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@Matt I've long given up on reading him.
 
It's the same person who asked the question about that number, you see...
 
リンさん、突然なんですけど。。。日本語の質問を聞いてもいいですか。
I have a word that I don't know the meaning of and your Japanese seems pretty solid.
 
いいですけど、japanese.stackexchange.comのことを知りませんか
 
Yes I do but I feel silly about asking a one word question.
 
QED
12:28 PM
How do you type that?
 
@QED: With suitable software, of course.
 
I switch my keyboard settings to Hiragana, write the word then press space to get Kanji.
 
QED
what do you mean write the word?
you type romanji?
 
No Hiragana.
 
You actually type hiragana? :o
 
12:29 PM
@QED Romaji is Japanese words written with Roman letters. That certainly ain't romaji.
 
QED
so letter 'h' (for example) is mapped to の and stuff?
 
@ZhenLin I don't know any faster way. : )
 
@Matt I type romaji...
and as far as I know most people, even in Japan, type romaji.
 
That makes no difference with respect to the amount of typing...
 
QED
how do you type it Matt?
 
12:31 PM
Ok, here is an example:
If I want to type "Konnichiwa" then I switch to Hiragana input and type こんにちは
this gets uderlined
then I press space
and it converts it to 今日は
: )
 
QED
cool but how do you type こんにちは?
is it one symbol per key on the keyboard?
 
So after switching keyboard settings the key sequence is:
konnnichiha
 
QED
ah I see
 
@ZhenLin Honto? Boku wa wasurenai...
 
QED
That's interesting
It seems I only have chinese support
not that I can use that
 
12:36 PM
@ZhenLin Do you know what 実況中継 means? I thought it might mean "live broadcast" but then that would be 生放送。。。
If not I'll ask it on JSE.
(Seeing as you're probably sort of a senpai I decided that I better be polite when talking to you in Japanese ; ))
 
Sure, but 中継 is the kind of word that they'll put on the TV screen during a live broadcast
whereas 生放送 is the kind of word used to refer to a live broadcast.
 
Ooh!! : ) So they do mean the same. Nice! Thank you!!
How long have you been learning Japanese?
 
I think it's been about 12 years? Not that I have that much to show for it.
Oh, and technically 中継 means "relay broadcast".
So there are subtle differences in nuance.
 
What does "relay broadcast" mean as opposed to "broadcast"?
 
See here.
 
12:47 PM
So Japan is not on the list there but as far as I can gather it means that in areas that are too far away from where the thing is produced they use a repeater to send the signal across a longer distance.
So from the point of the person who is watching they don't know whether it's a relay or just a normal broadcast.
 
Well, more to the point, the live feed usually doesn't come from a broadcasting centre.
Instead it is relayed via the broadcasting centre
 
So for example film it in the stadium, send it to the broadcasting centre via cable and then send it from there via satellite?
 
I actually have no idea how the relay is usually done, actually.
 
Me neither : ) But I'm glad I could ask you rather than posting a non-interactive question on JSE : )
Thank you!
@ZhenLin I like modesty.
 
Ah, it's not modesty. Just because I've been studying it for a long time doesn't mean I've learned much!
I still can't read a newspaper. I can barely read Wikipedia unassisted.
 
QED
12:53 PM
This handwriting recognition isn't great.. i.imgur.com/dPZ1R.png
 
You wrote backwards...
 
QED
今日は
oh he left
 
1:38 PM
@QED Start in the middle, draw a straight line to the bottom left and then without lifting up the pen draw an almost-circle.
 
QED
ah it can recognize that
 
@tb Is what Johan says true?
 
@Matt I would say so, yes.
The point is: the conclusion is correct but the argument seems wrong.
The choice of $\delta_x$ says that for all $y$ with $d(y,x) \lt 2\delta_x$ you have $|f(y)-f(x)| \lt \varepsilon$.
 
@tb Let me figure it out, ok? : ) I really need to finally be able to do analysis I.
 
@Matt okay. It's a slightly subtle thing and as I said, I'm not sure if it's a genuine mistake or not.
 
QED
1:48 PM
What do you mean by wa is a normal letter?
 
It's just a quirk of Japanese spelling.
 
わ=wa is a Hiragana letter.
 
QED
okay
this is complicated. .. :)
 
は=ha also is a letter but it can also be a particle.
 
Huh... that character QED had backwards was no, right?
 
1:50 PM
@ZhenLin And I wish my handwriting wasn't fugly. : )
@JM Yes.
 
@Matt Practice practice practice!
 
Indeed.
Now I'm going to leave this chat for a little while again. brb
 
QED
bye
 
2:03 PM
Hello @all .. I have a question about rate of convergence! if $g'(a) \neq 0$ then the rate of convergence is 1, if $g''(a) \neq 0$ then it's 2 and so on. Am I right?
 
Convergence of what?
 
Wth! MathSciNet asks me to login!
 
@JonasTeuwen You're on your university's IP, no?
 
Yes. In the top right corner it says "Technical University Delft".
 
@JonasTeuwen It's about "the Newton–Raphson method" in numerical analysis,
 
2:08 PM
@JonasTeuwen Something you should be bugging the IT or library guys, then.
 
@tb I'd much rather discuss it with you than with Johan. Are you still here and have a spare minute?
 
"Rate of convergence"
 
@JM I'm too shocked to call them. I'll grind some coffee first.
 
@Gigili Well, multiple roots do retard the convergence rate. If the root is simple, the rate is quadratic.
 
I posted my false reasoning in a comment and I'll be back later to check for responses.
 
2:09 PM
Computer says no.
 
@Gigili If the root you are trying to approach is multiple (i.e. the root is also a root of the derivative), then you get the linear convergence.
 
This P. Auscher guy seems to be important in my field. I see him everywhere in the references!
 
In my book, it says somewhere what I said above, but somewhere else it says if $h(a) =0$ but $h'(a) \neq 0$ the rate of convergence it's at least two .. I'm confused if it's exactly two or at least two
 
@Gigili What are you tokkin' about?
 
@JonasTeuwen Newton-Raphson
@Gigili It's supposed to be exactly 2. I don't think I've seen an example of NR cubically converging...
 
2:13 PM
Oops sorry, I meant to ping @JM instead of you. @JonasTeuwen.
 
Would it be possible that we have a fifth degree polynomial for which we are not able to find closed form expressions for the roots in terms of elementary (?) functions, but we are able to solve the Newton-Rhapson recursion to obtain a root?
 
@JonasTeuwen Certainly. But you need a good starting point. NR is one of those "garbage in, garbage out" processes.
 
@JM Cute.
 
Wait... you meant approximations, didn't you? NR can't do exact solutions, except in the linear case.
 
Well, I mean that one would be able to find the limit of that sequence!
By solving that recursion relation.
 
2:16 PM
@Matt your argument in the comment is correct.
 
$x_n+1 = x_n - \frac {f(x_n)}{f'(x_n)}$
 
@JonasTeuwen The recursion is awfully nonlinear. I haven't heard of anybody doing explicit solutions except in the important case of computing nth-roots.
(i.e., solving the polynomial equation $z^n-a=0$)
 
Okay, but there is nothing that excludes the possibility that it is possible, right?
 
For that case, you have a sequence of rational functions whose limit is the root function.
@JonasTeuwen None that I know of. :)
 
My advisor told me to check out P. Auscher on ArXiv, but the guy has 32 papers there... Mmmm.
Why is there nobody here?
 
QED
2:27 PM
one for each day of the month
 
I'm not aware of any month with 32 days.
 
QED
that'll be that day light saving time screwing with everything
 
There's a question in the book, asking " suppose a is a root of $m$ multiplicity ( $m \ge 0$) from $f(x)=0$. if $h(x)= \frac {f(x)}{f'(x)}$ , show that $h(a)=0$ but $h'(a) \neq 0$ then get the result that convergence of ${x_n}$ is computed by $$x_n+1 = x_n - \frac {h(x_n)}{h'(x_n)}$$ , the rate of convergence is at least 2. " .. That's why I asked if it's exaclty two, so the question is wrong, right @JM?
 
@Gigili The idea is to take $f(x)=(x-a)^m q(x)$, where $q(a)\neq 0$. Try running through the usual procedure with that construction.
BTW: that modified NR you speak of is sometimes called the Schröder iteration.
Well, proving that the convergence rate is at least two is (supposedly) easier. Proving that it's exactly two is the stiffer challenge.
 
Aha, thank you. But why at least two?
Got it, thank you very much.
If $f^{(m)} \neq 0$, the convergence rate is $m$. Right?
 
2:41 PM
Every derivative from $f$ to $f^{(m)}$, you mean.
 
@JM Do you know the book "Beyond the quartic equation"? It mentions the elliptic functions thingie.
 
No use having nonzero second derivative if the first derivative's zero.
@JonasTeuwen I liked that book. :) You're reading it right now?
 
@JM When did you find the time to read all these books? :|.
Yes. I'm reading it right now.
 
@JonasTeuwen Before there was such a thing as math.SE... :D
 
@JM Why not? Didn't get your point there.
 
2:46 PM
@Gigili If a function has a double root, the function and its derivative are zero there, so the convergence rate is linear.
I gotta admit, I keep seeing all these shiny new books at, say, SpringerLink, but math.SE is too distracting... :D
 
You have a university connection?
 
@JonasTeuwen Well, I'm an alumnus, so I have to pay for the privilege...
 
Oh :-).
You pay to Springerlink or do the university?
 
Well I'm again stuck at this problem : Find the largest integer which divides 62, 132 and 237 leaving the same remainder? 0_o
 
@JM $(x-1)^2=0$ , $f(1)=0$ and $f'(1)=0$ but $f''(1) \neq 0$ .. The rate of convergence is two, no?
 
2:53 PM
Damn, I drank too much coffee.
 
@JonasTeuwen The university does. What I'm paying for is access to their library and its resources. So, SD, SpringerLink, Wiley...
 
Sorry if I asked too much questions =\
 
@Gigili The NR iteration for that ought to converge only linearly, unless you do the Schröder construction or something.
@Gigili No worries; I don't see that much numerics questions here, and it gets lonely... ;)
 
I have all these papers on my desk. Someone that doesn't know better would think that I know what I'm talking about 8-).
 
@JonasTeuwen Stacked neatly, or in a haphazard heap?
@MrAnubis I think you need the Chinese remainder theorem here.
 
2:58 PM
@JM Never even heard about it. Lemme google it :)
 
@JM The latter.
 
@JonasTeuwen Then yes, it's a good way to maintain the illusion you speak of. ;)
 
@JM That theorem sucks (so large :D)
 
@MrAnubis On second thought, there might be an easier route. Let me think...
 
@JM: are you the one voting to delete those duplicate threads? If so, why?
 
3:06 PM
@tb Which? I voted on the answer in the prime question, but not the question itself.
 
@JM There are three votes for deletion of questions that appeared today and I'm not sure why any one of those threads should be deleted, except maybe the repeated exponentiation thing.
 
Odd. But one of the votes here is mine.
 
@JM yes, I'm the second one there.
On a more enjoyable note: Did you see the links I gave here? You might like them.
 
Anyway how is HCF is useful to know? what is its application? any analogy?
 
@MrAnubis Have you noticed that they are all multiples of three?
@tb I haven't. Thanks for those! :)
 
3:14 PM
@JM hmm, no , lemme think again:)
 
That MathOpenRef site's nice. Too bad they don't have a cartoon of how to draw a heptadecagon...
 
Anyway how is HCF is useful to know? what is its application? any analogy? ( I know what is it exactly , but..)
 
@JM Hmm. Do you happen to have a reference where they explain how to solve polynomials using the $\theta$-functions?
 
@MrAnubis The thing pops up a lot in number theory. I don't quite know what to say about "applications".
 
Without requiring me to read a complete book on polynomials.
 
3:20 PM
@JonasTeuwen Just the quintic, or arbitrary-order polynomials?
 
Arbitrary-order. Well, actually nineth-order.
 
QED
I love how we get mathjax rendering in this chat
 
@JM analogy when to use HCF?
 
@JonasTeuwen Umemura is a primary ref, but I've completely no idea if there's newer work...
 
QED
もしもし。
 
3:23 PM
@MrAnubis I don't understand how you're using "analogy" here, as it's not the use of the word that I'm accustomed to...
@JonasTeuwen ...and as you can see here, Riemann theta functions are rather... elaborate.
 
The sum of two number is 528 and their HCF is 33. The number of pairs satisfy the above condition?
 
@JM There are two of those books available at the library, is this I?
@JM Holy monkies!
 
I've no clue where to start :(
 
QED
factor 528
 
@JonasTeuwen apparently it's II.
 
3:28 PM
Bleh, the library doesn't have it anymore since they have cancelled the mathematics library...
(they will bring the books to the central library, but I'm not sure when)
 
@QED Why did you tell me to factorize 528?
 
@JonasTeuwen Eep, sorry to hear. I'm afraid I don't know newer refs, as I've said. I'll only note that "Riemann theta", "Siegel theta", and "multidimensional theta" are effectively interchangeable terms, if you'll try doing your own searches.
 
@tb Lovely, thanks. But there is still a mistake in the OP that I haven't found yet, right?
 
I'm 100% gonna fail in maths :(
 
@JM Yes, that would be nice. The cartoon on Wikipedia isn't really understandable. I liked Richmond's construction as given in Stewart's Galois Theory. When I was in High School I drew that using Indian ink.
 
3:33 PM
@JM Thanks!
 
@Jonas: Can you explain to me whether the Dutch sound represented by <ij> sounds distinct from the English sound in <eye>?
 
@Srivatsan you there?
 
@Matt After your comment I don't think it really is a mistake. Let me look closely once again (but don't worry it really is a quibble if anything)
@Srivatsan: Just in case you haven't noticed: it's Kalle
 
@ZhenLin It certainly is different. We have "mijn" (mine), and it is pronounced completely different and I'm not even sure how to write that in English. I'll look it up.
 
3:36 PM
@Jonas See, every time I listen to a Dutch recording of the sound, it sounds exactly the same to me...
 
Hmm. Could you give a word where they pronounce it like that?
 
@tb Ah :-)
 
Nijmegen, Valentijn, Moerdijk etc.
 
@MrAnubis it doesn't have a unique answer, any pair of $(x,y)$ as $x+y=16$.
 
@tb Thank you. If it's not 100% right I can't be sure whether people are just assuming that I'm being imprecise while I'm still not 100% clear about it. Earlier you said Johan's comment was correctly pointing out a mistake...
 
3:39 PM
@tb You drew yourself a heptadecagon? Hardcore, man. :D
 
@ZhenLin Hmm, sounds a bit different. I found that the "ij" should be pronounced like the phonetic writing: (Ei/EI/eI/ei/Ai/AI/aI/ai).
 
Can't kalle pick a less annoying handle? Or at least a handle that doesn't boast of his "discovery"?
 
QED
it's not really something to boast about
 
Who is kalle?
 
@Jonas Yes, I know that <ij> is usually spelled [ɛi] in IPA, while the English sound is usually spelled [ai]... but does it sound completely different?
(because I honestly can't hear much of one...)
 
3:42 PM
@ZhenLin To me it sounds completely different, yes :-).
 
(but people would understand you if you would say it like "eye")
 
QED
@Srivatsan, It's not uncommon for one to write something meaningless and have someone else interpret it as profound
 
@Srivatsan Whoah! Great!
 
@Jonas Then perhaps I just need to listen more carefully. Thanks!
 
3:44 PM
No problem :-).
 
@QED All our efforts into trying to convince the OP haven't yet stopped them from boasting. I'm sure they are still proud that they shook the foundations of mathematics, not to mention shaking the math community.
 
QED
I don't think you need to convince them, they'll realize it when they get older
 
Let's ask his name so we can nominate kalle for the Abel prize.
 
@JM we had this horrible course called "Technisches Zeichnen" (technical drawing) which was all about not spilling ink on a huge sheet of paper with the worst possible tools. Out of boredom I decided to challenge myself...
 
QED
3:50 PM
"And dont flatter yourself" flagged as rude/offensive
 
I'm tempted to respond, but what's the point? Let me just enjoy my coffee.
@QED Thanks, QED.
 
Kalle Clown
 
QED
haha
 
Enjoying coffee is a good thing.
 
4:04 PM
I'm pooped. Later, you guys.
 
QED
bye
 
@JM What? You're going to bed so early? =)
See you later, JM.
 
4:23 PM
You guys suck! I should know better than listening to guys on the internet. You were all like, you better return the necklace to your ex, its hers! And I am all like, I do not really want to, but I guess I will. So I placed it in her mailbox, with a short note and a piece of her favourite chocolate. And now she hates me.
 
QED
she sounds like an idiot
next time you're going to give her chocolate give it to me instead
I wont hate you at all
 
@Srivatsan: just let it be...
 
@tb Is this guy looking for the bar resolution?
 
@ZhenLin I think so. I just asked in a comment...
 
I must admit, the bar resolution isn't entirely obvious. Comonadic resolution using free presentations is much more obvious...
 
4:36 PM
@tb Yeah, he is getting to me. I wanted to do something (and that never ends well, I know)...
 
but that was before I learned about simplicial objects, so I was still thinking about chain complexes.
 
QED
I always feel that Byron is looking at me disapprovingly
Maybe it's my atrocious spelling
 
Maybe because you don't end your sentences with a full stop. =)
 
QED
I do so!
 
4:42 PM
I'm just kidding obviously.
 
QED
I don't end messages here with full stop, but is it correct to do so?
(I just noticed I'm the only one that doesn't..)
 
I don't bother when I'm lazy...
 
@QED Oh yes, I meant only the two sentences immediately preceding my comment. In chat, it doesn't matter.
This chatroom seems to be on the formal side. Complete with full stops, capitalisation and whatnot.
What in the devil's land is this guy doing?
 
0
Q: Is there a function that will give me nth prime number?

PragmaOnceIs there a function, like f(n), i insert n and it outputs nth prime number? I have been trying to find a pattern between the prime numbers, 1st differences, 2nd differences, but i can't find it.

What?
Perhaps I should post $$ n^2 + n + 41 $$ Just for the giggles =)
 
@N3buchadnezzar Did you see my message? =)
 
5:05 PM
@Srivatsan Seems to have been deleted. What happened?
 
Good evening folks.
 
The title is innocent enough.
Hey Asaf.
 
What is going on?
 
Nothing interesting.
 
I have an interesting idea but sharing it from the iPhone would be difficult.
 
5:12 PM
@DylanMoreland Perhaps he (again!) thought he solved it.
 
I woke up early because my dog had decided that my hair was particularly delicious this morning. Trying to get a third faculty member to show up for my qualifying exam talk. I will probably read a lot of Silverman after that.
 
Master degree Dylan?
 
I think I get one after the qual is over, yeah.
 
It is pretty cool that your advisor is on MSE/MO... I guess
 
It's nice to have a lot of his spiels written down somewhere.
He talks very fast.
 
5:16 PM
How much does he know on the topic of your research?
 
Quite a bit, I should hope!
 
I bet this is nice.
No one in my university knows much about my research area...
 
I didn't pick the topic myself. I had some vague ideas but there are way too many interesting Langlands-type projects to do out there.
 
It can drive me nuts sometimes. At least I can present the arguments and sometimes my advisor catches mistakes or helps me to come up with solutions.
 
He had two projects in mind and I chose the one that had more geometry.
By which I mean algebraic geometry. Actual geometry is difficult.
 
5:20 PM
I hate geometry... :-)
I mean all kinds of geometry.
 
@AsafKaragila Being in logic is probably always going to be hard that way, right?
 
Not entirely true, but it can be harder to find a knowledgeable advisor in all the topics.
 
QED
You can't hate circles
 
Working under Shelah is a lot more how you'd expect grad school to be. He offers you topics and you work.
 
QED
@Srivatsan, now I can't stop noticing all the full stops!
 
5:23 PM
Personally I doubt I could do that. :-)
 
Argh! Full stop before smilies, that is over the top. =)
 
WiFi on the bus = Awesome.
 
@AsafKaragila That is cool. I wish we had Wi-Fi on the subway.
 
@DylanMoreland You even had Wi-Fi on your food? !
 
Also full. Stops. Are. Awesome. : .-.).
 
@ZhenLin Admittedly, not using $\wedge$ is a horrible crime, right?
 
Why doesn't he turn on the heating?? It's bloody freezing on this bus!
 
@Dylan Given that the book says "multiple integrals", and this whole thing is being developed to justify "a mechanical substitution procedure", I think introducing $\wedge$ would just be confusing...
 
Well that's probably also true.
 
@AsafKaragila How cold is it?
 
5:30 PM
But a question on why differential forms work this way could be okay. It has surely been asked before, though.
 
Cold enough for me to complain on this chat, but won't go to the driver to ask for heating.
 
But the page in the question already explains the justification!
 
@AsafKaragila You should come to live in Norway. Harr harr harr
 
I know. My girlfriend and myself are both hardcore metal fans and a visit, a long visit, to Norway is on the wishlist.
 
Metal is so awesome
I think there are about 7 well known metalbands, from my hometown ^^
 
5:34 PM
I used to write to a metal e-mag whose origin was in Norway. Imhotep.
 
Gorgoroth, Taake, Burzum, Immortal etc
 
Yeah I figured you're from Bergen.
 
It`s rather cool, living in Bergen.
 
Literally too.
 
@ZhenLin I suppose the reading comprehension is about as well developed as the orthography...
 
5:36 PM
@Srivatsan are you free?
 
@QED You are welcome. =)
@MrAnubis Not for long, actually. But tell me.
 
@tb I hope you'll be around in an hour or so. I need to pick your brain.
 
@tb And now the question has vanished. Hm.
 
I'm editing all my old answers; hopefully improving them.
 
Srivatsan: are you a free module?
 
5:37 PM
@AsafKaragila make it 1 1/2 - 2 hours.
 
Great!
 
@ZhenLin deleted by OP
 
@Srivatsan well I am unable to figure out this Q : The sum of two number is 528 and their HCF is 33. The number of pairs satisfy the above condition?
HCF and LCM is really hard :(
 
Which one is deleted? kalle's?
 
I should be home in 40 and I'll have time for dinner too.
 
5:39 PM
HCF?
 
@N3buchadnezzar HCF=GCD.
 
QED
@MrAnubis, did you factor 528 yet?
 
@QED Why factor the sum of the numbers?
 
@QED yes but I had asked you tell me why do I factorize it but you never answered me back :(
 
QED
look at the factorization...
 
5:41 PM
 
A short question, do you guys have any mathematical software in the background, or just a pen and paper?
 
@ZhenLin - No intuition behind this question?
 
@MrAnubis Actually, you should factor it, kind-of. This is a standard trick in GCD/HCF: If the GCD is d, then write the numbers as d*x and d*y. Where x and y are relatively prime.
 
Mostly pen and paper.
W|A sometimes.
 
@Victor: There is a very well-developed calculus of differential forms that the book is alluding to there. The motivation is explained on that very page.
 
5:43 PM
GCD = 11 * 3
 
@Srivatsan which means if I only know GCD I can easily find the two numbers like that?
 
Why factor the gcd?
 
@ZhenLin - Just to get the "right" answer?
 
For fun!
 
@Victor Yes.
 
5:44 PM
@ZhenLin - How do you know that is the right answer?
 
@MrAnubis No, you cannot find the numbers given only the gcd. But the GCD is kind-of extra noise that you can filter it out.
 
@ all: request for papers on covering systems applications other than proving conjectures related to diophantine equations etc.
 
Let me explain.
 
@Victor Because it gives the right answer when you actually need to calculate things.
 
@NikhilBellarykar I think you'll be better served in the main site. =) Not sure if anyone here can answer this question...
 
5:46 PM
@ZhenLin - is there proof for the jacobian thing?
 
@Srivatsan ok :)
 
@Srivatsan How do I find those x ,y ? any trick?
 
@MrAnubis No no, no tricks.
 
@Srivatsan I'm waiting for explanation :)
 
@MrAnubis There's too much going on in the chatroom, that's why I cannot focus =).
 
5:47 PM
@Victor Yes. Go read Rudin's Principles of Mathematical Analysis.
 
QED
@Srivatsan, It's just a hint
He said he didn't know how to start so I said factor the number
 
@ZhenLin - which page in which edition?
 
@QED Right. Makes sense.
 
QED
I'm trying not to give the whole solution away
 
@Victor Does it matter? You need to read a lot of the preceding material before you will be able to understand the proof.
3
 
5:50 PM
@ZhenLin - okay, thanks
 
@Victor : take Zhen Lin's advice seriously! And if I may ask you a favor: before you ask a question on main please think about it for at least two entire days.
2
 
QED
Maybe I shouldn't have upvoted it
I wouldn't normally upvote it.
 
6:13 PM
Analysis is quite pretty.
 
I am watching Ducktales
That show is still great
 
QED
I just remember the music
life is like a hurricane
 
scrooouuuggeeee
 
@JonasTeuwen Out of nowhere? :)
 
@Srivatsan Yes.
 
6:28 PM
I mean, were you talking about analysis sometime back?
 
@Srivatsan his name is Jonas Teuwen and he's an al... er... analyst
 
@tb But he might get some recruits if he explains the context better... =)
 
6:56 PM
@MrAnubis: I told you what to do, seems you want solutions from special users.
 

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