Mathematics - 2014-05-19 (page 4 of 5)

Sawarnik

6:00 PM

@JasperLoy You are.

Sab ಠ_ಠ

I heard that Russians are maths geniuses

Ted Shifrin

If you can compute $\chi$, @Mike, you need some complex. Then It should be algebra with universal coeff theorems.

G.T.R

@N3buchadnezzar I'm a Latex noob (I know only basic typesetting). Can you provide a link/resources to learn to write math papers with Latex ?

Ted Shifrin

Howdy @Jasper

Sab ಠ_ಠ

I know only basic stuff too @G.T.R but you could find loads of resources off google

N3buchadnezzar

6:01 PM

@G.T.R look at the stack site

user116900

@TedShifrin Hi!

r9m

@Sawarnik multiply both sides with $e^{f(x)}$ and integrate from $1$ to $5$ .. it should give rise to a contradiction ..

user116900

@G.T.R tobi.oetiker.ch/lshort

Sawarnik

@r9m I have a solution .. using the MVT :)

Sab ಠ_ಠ

^

@Sawarnik It's a corollary of MVT which uses contradiction :P

r9m

6:02 PM

@Sawarnik okay .. nice :)

Sawarnik

@Sabಠ_ಠ Umm what?

G.T.R

@JasperLoy hmmm this looks very good, thank you

Mike Miller

@Ted Okay, I'll try that. I'm not sure I have much hope though - Tor is nasty.

N3buchadnezzar

@G.T.R But for reals latex.stack.exchange has everything and mpre

G.T.R

@N3buchadnezzar I'll ask specific questions there when I get stuck (I surely will :P)

N3buchadnezzar

6:05 PM

They have questions about starting using latex as well ;)

robjohn

@N3buchadnezzar for integer $a$ probably, but I would doubt it for non-integer values

user116900

I am wondering if my thinking is right here math.stackexchange.com/questions/801829/… please comment!

Ted Shifrin

@Mike: Not when you're doing fields!!

@Sab: Email sent to you.

Pedro Tamaroff

@TedShifrin Mike is doing what? =D

Ted Shifrin

heya mr @Pedro

He's being TORrid @Pedro

Pedro Tamaroff

6:14 PM

@TedShifrin How is it going?

Sab ಠ_ಠ

Thanks a lot @Ted :) I'll do them all :D

Ted Shifrin

Well, some are perhaps not appropriate for your course, @Sab. In particular, I didn't send the first exam, which is all the beginnings of Spivak (induction, algebra properties, definition of function).

Sab ಠ_ಠ

I see some questions which look like geometric proofs.

ಠ_ಠ

Hi @Ted

Ted Shifrin

hi mr eyeglasses

Didn't Alex change your name for you the other night?

Sawarnik

6:15 PM

@Sabಠ_ಠ Did you try my questions?

@eyes Hi.

Sab ಠ_ಠ

I saved them @Sawarnik I will be doing them later today

Sawarnik

Ok :D

Sab ಠ_ಠ

:)

Mike Miller

@Ted It works for char 0. It's horrifying for char nonzero.

Sab ಠ_ಠ

@Ted The numbers are a lot like what we get for our tests :D

This would be a great practice

Ted Shifrin

6:17 PM

@Mike: Do you know the Hopf Trace Lemma? I keep thinking that should be relevant.

I hope you enjoy @Sab :)

Sab ಠ_ಠ

I will Thanks again :)

Sawarnik

@Sabಠ_ಠ Do you use Hindi or some mixed variants at home?

Pedro Tamaroff

@MikeMiller Mike

Sab ಠ_ಠ

@Sawarnik not at all, it's kreol(a variant of French)

Sawarnik

@Sabಠ_ಠ is it the native language of there?

Sab ಠ_ಠ

6:20 PM

yap

I'm currently living in cape town though, so I speak English these days

Sawarnik

Hmm.

Mike Miller

@Ted I don't... I'm sure I should. Look it up in Weibel?

Ted Shifrin

It says that if you have any chain complex, and a chain map, the alternating sum of its traces on the chain level is equal to the alternating sum of the traces on homology/torsion. Prove it! @Mike

Mike Miller

Ooh, sounds fun... but what's the chain map? You want me to try to use UCT a a chain map?

Ted Shifrin

oh, take the identity for what you want (I meant chain map from the complex to itself)

Mike Miller

6:29 PM

@Ted I don't really get it. We need to change coefficient field somehow.

@pedro wat

Pedro Tamaroff

@MikeMiller Why did you ask about the abelianization of GL?

Ted Shifrin

So take the chain complex and tensor it with whatever field you want @Mike.

user116900

Wow, I am quite happy with the lhf I just answered.

Mike Miller

@Ted Oh, I suppose just tensor with the appropriate field?

Damnit.

@Pedro No reason.

Ted Shifrin

I haven't thought about this in 40 years, @Mike, but I think it gets somewhere.

Sawarnik

6:31 PM

@JasperLoy Why do you have to announce it in the chat always?

Chris's sis

@G.T.R did you manage to compute my limit? :-)

user116900

@Sawarnik It's called making small talk.

user116900

I wish people would upvote the questions more. Sometimes they only upvote the answer.

G.T.R

@Chris'ssis not off the top of my head (as Daniel would say)

user116900

@G.T.R Are you Gabriel? Changed your username?

r9m

6:33 PM

@Chris'ssis which limit ?

Mike Miller

@Ted Oh.... so what we want is the alternating sum on homology of this chain map to be zero. Then the Euler chars will be the same.

G.T.R

But I'll get back to you when I think enough about it @Chris'ssis. And yes @JasperLoy

Chris's sis

@r9m a limit I proposed and was given on a contest. Wait.

Ted Shifrin

People like this and this (same person) truly annoy me. No effort made to learn from the responses to the first question before making the identical error on the second. GROWL.

user116900

@TedShifrin You must try not to get upset so easily, bad for health.

Ted Shifrin

6:34 PM

Oh, I see @Mike ... you're using two different chain complexes: One and the tensored one. That should be interesting :P I hadn't thought of that.

Mike Miller

@Ted they're not here to learn.

Ted Shifrin

This is why I'm going to stop answering questions and disappear ...

Chris's sis

@r9m This one $$\lim_{n\to \infty} n 2^n \int_1^n \frac{1}{(1+x^2)^n} \ dx$$

Mike Miller

@Ted This isn't possible. My chain complexes were the singular complexes, for which trace I the ifentity map is nonsense.

user116900

@TedShifrin Yeah, write more books during your retirement, lol.

Ted Shifrin

6:35 PM

Trust me, @Jasper ... That has no effect on my health. Not after dealing with my mom's dementia almost did me in. Now I'm immune to everything :D

Yeah, you need finite complexes, @Mike.

Mike Miller

@Ted Sad stuff... same thing happened to my grandmother for years before she passed.

Ted Shifrin

Singular is worthless :D

Mike Miller

@Ted Aye, there's the rub! There's no reason my manifolds are simplicial!!

Ted Shifrin

Well, @Mike, it's been better the last few years, since the Alzheimers progressed. Of course, she has no idea who I am ... hasn't really for about 3 years now.

Mike Miller

So it truly is a complicated question.

Ted Shifrin

6:37 PM

So what manifolds have cellulations? Does Hatcher address that? @Mike

Mike Miller

@Ted I'm sorry to hear that.

Chris's sis

@G.T.R Is it a joke? I hope it is just a joke. :-)

user116900

@Chris'ssis There are no jokes in this chat, lol.

Chris's sis

@JasperLoy :-)))))

Mike Miller

@Ted Every smooth manifold is triangulable. As mentioned, they are also homotopy equivalent to a finite CW-complex. But I can't find a similar result for topological manifolds, which in general are not even PL.

Ted Shifrin

6:39 PM

So you really are in Kirby-Siebenmann land ...

@Jasper: Humor is not allowed in mathematics. A reviewer of my first book told me that.

Mike Miller

Yes, I thought it was a trivial question at first!

Ted Shifrin

I'll email a retired topologist colleague of mine who liked questions like that, @Mike, and ask him.

Mike Miller

@Ted Thanks! :)

Ted Shifrin

Just did. We'll see if he answers.

Studentmath

Prof @Ted I always loved humor in my teacher's marking.

Mike Miller

6:45 PM

@Ted I'll borrow a copy of KS and see if I can read the theorems. (I have no expectation of being able to read the proofs.)

Ted Shifrin

@Studentmath: My undergrad students in my grad diff geo course over 10 years ago gave me a lovely present at the end of the year. Hand-made wooden boxes with a dozen rubber stamps in there, with all the things I used to write on their homeworks. Things like SHOW!, GRRR!, Proof?, Good!, WHY?, etc. :P

It's quite recondite, @Mike. I still say you should look at Spanier and see what he says.

I'm not at school today, or I would have.

Studentmath

Haha, awesome

Mike Miller

@Ted I did. I didn't see anything like what I wanted.

@Ted Omg, I could use those. I'm going to order a set for myself.

Ted Shifrin

I am too lazy to use them, as it isn't convenient to keep a red stamp pad next to me while I'm grading ... :P

user116900

I keep thinking naslundx is Eric Naslund, lol.

Ted Shifrin

6:49 PM

I've forgotten all the other things they said. And the hand-carved boxes Rolf made (he's now a professor in NY) are wonderful :P

Mike Miller

That's a really awesome gift.

user116900

Is Spanier the most comprehensive book on AT?

Mike Miller

Your grad students didn't like you as much, I guess, @Ted

Ted Shifrin

Actually, the only students left in the second semester were the undergraduates. Seriously. And only Rolf got a Ph.D. in math. Others became lawyers, divinity scholars, economists, etc.

The class I taught this past fall was 9 grad students. You're probably right that several of them didn't like me; but they never turned in any homework.

Mike Miller

Why not?!

Bizarre... that's how I learn best.

Ted Shifrin

6:52 PM

To get grad students to take advanced courses here, we have to offer them automatic A's (in my case, B's) if they enroll. I hate it.

They didn't care about learning ...

The five who did worked a lot of homework and came to office hours to talk to me.

@Mike: My favorite teaching eval from diff geo this spring was this: "wonderful wonderful class, I cannot express how much I loved the material and how Dr. Shifrin presented it. I can't imagine what we will do without him after next year :(." :D

Mike Miller

Garbage. You should take classes that will - however tertiary - help you in your future endeavors. (And I would guess that diff geo is good for any mathematician.)

Ted Shifrin

When you get to LA, @Mike, you'll discover that attitude is not universal.

Mike Miller

it's bizarre! And if you are taking a class, you should darn well learn it! Or else what's the point?!

@Ted I share that student's opinion :)

user116900

Oops, I am guessing that Mike got into UCLA...

Ted Shifrin

There are a half dozen universities in LA, @Jasper!

You have no basis on which to share, @Mike :D

user116900

6:55 PM

@TedShifrin Well, I only know 1, lol.

Pedro Tamaroff

@MikeMiller @TedShifrin

Ted Shifrin

Cal Tech, USC, UCLA, Pepperdine, and several others slightly further out.

Mike Miller

@Ted I don't need much of one.

@Jasper I'll put my affiliation on my profile when I become a student... patience :)

@Pedro

Studentmath

Anyone with some knowledge over NSD in probability?

user116900

I will try to apply to UCLA if I do apply for grad school, then I can see robjohn, lol.

Ted Shifrin

6:57 PM

No, robjohn has left UCLA ... He's now corporate :P

Pedro Tamaroff

@MikeMiller A minimal prime in $V(\mathfrak a)$ corresponds to an irreducible component in ${\rm Spec}\; (A/\mathfrak a)$ amirite?

I think I am.

Just checking.

The component being $V(\overline {\mathfrak p})$.

Mike Miller

A minimal prime? Not a maximal prime?

Anyway, you know more than me, so I can't answer.

Chris's sis

@r9m let me know if you find a nice solution there.

r9m

@Chris'ssis i'm on it :)

Chris's sis

@r9m OK. Have fun! :-)

Studentmath

7:15 PM

Emrpf, I have a feeling that $0.84+0.01*\frac{0.0005}{0.0028}=\frac{20.208}{x}$ should get me to x=24. But every calculator leads me to x=24.061, but it rounds it up oddly..

Any way I can show up it is indeed 24?

Pedro Tamaroff

@Studentmath Your calculator might be retarded. Maybe.

Mike Miller

Consider voting to reopen this question.

Studentmath

Or maybe I am, should consider it too.

user116900

@Studentmath Don't use a calculator.

Studentmath

Pedro, that freak deer show is creepy!

user116900

7:19 PM

@Studentmath Also, check its rounding off settings.

Studentmath

It's a bit hard to solve it without one @Jasper. But I think the rounding off settings are at the sixth digit or so. And this has infinite digits.

It's 24, I know it for sure. Will just state it and hope no points are taken off.

Pedro Tamaroff

@Studentmath It's good.

user116900

@Studentmath Just observe that 28 times 3 is 84.

Pedro Tamaroff

@Studentmath Just use usual fractions.

Studentmath

Jeez I am retarded.

Thanks @Jasper @Pedro

I am just like all the recent k-12 graduaters, sucked up into the calculators... Pedro, it's a -kids- show about a post-acopocalyptic world with creepy tragic characters

user116900

7:22 PM

@Studentmath I am not really thinking here, just throwing in a random observation.

user116900

I still suspect that anon=Andre Nicolas, lol.

Pedro Tamaroff

@Studentmath Yeah, kids be different these days.

user116900

@PedroTamaroff The increasing use of computers is making them more stupid.

Mike Miller

Anon's just a kid.

user116900

So might AN, lol.

user116900

7:25 PM

They even start with the same letter, lol.

Daniel Fischer

@MikeMiller You may now post an answer, if you have one ;)

I wish I did.

@DanielFischer

@PedroTamaroff

@MikeMiller Yes?

7:34 PM

@PedroTamaroff Que?

user116900

@all

Mike Miller

@PedroTamaroff I wanted to ping someone too, but I didn't want to ping anyone I liked, because it might annoy them.

user116900

Why no response?

Pedro Tamaroff

@DanielFischer Hello. I've been wondering about this for some days. Suppose $f_i=\sum a_{ij}e_j$ in $\Bbb Z^n$. Let $K$ be the submodule generated by the $f_i$. Let $d=\det (a_{ij})$. Suppose $d\neq 0$. Then using Smith's Normal Form one can see $\Bbb Z^n/K$ has $d$ elements.

Alyosha

@BarlakaSen In the comments section here cp4space.wordpress.com/2012/08/29/elliptic-curve-calculator, it's noted that 'For a real number x, the numbers {…, 0.1x, x, 10x, 100x, …} are all represented by the same point on the elliptic curve. It’s a consequence of the Weierstrass P-function being doubly periodic — everything wraps around.' Do you have any idea what this means and why it's true?

Pedro Tamaroff

7:37 PM

I am trying to prove something similar for $A=\Bbb Z[i]$ and $I=(a+bi)$; that $A/I$ has $a^2+b^2$ elts.

And I should use SNF too.

Daniel Fischer

@PedroTamaroff The simple way is to note that $(a+bi)$ is generated by $a+bi$ and $-b+ai$ over $\mathbb{Z}$.

Pedro Tamaroff

@DanielFischer Right, I should look at $\Bbb Z[i]$ as a $\Bbb Z$-module?

I see.

Daniel Fischer

@PedroTamaroff One possibility.

Pedro Tamaroff

So I am looking at the submodule generated by $(a,b)$ and $(-b,a)$ in $\Bbb Z^2$.

Daniel Fischer

Yup.

Pedro Tamaroff

7:40 PM

Ah, that's it. Thanks, @DanielFischer.

Matt N.

@DanielFischer Congratulations!

Daniel Fischer

@MattN. Thanks.

Matt N.

@JasperLoy Ello. Here iz response.

Pedro Tamaroff

@DanielFischer I added it to this.

Matt N.

@DanielFischer I'm very close to getting an algebraic topology badge. The sad thing is I don't actually know much algebraic topology.

user116900

7:48 PM

math.stackexchange.com/questions/801934/… Why does Don say it is a parabola, it can be anything.

Daniel Fischer

@MattN. So what? I have a gold C++ badge on Stack Overflow.

(People way too often double-tag C and C++)

user116900

@MattN. It's OK. There is one user with more than 100k even though they are not too good in math, lol.

Matt N.

@DanielFischer I wouldn't know. I try to stay away from C++. I like C and Objective C. And Python.

@JasperLoy Don Antonio?

user116900

@MattN. Nope, not referring to him.

Daniel Fischer

@MattN. Not yet above 100K.

N3buchadnezzar

7:51 PM

100 rubiks cubes solved in 58 minutes

wop wop

user116900

I can't say more in this chat.

Matt N.

@DanielFischer A friend of mine got endorsed for Perl (on linkedin) even though he doesn't know any Perl by a graphic artist he used to work with. I think if someone endorsed me for Perl I'd endorse them for MS Paint.

@JasperLoy What's a vertex of a graph?

user116900

@MattN. Hmm, actually I don't know.

Matt N.

@JasperLoy And why do you think that graph is not a parabola?

user116900

@MattN. I mean one cannot assume it is of the form ax^2+bx+c.

Matt N.

7:53 PM

@JasperLoy Ok. And why not?

(I'm not disagreeing with you, merely curious)

user116900

@MattN. Because there is no reason to, I think. It could be given by some strange function.

user116900

That's all, I have no other reasons.

Matt N.

I imagined the lines continue (in the obvious smooth way) towards minus infinity.

user116900

For example, it could be part of a sine curve...

Matt N.

Given the level of the question I would assume not...

It looks like high school maths.

user116900

7:56 PM

I would closevote it as not a real question.

Matt N.

It seems real to me.

user116900

But assuming it is a parabola, then Don is obviously correct, lol.

Matt N.

Yes : )

Now I'm wondering who the user is who can't do maths.

But given the narrow choice there is but one possibility.

user116900

...

user116900

No comment.

Matt N.

8:00 PM

Yeah. I know. But who cares.

user116900

Hehe.

Matt N.

I should go. Got to feed the cats and stuff.

user116900

C u

user116900

I am going to bed.

Mike Miller

@TedShifrin My question was super trivial. I'm annoyed.

Chris's sis

8:04 PM

Oooo, what a great question I created!

Pedro Tamaroff

@Chris'ssis Yes, everything your create is awesome.

3

Chris's sis

:15609078 Tired here. (4 hours of sleep?)

Sawarnik

@Chris'ssis Then go to sleep! Ok.

Chris's sis

@Sawarnik Later ...

@PedroTamaroff Yes, it's true.

Balarka Sen

The guys in homotopy theory chat really know their mathematics.

@Alyosha First, my name is not Barlaka. About the comment, I have a shrewd idea, yes.

Pedro Tamaroff

8:13 PM

@BalarkaSen Bakarla?

Balarka Sen

No.

Pedro Tamaroff

Blakara? Barlaka? Bakalar?

Daniel Fischer

Ralabak

Balarka Sen

Hey, cut it out.

Pedro Tamaroff

Your name is super shuffable.

Kablara.

Mike Miller

8:14 PM

Ralabak sounds like it could be a good supervillain name.

I'm imagining a wizard-type dude, calls himself "The great Ralabak."

Balarka Sen

@MikeMiller Yes, Like Lord Voldemort.

Daniel Fischer

@PedroTamaroff Yours can only generate "O Derp"?

2

Mike Miller

Nah.

Pedro Tamaroff

And when the name is said, it should be "RrrrrralabaKK."

@DanielFischer Ordep, too.

Mike Miller

Who's the guy that always called you Ordep?

Pedro Tamaroff

8:15 PM

And Epord?

@MikeMiller I think it was Anthony.

Sawarnik

@BalarkaSen Bakra-la? (:P)

Balarka Sen

@DanielFischer Free group on a single generator.

Mike Miller

@PedroTamaroff No, it was some kid. Anthony is cool.

Pedro Tamaroff

@MikeMiller Dunno then.

Balarka Sen

@Sawarnik Arnik Saw. Some device to cut open trees.

Pedro Tamaroff

8:21 PM

nypost.com/2014/05/18/…

Balarka Sen

@Sawarnik is annoying sometimes.

Sawarnik

@BalarkaSen Pedro would be thinking, why don't you write ' he is annoying everytime'.

Balarka Sen

@Sawarnik Write? Write what?

Ah.

@Pedro You think @Sawarnik is annoying?

Pedro Tamaroff

@BalarkaSen Don't drag me into this.

Balarka Sen

Sure. I'll better get out too.

Sawarnik

8:26 PM

:D And me too.

Balarka Sen

Hello professor @TedShifrin

Ted Shifrin

@Mike: Pray tell.

Hi @Balarka

Mike Miller

@TedShifrin I'm no longer convinced. The answer is here. He says roughly what I already knew and then vaguely talks about a universal coefficient theorem argument.

Ted Shifrin

Well, I mumbled that much.

Mike Miller

I think I know the argument he means (Hatcher uses an argument like that on p249 to show that the Euler char didn't depend whether or not we took coeffs in $\Bbb Z$ or $\Bbb Z_2$.

But I'm chasing the details and I don't think it will work.

I think the fact that the latter field is $\Bbb Z_p$ is key.

Ted Shifrin

8:31 PM

His ANR argument to get a finite simplicial complex was the part that escaped me.

Mike Miller

@TedShifrin Yes, I'm annoyed I didn't realize that. But it's not a def. retract. of a simplicial complex. It's a retract.

All that gives us is that $H_i$ is finitely generated. Nothing more.

Chris's sis

Newly created & marvellous (it's required an elementary proof without integrals) $$\lim_{n\to\infty} \frac{\displaystyle \binom{n}{1}-\frac{1}{2^2}\binom{n}{2}+\frac{1}{3^2}\binom{n}{3}-\cdots+(-1)^{n-‌1} \frac{1}{n^2}\binom{n}{n}}{\log^2(n)}$$

I love this one.

Ted Shifrin

@Mike: See Ex 1 on p. 267.

Mike Miller

@TedShifrin The proof I have in mind only seems like it could possibly work for finite fields.

Ted Shifrin

I know I did this in grad school.

Mike Miller

8:40 PM

@TedShifrin You're right. It's an exercise.

So the proof is not like one he gives earlier (with Coho-UCT), but rather one with Homol-UCT.

Ted Shifrin

Yes, I said UCT first thing, hours ago ;)

Mike Miller

@TedShifrin I know. And I said "bah!" because Tor.

@TedShifrin Okay, I proved it. I should have just not been a dumb baby.

Balarka Sen

@RandomVariable Can you do me this integral complex analytically? $$\int_0^\infty \frac{\log^3(1+z)}{z(1+z)} dz $$

Sawarnik

@BalarkaSen Is there any function such that $f(x+y^2)-f(x)\ge y$? It won't be differentiable, and is increasing seems obvious, though.

Balarka Sen

@Sawarnik For all $x, y$?

Sawarnik

8:46 PM

@BalarkaSen Yes, and they are real.

Balarka Sen

@Sawarnik I have no idea.

Ted Shifrin

Dumb baby? That's star-worthy! @Mike

Mike Miller

@TedShifrin Well, you're free.

Chris's sis

Ooooo, I go further and claim this ...

Ted Shifrin

So what obvious thing were you missing? @Mike

Chris's sis

8:49 PM

@r9m don't miss this new one (elementarily done) $$\lim_{n\to\infty} \sum_{k=0}^{\infty} \frac{\displaystyle \binom{n}{1}-\frac{1}{2^k}\binom{n}{2}+\frac{1}{3^k}\binom{n}{3}-\cdots+(-1)^{n-‌‌1} \frac{1}{n^k}\binom{n}{n}}{\log^k(n)}=e$$

Ted Shifrin

Oh, the Tor stuff you were worrying about telescopes away?

Mike Miller

@TedShifrin I just haven't done a lot of Tor computations like I have Ext, and I seem to recall them being more difficult.

@TedShifrin Yes - it gives nontrivial summands in $H_i$ and $H_{i+1}$. But the number of such summands is equal.

Ted Shifrin

yeah.

Mike Miller

I'm quite frustrated I didn't just see that. I need to do more Tor computations.

Ted Shifrin

I remember this from 40 years ago?

Mike Miller

8:51 PM

@TedShifrin Now you're just poking fun at me.

Ted Shifrin

No, just saying I'm old and haven't forgotten everything ....

Mike Miller

OK. Well I'm going to go do some math.

Thanks for your time.

Ted Shifrin

Ok. I'm going to NCAA tennis semifinals.

Balarka Sen

@TedShifrin Are you familiar with Arnold's proof of insolvability of quintics?

Nevermind @RandomVariable. I just did it.

Random Variable

@BalarkaSen Using contour integration?

Balarka Sen

9:03 PM

@RandomVariable No. Real analysis. But I'll do it contour integrally too.

Right now, I think I need some sleep.

N3buchadnezzar

@RandomVariable What analysis books do you like / preffer / use?

Random Variable

@N3buchadnezzar Complex or real?

N3buchadnezzar

I'd guess both

Sawarnik

@BalarkaSen Too early for you.

N3buchadnezzar

@Sawarnik Hush, my son.

Random Variable

9:11 PM

@N3buchadnezzar In terms of complex analysis, I keep going back to Gamelin or Ahlfors. But I've been wanting to get the book from Stein and Shakarchi since they have a chapter on elliptic functions.

N3buchadnezzar

=) I've just used gamelin, I can see that it is good on the contour analysis bit, but overall I have not been a fan.

@RandomVariable Stein should be good, but perhaps a bit too technical ?

Random Variable

9:28 PM

@N3buchadnezzar Usually when I'm confused about something related to complex analysis, I ask @robjohn like 5000 questions rather than turn to a textbook. He's probably sick of me. :)

N3buchadnezzar

I do the same:p

Balarka Sen

@Sawarnik I know. I realize that sooner of later.

@RandomVariable I fancy a whole book on elliptic functions.

It's pretty broad of a branch.

Oh, and by the way, that integral there can be converted to usual zeta-gamma one by making a logarithmic sub, @RandomVariable

Random Variable

@BalarkaSen Where did that integral pop up?

Balarka Sen

Dunno. Someone gave me.

N3buchadnezzar