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user19161
12:00 AM
@dreamcrash It's hard to navigate in a world of formulas!
 
@robjohn undeleted
 
@dreamcrash hang on, I will look
 
thanks a lot
@WillHunting indeed
 
@BenjaLim I will look into the isoperimetric question.
 
@robjohn you mean answer it?
@robjohn I deleted it because it attracted no attention.
 
12:02 AM
@BenjaLim yes. I will try to answer it. I have done things like that before.
 
thanks!
@robjohn By the way I am just digging up stuff from the grave with that question
 
@BenjaLim You mean it is an old question?
 
well yes
 
what isoperimetric
does anyone know some nice things about the theta function?
I think it's holomorphic on $\mathbb C$
$$\theta(x) = \sum_{-\infty}^{\infty} e^{-\pi n^2 x}$$
this one
 
user19161
@ben How many brownies did you make?
 
12:08 AM
:(
 
@Argon, are you interested in the Pochammer continuation of Beta?
that is weird!
 
@cassandra0 It scared me... Did you make any progress?
 
@Argon, I keep forgetting to look it up in the library
 
@cassandra0 It is one crazy contour!
 
12:12 AM
:D
hmm why do bernuolli numbers comefrom zeta
 
user19161
I think I will make some brownies myself now...
 
25
A: Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$

Qiaochu YuanI have two favorite proofs. One is the last proof in Robin Chapman's collection; you really should take a look at it. The other is a proof that generalizes to the evaluation of $\zeta(2n)$ for all $n$, although I'll do it "Euler-style" to shorten the presentation. The basic idea is that meromo...

@WillHunting Yum
 
12:45 AM
I don't quite understand it
 
1:00 AM
Can anybody give me a quick hint on inverting the complex function 2z + 1/z ? I feel so stupid, I need it for a conformal mapping problem.
I know it = 3x + iy, but i dont see how that helps me invert.
 
2z + 1/z = y then 2z^2 - yz + 1 = 0
so z = quadratic formula of y
 
doh! what a great idea :-D
 
user19161
@Argon LOL
 
thanks cassandra0
 
:) the reason I know this is because phi - 1 = 1/phi
 
1:03 AM
haha you already know where im going
why do we choose the root with + discriminant and not - discriminant
 
"we"?
 
hey in the end aren't all of us in this together
 
I never made any choice about + vs - sqrt(disc). That was all you. Why don't you tell us why you picked it? :armchair:
 
1:19 AM
@anon, I should have said, why does my textbook say this is the inverse function?
 
well both are "the" inverse
 
there is more than one inverse, just like flipping y=x^2 across y=x gives you two different functions (y=+/-sqrt(x)). the text might have just picked the more obvious one arbitrarily.
 
right... was not sure if things worked the same way in complex land
But that makes sense, so the - so if one maps a region to the exterior of the disk, the other one is a perfectly good mapping too?
 
well basically
sqrt(-1) = i
let me draw a diagram
 
right, and z^2=i has root -i as well
*z^2=-1
 
1:33 AM
well I can't draw it but basically imagine you slice from 0 to infinity and then take hold of the bottom part and bend it around so that -1 gets moved to 1 and everything to its square root
 
gotcha
ty
 
so the thing with that is 0 to infinity was arbitrary
you could slice in any direction at all
 
I was gonna ask if it had to be along the real axis
 
what does it matter? Well consider sqrt(1 + i/100) and sqrt(1 - i/100)
do you see how their square roots are really far apart? (the first is near 1 and the second is near -1)
even though the numbers they are the square root of are close
 
algebraically yes... geometrically i am not seeing it
 
1:41 AM
that's just one square root function
the red line is where I made the cut
that's taking the points z on the complex plane to the image of sqrt(z)
does it make sense?
 
so it does what Id expect it to do in QI
 
what's that
 
The upper right quadrant
 
1 + i/100 would be near the pink bit and 1 - i/100 would be near the blue right?
 
Where are we gluing the two pieces together
 
1:47 AM
we aren't glueing anything
and there is only one piece
does it not make sense?
 
Im sure it does, but not to me...
i get that the pink/blue pts were close to each other before the sqrt was applied
 
it's the complex plane and then |-- sqrt --> [where everything went]
 
or that they could be arbitrarily close like 1 + i/epsilon, 1 - i/epsilon, and still end up about a distance two apart
 
I drew some colored shapes on it so you can see where everything went (if it was all white it'd be impossible to tell)
yes exactly
 
epsilon i
 
1:52 AM
because they are on other sides of the cut
 
so it rotates what is in the Re > 0, Im < -0 quadrant by -pi?
 
lol
 
it's more like a hinge
the complex plane gets squeezed in the hinge
 
@Melvin Sorry, I fell asleep. :P
 
2:01 AM
here's a different way to do it
this is what my calculator does
 
and the black lines in the second picture show what happened to the imaginary axis after the transformation?
 
yes
 
the two diagonal lines
and they meet at a right angle?
I guess it makes sense, sqrt i = 1/sqrt(2) + i/sqrt(2)
 
yes
notice in the first diagram sqrt(i) = -1/sqrt(2) - i/sqrt(2)
but these two do match up in QI
e.g. sqrt(1+i) is the same in both
 
right...
but not 1-i, right?
 
2:11 AM
yeah
 
ty much
 
so there are loads of different sqrt functions depending on which direction you make the cut and how you hinge out from it
but the cut always comes from zero
because zero is the only point which doesn't have two values
sqrt(0) = 0 that's all
if you think about inverse of something like z^2-4 that gets interesting because there are two places that a cut needs to come from (-2 and 2)
 
user19161
Haha @amwhy I posted my modulo answer 5 seconds later than yours so I deleted it!
 
@WillHunting Ahhh...don't you hate that! no need to delete!
 
user19161
2:26 AM
Also just now that new guy posted his distance speed time answer a minute earlier than mine so I deleted it!
 
user19161
There's no point in having two answers essentially the same.
 
user19161
If they are significantly different, fair enough.
 
@WillHunting You're right. I wish some of the very high rep users would realize that before posing an answer 30 minutes after mine, saying exactly the same thing!
 
Im downloading debian
 
user19161
@amWhy Sometimes it's because they forgot to read the other answers as they are too tired looking at the stupid screen!
 
user19161
2:28 AM
@Khromonkey Good luck installing it if you are a beginner. You will run into many problems...
 
why are you discouraging him from using debian
 
@WillHunting That could be. It's just frustrating when moments after they post, they get a handful of upvotes.
 
user19161
Also @amwhy today I learnt about the phrase "making brownies in the bathroom" LOL.
 
user19161
@cassandra0 Not discouraging, just preparing him...
 
@WillHunting I never heard that!? What does it mean?
 
user19161
2:30 AM
@amWhy Are you serious you don't know? =)
 
@WillHunting Never mind, I got it!
 
user19161
@amWhy HAHAHAHAHA
 
@WillHunting That's a new one to me. It beats some other not-so-pleasant ways of expressing the same!
 
user19161
@amWhy I just thought of a variation on that theme: making lemonade in the bathroom. HAHAHAHA
 
@WillHunting Hey, quick wit. I like that.
 
user19161
2:34 AM
@amWhy By the way, brownies go well with lemonade!
 
@WillHunting Yuck! sweet then sour. No thanks. I'll take milk with my brownies.
@WillHunting But in THAT sense, how true!
 
user19161
@amWhy Yeah true too, otherwise the lemonade will be too sour...
 
user19161
Over and out!
 
@WillHunting Where you going?
 
@amWhy Hi!
Thanks for the edits and stuff!
@WillHunting Good night!
23 hours ago, by Will Hunting
@Argon See you in your dreams.
 
2:45 AM
@Argon Edits just to get your attention :-) ... and a couple well-deserved upvotes.
 
@amWhy Hahaah!!! Thanks again!
 
@Argon My pleasure
 
:)
@amWhy What types of math do you like especially?
 
@Argon Math logic, set logic, philosophy of math, abstract algebra, combinatorics, topology, analysis $\to \infty$
 
@amWhy Hahaha!
 
2:50 AM
@Argon It's really hard for me to limit myself to one narrow branch. I like the connections between different branches of math, how they relate, etc.
 
@amWhy Well, some stuff I don't know!
 
@Argon You're in good company! =)
I think I'm going to head to bed soon...and my kittens are climbing on my laptop! I'll see you soon!
 
@amWhy Good night
 
@Argon g'night, sleep tight.
 
user19161
I am back.
 
3:01 AM
@WillHunting Hello 'dere!
 
user19161
@Argon I came back to see Aaron!
 
@WillHunting Yay! And I came to see Willy.
@WillHunting Did you go for a walk?
 
user19161
@Argon Nope, maybe this Friday...
 
@WillHunting Got to get down on Friday; everybody's looking forward to the weekend.
Fun, fun, fun, fun.
Good night all!
 
3:40 AM
Spivak's Calculus is weird...
@JayeshBadwaik Sometimes I feel this kind of conflict, I guess that "no single rule applies to all things" is a rule on a higher class and "including this rule" couldn't be applied to it.
 
@GustavoBandeira Maybe it is a rule schema over all rules, so that the truth of this proposition follows from the truth of it applied to each individual rule.
 
4:26 AM
@GustavoBandeira Weird? As in?
 
4:39 AM
@JayeshBadwaik Well, the beginning of the book has some obvious math properties.
They told me the book is hard as shit, when I've read these properties, I felt intimidated: I started to think that there may be some kind of hidden secret.
Anyone use Lyx?
 
@GustavoBandeira Obvious is the hidden secret here. :P
It is supposed to be an introduction to proof, better to start from axioms. Have you done of the construction of real numbers yet?
 
Oh, not yet. I'm trying to hurry in the functions chapter.
But I've already read some things from Rudin's book.
It starts on the proof of the irationality of 2, isn't it?
 
Yes. Actually, rather, its not proof of irrationality of 2. Rather it is the proof of "gaps" in the rational number line.
It is my opinion that if you study the construction of real numbers given at the end of the book first, you might understand the need for stating the obvious properties even though they are easy.
 
Oky.
Thanks for the advice.
 
@GustavoBandeira I used LyX until three years ago. But then, while writing a long course project report, I was frustrated with it, and finally took up hardcore LaTeX writing.
 
4:48 AM
It has an export tool.
I'm trying to adjust the exporting so that it can use more space on the page.
Let me show you.
 
Hmm, export to PDF?
 
hmm. and you want less margin?
or less space between the lines?
 
Less margin
This document could easily fit one page
But there's a lot of margin
 
Go to Document -> Settings -> Page Margins. Adjust your margins there.
 
4:58 AM
@JayeshBadwaik Thanks!
 
@GustavoBandeira welcome. :-)
 
I hoped (dunno if this word exist) it was something near the export in the file menu. =/
 
Hmm, the margins are the property of the LaTeX document, not of the compiler/interpreter/converter, hence, it is in the documents section.
Also, another option with slightly less control but more automation is using the package fullpage. In the Document -> Settings -> LaTeX Preamble, if you add the line \usepackage{fullpage} the margins will be reduced to 1.5 cm on each side, however, I myself don't use that a lot.
 
Got it.
 
5:15 AM
hii
 
@anilorap hello.
 
:)
 
hello
 
wassup?
 
Anyone know how to convert a loop into a closed form expression when floor division is involved?
 
 
1 hour later…
6:44 AM
Can non-usa students take the amc12?
 
@MaoYiyi I don't think so, the eventual purpose is to select the students for the US team to IMO. I guess you should at least have some sort of green card/ citizen status there.
 
 
1 hour later…
8:06 AM
between 5:00 and 6:00 UTC, there was a color shift from bluish to reddish gravatars. Interesting
 
@robjohn Good morning, mean square : )
 
Is that known as the ether-net red shift?
Because the internet is expanding :-)
 
8:47 AM
@aDangerousIdea Is the internet warped near MSE?
it certainly is near this chatroom :-)
 
@robjohn Indeed.
This chatroom certainly has a way of "sucking people in." ;-)
 
@robjohn was Edmonton a good place for mathematics sometime in the last fifty-sixty years?
 
@JayeshBadwaik I don't know; why do you ask?
 
@robjohn I see a fair number of mathematical jokes about an arbitrary mathematician from edmonton.
 
@JayeshBadwaik I must have missed those
 
8:57 AM
@robjohn hmm.
 
user19161
@JayeshBadwaik Unheard of.
 
user19161
@JayeshBadwaik Unheard of.
 
@WillHunting Doubled Posting. Chrome problems?
@WillHunting hmm.
 
user19161
@JayeshBadwaik Look carefully.
 
@WillHunting ahh, my bad,
 
user19161
8:59 AM
QED.
 
We forgive you :-D
 
:D :D
Q: What does a mathematician present to his fiancée when he wants to propose?
A: A polynomial ring!
 
user19161
Lame!
 
user19161
I experienced some automatic upvoting reversal when a fan gave me too many upvotes yesterday on English. 7 was too many. I don't exceed 5 when I upvote someone else.
 
@WillHunting were the votes reversed by the voter or by a moderator?
 
9:02 AM
or both
 
user19161
@robjohn By an automatic script.
 
or neither
 
@WillHunting that's what I meant by moderator
 
user19161
It runs at GMT 0300 every day.
 
A robotic moderator
 
9:03 AM
@WillHunting what is the limit that it looks for?
 
user19161
It happens now and then because there are new users who serial upvote and don't know the forces at work.
 
this is slightly insulting to philosophers
Q: What is the difference between a mathematician and a philosopher?
A: The mathematician only needs paper, pencil, and a trash bin for his work - the philosopher can do without the trash bin...
 
user19161
@robjohn I am afraid only the SE staff know the exact code, but from my experience 5 at a time once a day in addition to ordinary voting is OK.
 
@JayeshBadwaik I don't get it?
Why would it be slightly insulting to philosophers?
 
user19161
@JayeshBadwaik Don't get it.
 
9:07 AM
@WillHunting hmm, explanation would take out all the fun. :-)
 
user19161
Also, now that I have 25k on English and 2k on TeX I will retire there, though if things get deleted, I will have to top up now and then!
 
@JayeshBadwaik It could be slightly insulting to mathematicians.
 
@aDangerousIdea Naah, I assure you it is otherwise.
 
user19161
So I will be on Math full time from now till infinity, yay!
 
this is much much better, especially since my new found love of the killer phrase in the joke


A mathematician is asked by a friend who is a devout Christian: "Do you believe in one God?"
He answers: "Yes - up to isomorphism."
 
user19161
9:11 AM
Also, I love my latest answer on English, about "paid" and "payed". The surprising thing is that the latter actually exists!
 
user19161
@amWhy It is interesting that you have kittens.
 
From zero to infinity: what makes numbers interesting.
4 mins ago, by Will Hunting
So I will be on Math full time from now till infinity, yay!
 
user19161
From zero to hero; from victim to victor; from tragedy to triumph. QED.
 
From QED to QED
 
user19161
From PDE to ODE to OED to QED.
 
9:15 AM
From OPP you know me?
 
user19161
From A to Z, from 0 to 9.
 
@JayeshBadwaik What do you think of Mahatma Gandhi's philosophy?
 
@aDangerousIdea which one?
 
About God.
 
From what I know and have read, I don't think he had any particular philosophy about god. He was tolerant of what other's views were, though.
 
9:27 AM
Yes, tolerance was very important to him.
From what I've read.
From what I've read, he said that most people who believe in God believe that God is Truth, that is there is Truth in God. @JayeshBadwaik
 
You might have read his autobiography "My Experiments with Truth" wherein you might realize that due to different circumstances, incidences etc in his life, the scale which he used against which he measured all his activities was the scale of truth. I guess, it was in that sense he meant that God is Truth.
 
I also read that at the age of 50 or so he realized how to be tolerant towards those atheists who do not believe in God. To them, he believed Truth is God, that is there is God in the Truth.
 
No, I do not think that can be correct, he was an atheist since a very young age IIRC. He was, at least, not religious. So, I do not think that is correct.
 
9:53 AM
@JayeshBadwaik Point number 5 here.
But two years ago, I went a step further and said Truth is God. You will see the fine distinction between the two statements, viz. That God is Truth and Truth is God. And I came to that conclusion after a continuous and relentless search after Truth which began nearly fifty years ago.
 
I do not see how this justifies "I also read that at the age of 50 or so he realized how to be tolerant towards those atheists who do not believe in God. "
 
If a=b, then b=a.
 
Hindu scriptures are in themselves religiously tolerant. I have personally never come across any kind of hindu scriptures which promote religious intolerance explicitly. They do promote other kind of intolerances though.
@aDangerousIdea No. I disagree. The relation "is a" is not equivalent to "equality"
 
"is" is not equivalent to "equals"?
 
naah
Also, you must understand many of his thoughts from his own life experience POV, from his times, which were significantly different in terms of social fabric. (and the difference if not simply due to the change of technology etc, there is a fundamental difference between the India of then and India of now, change which cannot be accounted to the progress of time alone.)
@aDangerousIdea "is" can be used as an adjective.
 
10:04 AM
But in this context it is not used as an adjective, no?
 
It is.
"You will see the fine distinction between the two statements, viz. That God is Truth and Truth is God."
 
God is Truth.
God = Truth
 
no...
 
1 = 1
one = one
 
1 is a number .
1 $\neq$ number
 
10:07 AM
an idea
 
Can all elementary functions be expressed as either as a fourier series or a cosine series?
 
Define elementary?
 
@JayeshBadwaik See wikipedia ;)
28
Q: What makes elementary functions elementary?

rarIs there a mathematical reason (or possibly a historical one) that the "elementary" functions are what they are? As I'm learning calculus, I seem to focus most of my attention on trigonometric, logarithmic, exponential, and $n$th roots, and solving problems that have solutions which are elementa...

 
@N3buchadnezzar Hmm, if the function is piecewise continuous, you have a fourier series for it.
So, since an elementary function is everywhere continuous, yes, a fourier series exists I will say.
 
I was thinking about $T(x) = e^{-1/x^2}$
For a counter example, but I am not quite sure.
 
10:11 AM
wait.... Lipschitz continuity is also required
@N3buchadnezzar See this
 
Oh yeah, there needs to exists a contraction for the operator. Hmm
 
@aDangerousIdea also, there is a distinction when hindus talk about "God" in general as opposed to a particular god such as "Vishnu", "Mahesh" etc. The term "God" in the former usage is a reference to "Bramha" or "Bramhan" which can be interpreted to mean the universe itself.
 
@JayeshBadwaik I agree, this ultimately reduces to semantics :-)
 
As most of philosophy...
...that's why they don't throw anything away.
 
10:20 AM
:P
right..
 
1 hour ago, by Jayesh Badwaik
this is slightly insulting to philosophers
Q: What is the difference between a mathematician and a philosopher?
A: The mathematician only needs paper, pencil, and a trash bin for his work - the philosopher can do without the trash bin...
 
Lame
Wow $e^{-1/x^2}$ actually have a fourier series, although it is ugly.
It does not have a taylor series though, since it is not analytic.
 
10:48 AM
@JayeshBadwaik $$ \sin(2z) = \frac{1}{2i}\left( 2z - \frac{1}{2z} \right) $$ ? =)
 
@N3buchadnezzar How? What happens at $z=0$?
Or do you want solutions for the equation?
 
@JayeshBadwaik No, I want to write $sin(2z)$ on complex form using $z$ and $\overline{z}=1/z$
 
\begin{align}
\sin(2z) = \frac{e^{i2z} + e^{-i2z}}{2i}
\end{align}
 
 
1 hour later…
12:01 PM
Hi all
 
12:17 PM
Hi
 
12:39 PM
hi
 
What is the notation for an nxn zero matrix?
 
0
A: Question on Congruence

AmrLet $r-s=k(p-1)$ where k is a non-negative integer (assume wlog that r>=s) Using Fermat's last theorem we know that $x^{p-1}=1$ (mod $p$) therefore $x^{k(p-1)}=1$ (mod $p$). Thus, $x^{r-s}=1$ (mod p). Now multiply by $x^s$ to get: $x^r=x^s$ (mod p)

hehe
 
1:20 PM
hello!!!
@sonicboom why not just 0?
 
@JayeshBadwaik That page has an error on it.
 
hi @cassandra0, do you remember yesterday's speech about $\sum\limits_{n=1}^{\infty} e^{-n^2 t}$ ?
 
@JayeshBadwaik It says that a function of bounded variation must have a finite number of extrema on any given interval.
 
@Nimza, yes
$t = \pi x$
 
@JayeshBadwaik However, $x\sin(\log|x|)$ is a counter example
 
1:28 PM
@cassandra0 why integral $\int\limits_{c-i\infty}^{c+i\infty} x^{-s} \Gamma(s) \zeta(2s)ds$ does converge?
 
Any tricks for making this partial fraction decomposition easier?
 
@Nimza, I don't recognize that integral
 
$$ \frac{1}{(s+2)^2(s^2+1)} $$
 
@cassandra0 $\zeta(2s) \Gamma(s) = \int\limits_{0}^{\infty} \sum\limits_{n=1}^{\infty} e^{-n^2 t} t^{s-1} dt$. So I take the inverse Mellin transform
 
ah
I can't think how to show the integral converges. I think gamma gets large as t increases in s = c+it, but x^{-s} is constant and maybe zeta small
 
1:35 PM
it's interesting
@robjohn do you know how to show that $\int\limits_{c-i\infty}^{c+i\infty} x^{-s}\Gamma(s) \zeta(2s)ds$ converges for $x>0$?
 
@Nimza, what will we get from convergence inverse mellin transform?
 
@cassandra0 representation $\sum\limits_{n=1}^{\infty} e^{-n^2 x} = \int\limits_{c-i\infty}^{c+i\infty} x^{-s} \Gamma(s) \zeta(2s)ds$
 
@Nimza $\Gamma(z)$ dies exponentially off of the real axis...
 
oh! nice
oh I was wrong I thought it blew up
 
just a sec...
 
1:38 PM
@robjohn and what about zeta?
 
@cassandra0 Take a look at this answer
 
Goodday!
 
@robjohn that's a very cute estimate! I like it!
 
hello sos440
@robjohn, in polynomial decay is n related to y?
 
Hello~
 
1:44 PM
oh I understood it now,
just applying the first result n times
 
@sos440: Greetings
 
@Chris'ssister Goodday!
 
wow @robjohn that exponential decay is really cool!
 
@Nimza I think that $(25)$ on this page gives boundedness on the imaginary axis
 
how is Gamma(x+n)/n! < (n+x)^x
 
1:49 PM
@robjohn thank you, if I'm not mistaken, it follows also from $|\zeta(x+iy)| \leq |\zeta(x)|$ for $x>1$ (this inequality is true, right?)
 
@Nimza I am not sure about that, but even if that is true, we are looking at $x=0$
Oh, I see the $c$
@Nimza Yes, that is true for $\mathrm{Re}(z)>1$
 
@robjohn nice, thank you
 

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