log thing?

Yep

@DiscipleofBarney Do you not know this stuff either?

I don't really know much complex analysis, just basic exponential stuff. I am guessing the log stuff would come from working out the inverse of sine in the exponential form but I don't know. @Incurrence (I think I have told you already I don't really know complex analysis)

@DiscipleofBarney Oh okay, that's fine

You can probably start logging what you have

I am trying to show that $\left|\int_{\Gamma_n^B} \frac{e^{-z}}{z+1} dz \right| <\frac{1}{n}$ where $\Gamma_n^B = \{t-in : 0 \ leq t \leq n \}$

Would someone be able to help me with this?

I can only show it to be less than 1.

I can say that exp(-z) <= 1 (by writing z=t-in and then using the estimation lemma)

And I've written |z+1| > |1-n|

But this gives me too high an upper bound.

I can't find a way of getting 1/n

Hello!!!

Is anyone familiar with DFS?

This is the algorithm:

I applied it several times at the following graph:

in order to get all the possible results of DFS

and I wrote the nodes in decreasing order as for the finishing times

I should I also get the result {e, a, b, d, c} but I do not get it. Do you have an idea how we could get this result?

@DanielFischer @TedShifrin would you be able to help me with this question?

@Henrik Do you maybe have an idea?

@Incurrence Hi AC.

@evinda Aha!

Hi

I have only tomorrow to study for complex and I suck at it

@WillHunting Aha! @Incurrence Hi

And I have algebra to do by the day after that

@Incurrence It's an exam?

Yep

@Incurrence Good luck, I will pray for you.

Pray lots :)

Why are there so many in this room?

Is anyone able to help with contour integration?

@DanielFischer Are you familiar with DFS?

@WillHunting That's what I said at 31, now we have 36

@Incurrence I think maybe they are here to see you, the future Fields medallist.

@evinda Superficially.

@WillHunting lmao, don't say silly things when there are 36 people watching

@DanielFischer Did you see my question above?

@DanielFischer and mine? Do you know how I can show that that integral is less than 1/n?

@Incurrence It's alright, I think 90 per cent of them have put me on ignore already.

I'm looking.

@user112495 On $\Gamma_n^B$, you have $\lvert z+1\rvert = \lvert (1+t)-in\rvert = \sqrt{(1+t)^2 + n^2} > n$. If you use that estimate, you can get the desired bound for the integral.

@Incurrence If you always lmao, how come you still can lmao, since it has already come off?

@WillHunting lmao

@Incurrence I thought the 90 per cent statement would have been starred by now. The reason it is not starred is because it is true, lmao.

@WillHunting You should go to my chat room if you want to talk, I don't want to interrupt the chat with non-math when there are so many here(36)

@Incurrence I recommend you do Bredon, which contains point set, differential and algebraic topology all in one book. Forget Munkres and Hatcher altogether.

@WillHunting What why? Munkres seems really good, and is apparently the standard

(and I have to use it for my current functional course)

@Incurrence Take a look at Bredon. It's really good too.

@evinda You can't get that result from a DFS on that graph. Either you look at b before you look at e, then b and d are both explored before e, or you look at e before b, then all of e's descendants are visited before b, and that means b is the last node visited.

I'll check it out

@Incurrence Title is Topology and Geometry.

@Incurrence Who cares what is standard? If everyone uses the standard, newer books will never have a chance.

@WillHunting That is true, but usually they are the standard for a reason

As in, I should do the standard, and then do other textbooks

@Incurrence Yes, the reason is because when they were written long ago, they were the only ones available.

@Incurrence No, you should do what you consider best for yourself.

Brb

@DanielFischer but when you integrate 1/n between 0 and n, that just leaves you with 1 rather than 1/n.

@user112495 Well, don't integrate $1/n$. There's more in the integrand, you can use that.

@DanielFischer Hi Daniel, how is your health these days?

@WillHunting As usual. Ups and downs.

@evinda Oh, wait, in decreasing order of finishing time? Then that list is "First visit c, then d, then b, then a, finally e". That's possible, you just need to feed the nodes in a correct order to the outer loop.

@DanielFischer So at the DFS , suppose that we start from the node c. Then, it has no oufoing edges, so the discovery and finishing time will be 1 and 2, respectively. Then, we choose to visit d. Again, it has no outgoing edges, so the discovery and finishing time will be 3,4 respectively. Then we choose to visit b. Then, can we visit a, although it has incoming edges, and then e? So can we choose at each step any vertex we want, independently if it has incoming edges or not?

@evinda You can pass the node list to the outer loop in any order you want. The edges only matter for the visiting in the inner loop (and affect when a node is coloured, if a node has incoming edges, it can be coloured before the outer loop gets to it).

And all these possible outputs that we get are all the possible topological sortings that we could get? @DanielFischer

@evinda Need not be a topological sort. The one listed above isn't one, for example.

@DanielFischer And what if we want to find only the outputs that are also topological sorts? Would I have to add a condition at the algorithm pastebin.com/uZMnTguW ? :/

@evinda Wait, once more been confused by the strange thing that the DFS output is in decreasing finishing time order. So when you want to check whether the output gives a topological sort, shall that be in left to-right order, so every node is listed left of all its descendants? Or in right-to-left order?

@evinda So left-to-right. Then indeed your output is a topological sort, since a descendant is always finished before its ancestor in a DFS. Can you get all topological sorts that way? Yes.

Hi @JulianRachman really long time nooo seee!!!!!

@DanielFischer So even if we want to find a topological sort, can we visit at each step any node, even if it has incoming edges? So is the algorithm I sent you right? Does it return all the possible topological sorts?

Is he here @Sayan?

Hi @Incurrence

@Rememberme Hello

@Rememberme I think Julian is just afk here

I can see him.....

@Incurrence anyways where are you from?

@Rememberme I swear I have told you, Australia

@Rememberme Aren't you Sayan?

@Incurrence I am sayan i might have forgotten

@Rememberme It's alright, I can't find it in the transcript

@Incurrence you were blacklisted? How and why

@Rememberme Can't talk about it lol

@Kaj Are you busy these days, or the rooms isn't Mathy enough anymore?

Ok fine