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11:00
@JasperLoy I know exactly what you mean. They know almost nothing but they like to be seen as gods. The last time I visited one I was telling her how to prescibe me a certain medicine, gave me a wrong dosage.
@Committingtoachallenge 9 years in Medicine? That is a lot.
Yep, takes a long time
It takes a longer time to attain Nibbana, maybe a few trillion years of lives.
I reached nibbana when I was taking a nap once(jokes of course)
@r9m good. I've always thought you are a professor (assistant?)...
11:03
@r9m which inst, if you don't mind sharing?
@Chris'ssis I am an assistant janitor at MIT, LOL.
We already new that @Will xD
@JasperLoy lollllllllll :-)))))))))))))
What is the feeling I can have when someone that is bragging to me that he has a master's degree in mathematics, he was the best one in school, but he is not able to compute $$\lim_{x\to 0} \frac{1-\cos(x)}{ x^2}$$ without l'Hopital?
What is the feeling I can have?
@Chris'ssis That'd be pretty sad :P
@BalarkaSen True, too sad.
This is a real story in an interview I had. One can come to me with a diploma from Harvard, but if he/she failed such a question ...
11:12
It's somewhat similar to what students in CMI studying algebra, as I have heard.
@Chris'ssis Is there any other way to do it other than series expansion, though?
Do you enjoy integral transforms @Chris
@Committingtoachallenge These days I like some more the series like Au-Yeung's series :D
@DanielFischer
11:28
@BalarkaSen ?
Hi @IceBoy.
@r9m I'm done with this one
$${\large\int}_0^1 \frac{ \ln^2 (1-x)\,\operatorname{Li}_2\left(\frac{1+x}2\right)}xdx =\frac{81\, \zeta(5)}{32}+\frac{ 5\pi^2}{16}\zeta(3)-\frac{\zeta(3)}8\ln^22+\frac1{15}\ln^52\\-\frac{\pi^2}{ 18}\ln^32- \frac{\pi^4}{15}\ln2 +2\operatorname{Li}_5 \left(\tfrac12\right) + 2\operatorname{Li}_4\left(\tfrac12\right)\ln2$$
@Daniel $(X_i, d_i)$ be metric spaces I want to prove that the open sets of $(\prod^n X_i, d)$ and the open sets of $(\prod^n X_i, \bar{d})$ are the same (i.e., the two spaces are topologically equivalent), where the metrics are defined as $d(\mathbf{x}, \mathbf{y}) = \text{min} \, d_i(x_i, y_i)$ and $d(\mathbf{x}, \mathbf{y}) = \sum_{i=1}^n d(x_i, y_i)$. It is clear that open sets of the latter are the open sets of the former, but not sure about the other case. Walkthrough needed.
@DanielFischer LOL!
@Chris'ssis Why does Cleo always recieve $\bf\zeta$?
@Committingtoachallenge I'm not sure what you mean.
Why does $\zeta$ always show up?
Is it particularly common in these advanced integrals?
11:35
@Committingtoachallenge Yeah, in this kind of integrals you usually meet zeta all over.
@Committing Conjecture : A positive proportion of the integral with log-gamma integrand is a rational combination of zeta values.
Have fun even rigorising that before trying to prove it.
@BalarkaSen You mean $\max$, not $\min$, presumably. It isn't a metric otherwise. Show that a ball of radius $r$ in one metric contains a ball of radius $\rho$ in the other.
Does Algebra or Topology interest you at all @Chris
Seriously though, @Committingtoachallenge, that's currently a study of transcendental number theory. Google [ring of periods].
Thank you for sharing that @Bala
11:36
@DanielFischer Yes, I mean max.
OK, let's see.
Oh, and does this look good, @DanielFischer?
2 hours ago, by Balarka Sen
@GustavoMontano We have the two metric space $(X, d)$ and $(X, \bar{d} = d/(1+d))$. We want to prove that the open sets of the two are the same. Consider the open sphere $S_r(x_0)$ on $(X, d)$. $r > d(x, x_0) > d(x, x_0)/(1+d(x, x_0)) = \bar{d}(x, x_0)$ so we get an open ball $S'_r(x_0)$ in $(X, \bar{d})$. Conversely, $S_r(x_0)$ be the ball in $(X, \bar{d})$. Then $\bar{d}(x, x_0) < r$ implies $d(x, x_0) < \frac{r}{1-r}$.Pick $r_1 = r/(1-r)$ which gets you a ball $S_{r_1}(x_0)$ in $(X, \bar{d})$
I have proved that each ball in one space is a ball in the other.
But the book says that there is one exception. I can't figure that out.
@Committingtoachallenge These days I work on a book that is going to contain a collection of integrals, series and limits only. So, at the moment I'm only intersted in this area.
Looks fine. The exception is probably $r = 1$ (or $r \geqslant 1$) for the $\overline{d}$ metric, which is the entire space, and would be the ball of radius $\infty$ in $d$. If you have qualms about balls with infinite radius, you make an exception for that.
Ah. But then it is all of $X$, which is trivially open.
@Chris That sounds fun, how many pages?
More than 500
11:44
Already or is that the plan?
r9m
r9m
@Chris'ssis wat ?! I'm a student UG :P
@BalarkaSen :P LOL
@Committingtoachallenge work in progress here
@r9m The name of the inst is ":P LOL"?
r9m
r9m
@Chris'ssis Nice !! 0:-)
@BalarkaSen yes ,, that is right :) we all laugh out loudly .. LOL
@r9m I'm afraid of the day when I'll solve all these questions too easily ... no more pleasure ... :-(
11:49
@BalarkaSen. I like it!
Works great on the definitions.
Great work :).
Thanks.
@DanielFischer Let $d(\mathbf{x}, \mathbf{y}) < r$. Then $\bar{d}(\mathbf{x}, \mathbf{y}) < n \cdot d(\mathbf{x}, \mathbf{y}) < n \cdot r$. So pick up a ball of radius $rn$ on $(\prod X_i, \bar{d})$?
That does the work, right?
@BalarkaSen What are you trying to do? You should start with a ball $B^{d}_r(x)$, say, and then try to find a $\rho > 0$ such that $B^{\overline{d}}_\rho(x) \subset B^{d}_r(x)$. And with the roles of $d$ and $\overline{d}$ interchanged. If you start with a radius of $r$, you will not get a radius of $n\cdot r$.
@DanielFischer Well here is your ball $\{x : d(\mathbf{x}, \mathbf{y}) < r\}$ of radius $r$ on $(\prod X_i, d)$ centered at $y$. Then you have a ball $\{x : \bar{d}(\mathbf{x}, \mathbf{y}) < rn\}$ in $(\prod X_i, \bar{d})$ of radius $rn$ centered at $y$ inside which the previous ball is sitting, no?
@BalarkaSen You're doing it backwards. Given a ball in one metric, you need not find a ball in the other metric containing it, you need a ball that is contained in the ball you start with.
@DanielFischer. What is your favourite book on Real Analysis?
11:59
I want to show that open sets in $(X, d)$ are the same as the open sets in $(X, \bar{d})$. @Daniel. So why isn't that going to work?
@r9m due to my research I finished this one in less than 2 min $$\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x} \ dx$$
@GustavoMontano I guess Rudin (RCA).
@GustavoMontano Folland's Real Analysis is also a good book.
Great, I shall check it out.
12:01
Did I ever ask you if you liked Fundamental analysis by rudin @Daniel
I feel the need to collect books of Mathematics.
I need to buy this book. Pdf doesn't do justice.
Better study than collect @Gustavo
And the need for other books would naturally come
Yes, I have already finished Real Analysis, did it a while ago.
@BalarkaSen You want to show that every $d$-neighbourhood is also a $\overline{d}$-neighbourhood and vice versa. You need to place a $\overline{d}$-ball inside any given $d$-ball (and vice versa).
I would like to read some more.
Refresh myself.
12:02
@Committingtoachallenge You mean Functional Analysis, I suppose. I do. Though Meise/Vogt I like more.
@DanielFischer The vice-verse parts are trivial. Do you agree? As $d(\mathbf{x}, \mathbf{y}) < \bar{d}(\mathbf{x}, \mathbf{y})$.
Is there one you like more than baby rudin?
@Committingtoachallenge You keep misquoting the title.
I have done it more than once @Jasper?
@DanielFischer. I hope I am not asking for much, but could you find a link to where I can buy the book.
12:03
@Committingtoachallenge Yes, look at your profile.
I am seeing a lot of things and I'm not sure which is the right one.
@Jasper Wow haha
@GustavoMontano Have you checked your local bookstores? That's where one usually buys books. They can order it, if it is not out of print.
@GustavoMontano Check amazon.com, abebooks.com or gen.lib.rus.ec for normal, international and electronic versions respectively.
@BalarkaSen $d \leqslant \overline{d}$ says it's trivial to find a $\overline{d}$-ball contained in any given $d$-ball: take the same radius.
@GustavoMontano I meant "Real and Complex Analysis". PMA is fine too, as far as I remember, but I've only flipped through it once a quarter century ago.
Oh, RCA!
I know what you're talking about now. Great ^_^.
@gus You should really take a look at my list of 12 holy math books.
Please share!
(Love the nickname - it is common :) )
@Daniel Is there one you prefer more than Rudins Principles of M.A?
12:10
Oct 6 at 4:00, by Jasper Loy
Marsden and Weinstein: Calculus I, Calculus II, Calculus III; Cohn: Classic Algebra, Basic Algebra, Further Algebra; Rudin: Mathematical Analysis, Real and Complex Analysis, Functional Analysis; Lee: Topological Manifolds, Smooth Manifolds, Riemannian Manifolds
2
Great. I'll note this down now.
@GustavoMontano Please star it if you like it so that more people will know about it and benefit. I thought about this for 5 years before coming up with this list.
Note that I do not earn anything from promoting these books, lol.
Do the stars join to the copied post from Oct 6, or just that one?
12:15
Just that one.
Right. Then for the nontrivial part, we want to find a $d$-ball contained in any $\bar{d}$-ball, right, @Daniel?
I have left many copies of this in this chat. Many are sick of it already.
I have also changed the list slightly from time to time, but it is final now for all eternity.
I notice you now credit Weinstein
Well, two authors wrote the book. If I used one it was for brevity.
I have also trimmed the titles. Some are not the full titles.
@r9m I finished this one less than 1 min
$$\int_0^1 \frac{\log^3(x)\log(1+x)}{1+x} \ dx$$
12:17
Me too Chris
@Committingtoachallenge to do things like that you need years of training and very hard work (even so, there is no guarantee you can do things like that).
I got $\cfrac{Li_2\zeta(3)}{\pi*e^3}$
@r9m in less than 1 min $$\int_0^1 \frac{\log^5(x)\log(1+x)}{1+x} \ dx$$ I skipped 4 power on purpose
@DanielFischer Looks like $r/n$. For any given $\bar{d}$-ball of radius $r$, a $d$-ball of radius $r/n$ is contained in it.
@BalarkaSen Yep.
12:22
YES
@DanielFischer Assume $d(x, y) < k$. Then $\bar{d}(x, y) < n \cdot d(x, y) < n \cdot k$. Picking $k = r/n$ does the trick.
Thanks for being of great help @DanielFischer
Hey can you guys check this out : math.stackexchange.com/questions/976439/…
Im struggling so hard....and this person who's commenting telling weird stuff I ddint kno. Can you guys check this
@ShanikaSamarasinghe $X=x \lt t$ makes more sense does it now?
Probability UPTO that point, instead of probability greater than?
@Committing to a challenge, you mean probability less than t?
But normally it's $P(X>x)$ is the cumulative distribution function
Ok look ummmm
The pdf of a negative exponential distribution is defined as :
You mean $P(X \gt x) = 1 - P(X \leq x)$, where $F_X(x) = P(X\leq x)$
12:37
Oh, I have confused the two. Thank you so much
I really appreciate it
That is alright :)
Thanks again :) :')
Are you all going to watch the apple event Tonyt ? :)
What apple event is that?
The tech company :) They are announcing the new IPads and iMac
@Committingtoachallenge ...
12:47
What @Chris
Oh yeah I got that aswell
@Committingtoachallenge aha, in your dreams?
@Committingtoachallenge ;)
12:49
A hand wrote them out in blood, I wrote them down, some call it magic
What what what :D
Why why why :D
Oh yesterday I saw this question that said something
It is a reference to Srinivasa Ramanujan: "While asleep, I had an unusual experience. There was a red screen formed by flowing blood, as it were. I was observing it. Suddenly a hand began to write on the screen. I became all attention. That hand wrote a number of elliptic integrals. They stuck to my mind. As soon as I woke up, I committed them to writing."
It seems that there is this partial derivative or something which is unsolvable....
12:52
What is it?
Riemman hypothesis.....by looking at it, it doesn't look that hard
Oh yes, not solved yet
7
Q: Can you solve this captcha?

VividDI found the following problem in a captcha: What does it mean, and what would the solution be ?

But why ?
$\mathcal{O}(x^{\frac12-\epsilon})$ isn't nice to work with
@Chris Can probably solve it in 1 min though ;P
that's not the problematic term...
12:57
I once checked the Rieman hypothesis , and it had so many paragraphs about the problem. I didn't know even what the problem was :p it was today I saw this post, and I was like oh it's jus one line thing.
@Mile Miller , what do you mean? :)
the $\pi(x)$ term is the one we know the least about... not the others
@Mike Mertens function $M(x)=\mathcal{O}^{\frac12 - \epsilon}$ is equivalent to the R.H. for all positive $\epsilon$
@Commiting to a challenge You mentioned can Chris solve it in 1 min. Didn't get that (I assume that was a joke)
Who's Chris?
Are the following exercises correct? puu.sh/cerxI/e225170677.png
13:05
3 hours ago, by Chris's sis
@Committingtoachallenge This is one of my best proofs I've ever created. Of course I care of it. And it's even more than this, it's the best proof in the world in my opinion. How to give up such a creation?
That person ^
@robjohn you're too silent these days, I miss some proofs of yours. For some time I try to find a strategy to compute $$\sum_{n=1}^{\infty} \psi^{(2)}(n) \psi^{(3)}(n) \psi^{(4)}(n) $$
@Committing Yes, and the problem with proving that is understanding Mertens function :P
@Comming to a name , Oh Ahhh lol. So just like that, chris knows the solution to the rieman hypothesis too?
:p
@ShanikaSamarasinghe No, I don't. That comment was about the solution to the famous Au-Yeung series.
Careful @ShanikaSamarasinghe, Chris doesn't joke
13:10
Hmmmm ok. Maybe work on that ? @Chris's sis
She also doesn't talk about anything outside of Mathematics
@Committing to a challenge , oh WOW. The right person to solve it then :)
@Chris's sis Hi :)
@ShanikaSamarasinghe Hi
@Chris's sis What do you think about Riemann hypothesis. Have you ever tried to solve it ? :)
@ShanikaSamarasinghe I received some proposals from some mathematicians some time ago, but I refused to work on it.
13:16
@Chris's sis But why ? Who doesn't love that Eureka moment
Gauss refused to work on Fermat's last theorem too.
Exactly and that was solved finally :) @Ice Boy
@ShanikaSamarasinghe I think out there are many much much better than me to do that. Besides that it would take some years of learning of that stuff since I don't come in place with a background in mathematics.
@Chris's sis , don't you think everyone has that feeling?
@Committing to a challenge would you like to commit to a challenge? ;)
@Chris's sis aren't you a mathematician?
@ShanikaSamarasinghe Maybe
13:21
I am but a lowly Mathematics student, my current challenge is to complete a specific list of $9$ textbooks by 03/10/2015(DD/MM/YYYY)
@ShanikaSamarasinghe No, I don't have any background in mathematics. I'm an accountant.
@Committingtoachallenge Good! I am only a banana.
@ShanikaSamarasinghe That word is not well-defined.
@Chris's sis I asked because You said that you received proposals from mathematicians, so isn't it funny that an expert asking someone else (who doesn't have a math background) to solve a problm?
13:23
@Chris'ssis I don't have any ideas on that right off... I'd have to play around with it a bit.
@robjohn OK
@Jasper "chatting bananas" that's not well defined either :p
@ShanikaSamarasinghe Not weird at all. Nobody knows how to solve every problem.
It is very interesting that there are now two Supermans and two Mean Squares in this chat, this is a miracle.
3
@ShanikaSamarasinghe lol, maybe they only made fun of me ... I never know since I didn't accept that.
@Jasper Loy, I was jus kidding. Yes but mathematicians don't just go and propose ordinary people to solve things they can't.
@Chris's sis I guess :)
@Committing to a challenge , Chris is a nice person :)
2
I mean Chris's sis
@Chris'ssis Sorry, things offline have been taking a lot of time. I've not been able to get as many questions answered recently either.
@Jasper Loy , how dare you forget that im in the chat too :p
@ShanikaSamarasinghe Sorry, you are only a papaya.
Now I have to go to the park... bbl
8
A: Evaluating $ \sum\frac{1}{1+n^2+n^4} $

robjohnFirst, we have $$ \begin{align} \frac1{z^4+z^2+1} &=\frac1{12}\left( \frac{-3-i\sqrt3}{z-e^{\pi i/3}} +\frac{3+i\sqrt3}{z-e^{4\pi i/3}} +\frac{3-i\sqrt3}{z-e^{2\pi i/3}} +\frac{-3+i\sqrt3}{z-e^{5\pi i/3}} \right)\tag{1} \end{align} $$ Let $\gamma$ be the rectangle $$ [-1-i,1-i]\cup[1-i,1+i]\cup[1...

13:30
Later
This one reminds me of an old generalization of mine you answered. I need to find it.
@Jasper Loy shush banana
@robjohn OK
Ok here's a question to everyone
Favourite Mathematician of all time ? :)
Srinivasa Ramanujan
I like the story of Galois aswell
I like my own story, even though I am only a banana.
Leonhard Euler
;)
Who else is there in this chat ? searching for answers
@Chris's sis why did you remove? :)
I changed my mind.
@ShanikaSamarasinghe by mistake
13:36
Ice deletes 23.833% of his posts from here
top lel
@Chris's sis okay :) Ive heard his story a natural genius
Define "favorite"?
@Ice Boy Favourite means UMMM the work and life of that person..you get it ?
@ShanikaSamarasinghe Indeed.
13:40
@Ice Boy you know the difference between best actor and favourite actor right? im asking FAVOURITE, which means it's personal
@Chris's sis Yep
well, well, we have a new result in place
How do you call a map that is the identity but the codomain is greater than the domain ?
So $f:X->Y : x \mapsto x$ and $X\subset Y$, $X\neq Y$
@Kasper "The inclusion map of $X$ into $Y$"
oh that is the word ! thanks
I don't get worlframalpha.
$$\frac{\frac{k^4}{4n^2}}{\frac{kn}{n}}$$
Shouldn't it be $k^4/4kn^2$?
Am I that stupid that I forgot all math?
13:50
Allow me to promote my latest OP to you in case there's someone here can answer it. Thanks. math.stackexchange.com/q/976608/133248
$\frac{\frac{k^4}{4n^2}}{\frac{kn}{n}} = \frac{k^3 }{4n^2 }$
Yeah, I know.
Wolfram claims otherwise.
I swear.
No?

http://www.wolframalpha.com/input/?i=does+%5Cfrac%7B%5Cfrac%7Bk%5E4%7D%7B4n%5E2%7D%7D%7B%5Cfrac%7Bkn%7D%7Bn%7D%7D+%3D+%5Cfrac%7Bk%5E3+%7D%7B4n%5E2+%7D&dataset=
It says true @Studentmath
It works there aswell?
It says $k^3/4n^4$.
Oh I see sorry, yeah it did a/b/c/d in that order
Yeah
Holy shizzle I think I just managed to prove what I've been trying for a week
Really? That is great to hear!
13:57
I hope I didn't make a mistake along the way
But this is great
@Anastasiya-Romanova does it work to let $x\mapsto \pi/2-x$ and then change the variable and map all onto $[0,1]$? Or you can do it straightforward.
@Anastasiya-Romanova wait, there is a problem with the integral you got ...
@Chris'ssis Let me try it. It doesn't cross to mind
31
Q: A problem I can't solve from my childhood to now.

Tharindu When I was a child I was given this problem to send a wire from electricity, water, and internet to each of the houses, all three houses must have all three wires connected without being crossed over each other (wires can't meet and there is no such thing as a wire going on top of another wire)...

What do you guys think about my solution ? :)
Scroll down to see my solution, if you like it. Upvote :)
@Anastasiya-Romanova I'm afraid you wrongly applied teh differentiation.
@ShanikaSamarasinghe How can I post the chat like that?
14:10
@Anastasiya-Romanova sorry didn't understand :)
What do you mean :)
@Anastasiya-Romanova do you read me? You can do that on your own, you need no help. Just correct your work.
@Chris'ssis Yes. I get it where is wrong with the differentiation
@ShanikaSamarasinghe I mean how to post link problem like that in the chatroom?
@Anastasiya-Romanova , copy link---> paste--->send :)
I don't know darling, I didn't do anything magical . Just copied and pasted
2
Q: Prove $\int_{0}^{\pi/2} x\csc^2(x)\arctan \left(\alpha \tan x\right)\, dx = \frac{\pi}{2}\left[\ln\frac{(1+\alpha)^{1+\alpha}}{\alpha^\alpha}\right]$

Anastasiya-RomanovaWhen I showed to my brother how I proved \begin{equation} \int_{0}^{\!\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, \mathrm{d}x=\pi\ln\ln2 \end{equation} using the following theorem by Mr. Olivier Oloa \begin{equation}{\large\int_{0}^{\!\Large \frac{\pi}{2}}} \frac{\cos \left(\! s...

@ShanikaSamarasinghe Ahh, I get it. Thanks
@Anastasiya-Romanova :)
14:29
Hellu :D
@Committingtoachallenge Fake -__-
Picture
14:32
Da mean evil square
Hi all
@Chris'ssis How was it generalized?
Does anyone know the probability that I will be alive in 20 years time?
@ShanikaSamarasinghe 0.7
$0$ @ShanikaSamarasinghe
14:35
Where are you from? @ShanikaSamarasinghe
@The Game Thank you. How did you know ? :)
@ShanikaSamarasinghe uh ? that was a joke
It's a known problem
'probability that tom is alive in 20 years is 0.7'
Does anyone know if the symbol [ ] is used to denote the scalar product in any book?
@Alex Where do you think Im from?
The book will give notation somewhere surely @Alex
14:37
@ShanikaSamarasinghe The internet
:P
@Committing to a challenge . Why u so mean? :'(
When was I mean @Shani?
@Shanika India or Durban, South Africa.
@The Game haha nice one :)
@Alex Yeah im a Ebola patient
@Alex where are you from?
@Committingtoachallenge I'm asking if anyone has encountered this notational use.
14:39
@Alex I have never seen it in a book, but I have seen it use on stack exchange before
@Committingtoachallenge Damn that doesn't cut it, I don't feel like changing notation in thesis.
@Committing to a challenge mean for using probability to tell I will die
@ShanikaSamarasinghe Sorry, but I did the calculations
@Shanika Where do you want me to be from? The place from whence you think Ebola inhabits.
@Committing to a challenge okay -.-
14:42
@robjohn Oh , it was
@Alex why did you ask where I am from in the first place? :)
@robjohn I think you can also use a similar strategy to your question.
I am sorry @Shan Can you forgive me?
You will live with proability $1.01$
@ShanikaSamarasinghe I thought you were from India and if you were then I would have had a follow up question.
@ShanikaSamarasinghe you live with a probability -1. Get out, ghost :P
14:44
@ShanikaSamarasinghe Where are you from?
@Alex Im Indian living in Western Australia :) dont tell i will bit by snakes
@Commting to a challenge :) Thank you :p
@The Game oh gosh....probability is defined from 0 to 1.
@ShanikaSamarasinghe That's nice, are you going to watch the rugby this weekend?
Chance to reproduce ;)
@ShanikaSamarasinghe Quantum quantum :D
(Asexually of course)
(referring to my $1.01$)
14:48
@Committing to a challenge what do u mean? :p
@Alex, what was the follow up question. And no Im not a fan of rugby, plus my exams are near
@Alex I don't even know which rugby match there is :) or if there is any
@Alex tell me the probability that you had in mind for me :)
@ShanikaSamarasinghe It's New Zealand vs Australia :)
sheep VS kangaroos
@The Game sheep vs kangaroos? .....quantum? im not a particle
@The Game , sheep?
@ShanikaSamarasinghe I have an indian girlfriend by the way. It's quite a good product...indian girls.
They call New Zealand as sheep? O.o
14:51
New Zealand <-> Sheep
@ShanikaSamarasinghe Did you watch the movie Wolfcreek?
They WorkSheep Sheeps
:P
@Alex, really ? :) what do you mean by product? Where are you from? Btw when are you going to answer my question
@ShanikaSamarasinghe Yes really. Product probably not a good choice of word. Rather they are nice things...indian girls. Does your mother make good rotti?
@Alex Nope I haven't :) I watched ninja turtles last week
14:54
Why are you talking about India ?
@ShanikaSamarasinghe It's about murders in the Australian outback. Loosely based on actual events of Ivan Milat, serial killer.
@Alex rotti is mostly pakistani ;) my mom knows to make the best pani puri...have u heard of that food?
@The Game , they are asking questions and I am answering :)
9 mins ago, by Shanika Samarasinghe
@Alex Im Indian living in Western Australia :) dont tell i will bit by snakes
Ooooh I hadn't seen this one
Ok
@Alex you call your gf a product? Show this to her, if she's truly indian, lol trust me your going to have a bad night
There's no Ebola in India though
Only Environment Pollution in the State of Andhra Pradesh
14:56
@The Game I know obv....
@ShanikaSamarasinghe :) Was a bad word choice. But indian girls are quite nice things.
@Chris'ssis :O :O :O
@The Game and im not living in India.....so when I said I have Ebola (even for joke) didn't mean it's there in India
Those coefficients
@TheGame yeah ...
14:57
@Chris'ssis Looks like chemical shrodinger equations :P
@TheGame :D
@Alex things?
$\pi\dfrac{17\sqrt{24}}{2e}$
Gimme 10 mins guys
That's $10

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