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3:00 PM
@Chris'ssis Non-math question, what do you think about this glasses?
 
let me check something interesting now ...
 
@Chris'ssis Does that look cute if a girl wear it?
 
@Venus They look interesting.
@Venus Yeah, I think so. :-)
 
@BalarkaSen I don't think they're homotopy equivalent. A continuous map from the earring to the wedge can only meet finitely many loops in the wedge.
 
Girl's concerns >.>
 
3:03 PM
@Chris'ssis I plan to ask my BF to buy it for X-mas
 
@Venus To buy the girl ? :D Are the glasses extra ?
:P
 
@Venus That's not bad at all. :-)
 
@Hippalectryon I don't get it your question
 
@DanielFischer In any case, YIKES. That's a crazy covering space.
 
@Venus It's just a joke :c nvm
 
3:05 PM
@Chris'ssis What did you mean by not bad at all? It sounds it's totally bad to me
@Hippalectryon I get it now haha
 
@Venus lol, why? Do you want more? :-)
 
@DanielFischer They're not even a little bit homotopy equivalent; the wedge has countable fundamental group, and the earring has uncountable.
 
Xmas = chocolate :D
 
I seem to recall the earring even has nontrivial $H_2$...
 
@Chris'ssis The answer should be good or bad. Your answer makes me confused
 
3:06 PM
@BalarkaSen Yes, it's pretty un-pretty ;)
 
@Chris'ssis I saw Annastasiya wear it & she looks cute
 
@Venus Well, it's good, that I meant. :-)
 
@Venus It's Anast not Annast
Anastasiya-Romanova
 
@MikeMiller I am not sure why so. Intuitively, you can "chuck out" the circles inside to pull it to outside and you'll get a bunch of circles wedged at a point.
 
@Chris'ssis OK, I will start to browse online catalogue then
@Hippalectryon Sorry, my bad. You seems care with her :P
 
3:08 PM
That doesn't sound very continuous.
 
@Venus She's in my gold users list, with @Chris'ssis. They produce such nice Q&A :D
 
Might be @Mike. Hrmph. I forgot continuity in this mess.
 
@Venus :D
 
@Balarka Try to prove what I said about the fundamental groups involved.
 
@Hippalectryon lol, gold users? Platinum users don't you have? :-)))
 
3:10 PM
@Chris'ssis Unfortunately not
 
@Hippalectryon OK :-)
 
I was scared away by your $H_2$, @Mike. But if you say so...
 
@Hippalectryon You like her, like emo too. It's kind of a sign ^^
 
@Balarka The first thing I said is completely unrelated to the letter $H$.
 
Ana might not be Anna but @Hipp is still Hipp, not Hip.
 
3:11 PM
@Hippalectryon How did you make a feature like that?
 
@Venus >.> What kind of reasoning is that (ノಠ ∩ಠ)ノ彡( o°o)
@Venus It's a chrome extension
 
@Hippalectryon Just kidding :D
@Hippalectryon Where I can get it?
Link please
 
On the deadly flower's profile
 
@BalarkaSen But she want us to call her Anna, not Ana
 
3:14 PM
@MikeMiller it's evident that it's homotopy equivalent to $\Bbb R^2$ with $(0, 1/n)$s chucked out, so I guess I'd start from that.
Wait
 
Path-ological spaces are confusing =(
 
Is that evident?
 
No, @Mike.
 
Best book for abstract algebra?
 
3:14 PM
That's why I said "wait".
@GustavoMontano Artin
But Dummit-Foote is also good @GustavoMontano
 
OK. Don't try to give it other forms, just use the standard definition...
 
@Hippalectryon I have no chrome, haha
 
I want to buy a good book on abstract algebra to read over the holidays.
 
Will have a look at these, thanks @BalarkaSen.
 
3:16 PM
It's not @Mike
 
@Venus What do you use ?
 
$\Bbb R^2 - \bigcup_{n} (0, 1/n)$ is deformation retractable to consecutive wedge of countably many circles, homotopy equiv to $\Bbb C - \Bbb Q$
 
Mozilla
 
Hahaha, @WillHunting's starred comment made me laugh!
 
3:18 PM
@GustavoMontano Look at what?
 
The books.
 
Which books?
 
Artin & Dummit-Foote
 
I've studied those @GustavoMontano. Not sure what you mean.
Oh darn
I misread "Will have a look" as "Will you have a look"
 
Hahaha :)
 
3:21 PM
This happened to me twice a day, probably I should shut up and think about the Hawaiian earring.
 
Did they change mathjax?
 
@Balarka The countable wedge of circles is not homotopy equivalent to $\Bbb R^2\setminus $\Bbb Q$.....
 
I mean consecutive wedge.
 
And I've my doubts it's homotopy equivalent to $\Bbb R^2 \setminus \{1/n\}$ either.
 
a bunch of circles arranged sidewise tangentially.
 
3:23 PM
There's no such thing as a consecutive wedge.
 
@UserX ?
 
Oh, I see. That's still wrong.
 
@Hippalectryon it renders differently now...
 
oh i am thinking about $\Bbb C - \Bbb N$
grr pathological spaces.
$\Bbb Q$ is dense in the x-axis.
 
@UserX I don't see any difference
 
3:26 PM
It doesn't become grey while rendering
 
Yes, @BalarkaSen, and that $\Bbb C \setminus \Bbb Q$ has uncountable fundamental group is easy to see. Cardinality again...
 
Oh that
Yeah then @UserX
 
13
Q: Mathjax 2.5 alpha

Geoff DalgasWe have deployed the latest Mathjax 2.5 alpha on Math.SE. As with any beta release please post an answer to this question if you find any issues. We will be monitoring this thread closely along with the authors of Mathjax to ensure this release goes smoothly. This particular release has a new f...

 
What's the difference between alpha and beta versions?
 
How do I enable fast rendering?
 
3:27 PM
Wait, is there such thing as a beta version? I am assuming there is.
 
@MikeMiller By reading the post
 
@MikeMiller it's the default I guess
You can turn it off if you want
 
@UserX no
 
I have a simple question for you ... did you ever see this integral in closed form $$\int_0^{\pi/2} \arcsin(\tan(x)) \ dx$$?
 
@UserX It means you should buy a new laptop. It's holiday month, sales are everywhere :D
 
3:28 PM
@Venus @Hippalectryon ^^^
 
@Venus I'm on my phone
 
That didn't work when I tried it at home... bizarre. Maybe i'll clean my cache. Pleasant comment regardless.
 
98% of my answers and browsing are done with my phone
 
@Chris'ssis I don't have a good memory for that :/ BUT don't give me the closed form. It doesn't seem that awful.
 
@Chris'ssis The answer is NEVER :D
 
3:29 PM
I can't do it @Mike it's too complicated for me
I just can't think about these spaces.
 
Ask your guru @Chris'ssis :-)
 
OK, let me know when you change your mind.
 
@Venus I did it on my own.
 
i thought you'd reveal the solution @Mike
 
No.
 
3:30 PM
heaves great sigh
 
Yikes, fast preview is ugly. I guess that's why it's a preview.
 
@Venus Don't understand me wrong, but I'll always appreciate someone that invested time and effort in me. He is and it remains a guru to me, but this doesn't mean I do not appreciate the others. I don't like when others want to think for me and speak about me as if they knew what my options are.
@Venus I love your answer, M.N.C.E.'s answers, you're on top, that's clear.
 
Hello!! What does it mean <<prove the correctness of the algorithm>> ??
 
@Chris'ssis I never get you wrong, seriously. I'm on your question anyway. That was only a joke
 
@Venus OK :D
 
3:35 PM
@Chris'ssis I never get pissed off by anyone on internet. I'm always cool ^^
 
@Venus That's great!
@Hippalectryon Did you manage to finish those series? Do they match numerically with the integral? I'll check the details when I return. I need to buy some food for my pets.
Back in 20-30 min.
 
@Chris'ssis I didn't work on them any further. I don't have mathematics, and I usually make stupid mistakes, so I thought it would be better if you could proofread it first.
 
@Hippalectryon OK. I'll finish the job there (soon).
 
@Chris'ssis Make sure I didn't make a mistake in what I sent you though
 
@Hippalectryon Sure. That's clear.
 
3:39 PM
It would be a shame to compute it, then realize it was wrong
 
@MikeMiller It is clear that $\pi_1(\mathfrak{H})$ has a copy of $\prod \Bbb Z$ inside, but I don't see any other elements though.
 
"It is clear" is not a proof.
You use those words too much.
Anyway, i've gotta teach. See ya.
 
OK, I am gonna use "if you don't believe me, I'll punch you" from now on.
 
Anyway, if you can prove that group embeds, you're done... since $\prod_{\Bbb N} \Bbb Z$ is uncountable
 
3:56 PM
@MaryStar It means "prove that the algorithm does what it is meant to do". Whether that is sorting an array, factoring a number, whatever.
 
@Chris'ssis $$\int_0^{\pi/2} \arcsin(\tan x) \ dx=2{\large\chi}_2\left(\sqrt{2}-1\right)-\frac{\ln\left(3-2 \sqrt{2}\right)}{2}\sinh^{-1}(1)-\frac{i\pi}{2 }\sinh^{-1}(1)$$
 
@Venus Nooooo :c
 
@Hippalectryon Hohoho...
 
31 mins ago, by Hippalectryon
@Chris'ssis I don't have a good memory for that :/ BUT don't give me the closed form. It doesn't seem that awful.
I wanted to try it before :C
 
It came from Santa
 
4:00 PM
Santa is mean
 
@Hippalectryon Didn't you know Santa? I mean Santa Claus
 
I know him :c
He never gives me what I want
 
@Hippalectryon Coz you're not a good boy *kidding
 
I always get it every X-mas
 
4:20 PM
OK OK OK, that one was far too easy
clap, clap, clap...
 
@Chris'ssis Which one ? The arcsin(tan) or arctan^3(sin) ?
 
Let's see this one ...
@Hippalectryon $\arcsin(\tan(x))$
$$\int_0^{\pi/2} \arcsin^2(\tan x) \ dx$$
@Venus Good job.
 
@Chris'ssis I'm honored it came from you, but that was not easy for me
 
@Venus How would you rate it as difficulty (on a sclae from 1 to 10)?
 
@Chris'ssis Do you always follow fashion trend?
I think it's 7
 
4:30 PM
Girl's stuff >.>
 
@Venus What do you mean by "fashion trend"?
@Hippalectryon :D
 
@Chris'ssis Fashion like clothes, scarf, hat, etc
 
@Venus Yeah, I know, but what is the meaning in this context? Do you refer to integrals or to clothes, scarfs and hats (and so on)?
@Venus I don't follow the trends posed by the others, never. I define for me the fashion trends as I like.
 
it's just a personal question. Please don't always talk about math
 
@Venus hehehe, I was thinking of math. :D
 
4:34 PM
@Chris'ssis You're too serious :D
 
@Venus You got me here! :-)
 
What kind of pet do you take care of?
 
@Venus Dogs
 
More than one?
 
@Venus I had $7$ some time ago, but now I only have $3$.
 
4:36 PM
@Chris'ssis What??
I envy you. I can keep dogs here T_T
 
plot twist : @Chris'ssis can't integrate, she just steals her dog's works
:D
 
@Venus Yeah, I had $7$ dogs! I often prefer to talk to them, and they answer back to me barking. :-)
 
Knock knock!
 
@Hippalectryon :-))))))
 
@Chris'ssis Owww, that's very cutteee!!
 
4:38 PM
@Venus yeap :-)
 
@robjohn @UserX Any Idea about how much votes are required to trigger serial voting script?
 
@Integrator couldn't say
 
@robjohn okay!
@robjohn Any update about mod-elections, I need badge!
 
Anonymous
Anybody has a solution book to Linear Algebra and its Applicatins?
 
Anonymous
Thers's one guy on the web who claims he has solution books to many books,but demands money.Can I buy the book from him?
 
4:48 PM
@AshwinGokhale I don't trust anybody!
 
@AshwinGokhale Never trust a man :D
 
@Integrator you need a badge? any badge in partcular?
 
Anonymous
@Committingtoachallenge Are you taking any Physics courses?
 
@Integrator I get my first gold in 5 hours
 
5:06 PM
I still need the Self-Learner Badge ;_;
 
@UserX Which one? Electorate? Fanatic?
@UserX I think Fanatic!
@Venus I'm just a kid :P
 
Grrr... I still only need one more upvote...
 
@Hippalectryon I think something is wrong there since numerically the last series does not match the integral.
@Hippalectryon I'll go my way ... from scratch.
 
@teadawg1337 I could have up-voted you but, I've reached my voting limit for today, and I'll need to spend my votes more wisely to earn a Sportsmanship badge! So, will help you after two days! @ least!
 
@Integrator fanatic. I won't probably get the upvoting ones soon as I'm too picky on my upvotes
 
5:28 PM
@Hippalectryon
2 ((-1+Sqrt[2])^(-3+2 k+2 l+2 n) (1/(3-2 k-2 l-2 n)+1/(-1+2 k+2 l-2 n)+1/(-1+2 k-2 l+2 n)+1/(-1-2 k+2 l+2 n)))/((-1+2 k) (-1+2 l) (-1+2 n))
@Hippalectryon The calculations are not hard but pretty tedious, boring.
 
@Integrator I see. I never thought that election badges (which are gotten quite easily when there is an election) were that important.
@teadawg1337 for?
 
@MikeMiller oh you just want me to prove that $\pi_1(\mathfrak{H})$ is uncountable? that is pretty obvious i'll punch you if you don't believe that's true.
@Venus everyone are too serious in here.
 
I do believe it's true, but I assume you're trying to learn some math, and you don't do that by asserting things without proof.
 
@MikeMiller i was just joking.
 
ok. ping me when you have a proof.
 
5:43 PM
@BalarkaSen I'm too serious to answer this one too
0
A: Proving $\frac{200}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}=25$

VenusWe will evaluate $$S=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh\left(\frac{\pi}{2}(2n+1)\right)}$$ Since the series is alternating, use $f(z)=\pi\csc(\pi z)$ and we have $$\oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh\left(\frac{\pi}{2}(2z+1)\right)}\,dz$$ The poles are at $z=-\frac{1}{2}, \;\ z=...

 
@Venus i don't see the context. i was just asserting that mathematicians in MSE chat are serious in general.
come along to the dark side. don't try to resist :P
 
@BalarkaSen I didn't see the chat because I was too busy to answer a question
 
eh? I was replying to this :
1 hour ago, by Venus
@Chris'ssis You're too serious :D
 
@BalarkaSen How do I even know that? You just mentioned my name, so I replied it
 
@Venus Click on the small arrow on the message.
^ do you see a small grey arrow?
 
5:52 PM
I created a cryptographed code. Anyone up for deciphering it?
 
What arrow? I can't see anything
 
@Integrator maybe you're interested
 
@Venus this is an example of a message with no arrow
@Venus this is an example of message with a arrow
 
@Integrator Your comments about election / badges / Sportsmanship suggest you took Jyrki's comment too seriously.
 
@WarmFuzzies I like the new name. How do you change so frequently? I thought there was a monthly limit.
 
5:54 PM
wait @WarmFuzzies = care bear?
 
@BalarkaSen I get it now. I thought that arrow was meaningless
 
@Venus it's pretty useful.
 
@MikeMiller Join new SE site -> change name there & copy to the rest of network. A known workaround.
 
Also = Raff, Rafflesia Arnoldii, this is much healthier, 900 sit-ups per day, Post No Bulls...
@WarmFuzzies Ah, so one day you'll run out of free passes.
 
@BalarkaSen How to type a chat contains arrow like that?
 
5:56 PM
No, because I can also delete those accounts. And new sites come up almost every week.
 
Ah, I didn't know you could delete accounts.
 
This is super easy on a site where you never voted or posted: there is a delete button on the profile page.
 
@Venus mouseover a message you want to reply to, and clickon the downarrow that appears. you'll get a drop down list containing "reply to this message". click on it.
 
@BalarkaSen This!
It's so easy Haha
What kind of magic that I haven't known yet in chatroom?
 

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