00:00 - 20:0020:00 - 00:00

12:01 AM
Let's see if this works. On main you can use AMScd to make commutative diagrams. $$\require{AMScd}\begin{CD}A @>f>> B \\ @VVV @VVV \\ X @>>g> Y\end{CD}$$
Ah, it works in chat too.

Ah think I found my answer in an old paper of Khovanov's
He calls the cubes skew-commutative

Indeed, looks promising.

12:21 AM
\begin{align} \prod_{k=0}^\infty\sqrt[\Large2^k]{1+\frac{2}{2^{2^k}+2^{-2^k}}} &=\prod_{k=0}^\infty\left(\frac{\left(2^{2^{k-1}}+2^{-2^{k-1}}\right)^2} {2^{2^k}+2^{-2^k}}\right)^{1/2^k}\\ &=\prod_{k=0}^\infty\left(\frac{\left(1+2^{-2^k}\right)^2}{1+2^{-2^{k+1}}}\right)^{1/2^k}\\ &=\prod_{k=0}^\infty\frac{\left(1+2^{-2^k}\right)^{1/2^{k-1}}}{\left(1+2^{-2^{k+1}}\right)^{1/2^k}}\\[9pt] &=\left(1+2^{-2^0}\right)^{1/2^{-1}}\\[24pt] &=\frac94 \end{align}

12:32 AM
@robjohn Hello!

@Charlie hi, how are you?

@robjohn I'm fine, and you? :)

@Charlie pretty good. Getting ready to take the dog for a walk.

@robjohn great!
btw, I like when @anon explains non-math stuff. It's funny.

12:46 AM
@PeterTamaroff I hear a ping, but I don't see who pinged

@Charlie Should I call you sillypantalones to wear your mysterious anger away?

@PeterTamaroff I'm listening to things...

:P'

@Charlie You are a mystery.

12:55 AM
@PeterTamaroff yes, I am
@PeterTamaroff what made you think it?

@Charlie Common sense.

@PeterTamaroff hmm

1:11 AM
Aren't we all? @Charlie, you feeling better?

@TedShifrin Yes, very, Thanks for asking! and how are you?

Just fine, thanks.

@TedShifrin :)
@PeterTamaroff I couldn't find
i was looking for one specifically
After reading 40 pages of starred messages...
look what i've found, @PeterTamaroff:
Mar 16 '12 at 16:36, by anon
@Kanna: When people downvote your posts arbitrarily just imagine them in their underwear.

1:27 AM
still wonder how that user knew what bdate I have on the mse system

@anon maybe he's looking over your shoulder right now
@anon how are you
?

pretty good

@anon good to know

1:42 AM
@TedShifrin Hey are you around?
I have a question
@anon Hey

yea

I have a question
consider $G = \{\left(\begin{array}{cc} y & x \\ 0 & 1 \end{array}\right)$
with $y > 0$
as a topological space just the upper half plane in $\Bbb{R}^2$ @anon
The book I am reading says that the left invariant measure on $G$ is given by $y^{-2} dydx$
@anon what does $y^{-2}$ there mean?
I assume $dydx$ is just the ordinary Lebesgue measure

y^-2 means y^-2
so if you wanted to measure a set A, you would integrate y^-2 dydx over that region

Hmmmm
Ok if $A$ is the unit square with vertices $(0,1),(0,2),(1,1),(1,2)$

for example an invariant measure on the topological group $\Bbb R^\times$ is $dx/|x|$, so $\mu(A)=\int_A\frac{dx}{|x|}$ for any measurable $A\subseteq{\Bbb R}^\times$
you can check by substitution that satisfies $\mu(cA)=\mu(A)$ for any real $c\ne0$

1:49 AM
@anon Is that something similar to the "Haar invariant measure"?

@user38268 then that would be $\int_0^1\int_1^2y^{-2}dydx$
@PeterTamaroff same thing

Hmm this dx thing is confusing me because I thought to define the integral we have to define a measure first.

@anon Oh. =)

@user38268 we're defining one measure in terms of another

@anon You're saying from the Lebesgue measure we define the new measure above.

1:50 AM
yes

@user38268 They just defined it, didn't them?

for example one may write $d^\times x=\frac{d^+x}{|x|}$ or such things to distinguish if desired

@anon Why the absolute value? I heard people talk about how the $\Gamma$ function is $t^{x-1}e^{-t}$ w.r.t. the Haar invariant measure.
Or is it just the case $\int_0^\infty$ in $\Gamma$ leaves it $>0$?

the topological group $\Bbb R^\times$ includes negative numbers.

@anon So to see left invariance of my measure I need to know $\int_{gA} y^{-2} dxdy = \int_A y^{-2} dx dy$ for any $g \in G$

1:53 AM
correct

goodnight all

night
@PeterTamaroff so yes, if you restrict yourself to positives the |-| is superfluous

@TedShifrin Hey.

Rehey

I'm rading your book at the moment. I should get a schedule to study! =D

1:59 AM
Not to mention your actual classwork :)

@TedShifrin Well, that I work on, mostly at uni, worry about it there. It isn't too challenging, yet.

Ah ....

@TedShifrin I guess I will have to worry more when I start studying stuff I don't know about Linear Algebra, dunno.,

My students are freaking out with closed sets, interior, and frontier points.

But then again, I am finding stuff easier than it was some months ago, so maybe it will be just fine.
@TedShifrin Why? Did you show them metric spaces before topological ones?

2:02 AM
We're only in $\Bbb R^n$ in my book. Hard enough.

@TedShifrin Oh. Why are they freaking out?

Did you sort out Jean-François?

@TedShifrin Dunno, he went away.

Ah. Learning logic and language is hard for most.

@TedShifrin Sure. But maybe there is something they don't grokke or feel is unnatural?

2:08 AM
Evening!

Today a physics student was freaking out about the theorem that given a set of generators in $V$ and another in $W$ of the same size, there is one and only one linear transform with $T(v_i)=w_i;i=1,\ldots,m$.

Then I threw weirdness of limits in $\Bbb R^2$ at 'em. Start derivatives tomorrow ...

(I mean base in $V$, gens in $W$)

Hmm, two people randomly answered some old question of mine today. I wish I had that a couple of months ago.

I tried to tell him it was something "natural", made him think how base make up everything, and uniquely, thus a LT is fully determined by its values on the base, and such.
@FernandoMartin Hey.
@FernandoMartin Which one?

2:10 AM
Some stuff I wanted to know about categories
12

Has anyone come across a cheat sheet containing basic properties of the most well-known categories (i.e. does it have (co)products, (co)equalizers, (co)limits, etc?)?

Well, yeah, linear maps are abstract for most physics students. Wait for quantum mechanics :)
Count me out @Fernando.

Haha, it's okay, I needed that for revising for an exam I took a while ago; it's no longer needed.
But it's weird that someone got to answer it after such a long time.

Sometimes things go unnoticed ...

@PeterTamaroff: Could you find a hostel?

@FernandoMartin Yes, Freedom.

2:15 AM
That's the same we're staying in.
Did someone cancel their reservation?

Where are you traveling?

(is that correct? is it 'their'?)

@FernandoMartin Wouldn't know. I am sharing a four person's room with a girl named Yamila.
@FernandoMartin (yes ;) )

@PeterTamaroff No clue
Well, that's good
@TedShifrin: There's some national inter-university meeting of sorts

@TedShifrin It's the Annual meeting of the Arg. Math. Association, and we're going to the "students" part of it.
(at least me)

2:18 AM
Awesome. I just was asked to give a lecture for undergrads at the southeast Math Assn of America meeting this spring. Have to come up with something ...

@PeterTamaroff I thought the courses were longer
There's a lot of free time actually

@FernandoMartin Micro-tourism!

Haha, yup. I've never been to Rosario before.

@FernandoMartin Neither have I. If the weather is still shitty like this, I have some good games, an XBOX controller and an i7 =D

Hahah
It was a total pain to go to class today.

2:21 AM
@FernandoMartin Yeah. Lights went out about 4:30 pm.

Oh. Didn't know that, I arrived at 5
Well, gotta go
See you!

@FernandoMartin Bye byes.
@MarianoSuárez-Alvarez Hey! Fernando just left.

Hi
Fernando?
Ah

@MarianoSuárez-Alvarez Did the power out affect you today?

2:25 AM
@MarianoSuárez-Alvarez Oh. I thought of the CS guys.
@TedShifrin I can't find in your book where you define orientation (for Green's theorem.)
@MarianoSuárez-Alvarez I'll have to get a different hostel for Friday, since Freedom is full. =/

Counterclockwise, with region on your left.

@PeterTamaroff Ouch
when are you leaving?

@MarianoSuárez-Alvarez I don't know. Maybe I should leave Friday night? When does it end?

Can someone give me the reasoning for why the reputation incentives on this site are biased towards answers over questions?

official activities end on Fri

2:28 AM
@Alexander Biased?

there is a lunch on saturday, I guess

@MarianoSuárez-Alvarez Ah. I guess I will figure it out.

@PeterTamaroff I guess I just mean that an answer point is 2x a question point and there are many more badges for answers than questions.

@Alexander Maybe one shouldn't worry that much.
@TedShifrin "...with region on your left." What do you mean by that?

@PeterTamaroff I think you're right. It doesn't really bother me, I was just curious as to whether there was some philosophical reason.

2:31 AM
@Alexander, the idea is that what makes the site valuable is the answers
a place where all people do is ask questions is not that useful!
on MO originally votes on questions and on answers were all 10 pts, iirc
when we moved here, reputtion was recalculated according to the new rules
and I don't think it changed anything significantly
people with a significant number of good questions has an even more significant number of answers
Here, I would say that the most prolific answerers have very few questions

@MarianoSuárez-Alvarez Yes, that's true. But some people are here just to teach, say Brian M. Scott. =)

@MarianoSuárez-Alvarez That makes good sense. My original thinking was that a good question causes much more activity in the site users than a good answer, but I can see both sides of the coin.
*from the site users

@Alexander, but you want to keep those that are good at answering hooked, too
more so than those asking
because otherwise who will the askers ask?
people who can answer questions are rarer than those who can ask questions, at least at the level of M.SE
on MO it is a bit different: asking a good question at that level is also rare

@MarianoSuárez-Alvarez Yes, I can see that. Also, it appears that there are very few dedicated question askers. The answerers form the bulk of the community

I suppose I need to think of some good questions to ask. Even better if I don't know the answer. :)

2:44 AM
@TedShifrin I have a question.

LOL ... I didn't answer the last one, but you can figure it out.

We have seen that whenever $ω = df$ for some function f, it is the case the integral along any closed curve of $ω$ is $0$. So certainly we expect that the size of $d\omega$ on a region will affect the integral of $\omega$ along the boundary of that region.
@TedShifrin (The one of $\Bbb C^3$?)

No, I was meaning your question about "on the left."

@TedShifrin Oh.
@TedShifrin I don't understand what you mean by that.

Walk around with your body along +z-axis. Put your left arm out. It should point into the region. I can say this all purely mathematically. That is done in 8.6 for Stokes's Thm and boundary orientation on manifolds with boundary.
Walk around the boundary, I mean.

2:50 AM
$$n^{x}+(d(n)-1)n^{x/2(1-\frac{1}{d(n)-1})}\leq \sigma_x(n)$$
n>1

@TedShifrin Can I get the mathematical def?

You want the outward normal to the bdry followed by tangent to the bdry to give the standard orientation on $\Bbb R^2$.

@TedShifrin Right, what I guessed. Though it isn't immediate to see, maybe, how a surface is oriented?

Green is only for regions in the plane.
You need Stokes in general.
Oh, so 8.5. I lied.

@TedShifrin Tsk tsk!

2:57 AM
Gewiß.

@AlexYoucis It was "the grave diggers" it seems.
@TedShifrin "Surely"?

is the product of two independent variables also independent

For sure :)
Certainly.

why

Sorry, @Elp, I wasn't answering you.

3:02 AM
@TedShifrin bummer
@TedShifrin well if this questions piques your interest:

If X  N(0; 1), Z is independent of X with P(Z = 1) = 1 - P(Z = -1) =
0.5, and Y := XZ, then X and Y are uncorrelated but not independent.
True or false

I'm no expert on probability ...

the wikipedia article does even discuss independence of the product. en.wikipedia.org/wiki/Product_distribution
and bummer again

Seems unlikely that $XY$ and $X$ would be independent.

i agree but need to explain
I'm going go with that the joint pdf can't be broken up into discrete things

I'm not sure I've ever voted to reopen a question in M.SE.

3:21 AM
@Thomas: What's your point?

No point, just an observation.

Oh, ok ... I'm getting more and more annoyed by both questioners and answerers, so I might vote to close me.

I vote for you to stay open to suggestions :-)

You're out of order, @cyber! :)

@TedShifrin :)

2 hours later…
5:22 AM
The Civic Duty badge wasn't too painful to get. I don't, however, know if I'll ever be able to get that Electorate badge. Certainly I'm much closer to Copy Editor.... These badges are strange.

5:32 AM
Greetings
@robjohn congratulations! I'm glad you exist! :-)
I've rarely managed to find someone that really enjoy calculus questions.

5:55 AM
Jay Hanlon on September 16, 2013

Stack Overflow officially launched on September 15, 2008. In five short years, you’ve answered over 5 million questions on more than 100 sites, and helped hundreds of millions of people find the answers they needed. Today, we want to celebrate how, together, we changed one small corner of the Internet for the better.

We want to hear your stories about how someone on Stack Exchange helped you.

Before it went into beta, stackoverflow.com had a comic on the landing page that came to symbolize what we were setting out to do: …

@cyberskull They missed me here " People came because they wanted to help other people, because they needed to learn something new, or because they wanted to show off the clever way they’d solved a problem."
@cyberskull I've come to share beautiful things. Surely, I also learned a lot of nice things. :-)

6:18 AM
@robjohn crazy the last question, isn't it? :-) How would you rate it as difficulty?
@robjohn I wonder if it's a proper question for a high school exam.
The students I talked to all failed to come up with a solution. (unfortunately)
@robjohn well, if one has in mind the telescoping idea, all becomes very easy (or easy).

7:08 AM
@Chris'ssis :D

@cyberskull :D Great one, how are you this morning? :-)

@Chris'ssis Fine thanks, how are you greatest one?
:-)

@cyberskull Fine here too! Thanks! I'm writing down some proofs and listen to some music.
:-)

@Chris'ssis cool

@cyberskull you're the greatest! :-)

7:12 AM
@Chris'ssis I'm just a banana :-)

@cyberskull hehehehe :-) Then stay away from monkeys! :D

7:40 AM
@cyberskull You're a banana too?

8:03 AM
@Chris'ssis like many puzzles, once you know it key, all the pieces fall together.
I'm afraid that someone has massively upvoted my posts again. It really messes with your reputation when serial upvoting gets corrected. I'll have to see what happens.

@robjohn how many upvotes? Maybe not so many ...
@robjohn hmmm, OK.

8:59 AM
@robjohn I'm thinking to create some nice infinite double series that can be easily done by squeeze theorem. Do you have such a double series in mind? I simply don't remember any example of this sort.

9:13 AM
@GustavoBandeira We come in bunches :-)

1 hour later…
10:24 AM
hi guys

10:41 AM
hi guy

how is it going

fine thanks, how are you?

fine too

:-)
1 hour later...

2 hour later…
1 hour later…
0 hour later
...

10:45 AM
:D

good bye

11:32 AM
later pal
9000 hours later...

11:54 AM
@cyberskull sounds messy

hehe, I just found a clever way to prove that a certain double series converges. :-)

1 hour later…
12:56 PM
hi

Can someone help me with this ? math.stackexchange.com/questions/496378/…

@Carpediem what is wrong with the hints already given?

@I don't understand them

@Carpediem which part?

@TobiasKildetoft I am unable to make the connection with the problem

1:05 PM
@Carpediem you don't see how to connect the fact that a matrix is invertible iff the columns form a basis to this problem?

@TobiasKildetoft I know that the rank of a matrix i.e. Rank of the family of colums equals the format of A iff A is invertible

@Carpediem is the problem that you cannot connect the fact to the problem or that you cannot prove the fact?

@TobiasKildetoft I don't understand how the above property is enabling us to determine the number of bases of V

@Carpediem each basis of $V$ determines a matrix by having the basis vectors be the columns

$$\Rightarrow \cdots \cdots$$

1:33 PM
@robjohn :D

1:54 PM
I need to show that the set of nxn matrices of trace equal to 0 is supplementary to a subspace of $M_n(\mathbb{K})$.
I wanted to start by determining a set such that the intersection with the above set equals {0}
So I need to find a condition in addition to tr(A)=0 which would imply A can only be the null matrix

quora.com/Mathematics/What-do-grad-students-in-math-do-all-day the top answer is a great read for anyone that has been, or may one day want to be, a mathematics grad student.
4

Can anyone help ?

2:37 PM
@DanielRust Yes. I've seen it.

2:53 PM
@Carpediem what does "supplementary to a subspace" mean?

complementary

@anon algebraic geometry is hard man

@Carpediem Here's a hint: in F^n the set of vectors (a,b,...,c) with 0=a+b+...+c=(1,1,...,1)*(a,b,...,c) is complementary to the subspace generated by (1,1,...,1). Now think about F^(nxn)

@anon
m

eh?

3:15 PM
in Tagging, yesterday, by Martin Sleziak
There is one question tagged . Do we want this tag?
Ooops! You can ignore what I wrote above. I did not notice that Asaf has already removed it.
Sorry.

np, thanks for the news flash :)

4:11 PM
good evening folks
anyone who can suggest a good complex analysis book?
@DanielFischer for example

user87637
@DanielR Eberhard Freitag's Complex Analysis 1 and 2.

@DanielR Did you have a look at questions tagged complex-analysis+reference-request?
For example, Complex Analysis Book.

thanks guys
@MartinSleziak I realize now that my question is like being in Manhattan and asking for the road to New York.

4:26 PM
It's still possible that some people here in chatroom could tell you their personal experience with certain book.
That would be certainly useful.
I know very little about complex analysis, so you can't expect from me anything better than giving you a link to a question on the main...

1 hour later…
5:40 PM
How to evaluate $$\int^1_0\log^i t^{-1} dt$$

What is $\log^i$ ?

natrual log of x raised to the power of i the imaginary unit

@Alizter is that $$\int_0^1 \left(\log t^{-1}\right)^i dt$$

@DanielR Yes, sorry about my unclear notation

what happens when $t = 0$? looks a bit fishy...

5:45 PM
it was my attempt at evaluating $i!$ or $\Gamma(i+1)$ or $\int^\infty_0x^ie^{-x}dx$

ah, well it seems to converge, so everything is fine obviously
the answer is $(-1)^i \Gamma(1-i)$ but i have no clue how to get there :)

overall I am trying to prove that $$|i!|=\sqrt{\pi\text{csch}\pi}$$

6:06 PM
@Alizter $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z}$. Insert $z = i$, you get $\Gamma(i)\Gamma(1-i) = \frac{\pi}{i\sinh \pi}$. Multiply by $i$ to get $\lvert \Gamma(1+i)\rvert^2 = \Gamma(1+i)\Gamma(1-i) = \frac{\pi}{\sinh\pi}$.

1 hour later…
7:16 PM
@DanielR: I really liked Remmert's books on Complex Analysis

@robjohn I plan to compute (26) elementarily (POTE (problem of the evening)) mathworld.wolfram.com/InfiniteProduct.html
@robjohn I mean I won't touch any form of the gamma function. (I wish myself have some luck - - 99.99% I'll be successful)

@Chris'ssis Is that the one for today?

@robjohn it's for this evening only. :-)
(I already attended others)

Did I miss the others?
I was away working on a nasty change of variables problem

@robjohn no. I didn't post other problems.

7:22 PM
so do you mean (24) of (25)?

@robjohn no. Just $(26)$. By the way, you should like to take a look here mathematica.stackexchange.com/questions/32513/…

$$\prod_{n=2}^\infty\left(1-\frac1{n^4}\right)$$
Oh, I thought you wrote (25)

@robjohn $$\prod_{n=2}^\infty\left(1-\frac1{n^6}\right)$$

So (26) is the POTD

@robjohn Yes! :-)

7:25 PM
@Chris'ssis: How did you come up with that double series?

@FernandoMartin personal research

Yep, but in what context?

@FernandoMartin in the context of creating things and doing some research. I just asked myself questions...

@Chris'ssis Note that (27) on that page is wrong. It is 0. They meant to start at $k=2$

@Chris'ssis That's ok

7:27 PM
@robjohn it seems so.
@robjohn Another beautiful thing I did today was to prove the convergence of $$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{kn(k+n+m)^2}$$ in an elementary way.

@FernandoMartin Thanks! Do you mean Theory of complex functions or is there one entitled Complex Analysis?

@Chris'ssis The convergence or the value?

@DanielR: There are two Springer books by him
Let me check their names...
There's the one you said, which is the first one in the series, and the other one is called "Classical Topics in Complex Function Theory"
The second book is considerably more advanced than the first, I just covered the first chapter of it.

@FernandoMartin Right, I'll check them out, thanks! They seem to be for graduate studies, and may be a bit too advanced for me.

@DanielR: Is it for a first course in CA?
I'm an undergraduate as well and I used them for my first course, they're pretty accesible

7:34 PM
@FernandoMartin No, it's for self-studies. Or "for fun" is closer to the truth.

I see
CA is pretty fun
One thing I really liked about Remmert's books is that they give the historical context in which stuff was discovered

that sounds good

@robjohn the convergence only. I know to compute it.

@Chris'ssis How can you compute it without knowing it converges?

@robjohn I meant that I know to prove the convergence and then to compute it. But my requirement above was only related to the convergence.

7:38 PM
@Chris'ssis but you know the value?

Surely.
@robjohn $\pi^4/45$

@Chris'ssis If you can compute that and it is all positive terms, how could the convergence be at issue?

@robjohn I only imagined a question that sounds like that "Prove the convergence of the triple series ..."

@Chris'ssis The convergence is pretty easy.

@robjohn sure, it's not hard.

7:46 PM
That was the problem that you posed on Saturday, was it not?

@robjohn I don't remember exactly, but I think so. OK. Let me send you something (related to the version of the double series of the above problem I finished today). ok?

okay

How would you pronounce Tychonoff?

tick an off

My prof says something like ti-HON-off
Russian names are complicated.

7:51 PM
@robjohn sent it. Let me know when you've got it.

got it

hi guys
how are you ?
how are you ? @Chris'ssis

@what'sup hi. Not that bad. How about you?

i'm fine thanks

@what'sup that's great.

7:59 PM
:-)

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