Mathematics - 2014-10-21 (page 2 of 6)

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3:00 PM

@Anastasiya-Romanova Schizophrenia is characterised by hallucinations and delusions, not split personalities. You can read Wikipedia for more information.

my impression that split personality disorder doesn't really exist, at least not in the sense that popular culture would make one believe

but then pop culture's relation to mental illness in general is pretty much BS

Yes, pop culture makes people misunderstand OCD.

Anastasiya-Romanova

@JasperLoy OK

@DanielFischer: if you know a name/reference for that, i'd appreciate it. right now i'm tempted to put it up as a Q&A, but that's wasteful if it's already cited somewhere

@Semiclassical You can say it's an instance of Cavalieri's principle. But I don't know of a special name for this case.

3:08 PM

hmm

@JasperLoy Schizophernia is just too terrible to imagine. :|

that might be enough, really

@Sawarnik And OCD cannot be imagined either.

It sounds a bit less terrible.

@Anastasiya Can I ask one more thing?

the word 'terrible' seems not entirely helpful. mental illness can be horrible and painful (and i say that as someone who struggles with depression/anxiety). but it isn't a death sentence; it's something one has to live with, as best one can

3:10 PM

@Semiclassical You struggle with depression?

yeah

@Semiclassical Yes, horrific and painful.

@Semiclassical It doesn't seem so :O :O Why though?

b/c depression isn't something you can just look at someone and see

much less notice through the internet

or, to put it another way, depression \neq being sad and mopey all the time. it can manifest that way, but it's not reducible to such

@Semiclassical You seem cheerful, humorous, intelligent and ... that's why.

Anastasiya-Romanova

@Sawarnik Yes? What is the question?

3:13 PM

@Anastasiya-Romanova Why did you change your location to Russia :D

Holy cow, math.stackexchange.com/questions/984185/… has seven (undeleted) answers and over 300 views.

Yeah, expected the Dr. to be one of those.
Those are the questions which always get so many low-quality answers. And its not closed as duplicate.

Anastasiya-Romanova

@Sawarnik I change my location depends on my mood or someone who I admire

@Anastasiya-Romanova And you name too?

would rather not try to explain things in depth. but this environment != daily life. (plus depression is one of those things that's really hard to explain)

3:16 PM

@Semiclassical Ok :) I will talk to you on this someday.

Anastasiya-Romanova

@Sawarnik Yes

but, i do appreciate that remark :)

@Sawarnik If you know a dupe, give the link.

@DanielFischer I am surprised I have no answer there, lol.

Ack took $\pi/4$ as $90^\circ$ in the test today.

3:18 PM

@DanielFischer The dupe is simply any post that talks about division by 0 being not allowed.

@DanielFischer That's what I say despite getting the same question everyday, we don't have dupes.

coming in 1 min...

@Sawarnik The sad thing is that even if there is already a duplicate link plus a couple of close-votes, some people still hasten to drop their answers rather than close-voting too.

they want to get rep while the getting is good :P

one thing i find more troublesome when it comes to dupe is people replicating their own unanswered questions

you can't exactly call it a duplicate of the old question if it's not answered

@DanielFischer Many of these lhf attract poor answers, lol.

3:41 PM

What should be done about newly created beta-function? For more details and some background see here. (Previously there was a discussion on meta with the outcome that such tag is not needed.)

without knowing about the meta discussion, just looking at that tag i notice a problem: is it about the beta function as a special function, or the beta function as a pdf

@Semiclassical The tag-excerpt says special function.

right. but two of the four tagged problems are about the latter

which suggests that, among other things, it'd likely attract quite a bit of noise

Well, then they are certainly misstagged. (Or at least not tagged in accordance with the tag excerpt.)

right. my point is more that, if two of the four tags of a new tag are incorrect, it doesn't say great things about the likelihood of that tag to be used productively

3:46 PM

I'd say that people who are using ambiguous tags should read tag-info (tag-excerpt and tag-wiki). Although I did not know that this tag was ambiguous. (I did not know about beta distribution.)

i've heard of it, but i'm not a stats guy. (evidently the definition of the pdf involves the beta function, but it's not identical)

namely the beta function serving as a normalization constant

back.

@DanielFischer Yes :(

Some time back, I used to be like that too.

"Banach or bust" sounds like a functional analyst's version of "(Aut) Caesar aut nihil". — anonymous Apr 23 at 20:14

4:09 PM

@rehband Space filling curves works.

@DanielFischer LOL

Hi guys, quick question. Is there a symbol for arbitrary number (excl. letters)? In some maths I had earlier, we were working out the values of the discriminant of ax^2+bx+c, to see if there were 2 real roots, 1 repeated or no real roots.

@BalarkaSen I said the action on RP^1, not H or C

I had to work out: 0-4(12). However, the value doesn't matter, as I know it's negative, and that's all I care about. What I'm asking is, what would the appropriate replacement of "a" be in this equation? 0 - a < 0

@DanielFischer I think this also works : $S_r(x)$ be any open sphere around $x$. As $x$ is a lim point of $A$, we have nonempty $B = A \cap S_r{x}$. But then if this is finite, there is $S_{r'}(x)$ such that $B \cap S_{r'}(x)$ is nonempty. If $F - B$ is nonnull, repeat this with $A$ and $S_{r'}(x)$.

This way you'll cancel out a class of disjoint subsets $\{A_i\}$ in $A$ with which some open sphere around $x$ doesn't intersect with and construct some other set $A_{n}$ disjoint from $A_i$s with which an open ball around $x$ doesn't intersect with. This must imply that there is an open sphere around $x$ which doesn't overlap $A$ at all, coming to a contradiction, no?

@anon CP^1 would have been more natural but similar happens. GL(2) acts on RP^1.

mmhmm, that's what Ted said right after my question

4:24 PM

I noted, just wanted to be explicit about how it relates.

@anon I am studying some topology.

4:40 PM

hmm, perfect timing ... @anon

Hello @TedShifrin

hello

@MikeMiller!

Did you unboycott me?

:P

perhaps

heya @Balarka @Mike ... @DanielF, surely you're not shocked that there are so many answers ...

Well, @Balarka, in five years he can mancott you. :)

4:45 PM

sure, @TedShifrin. he already oldcotted you.

He was not the first.

@Balarka: Don't you ever go to school?

LOL @Ted

It's 10:16 PM, @Ted. I'd be pretty surprised to think of a school which opens at the dead of a night.

Ello invites! Ello invites all around!

just seems like you're always here until 6 AM ...

4:48 PM

he goes to school, @Ted - that's just where he sleeps :)

@TedShifrin yes.

schools starts at 10:00 AM ends at 4:00 PM

wow, school starts before 8 AM in this country ... sometimes 7:30.

@BalarkaSen You've got different school timings.

@TedShifrin that's because i am not in that country.

@Ted but ends at 3

4:49 PM

cool

The schools affiliated to the central board here do start at 8 AM.

@ParthKohli i am not in those BS central boards.

@BalarkaSen WBBSE?

School here started for me at 8 and ended at 5

WBBSE

4:50 PM

@Studentmath :)

@Studentmath :O

Plus drive, it was like from 7 till 6..

Yeah

@BalarkaSen Yeah, I mean WBBSE is still not far from CBSE.

Hey, @Studentmath, you have any good questions for me to put on my exam next week?

@ParthKohli you're woefully wrong.

4:51 PM

I have a mean question to show you, but none in mind for them - what's the subject of the exam?

@BalarkaSen Why would you say that?

Um, binomial and Poisson, continuous random variables, normal random variable, Markov and Chebychev inequalities

I really like Brian's take on this afterwards. math.stackexchange.com/questions/982134/couple-probability/…

Hmm let me think

@TedShifrin A student asked for a one-on-one meeting so that they would pass their midterm... I told them when my and the prof's office hours were, they said they'd come. Of course, here I am, 50 minutes later.

@Studentmath: Of course, the $1/e$ for no letter in the right envelope has to show up. That's always what you get when you count the permutations with no fixed point (for large $n$).

4:53 PM

I have started think about $\overline{S_r(x)} = S_r[x]$ a little bit. Let $y$ be a point in $X$ which is a limit point of $S_r(x)$, then for every $r'$, $S_{r'}(y) \cap S_r(x)$ is nonnull. Let $z$ be inside this intersection, then $d(x, y) \leq d(x, z) + d(z, y) \leq r + r'$.

Not sure where this leads to though

The Peter Principle, @Mike.

@MikeMiller One-on-one meeting? My thoughts are running wild, lol.

@JasperLoy don't be crass.

Of course, @Mike, you should be in your office hours regardless. I just gave an exam today and 20 of 'em will be in my office hours 2-4:30, so you can come help me.

@Ted yeah, that's true..

4:54 PM

@Ted I'm still in my office, waiting, for the next five minutes.

@Mike: For the first time in my teaching career (about 40 years, all told), I had a student show up for Exam #2 (I've already withdrawn him failing, but the Registrar hasn't entered it, so he doesn't know). He's done 0 homework, come to 3 classes (one class and two exams) and failed Exam #1. WTF?

oh we want every $z$ in $S_{r}(x) \cap S_{r'}(y)$ such that $d(x, y) \leq r$. Say, this requires $d(z, y) = 0$. bah.

@Ted, "Anything that works will be used in progressively more challenging applications until it fails'/

will think about this later

@BalarkaSen Quick, give an example where they're not equal.

4:56 PM

OK, I have to grade exams. Need them done tonight, and I have a concert to go to.

@Ted this one I really like, it's really small and simple:

ok

@TedShifrin that sounds fun.

first midterm for my class is next week

@TedShifrin that's the first time someone has shown up for your exam 2 in 40 years?

Let $X$ be the standard normal RV, compute $E[e^{-2X}]$

4:57 PM

@anon hahahaah

smacks @anon

I did take MC on some exam days when I did not feel too well.

@Parth

@Studentmath, I can't put that on an exam, as they haven't had to do similar problems. But I can put it on next homework :P

I just had to redo the course another semester, that is all.

4:58 PM

@Sawarnik Hi.

@Ted yeah, it's nice

Only if you know how to complete the square, @Studentmath :P

@ParthKohli Geometry?

well, an hour has passed... at least I had a student here for the first half that got something out of it.

Old John has not been coming to this chat.

4:59 PM

@TedShifrin Do you know this one ? math.stackexchange.com/questions/984473/…

What's the best mathematical pickup line? Anyone?

@TheArtist Best not to use pickup lines. They are all stupid and ineffective.

@JasperLoy Says the expert.

@Hippa: I vote no.

@JasperLoy pickup lines work . (proof : youtube vedios)

5:01 PM

Also, for $n$ independent geometric RVs with parameter p, using chebyshev show that $P(\bar X_n \le 2/p)\ge 1-\frac{1-p}{n}$

@TedShifrin :c

That's all I have in mind

@TheGame You're looking for an easy example or a proof of existence, so you won't accept a proof of nonexistence?

@TedShifrin Then it's probably even harder to prove xD

@MikeMiller I edited it

@ParthKohli Comments the expert.

5:02 PM

Thanks, @Studentmath :) I'll steal some.

@The Game what do you mean ? :D

@TheArtist Wrong ping :)

@Hippa: Can you give me such a basis for $n=2$? I believe I can prove you cannot.

@Ted sure :) if you find anythnig interesting to throw at me, will be glad

@The Game Im a tubelight :p can you explain it better?

5:04 PM

hi

@TedShifrin can the n-sphere carry nontrivial bundles of rank less than n?

Anyone ? Best math pickup line? :)

@nick Answer's length does not matter.

@TheArtist I wanted to write @TedShifrin and with auto completion I sent @TheArtist

@thegame Are you going to change your username?

5:06 PM

how to prove$\int_0^1 t^t (1+\ln t)\ dt = 0$

@JasperLoy Can't yet

@TheGame OK. What are you going to change it to? Let us know in advance.

I found a nice pickup line. Who wants to hear it?

none of us, clearly

@JasperLoy tomorrow

5:08 PM

@MikeMiller The proof is not clear.

@Nick

Im going to send it anyways. Here it is : I wish I was your derivative so I could lie tangent to your curves.

@Mike: Depends on $\pi_{n-1}(GL^+(k)) = \pi_{n-1}(SO(k))$ (since every bundle will be orientable for $n>1$). But we should know this from the exact homotopy sequence.

@TheArtist It turns me off.

^^ Would this work? Imagine if your the girl? Will you be impressed?

5:10 PM

@TheArtist @Khallil said something like that once.

@Hippa: You haven't answered me.

@JasperLoy :D

That is a line that's more than 60 years old.

@JasperLoy :/ oh ok :/

@TheArtist To impress a girl, be original, and be yourself.

5:12 PM

@Sawarnik who's that ?

@JasperLoy :D

Hi, I have to prove that any group or order $2$ is isomorphic to $(Z/2Z)^3$, I think we can look at vector space but my professor said that I have to check all the axioms wich is a bit long. Any ideas please?

@Sawarnik I sound like a real expert, LOL.

@Jasper Loy , you mean quiet? You should start a convo right? That's the pickup

@JasperLoy: To impress a girl, call in Hugh Jackman as your wingman.

5:13 PM

@JasperLoy :D

@TheArtist I did not tell you to keep quiet, lol.

@MarcGato edit your comment until your question is fixed

@Jasper Loy , you said to be original and yourself

I am reminded of how I was so not myself in high school. It is so silly.

@TheArtist I did not say be quiet, lol.

@anon edited, thanks.

5:14 PM

@Sawarnik How are you?

@MarcGato still wrong. do you mean exponent 2 and isomorphic to (Z/2Z)^n ?

Impressing a girl involves two steps:
Step 1: Be attractive.
Step 2: DON'T be unattractive.

Anastasiya-Romanova

I'm so disappointed with most of users here... I wanna cry (╥︣﹏᷅╥᷅)

@Anastasiya-Romanova Why?

@anon No, is $(Z/2Z)^3$ (it was in exercise of abelian group of order 8 with no element of order $4$, we said that $G$ has an element of order 2)

5:18 PM

@Anastasiya-Romanova No, it's just Jasper.

@MarcGato see, that's the kind of thing you should say. because "any group of order 2 is isomorphic to (Z/2Z)^3" is clearly, trivially, obviously wrong.

@ParthKohli Why?

Oh a girl :D now we can discuss this deeply

What do you think @Anastasiya-Romanova

Oh @anon right, sorry. A group of order $8$

Anastasiya-Romanova

Answering difficult OPs yet I got less upvotes （つ︣﹏╰）

@TheArtist Think of what?

5:20 PM

with $G$ has no element of order $4$.

@Anastasiya-Romanova what is the best mathematical pickup line that a boy can use ?

@MarcGato show the group is abelian (by showing all elements have order|2), pick a & b nontrivial in it, then pick c in G\<a,b>, then show G=<a,b,c> is isomorphic to (Z/2Z)^3

@Jasper Loy is good. He's giving me valuable advice. Don't tell anything against him over here

@anon Thanks! I will try right now :)

@TheArtist Are you some kind of artist? What art form do you like?

5:24 PM

@ParthKohli Just fine :(

@TheArtist lol

@Jasper Loy , Painter , tattoo artist, and a photographer

@Sawarnik what?

@TheArtist I see. Sometimes I see myself as an amateur mathematician, musician and mystic.

@JasperLoy Perhaps its the movie.

Anastasiya-Romanova

@TheArtist I've never picked up line to a boy

@Anastasiya-Romanova Gimme the link.

Anastasiya-Romanova

5:26 PM

@TheArtist Maybe: "There are infinite numbers between 0 and 1. There’s .1 and .12 and .112 and an infinite collection of others. Of course, there is a bigger infinite set of numbers between 0 and 2, or between 0 and a million. Some infinities are bigger than other infinities.… I cannot tell you how grateful I am for our little infinity. You gave me forever within the numbered days, and I’m grateful."

It always happens :(

How about this : My love for you is like a fractal - it goes on forever.

@robjohn maybe it's worth saying you actually use the mean value theorem for integrals in $(4)$ since some people will be in trouble with that (although it's completelty elementary)

7

A: Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$

$$ \begin{align} \lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{c\,t^{b-1}}{1-t^c}\right]\mathrm{d}t &=\lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{t^{b/c-1}}{1-t}\right]\mathrm{d}t\tag{1}\\ &-\lim_{d\to1^-}\int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\tag{2}\\ &=\lim_{d\to1^-}\int...

No, this one: Ey, is your dad in jail? 'Cuz if I were your dad, I'd be in jail.

I see many students that have no idea about MVT for integrals that is pretty sad.

5:29 PM

This chat has become a toilet bowl.

@JasperLoy lol true

@TheArtist this guy, chat.stackexchange.com/transcript/36?m=14976657#14976657 you can get many more from his chat history

@Chris'ssis I just bound it above and below by multiples of $\int_d^{d^c}\frac{\mathrm{d}t}{1-t}$. Multiples that tend to $1$. I don't think I am using MVT.

@Anastasiya-Romanova This is really nice. Thank you a lot. Only if I could remember this in my head and make it to the last word (coz it's too long)

@ParthKohli Silly.

@Sawarnik hahaha

5:30 PM

unstar it please.

@TheArtist See what @Sawarnik said above

@robjohn OK, but for me it was natural to think of MVT. Anyway.

Anastasiya-Romanova

@TheArtist It doesn't work for me. It seems understandable. This one is good: "Hi, I hear you're good at algebra.....Will you replace my eX without asking Y?"

@Sawarnik Ahan I see

@Chris'ssis I am sure that it can be shown using the MVT, but that was not my thinking at the time.

5:31 PM

@TheGame Can you untar it?

@Sawarnik Why ?

@TheGame It seems odd.

Nah it's ok

@robjohn MVT is simply straightforward, no need for inequalities, but it's OK.

ok :D

5:32 PM

@Sawarnik aha chat.stackexchange.com/transcript/36?m=17917221#17917221

@TheArtist Or it's more twisted than anyone can imagine...

Anastasiya-Romanova

@Chris'ssis It's not completely elementary to me. Mr. @robjohn looks like Cauchy PV in residue method

Mean value theorem

In mathematics, the mean value theorem states, roughly: that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval. More precisely, if a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that A special case of this theorem was first described...

@robjon what do you mean?

@TheArtist Girl: But it may have a finite area :P

5:33 PM

OK, let me find the problem I received some months ago from a friend that is matehmatician ... (to emphasize the power of MVT for integration)

@TheArtist Can I be Jasper? You misspelled robjohn.

@JasperLoy Why not a trash can?

@Sawarnik Same thing.

Has anyone else here ever listened to Explosions in the Sky?

@ParthKohli Actual explosions?

@Sawarnik The "e" is uppercase.

5:35 PM

@Anastasiya-Romanova I don't have an eX :p

Alright :P

Anastasiya-Romanova

But I think, in my personal opinion, the boys should never pickup lines girls using mathematical ways. Try love poems, it always works

@ParthKohli When did you turn 15?

@Sawarnik In July.

@TheArtist Be a real mathematician, don't even care about girls :P

3

Integrals are much prettier

5:36 PM

Now, I ask you this: why I was able to compute it without pen and paper? $$\lim_{p\to\infty} \int_{p-7}^{p+9} \left(\frac{2x+1}{2x}\right)^x \ dx$$

@Parth Ok, bhaiya :P I will, on 29th this month. Balarka on January perhaps. Not sure about V-Moy.

@TheArtist fractals are infintely twisted and convoluted... just making a joke.

@Chris'ssis Cuz you've spent $n$ years if your life doing that kind of stuff ?

:D

@TheGame lol...very true

@Sawarnik Who is V-Moy?

5:37 PM

@TheGame Not true. I think Grothendieck had kids with 4 women?

@Anastasiya-Romanova I found a similar one to yours : I don’t like my current girlfriend. Mind if I do a u-substitution?

@ParthKohli @Anastasiya-Romanova

@TheGame LOLLLL :-)

@robjohn Ah it didn't come to my mind your referring to that

@Sawarnik They're 14? And they're answering integrals?

5:37 PM

@The Game lolzzzzzz

@ParthKohli Yes :/ :/

@The a Game you don't care? ;) be honest ;)

Anastasiya-Romanova

@TheArtist "Are you a $30^\circ$ angle? Because you’re acute-y. "

@TheArtist ask @Anastasiya-Romanova

@Sawarnik Great, I didn't really need self-esteem issues at this time of the night. Thanks anyway.

5:38 PM

@TheArtist Didn't you see the little grey arrow on the right?

@ParthKohli lol, why?

@TheArtist that was where my comment linked... you are aware of comment linking (using the arrows at the right of each comment) in chat? It makes chat understandable, even with multiple conversation threads going.

That would get flagged, I guess

Yes I got my latex code to work

@TheGame Mine?

@Sawarnik Not yours >:c

5:39 PM

@Sawarnik and @robjohn no there is no such thing in the right O.o

@TheArtist Have you noted the one I posted ? :P

@The Game ask what? :p

Less borderline : By looking at you I can tell you're 36-25-36, which by the way are all perfect squares.

@Anastasiya-Romanova nice one (thumbs up)

@The Game you removed it before I could see. Send again ;) and what should I ask? I asked you if u don't care :p

Well...

$$\lim_{p\to\infty} \int_{p-7}^{p+9} \left(\frac{2x+1}{2x}\right)^x \ dx=16\sqrt{e}$$

5:42 PM

@TheArtist I don't care at all :c

@TheArtist and I'm 16 :P

Ah, I got disconnected..

@TheArtist If you hover over someone else's comment, the right hand side should look like this:

user image

@Sawarnik @TheArtist There are so many borderline pickup ones xD

@TheArtist If you click on the arrow, it will link the comment you're editing to that comment.

@TheGame I'm a perfect square too.

5:45 PM

@TheArtist to see where a comment is linked, click on the arrow at the left of a comment.

And again..

Anastasiya-Romanova

@TheArtist You know what my big bro said when I asked about math pickup lines? He said, "No need math, use money instead" ROFL

@TheGame lol, I know :P

@Sawarnik Hey, head over to the Root.

@TheGame Nice though. But if it turns out to be wrong....

@ParthKohli There in 2 mins...

@TheGame Didn;t see:(

5:46 PM

I need 60 points to reach 2000, then I will retire, lol.

@Sawarnik :c

@Anastasiya-Romanova Mine slapped me and told me to concentrate on studying.

@Anastasiya-Romanova lol

You said that he would kill you or something for that? :P

@ParthKohli How?

:D

lol

Anastasiya-Romanova

@Sawarnik If I had a BF

5:49 PM

@Sawarnik Square - (n; adj) A person who is regarded as dull, rigidly conventional, and out of touch with current trends.

But then he is giving you advice? :P
And anyway, doesn't he have gfs himself? :P

@Sawarnik destructingmessage.com/msg.php?k=mpxvo024x2

@ParthKohli Oh yes, you are dull.

@TheGame Its self-destructed.

@Sawarnik I read it. Hahaha.

@Anastasiya-Romanova lol you actually asked your bro? Maybe hes right, But math pickup lines are fun :p

5:50 PM

@TheArtist They are, sometimes.

@robjohn No I don't see anything like that. Any pickup lines? :D

Anastasiya-Romanova

@ParthKohli Be Peter Parker, he is smart & charming, but more important be Tony Stark, he is not only smart & charming, but also rich LOL

@TheGame I have something nice to share with you ...

@The Game ahaannn I see

Anastasiya-Romanova

@TheArtist Ya, he was in my room

5:52 PM

@Anastasiya-Romanova I prefer to be Will Hunting, lol.

@Sawarnik what's the country? :p

@ParthKohli You have big bro?

@Sawarnik So you're stalking them too. Argh.

@Anastasiya-Romanova your bro has a gf? :p

@JasperLoy Good will or last will?

5:53 PM

@ParthKohli Oh, its too tough to resist the temptation :(

@ParthKohli but not it the way i did with you sorry :P

Anastasiya-Romanova

@JasperLoy I prefer loving Bruce Wayne to loving Will Hunting

@Anastasiya-Romanova I like Matt Damon, and also Justin Bieber, lol.

@Anastasiya-Romanova because Bruce Wayne has about $7 billion dollars? :p

Anastasiya-Romanova

@TheArtist A fiancée

@TheGame $$\int_0^1 \frac{1+2t}{1+t+t^2} \sum_{k=1}^{\infty} t^{3^k} \ dt=1-\gamma$$

Anastasiya-Romanova

5:55 PM

@TheArtist Yup! He is rich

@Jasper Loy you like Justin bieber? :o I thought u were a boy :o

@Anastasiya-Romanova Yay, I m the Batman!

@TheArtist :P :P

@Anastasiys-Romanova What about Sheldon cooper? He's rich too :p

@TheArtist Yes, I am. But boys can also like Justin.

Anastasiya-Romanova

@JasperLoy I like Taylor Lautner

5:56 PM

@Chris'ssis $$\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m(2^n+2^m)}=6$$

@Anastasiya-Romanova Jesus. You remind me of this girl in my class.

@Anastasiya-Romanova Me too. I like Steven Strait. This chat is becoming a toilet bowl again, lol.

Anastasiya-Romanova

@Sawarnik Hahaha...

@ParthKohli Why I remind you to Jesus? lol

@Sawarnik in an open relationship with.....

Anastasiya-Romanova

@JasperLoy Who is that?

5:58 PM

@Anastasiya-Romanova lol

@TheArtist with... continue?

@Alizter This is a question for kids ...

@Anastasiya-Romanova Do a google image search, lol.

$$\LARGE \text{Why?}$$ ... $$\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m(2^n+2^m)}=\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{m+n+mn}{2^n(2^n+2^m)}$$

Anastasiya-Romanova

@JasperLoy Did you say this: To impress a girl, be original, and be yourself.?

@Anastasiya-Romanova Ya, he did. 46 mins ago.

5:59 PM

@Sawarnik is in an open relational with A_ _ _ _ _ _ _ R_ _ _ _ _ _ :p

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