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Peter Sheldrick
Peter Sheldrick
5:00 PM
@Huy, the flaw, and the appeal probably, of that video is that it's very high-school-math-ish
high-school algebra and such
 
Huy
Huy
It saddened me that my friend who has a MSc in EE didn't notice the mistake ._.
 
skullpatrol
skullpatrol
high schools do not teach students that integers add up to fractions
 
Peter Sheldrick
Peter Sheldrick
my point was they learn the most basic rules of algebra
 
skullpatrol
skullpatrol
integer + integer = integer
is that basic?
 
Peter Sheldrick
Peter Sheldrick
um infinite sums?
 
skullpatrol
skullpatrol
5:06 PM
infinity is not an integer
 
Balarka Sen
Balarka Sen
@skull : As I have said numerously many times to numerous people, it depends on the sense you are talking about. If cauchy, then everyone has viral infection. If abel, ramanujan got serious mental disbalance problems, etc.
@skull Correct.
 
Peter Sheldrick
Peter Sheldrick
@skullpatrol, if you are curious there are a lot of fun threads to read about this
 
skullpatrol
skullpatrol
yes, they are fun :-)
 
Balarka Sen
Balarka Sen
@Peter Just to point out, in the Cauchy sense, see Riemann rearrangement theorem. A generalization to in turn divergent series are possible, though.
 
Peter Sheldrick
Peter Sheldrick
the point were it made click for me was this blog post terrytao.wordpress.com/2010/04/10/… but that is also linked in 9/10 of those threads
that was a few months back
to just refer to 'analytic continuation' at that point is a bit heavy handed - i like that he presents a real analysis alternative (that part clicked with me)
 
Balarka Sen
Balarka Sen
5:21 PM
1 + 2 + 3 + 4 + ... = -1/12 often crops up in Casimir effect... Just sayin'.
 
skullpatrol
skullpatrol
context, context, context please
Just sayin'
 
Peter Sheldrick
Peter Sheldrick
@BalarkaSen, yeah that's what Mr. Motl opens up with in his answer to the main math.se thread on the topic math.stackexchange.com/questions/39802/…
which is also liberally linked on the social media
@skullpatrol, you started with the non-sequiturs
 
Balarka Sen
Balarka Sen
@skullpatrol I am definitely not physicists, so I cannot provide further context, admittedly. I saw this one explained in Marty Green's blogpost once.
:P
 
Jayesh Badwaik
Jayesh Badwaik
@skullpatrol Hello skull, I'm fine. I see that you have come back to your original naming scheme?
 
skullpatrol
skullpatrol
@PeterSheldrick I know, I was being a numb skull
@JayeshBadwaik yep
@BalarkaSen we definitely need the Physics here :-)
 
Balarka Sen
Balarka Sen
5:27 PM
Right.
 
Jayesh Badwaik
Jayesh Badwaik
@skullpatrol How are you doing nowadays?
 
Amr
Amr
Hello Everyone
Does anybody know what the prerequisites for studying this book : amazon.com/Riemannian-Geometry-Geometric-Analysis-Universitext/…
 
skullpatrol
skullpatrol
@JayeshBadwaik Fine thanks, how are your studies going?
 
Balarka Sen
Balarka Sen
Okay, here's where I have read it : marty-green.blogspot.in/2010/03/…
 
Jayesh Badwaik
Jayesh Badwaik
@skullpatrol My semester has just began, and I'm slacking off now, have to gear up soon though.
 
skullpatrol
skullpatrol
5:29 PM
ic ic
 
Peter Sheldrick
Peter Sheldrick
@Amr, hmm pretty advanced... complex analysis? (not all of it, but an appreciation at least)
 
Amr
Amr
@PeterSheldrick Rudin's real complex analysis suffices ?
 
Peter Sheldrick
Peter Sheldrick
sure
'the hottest topics of contemporary mathematical research'
is it really that hot?
it's been around...
 
Amr
Amr
@PeterSheldrick No idea, but I am intrested to learn geometric analysis
 
Peter Sheldrick
Peter Sheldrick
that part is definitely hotter than 'Riemannian Geometry'
and to be honest i saw some 'geometric algebra' but 'geometric analysis' is actually new to me
 
Amr
Amr
5:34 PM
@PeterSheldrick What other prerequisites do you think I might need ? I suppose differential geometry
 
Peter Sheldrick
Peter Sheldrick
yup
 
Amr
Amr
Are functional analysis or PDE s needed
 
Peter Sheldrick
Peter Sheldrick
just reading the description... makes it sound pretty hot
 
Amr
Amr
:)
 
Peter Sheldrick
Peter Sheldrick
not sure about functional analysis
 
Amr
Amr
5:36 PM
hmmm ok
 
Peter Sheldrick
Peter Sheldrick
can't hurt...
 
Amr
Amr
Sure
but I have lots to self-study
 
Peter Sheldrick
Peter Sheldrick
physics from the sound of it
 
Amr
Amr
What do you mean ?
 
Peter Sheldrick
Peter Sheldrick
like, physics as a prerequisite
functional analysis is a bit technical and is used for example in numerics
but this sounds more like a theory extravaganza - not sure if they care much about actually solving stuff numerically
but i'm just guessing from the description
 
Amr
Amr
5:38 PM
@PeterSheldrick I don't understand what you meant by : "physics from the sound of it"
 
Peter Sheldrick
Peter Sheldrick
proper theoretical physics is probably a prerequisite from reading the description (from the sound of the description...)
 
Amr
Amr
ahh oh no
i thought this was a pure math book
 
Peter Sheldrick
Peter Sheldrick
so you have 'quantum field theory' in the description
that's not just quantum theory but also field theory (for example in electromechanics...)
 
Amr
Amr
i thought they would solve mathematically intresting problems that are also occur in QFT, but QFT wouldn't be needed as a prerequisite
this is just a guess
 
Peter Sheldrick
Peter Sheldrick
okay, find the book in a library and base any purchasing decision on that first
 
Amr
Amr
5:46 PM
@PeterSheldrick OK. Thanks a lot for your help. I will need to leve now to continue studying Rudin's book Bye
 
Peter Sheldrick
Peter Sheldrick
bye
 
skullpatrol
skullpatrol
later
 
Mike
Mike
@Huy No - what if you pick $A$ to be the complement of $D$?
 
Huy
Huy
@Mike: Right.
@Mike: Could you by any chance help me here as well? chat.stackexchange.com/transcript/message/13276321#13276321
 
Mike
Mike
The actual method to proceed isn't immediately obvious to me. Did you solve that one yet?
 
Huy
Huy
5:58 PM
@Mike: It is a theorem in our lecture notes including a proof, if you want I can copy it, it's not too long.
 
Mike
Mike
I'd like that. What do you mean by $\text{Gl}(X)$ in the link you just sent?
 
Huy
Huy
Linear, continuous and invertible, I think.
(with linear, continuous inverse)
 
Mike
Mike
And $\sigma(A)$ is the spectrum of the operator?
 
Huy
Huy
Exactly.
 
Mike
Mike
Ah, well by definition of the spectrum, $(A-\lambda)$ fails to be invertible.
Hmm... that's not true
Give me a second
 
anorton
anorton
6:03 PM
Something to downvote: http://math.stackexchange.com/a/643970/23353
Copy/pasting answers isn't good. I almost wonder if this is a spambot that is trying to build credibility. (I already flagged for moderator attention.)

I only mention it here because, if it is a spambot, we could see more behavior like this.
 
Mike
Mike
@Huy Do you have the bounded inverse theorem? I.e. that a bijective bounded linear operator has a bounded inverse?
 
Huy
Huy
It doesn't sound familiar.
 
Mike
Mike
Hm.
Well, if you had that, it's clear why your function must be either not injective or not surjective: because if it were bijective, it would have a bounded inverse.
 
Huy
Huy
Is the proof to that theorem simple?
I know that if it's bounded it will be continuous.
Oh.
I had the theorem of open maps.
As it is linear and continuous and bijective, its inverse is linear and continuous as well.
Hence bounded.
 
Mike
Mike
No, the proof I know uses open mapping theorem, which requires Baire Category theorem
 
Huy
Huy
6:07 PM
Or am I making a mistake?
 
Mike
Mike
Oh, if you have that theorem, you're golden.
 
Huy
Huy
Exactly.
I have the open mapping theorem and Baire Category theorem as well.
I guess maybe somewhere hidden in the lecture notes that "theorem" was mentioned but I overlooked it. But it's clear now.
 
Mike
Mike
Yeah, it's an easy consequence of open mapping. But open mapping is a lot to assume.
 
Huy
Huy
I don't know about that. But now I see why it is in the chapter of "principles of functional analysis". :P
 
Mike
Mike
Well, I just mean that Baire is a hard theorem, so that I wouldn't assume every course covers it.
 
Huy
Huy
6:10 PM
Oh, okay. I don't know, I actually liked the Baire part more than the part about reflexive and seperable spaces.
 
Mike
Mike
I agree.
 
Huy
Huy
The proofs were much less abstract, at least to me.
 
Mike
Mike
It's an awesome theorem, I think.
 
Huy
Huy
I thought Riesz' representation theorem was mind-blowing :o
 
Mike
Mike
Did you do the Riesz representation theorem for measures, too?
I think I like that one even more.
 
Huy
Huy
6:13 PM
I don't know about that one. We just did it for Hilbert spaces.
To be honest, I kind of lack knowledge about measure theory because I didn't enjoy it very much.
 
Peter Sheldrick
Peter Sheldrick
measure theory is the pits ._.
 
Mike
Mike
I mostly agree with you, I found a lot of measure theory very tedious and with no really exciting results, but I think you'll like this one
Let's start with a locally compact hilbert space X, and consider the continuous, compactly supported, real-valued functions $C_c(X)$
 
Kevin Driscoll
Kevin Driscoll
I don't understand these things yet, but I am starting a Coursera course on Functional Analysis in a week. Maybe in a few months I'll have it down.
 
Mike
Mike
Now, given a functional $\phi: C_c(X) \rightarrow \Bbb R$, with $\phi(f) \geq 0$ if $f(x) \geq 0 \forall x$, then there exists a unique regular measure $\mu$
With $$\phi(f)=\int_X f d\mu$$
 
Kevin Driscoll
Kevin Driscoll
I find the frequent use of compactly supported functions in math to be a double-edged sword. For some applications, like statistics, its totally useful. For others, like physics, it makes things a real pain.
 
user2584228
user2584228
6:18 PM
Hello there! I am studying some basic Trigonometry and I'm stuck in a problem and hope you guys can point me out how to continue. It says:

"Determine the three cube roots of (2-j)/(2+j) giving the result in modulus/argument form. Express the principal root in the form a+bj, where j^2=-1"

I first calculated (2-j)/(2+j), which gives 0.6 - 0.8j (a+bj form). Then I calculated the modulus (r^2=a^2+b^2, it gives 1 in this case). But now, how do I continue? I do not get the first part of the exercise (I know what a cube root is of course, but...).
 
Mike
Mike
In other words, any such linear functional can be represented as an integral with respect to some unique measure. I believe this theorem can be restarted with a complex codomain as well, allowing the measure to be a complex measure.
 
Balarka Sen
Balarka Sen
Hullo @Mike.
 
Mike
Mike
Hi
 
Huy
Huy
@Mike: That does look pretty powerful.
Where did you learn about it?
 
Mike
Mike
It's used in the proof of the existence of a Haar measure on a compact group. I don't know the proof for a locally compact group, but I imagine it also uses it
 
Huy
Huy
6:21 PM
I have never heard of such measures.
 
Kevin Driscoll
Kevin Driscoll
@user2584228 That sounds like a question for Main, rather than chat.
 
Mike
Mike
Oh. A Haar measure on a group is a regular measure that's left-invariant; i.e. $$\int_G f(g) d\mu = \int_G f(hg) d\mu$$
or rather that $\mu(S) = \mu(hS)$
For compact groups it's also right-invariant, but that's not necessarily true for only locally compact.
 
Huy
Huy
I see.
 
user2584228
user2584228
@KevinDriscoll I did not want to open a question for this as I only need to have quick guidance about the text. I can do it nevertheless if you think it is better.
 
Mike
Mike
It's important (to me) because having a measure (or ability to integrate) on a group is rather powerful. It lets p-adic analysis really come into its own, for instance.
 
Balarka Sen
Balarka Sen
6:25 PM
Interesting : Find two complex numbers x and y such that 1) if one of them is transcendental then the other is not 2) both have transcendental status yet undecidable.
Ah, I found one with a respectively weaker condition that (1) applies.
$\log(\pi)$ and $\log(\log(\pi))$
If the first is not transcendental then the other is not.
Ughh, that's too trivial.
 
Balarka Sen
Balarka Sen
Any ideas?
 
Huy
Huy
@Mike: Let $(x_k)_k$ be dense in $M$, $a_0 \in A \subset M$. For $k,l \in \mathbb{N}$ with $A \cap B_{1/2l}(x_k) \neq \emptyset$ choose $a_{kl} \in A \cap B_{1/2l}(x_k), a_{kl} = a_0$ otherwise. Then, $A \cap B_{1/2l}(x_k) \subset B_{1/l}(a_{kl})$ and $$A = \bigcap_{l=1}^\infty \bigcup_{k=1}^\infty (A \cap B_{1/2l}(x_k)) \subset \bigcap_{l=1}^\infty \bigcup{k=1}^\infty B_{1/l}(a_{kl})$$ hence $(a_{kl})_{k,l}$ dense.
 
Mike
Mike
Ah, that makes sense. I like that argument.
 
Huy
Huy
@Mike: How long ago did you learn functional analysis?
 
Mike
Mike
Fairly recently. I first took a class in the summer.
And then I had a (rather unsatisfactory) semester-long class that I supplemented with Rudin somewhat.
The summer one was very short and we covered things too quickly to get a real grasp on anything.
Our teacher tried to cover the entirety of functional analysis in about six lectures and then moved on to the representation theory of Lie groups...
 
Huy
Huy
6:41 PM
A very interesting approach.
 
Mike
Mike
I got nothing out of it, personally, except for the first couple weeks of functional.
 
MSEoris
MSEoris
7:07 PM
Is it true that the number of Generators of Z_a X Z_b = # of Gen of Z_a * # Gen (Z_b)?
 
Mike
Mike
are those intended to be arbitrary groups
actually even if they are, no
 
MSEoris
MSEoris
they were, but after some thought i think a,b must be rel prime
 
Mike
Mike
i'm not sure what you mean then
$\Bbb Z_a$ is always going to have one generator
 
MSEoris
MSEoris
?
Z_5 has every element cept e a generator
 
Mike
Mike
oh, I misunderstood
I thought you meant the number of elements needed to generate the group
 
MSEoris
MSEoris
7:12 PM
no the actual count of hte generators
 
Mike
Mike
well, the number of generators of $\Bbb Z_a$ is $\phi(a)$
and by the chinese remainder theorem if $\gcd(a,b)=1$, then $\Bbb Z_a \times \Bbb Z_b = \Bbb Z_{ab}$
 
MSEoris
MSEoris
ok thats basically my question/ answer
 
Mike
Mike
yeah
 
MSEoris
MSEoris
guess Ill go read the chinese remainder theorem
 
Mike
Mike
you then have what you want by multiplicativity of $\phi$
 
MSEoris
MSEoris
7:14 PM
neat. thanks for the pointer
 
Mike
Mike
no worries
and yeah it's definitely false in general, $\Bbb Z_2$ has one generator, while $\Bbb Z_2 \times \Bbb Z_2$ has no elements that generate the whole group
 
MSEoris
MSEoris
yeah that was a duh moment for me
i learned these things once upon a time and its all slowly popping back into my brain
alright, back to reading - thanks again - toodles
 
Ted Shifrin
Ted Shifrin
hi @Mike
 
Jasper Loy
Jasper Loy
Hi @ted, it's too late to say hi.
 
Ted Shifrin
Ted Shifrin
it is? @Jasper
 
Daniel Fischer
Daniel Fischer
7:25 PM
I hate it when I try to gently nudge somebody to solve the problem themselves, and then some dork jumps in and posts a complete solution.
 
Ted Shifrin
Ted Shifrin
That's what I've been bitching about for months, @Daniel.
You've noticed I've basically quit answering.
I've gotten angry a few times and had the offender retract his solution.
 
Jasper Loy
Jasper Loy
@DanielFischer That sounds like me, lol. I like to post more or less complete solutions, but those are elementary ones.
 
Ted Shifrin
Ted Shifrin
still, @Jasper, "elementary" askers need to be made to think, too, not just fed the answers
 
Daniel Fischer
Daniel Fischer
@TedShifrin There are still a couple of questions left where a complete answer is adequate. And sometimes, the nudging works, that's always a great joy.
 
Ted Shifrin
Ted Shifrin
Yes, I've actually gotten some serious thanks from people whom I nudged and worked with ...
 
Jasper Loy
Jasper Loy
7:28 PM
@DanielFischer Wow, now you sound you are really committed to this site! I don't get a great joy just from some website on the internet.
 
Ted Shifrin
Ted Shifrin
I'm gonna have to bitch out my undergrad diff geo class ... about half of them didn't leave enough time or make enough of a serious effort on homework .... and there's no way they'll pass at this rate.
 
Mike
Mike
hi @Ted
 
Daniel Fischer
Daniel Fischer
@JasperLoy It's a great feeling when teaching works.
 
Nimza
Nimza
Hi @robjohn, do you know if an estimate of the type $|\pi(z)| \leq C (1+|z|)^N \exp(B|\Im z|)$ holds for all $z \in \mathbb C$ for some positive constants $C$, $N$, $B$, where $\pi(z) = \frac{1}{\Gamma(1+z)}$?
 
Ted Shifrin
Ted Shifrin
@Mike must actually be working today ... sooo quiet ;P
 
Kevin Driscoll
Kevin Driscoll
7:31 PM
@Ted I was also having that problem last semester with some of my students. They'd do half the problem set correctly and then just not do the second half.
 
Mike
Mike
@Ted Doing homework for the finance class. I should still drop it eventually.
 
Ted Shifrin
Ted Shifrin
although you might learn something interesting, I doubt the marginal learning is high enough :P
 
Balarka Sen
Balarka Sen
Solved two old problems today.
Oh, hullo @TedShifrin
 
Mike
Mike
@Ted I intend to just sit in on the lectures, but I haven't dropped yet
I think I'll see whether I can pass the first exam without studying for even a minute before officially dropping.
 
Kevin Driscoll
Kevin Driscoll
@Mike whats the title of the course?
 
Ted Shifrin
Ted Shifrin
7:32 PM
frustrating, @Kevin. I warned them specifically the first few days ... And some of them are my advisees. Even the 8 I've taught before didn't all exactly ace the assignment. I figure that if they have to do 5 problems, only 1 of which is specified, and they get to choose the level/content of the other 4, it shouldn't be hard to get close to 100%.
 
Mike
Mike
@Kevin "Mathematical Finance"
 
Ted Shifrin
Ted Shifrin
That'll make the professor feel good, @Mike.
Our course is called Mathematics of Options Pricing.
 
Mike
Mike
@Ted Well, don't tell him then!
 
Kevin Driscoll
Kevin Driscoll
@Ted @Mike Is it all Black-Scholes and such?
 
Mike
Mike
(He knows I intend to drop it and invites me to continue sitting in)
 
Ted Shifrin
Ted Shifrin
7:33 PM
Then no problems,@Mike. Frank F. is the only person on the faculty i actually know personally.
 
Mike
Mike
@Kevin Not even that yet. We've only just defined Brownian motion.
 
Ted Shifrin
Ted Shifrin
Our course started with a "review" of probability for the 95% of the class that hasn't yet taken the probability class. It's not a prereq.
 
Kevin Driscoll
Kevin Driscoll
@Ted That does sound totally reasonable. I have to think about what a behavioral economist would say though
 
Ted Shifrin
Ted Shifrin
say about what, @Kevin?
 
Kevin Driscoll
Kevin Driscoll
@Ted About allowing your students to choose a large subset of the problems they get to complete
 
Ted Shifrin
Ted Shifrin
7:37 PM
@Kevin: I started doing this about 10 years ago. This is the unique course I teach that way. I divide the problems into three levels (flower, pinecone, and pyramid) :P Generally, the most advanced students do more of the latter two, and the C student who needs to stick more to computations does mostly flowers. The one required problem on each problem set and stuff done in class forms the basis for exams.
 
Jasper Loy
Jasper Loy
@TedShifrin Flower, pinecone and pyramid? Very exotic choice!
 
Ted Shifrin
Ted Shifrin
I thought you'd appreciate the poetry of it, @Jasper :) I have little pictures on the left :)
 
Kevin Driscoll
Kevin Driscoll
@Ted Well, that sounds like some solid results and a good way of catering to disparate students
 
Ted Shifrin
Ted Shifrin
It's not fun keeping the grading scorecard with 34 students, @Kevin. That's about double the size I usually have, but I doubt I'll have 34 at midpoint.
 
Kevin Driscoll
Kevin Driscoll
@Ted Fall of 2012 I had 41 students and it almost killed me. I think I averaged 20 hours/week grading their homeworks.
So I feel your pain
 
Ted Shifrin
Ted Shifrin
7:43 PM
Did you abolish your account and start over, @Kevin?
 
Jasper Loy
Jasper Loy
@TedShifrin Abolish is another fanciful word, I just use delete, lol.
 
Ted Shifrin
Ted Shifrin
Well, you can delete me if you prefer, @Jasper.
 
Kevin Driscoll
Kevin Driscoll
@Ted Nope. I've never had much rep/activity on the site. I usually only ask questions if I'm really stumped and I mostly don't know the answers to the questions asked.
Why do you ask?
 
Ted Shifrin
Ted Shifrin
Oh, hello @Balarka
ah, cuz you'd been awol for a while and I guess I hadn't noticed your rep before.
 
Kevin Driscoll
Kevin Driscoll
Oh yeah, I've been around 400 for ages. Just been focusing on research.
 
Alizter
Alizter
7:48 PM
What properties does a function need for $f(0)=0=F(0)$
 
Mike
Mike
err... assuming $F(x) := \int_0^x f(t)dt$, then all you need is $f(0)=0$
 
Alizter
Alizter
hmm @Mike $F$ is a primitive of $f$
 
Mike
Mike
then you need it to have $F(0) = 0$ :P
there's nothing more you can say than that
 
Alizter
Alizter
Nothing such as being strictly increasing or decreasing?
 
Mike
Mike
nope
 
Ted Shifrin
Ted Shifrin
7:51 PM
Assuming $f$ is continuous, saying $F'(0)=0$ will give you $f(0)=0$. Otherwise, I have nada.
 
Mike
Mike
of course, the primitives of $f$ are all $\int_0^x f(t)dt$ up to adding a constant everywhere
so that if $F(0) = 0$, it must necessarily be $F(x) = \int_0^x f(t)dt$
but that's the best I can give you
 
Alizter
Alizter
I am specifically looking at $\int x^{1/x} \;\mathrm dx$ it is a crazy mf
 
Mike
Mike
I have opinions on that sort of integral that I think are controversial in these parts
 
Jasper Loy
Jasper Loy
@Alizter What is mf?
 
Mike
Mike
@Jasper m other f ucker
 
Alizter
Alizter
7:55 PM
A rude phrase, hence abreviated.
 
Jasper Loy
Jasper Loy
@Mike LOL
It could also be mezzoforte!
 
Alizter
Alizter
It could also mean mitten fest
 
Mike
Mike
Luckily for our colored tongues, it is none of those
 
Alizter
Alizter
Looks at tongue wait a minute
 
Mike
Mike
@Ted I'm already sick of this class and I've only just started the homework
I guess I'll just make the drop official tomorrow. :P
 
Alizter
Alizter
7:57 PM
@Mike What class?
 
Nickolas
Nickolas
howdy
 
Mike
Mike
Mathematical finance
 
Jasper Loy
Jasper Loy
@Mike Don't wait till it is too late.
 
Mike
Mike
@Jasper I have two more weeks
 
Ted Shifrin
Ted Shifrin
well, @Mike, you'd probably be sick of diff geo, too ... you'd like my pinecone and pyramid problems but would be bored by the routine ones.
Why is $x^{1/x}$ controversial, @Mike?
 
Mike
Mike
7:58 PM
Pinecone? Pyramid? I like those difficulty ratings, though I'm not sure how they correspond to medium/hard
 
Nickolas
Nickolas
May I ask a question : At the concatenation of two strings $x$ and $y$ , I guess the two strings belong to the same alphabet $\sum$ ?
 
Ted Shifrin
Ted Shifrin
pinecones are prickly but not deadly :) pyramids are difficult to climb. Flowers are child-friendly :P
 
Jasper Loy
Jasper Loy
@Mike I think it's pretty arbitrary choice of pictures!
 
Mike
Mike
@Ted The function's certainly not controversial, but my opinion on the beauty of evaluating integrals is, likely
 
Ted Shifrin
Ted Shifrin
I just explained, @Japser.
Oh, @Mike, I'm totally on your side.
 
Jasper Loy
Jasper Loy
7:59 PM
OIC.
 
Mike
Mike
I have nothing against those who like them, but I may never understand it myself.
 
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