Mathematics - 2015-11-25 (page 1 of 3)

12:01 AM

but with beamer you can't do what I want to achieve, it is a different kind of thing, I want to show the audience how the different sheets logical connect with each other, which sheet implies the other sheet

Dan Rust

@MikeMiller yeah I'm silly for overthinking this whole problem....

Kasper

and then I want to zoom out, and show all the sheets at once

Mike Miller

@Ted: H-spaces are better than that: the action of $\pi_1$ on higher homotopy groups is also trivial. This is not true for say $\Bbb{RP}^2$.

L33ter

this is so cool

Dan Rust

@MikeMiller you may like a little result I found today... If $X$ is an aspherical finite CW complex then its fundamental group is torsion-free

the proof is cute

L33ter

12:04 AM

this fundmental group stuff is very elegant

I am using allen hatcher book along with munkres for my project

allen hatcher is actually very nice

I like it more

because it has more pictures

Dan Rust

some people don't like Hatcher, though I'm personally a fan

L33ter

I like it more than munkres it gives more intuition

PVAL

The exercises are alright. The expositon is useless.

Ted Shifrin

@MikeM: Geometry to the rescue! :)

Mike Miller

@DanRust: Yes, it's a nice fact. In particular this is true for knot complements.

Dan Rust

12:14 AM

@MikeMiller ah that's true

Mike Miller

Or infinite fundamental groups of 3-manifolds in general.

Hmm... Orientable.

Clarinetist

To Windows 10 or not Windows 10?

That is the question

Dan Rust

7>10>8

Clarinetist

Hmm, I haven't had any problems with 8.1, and I liked 7

robjohn

@OldJohn Hey, there. We haven't been around at the same time for a while.

PVAL

12:19 AM

@MikeMiller I take back my take back $\Sigma (2,3,5) #M$ where $M$ is any closed 3-manifold with $\pi_1$ infinite would be a counterexample.

Mike Miller

Ok, sorry. Any 3-manifold that doesnr have a spherical manifold in its prime decomposition.

morphic

12:43 AM

@Clarinetist Don't do it

user174558

@morphic Hi Bart.

user174558

@Clarinetist I think all versions of Windows are good, much better than Linux.

robjohn

@JasperLoy Ack!!! (holding a cross between me and Jasper)

@Clarinetist Don't do it!

Never knew that $$\int_0^1\int_0^1(xy)^{xy}\,\mathrm{d}x\,\mathrm{d}y =\int_0^1x^x\,\mathrm{d}x$$ I just proved it for an answer.

Sajindia

1:04 AM

@JasperLoy

user174558

I know who you are. Don't try.

Sajindia

@JasperLoy I need to talk to you on private

user174558

@Sajindia OK, here will do.

Sajindia

I tried to send you an email to jasperloy@outlook and @jasperloymathematics@gmail but it can't be sent

L33ter

can you have abelian fundmental group that isn't Z?

user174558

1:10 AM

@Sajindia That is because I have changed my email. I will not disclose it to you. You can talk here.

Sajindia

ok thank you.

morphic

@L33ter What's the fundamental group of $\Bbb RP^2$

L33ter

what is $P^2$?

morphic

Real projective space

L33ter

I don't know this space

morphic

1:11 AM

It's pretty important in topology; you should get very familiar with it

L33ter

can you explain it ?

user174558

See @morphic is so smart.

morphic

@L33ter There's several different ways to view it but one of them is the sphere with antipodal points identified

L33ter

I see

can we have the fundmental group being R?

Nitrogen

For any group $G$, there exists a space with fundamental group $G$.

Ted Shifrin

1:25 AM

I'm not sure I believe that.

Rehi @morphic

L33ter

Take any simply connected space $X$ and any space $Y$ with fundamental group $A$. Then $X \times Y$ has fundamental group $A$ as well... — Qiaochu Yuan 6 mins ago

@TedShifrin

nice

morphic

@L33ter This answer also by Qiaochu might be of interest math.stackexchange.com/a/36780/130018

L33ter

I have a question also lets say we have a group G

how can I get a space such that its fundmental group will be G?

morphic

@L33ter Read the link I posted

L33ter

ohh

niceee

Sajindia

1:36 AM

Could someone please tell me why $\chi(\phi)=[\sigma]$ (the part in green) in this theorem? postimg.org/image/madlpge2d

The Substitute

2:27 AM

In probability, where does the word 'moment' come from? Does it have anything to do with how the word is regularly used in English?

PerplexedGuest

3:50 AM

Hello all! :)

Chantry Cargill

@PerplexedGuest How's it going?

PerplexedGuest

Not too badly! Working on my research, have been for around 9 hours so far today. xD

Yourself? :)

Chantry Cargill

Not bad. Marking IT exams lol.

What's your research area?

PerplexedGuest

Ooh haha, you're a professor? :0

Self-avoiding walks. :)

I'm in undergrad, so this isn't like ... serious research.

Chantry Cargill

Secondary school teacher.

I'm sure it's serious research from your point of view.

PerplexedGuest

3:52 AM

Ah, that's cool. High school teachers have my respect, that's hard work.

Yeah, it's pretty rough from my side. xD

Chantry Cargill

I'd love to teach college or university eventually. I'm currently restudying a lot of the things I learned in undergrad so that I can move on to graduate work.

PerplexedGuest

That's my goal as well!

Though I am much more attracted to the idea of math research being my job. xD

Chantry Cargill

You seem to have the right attitude.

What year are you in?

PerplexedGuest

Hm.

Junior.

crocket

5:58 AM

Can anyone give an answer to the comment on math.stackexchange.com/q/1542404/231595 ?

0

Q: Did I solve a basic derivation problem correctly?

The following problem is from "mathematical logic" by ian chiswell and wilfrid hodges, 2007.

RustyStatistician

6:56 AM

Is there a way to draw attention to questions in other communities that would benefit from the mathematics community?

For example, this question has not gotten much attention and I think it's because its needs someone with better mathematical chops: stats.stackexchange.com/q/180618/95564

crocket

7:14 AM

Did I prove mathb.in/47200 correctly?

Paradox 101

11:00 AM

Can anyone explain how o expand $ {-\frac{1}{2}\choose {k}}$ into something in terms of double factorial?

Paradox 101

11:15 AM

Anyone??

user174558

11:47 AM

math.stackexchange.com/questions/1544087/… put on hold? Absurd.

user174558

It might not be phrased well in its original state, but certainly deserves to be well answered.

Tobias Kildetoft

@JasperLoy I disagree. That question is clearly asking for opinions, which is not allowed for good reason.

user174558

Hello Tobias, you have a nice name.

Tobias Kildetoft

thanks

user174558

It's one month to Christmas, whee!

I'm mostly just an idiot

12:09 PM

hi

is that sent?

why is there a line of dots over my first message? do these come in a set?

\help

/help

!commands

I'm mostly just an idiot

12:27 PM

Can I chat here about commutative algebra as I learn it? The commutative algebra room seems inactive

Tobias Kildetoft

@I'mmostlyjustanidiot Certainly

I'm mostly just an idiot

A friend recommend I learn from Miles Reid, so I have that text and I hope I am capable of progressing

Why does it say $\dim k[x_1,\cdots,x_n]=n$ but $\dim \Bbb Z[x_1,\cdots,x_n]=n+1$?

Tobias Kildetoft

@I'mmostlyjustanidiot because this is true?

in general, if $R$ is an integral domain that $R[x]$ has dimension precisely one greater than $R$

I'm mostly just an idiot

$k$ is a field, so is an integral domain, so then $k$ has dimension $n-1$?

Tobias Kildetoft

no, a field has dimension $0$

I'm mostly just an idiot

12:36 PM

characteristic = dimension?

Tobias Kildetoft

no

you need to look up the definition of dimension

I'm mostly just an idiot

ok

I thought since it was a 1-dimensional vectorspace, it had dimension 1

Tobias Kildetoft

@I'mmostlyjustanidiot Again, look up dimension

I'm mostly just an idiot

which type should I be looking at?

Tobias Kildetoft

@I'mmostlyjustanidiot Well, obviously the one that applies to all rings, as defined whereever you saw the claim for polynomial rings

I'm mostly just an idiot

12:40 PM

sorry about that tobias, I was in the welcome chapter, so it didn't define it

I'll go to chapter 1

krull dimension ok, I had honestly never heard of that

Yeah it turns out that the proposition I mentioned was meant to be motivation to read that book lol

Tobias Kildetoft

@I'mmostlyjustanidiot Yeah, it is not a trivial thing at all

I'm mostly just an idiot

@TobiasKildetoft Well that's good to hear :-)

N3buchadnezzar

1:07 PM

Does there exists a good approximation for $\sqrt{1-x^2}$ from $t$ to $1$, where $t$ is close to $1$?

I'm mostly just an idiot

What does that mean?

Huy

what does $t$ have to do with that expression?

I'm mostly just an idiot

Did you mean $\sqrt{1-t^2}$?

Huy

@N3buchadnezzar: en.wikipedia.org/wiki/Puiseux_series Maybe this is helpful.

I'm mostly just an idiot

how is that helpful????

@N3buchadnezzar can you expand on your question? It looks like if $x\to 1$ from above, we get $0i=0$, and from below, we get $0$

N3buchadnezzar

1:13 PM

@I'mmostlyjustanidiot Below of course?

I tried combining the maclaurin series of $\sqrt{1-x^2}$ with some splines. However it left some large error margins.

I'm mostly just an idiot

oh I don't even know what a spline is oops

morphic

Is it a spine and a spleen fused

Akiva Weinberger

2:00 PM

So let's say you lost your compass

But you have a marked ruler

And you want to draw a right angle

How would you do it?

Solution: Draw two intersecting lines, call the point of intersection $O$. On each of the lines, mark off two points each $1$ inch away from $O$. These points form a rectangle; join three to make a right angle.

(Remember, I said marked ruler.)

Someone figured out how to draw an equilateral triangle with just a marked ruler. It's a brilliant construction, but it's kinda long.

morphic

How do you measure an inch

Oh I thought you said you have a ruler that's not marked

N3buchadnezzar

2:29 PM

Semi interesting I guess? math.stackexchange.com/questions/1545872/…

N3buchadnezzar

2:46 PM

I am the dumbest dumbo dumb dumb. Just take $x \mapsto (1-T) u + \sin u$...

robjohn

@N3buchadnezzar if you use the proper splines (e.g. rational quadratic splines), you can get $\sqrt{1-x^2}$ exactly.

N3buchadnezzar

3:01 PM

@robjohn example? =)

robjohn

@N3buchadnezzar rational quadratic splines are segments of conic sections. Matching a circle is no problem.

See this section

QuickDraw GX used quadratic rational splines as a primitive. (Note the developers)

Brennan.Tobias

hello

I'd like to ask about an answer I found when researching something

this one: math.stackexchange.com/a/9674/262437

hes say consider a with a^(4k) = -1 (p), but how would we know there is such an a?

robjohn

@Brennan.Tobias This is in $\mathbb{Z}_p[x]$ where $p=4k+1$.

Brennan.Tobias

in that case we have x^4k = x^(p-1) = 1 by fermats little theorem

but -1 is written there

oops! it's 8k+1

thanks I see that part now!

robjohn

@Brennan.Tobias 8k+1? Ah, yes, in Robin's answer.

Brennan.Tobias

3:12 PM

oh I don't quite follow what you are saying

I'm looking at the answer by Robin Chapman

why would we have to use GF(p^2) in the p = 8k+3 case?

Dave

Hi, I'm writing an homework assignement and I want to simulate a production of a banana plantation. I would like the production to start low, grow up, and then I want it to reduce because the tree is too old.

I can't figure out what type of curve or what type of equation I need,

Can you point me out the right direction

Is this the only solution ? en.wikipedia.org/wiki/Piecewise

Piecewise

well, it seems that could be appropriate ? Gaussia function en.wikipedia.org/wiki/Gaussian_function

not sure..

Dave

3:38 PM

Friend just told me to use a negative quadratic function, what you think ?

Huy

3:49 PM

@Dave: There are infinitely many functions which "start low, grow up and then reduce". You'll need to restrict it further.

@Dave: Piecewise definition is one possible answer, some sort of Gaussian might work and a negative quadratic works too, if applied correctly.

N3buchadnezzar

4:02 PM

@Dave $\sin^2x$ ?

Huy

@N3buchadnezzar: are your banana trees perpetuum mobiles?

Dave

@huy I'm sure you understood what I meant, didn't you ?

Huy

yes

N3buchadnezzar

@Huy ?

Dave

@N3buchadnezzar How exactly am I going to use it ? I mean, I can draw it using google, but what is going to be my x value,

@N3buchadnezzar I'm looking to pass the tree age to the function and get a number of banana produced accordingly to it's age

N3buchadnezzar

4:08 PM

@Dave Yes?

Dave

@N3buchadnezzar Let's pretend the age of the tree is 4, how do I use the sin function to get a low number of banana

N3buchadnezzar

@Dave When is the sine function zero?

zero

And the next zero?

PI

4:11 PM

@MikeMiller Hey Mike, are you around?

N3buchadnezzar

Awesome. So since $\sin^2( \pi \cdot 1) = 0$ But we want to expand this so we could say $B(x) = \sin^2(\pi x/10)$. This function would first reach zero for year $x=10$.

Dave

y = 2^sin (40/100*3.14) * 70

I tried with age / 100 * pi

does it makes sense ?

*70 is number the minimum number of bananas

N3buchadnezzar

@Dave 70 would be the maximum number of bananas. Try to plot the function for various values.

Dave

yeah I think you are right

I want 500 as max number of banana

Mike Miller

@iwriteonbananas: Nice timing.

N3buchadnezzar

4:17 PM

@Dave Then you would use 500 instead. It seems you expect your trees to live to 100?

iwriteonbananas

Cool. I'm trying to prove that if $f:S^n\times S^n\to S^n\times S^n$ is a homeomorphism, then the induced map $H^n(f)$ is given by either one of the following matrices: $$\begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1\end{bmatrix},\quad \begin{bmatrix} 0 & \pm 1 \\ \pm 1 & 0\end{bmatrix}$$

Since $f$ is a homeomorphism, it restricts to a homeomorphism on each copy of $S^n$. So the restriction to one copy is a homeomorphism onto one copy in the codomain, and the restriction to the other copy in the domain is a homeo onto the other copy in the codomain, right?

N3buchadnezzar

@Dave Seems banana trees are not actuall trees at all ehow.com/info_8414744_long-banana-tree-live.html and have a life expectancy of about 6 years.

iwriteonbananas

I suppose that proves that either both diagonal entries consist of $\pm 1$ or the "other diagonal" entries consist of $\pm 1$

Mike Miller

@iwriteonbananas: A homeomorphism could do weirder things than swap the factors

In principle

iwriteonbananas

Right.

Dave

4:22 PM

@N3buchadnezzar hahaha yeah I just read on wiki that banana trees don't work like I'm using it, well whatever,

@N3buchadnezzar here's my final answer : y = sin (x/100*pi) * 500

Mike Miller

I guess probably this should follow from the ring structure?

Dave

@N3buchadnezzar thanks for the help I kinda figured it once you told me that 3.14 was the next zero point

iwriteonbananas

Yeah, we just did the ring structure.

Mike Miller

It has to preserve $a \cup b$, so...

N3buchadnezzar

@Dave Yeah. Like they kinda work like that, if you look at a collection of trees or a whole field. Also consider the $\sin^2x$ to be the probability distribution. Instead of the precise number of bananas... I guess @iwriteonbananas knows more about this than I do.

Dave

4:25 PM

wow wait a minute there's a guy that writes on banana in here, haha wtf

@N3buchadnezzar for some reason it didn't give me what I expected when I used 2^sin .....

user174558

@N3buchadnezzar No meat, no pudding.

N3buchadnezzar

@Dave ?

Dave

@N3buchadnezzar yeah I think I got it now you told me to use 2x not 2^sin

@N3buchadnezzar but when I use 2x instead of x, what does it change ?

N3buchadnezzar

@Dave Changes where the zero lies

Dave

Oh damn

@N3buchadnezzar Right.

iwriteonbananas

4:28 PM

@Dave Bananas are surprisingly easy to write on, I suggest you give it a try.

Dave

@iwriteonbananas interesting.

N3buchadnezzar

@Dave $500 \sin^2(40\pi/100) \approx 475.5282582$. Since if the life expectancy is $100$, the tree will be most productive at year $50$.

Dave

@N3buchadnezzar Yeah I think I figure it now,

@N3buchadnezzar thanks a lot for the help, was fun

@N3buchadnezzar have a nice day

user174558

@dave you are welcome

iwriteonbananas

@MikeMiller Let's call the generators of $H^n(S^n\times S^n)$ $\alpha, \beta$. Then $f^*(\alpha)=f^*(\alpha\cup 1)=f^*(\alpha)\cup f^*(1)=f^*(\alpha)\in \{\pm\alpha,\pm\beta\}$ since $f$ is a homeomorphism?

Mike Miller

4:32 PM

Yah. I guess you're assuming that the homeomorphism is orientation preserving

Hmm, wait. You probably want to make an assumption on $n$.

iwriteonbananas

I guess I should write $\pm \alpha$ and $\pm \beta$ instead

user174558

Mike, since you are ignoring me, I won't talk to you anymore. Bye.

iwriteonbananas

Ohhh damn it

Mike Miller

Well it needn't be 1 in the end unless it's orientation preserving

iwriteonbananas

I forgot to mention: $n$ is even!! Sorry.

Mike Miller

4:34 PM

Oh, cool

That helps us by commutativity probably.

iwriteonbananas

But where does that even come into play?

Hmm ok.

Mike Miller

a cup b = b cup a

iwriteonbananas

Right...

Mike Miller

so use the fact that a cup b maps to 1 to get a minor restriction on the matrix, then do the same for b cup a to get another restriction

Agawa001

@JasperLoy perhaps, if u learn how to not run ur mouth with foulish, ppl would unignore you again

user174558

4:36 PM

@Agawa001 Thanks, it is not me who talks rot here about others. I have said enough for those who care to figure out. Bye.

robjohn

@iwriteonbananas fountain pens, ball-point pens, felt pens, pencils, chalk, spray paint, chisels... with what do you write on bananas?

Sledge hammers leave a lasting impression...

user174558

@robjohn Hi Rob. Have a nice day.

iwriteonbananas

@MikeMiller Hold on...$a\cup b$ generates $H^{2n}(S^n\times S^n)$, right? What does that have to do with the matrix of $H^n(f)$?

robjohn

@JasperLoy howdy... I am looking forward to our feast tomorrow. :-)

user174558

@robjohn Is tmr Thanksgiving?

robjohn

4:41 PM

@JasperLoy It is

Andrew Thompson

Just finished diffgeo exam. Thanks for the help during this semester, @MikeMiller.

Mike Miller

@iwriteonbananas: The automorphism you get of $H^n$ needs to preserve the cup product.

user174558

@robjohn Ah, I have nothing to be thankful for.

Mike Miller

That places restrictions on the cup product.

@AndrewThompson: I don't think I gave any, but grats.

robjohn

@MikeMiller I have no cups that produce anything...

iwriteonbananas

4:42 PM

@robjohn I usually just stick to a good old ball-point pen.

Andrew Thompson

Oh, you gave plenty, I don't think you realize how much people on the internet bother you with stuff.

user174558

@Agawa001 Someone will almost certainly star this line later to discredit me, it's OK.

robjohn

@iwriteonbananas Okay. Just wondering.

user174558

@robjohn Will there be turkey tmr? I think I should not come here anymore, since I am not welcome.

robjohn

@JasperLoy Why would you say that? Just because some people are ignoring you does not mean that you are not welcome.

user174558

4:46 PM

@robjohn OK. Anyway, turkey here is rare.

robjohn

@JasperLoy It is not good to eat rare turkey... it is better to cook it well to rid it of things like salmonella.

user174558

@robjohn Have they restored your chatjax site in UCLA?

robjohn

@JasperLoy someone was looking into it, but I don't think it has been yet.

user174558

@robjohn Might be an act of terrorism. Did you know that Turkey shot down a Russian plane? It's on Wikipedia.

robjohn

@JasperLoy The page still returns an error.

iwriteonbananas

4:50 PM

@MikeMiller Yes, but I'm still not sure what you mean. The only cup products that we can plug into $H^n(f)$ are $a\cup 1, 1\cup a, b\cup 1$ and $1\cup b$, right? What's the above-mentioned restriction you want to put on the matrix?

robjohn

@JasperLoy It was supposedly flying in Turkish air. The US radar concurs.

r9m

@robjohn maybe this will be of interest to you! :) I couldn't manage anything useful :|

iwriteonbananas

I also proved that the projections $S^n\times S^n\to S^n$ induce an isomorphism $H^n(S^n)\oplus H^n(S^n)\cong H^n(S^n\times S^n)$ for what it's worth.

user174558

@robjohn Still, no reason to shoot it down even with 10 warnings, considering the Russians are there to fight the terrorists.

Mike Miller

@iwriteonbananas: Trivial restriction: the matrix has to have determinant one. You get that from preserving $a \smile b$.

Ainz Ooal Goal

4:54 PM

@JasperLoy It wasn't the first time at all they were infringing on their air space, and allowing foreign planes into your air space without a good reason is not a very good idea

Mike Miller

I haven't worked out whether the fact that it also preserves $b \smile a$ is interesting. I suspect it is.

Agawa001

@JasperLoy i dont belong to ur star-games

neither flags

user174558

@AinzOoalGoal Yes, but to kill people like that, sad.

Mike Miller

I think what you're going to get is that it's an orthogonal matrix which should give you the desired restrictions.

user174558

@Agawa001 I am not talking about you, of course.

Ainz Ooal Goal

4:55 PM

@JasperLoy Did they kill them ? iirc, the army didn't kill them, rebels did.

I'm pretty sure they did not want to kill anyone to avoid too much trouble

user174558

@AinzOoalGoal I mean the bomb could have killed those on the plane.

Ainz Ooal Goal

I'm no expert, but I suppose they're trained to eject quickly in those situations... idk

iwriteonbananas

@MikeMiller I guess I'm being dumb, but could you explain how we obtain that restriction?

Mike Miller

@iwriteonbananas: You need to actually write stuff down... if $f$ is given by [[a, b], [c,d]] then $f(\alpha \smile \beta) = (a \alpha + c \beta) \smile (b\alpha + d \beta) = ad+bc$, which is supposed to be 1. I guess that's not invertibility but whatevs.

Fiddle with what you know explicitly like this.

robjohn

@r9m I wonder if there is a reason to believe that there is a nice closed form.

@Paradox101 has anyone answered this?

iwriteonbananas

5:04 PM

@MikeMiller That's what's confusing me. $\alpha \smile \beta$ is in $H^{2n}$. We're trying to determine the matrix of $H^n(f)$. And $H^n(S^n\times S^n)$ is generated by $\alpha=\alpha\smile 1=1\smile \alpha$ and $\beta = \beta \smile 1=1\smile \beta$.

r9m

@robjohn I have no idea .. :{ It changes to $\displaystyle \frac{1}{a}\mathfrak{Re}\int_0^{\infty} \Gamma(1+ix)e^{-ax}\,dx$ .. but I'm not sure how to handle this form either :(

Agawa001

@AinzOoalGoal nothing fine is predictable, war flags are waving!!

we must as crippled humans, pray and hope the best situation fo both sides

robjohn

@r9m how do you get that?

JeSuis

Hi

@AinzOoalGoal Salut

robjohn

@Paradox101 $$\binom{-\frac12}{k}=\left(-\frac14\right)^k\binom{2k}{k}$$

Mike Miller

5:10 PM

@iwriteonbananas: I don't understand the confusion. You're askinf how I can write it as a matrix like this?

iwriteonbananas

@MikeMiller We want to determine the matrix of $H^n(f)$, so we check what it does with the basis elements $\alpha$ and $\beta$. $\alpha\smile \beta$ is not a basis element of $H^n(S^n\times S^n)$ though -- it lies in $H^{2n}(S^n\times S^n)$, right?

Mike Miller

The point is that your $f$ is a homomorphism.

Here's another way of saying what I'm saying. You have an intersection form [[0,1],[-1,0]] (I'm having some confusion about signs here, can't really figure it out, maybe you can.) The fact that $f$ is a homomorphism means that the automorphism it induces on $H^n$ preserves the intersection form. So find the matrices that preserve it.

I have to go... good luck

r9m

@robjohn $\displaystyle I = \frac{1}{a}\int_0^{\infty} e^{-x}\mathfrak{Re} \left(\frac{1}{a-i\log x}\right)\,dx = \frac{1}{a}\mathfrak{Re} \int_0^{\infty} e^{-x}\int_0^{\infty} e^{-(a-i\log x)y}\,dy\,dx = \frac{1}{a}\mathfrak{Re}\int_0^{\infty} e^{-ay}\int_0^{\infty} x^{iy}e^{-x}\,dx\,dy$

iwriteonbananas

@MikeMiller Ok, thanks. I'll try to see where I'm going wrong...

robjohn

@r9m looks good.

@r9m had you ever seen this before?

16 hours ago, by robjohn

Never knew that $$\int_0^1\int_0^1(xy)^{xy}\,\mathrm{d}x\,\mathrm{d}y =\int_0^1x^x\,\mathrm{d}x$$ I just proved it for an answer.

r9m

5:18 PM

@robjohn Holy **** .. no !!! :D

@robjohn (+1) at answer! :D Awesome!!!

JeSuis

@r9m I cannot find the answer you are talking about..

Ainz Ooal Goal

@JeSuis o/

JeSuis

@AinzOoalGoal par hasard isomorphisme d'anneaux tu aimes ?

r9m

2

A: Seemingly impossible double integral reduction

$$ \begin{align} \int_0^1\int_0^1(xy)^{xy}\,\mathrm{d}x\,\mathrm{d}y &=\int_0^1\int_0^yx^x\,\mathrm{d}x\frac1y\,\mathrm{d}y\tag{1}\\ &=\int_0^1\int_x^1\frac1y\,\mathrm{d}y\,x^x\,\mathrm{d}x\tag{2}\\ &=\int_0^1(-\log(x))\,x^x\,\mathrm{d}x\tag{3}\\ &=\int_0^1x^x\,\mathrm{d}(x-x\log(x))\tag{4}\\ &=\...

Ainz Ooal Goal

@JeSuis Les anneaux c'est pas trop mon truc généralement... mais dis quand même :P

JeSuis

5:24 PM

@r9m thanks!

robjohn

@r9m That integral is in turn $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$

r9m

@robjohn yes! Sophomore Identity :-)

user174558

Hi @Chris'ssistheartist!

JeSuis

@AinzOoalGoal je dois montrer que $\Bbb{Z}[i]/\langle p\rangle\simeq \Bbb{Z}[x]/\langle x^2+1,p\rangle $

robjohn

@r9m That double integral identity really surprised me.

Ainz Ooal Goal

5:26 PM

@JeSuis Ah oui en effet, c'est pas mon domaine xD

JeSuis

@AinzOoalGoal :D, t'es dans quel domaine ?

r9m

@robjohn yes! I had no idea there was a nice relation like that!

Ainz Ooal Goal

@JeSuis En PC on voit pas trop les anneaux et groupes quotients.

JeSuis

@AinzOoalGoal ah ok! vous avez bcp de maths qd mm non ?

Ainz Ooal Goal

@JeSuis Oui, mais moins qu'en MP de toute évidence :-) là on a commencé les probas

JeSuis

5:29 PM

@AinzOoalGoal clairement vu le niveau des ENS, je suis à la fac, je connais pas trop le système prépa. probas, j'aime pas!

Ainz Ooal Goal

Ca rime :D

Chris's sis the artist

@JasperLoy JASPER!!! Hi. How is it going?

user174558

@Chris'ssistheartist Hehe. My book arrived in the mail today.

Chris's sis the artist

@JasperLoy Great!

robjohn

@Chris'ssistheartist Good morning. Did you see the identity I mentioned to r9m above?

JeSuis

5:31 PM

@AinzOoalGoal :DD

Chris's sis the artist

@robjohn Hello. No, I didn't. Which one?

robjohn

@Chris'ssistheartist It's kinda freaky that the same integral that gives one of the Sophomore Dreams, also has a double integral identity like that.

Chris's sis the artist

@robjohn Interesting (+1). Do you think there is something similar in 3 variables?

robjohn

@Chris'ssistheartist I don't know...

Chris's sis the artist

@r9m did you manage to calculate my series (in the spirit of the art :-))?

@robjohn I might take a look of that variant.

robjohn

5:34 PM

@Chris'ssistheartist checking out the three dimensional equivalent.

Chris's sis the artist

@robjohn Yeap.

robjohn

@Chris'ssistheartist It doesn't hold by the same extension... too bad

Chris's sis the artist

@robjohn I didn't check yet, but I trust you.

robjohn

$$\int_0^1\int_0^1\int_0^1(xyz)^{xyz}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \ne\int_0^1\int_0^1(xy)^{xy}\,\mathrm{d}x\,\mathrm{d}y$$

Chris's sis the artist

@robjohn right.

@robjohn The key for the former one is to wisely use the derivative of $x^x$.

PerplexedGuest

5:39 PM

Hello all! :)

Ainz Ooal Goal

@Surb Is your profile picture set by yourself, or is it generated by MSE ? It looks like MSE's, but I've never seen one that blends colors like that before !

robjohn

@Chris'ssistheartist yeah... when I saw the $x\log(x)$ floating about, I was able to get the thing immediately.

Chris's sis the artist

@robjohn Without the proper approach these ones give some headache.

Ted Shifrin

hi @Perplexed, M le Méchant :), @robjohn

robjohn

@TedShifrin Good morning! Ready for T-Day?

Ted Shifrin

5:44 PM

@robjohn: Pretty much every year but 2 in GA I hosted 10-20 people. Tomorrow I'm going to friends' and bringing a dish :) You?

Ainz Ooal Goal

@TedShifrin o/ What's T-day ? Ted Shifrin day ? :P

Ted Shifrin

Set any laptops on fire lately, M le Méchant? :)

la journée du dindon :)

Ainz Ooal Goal

Aah ça, on ne le fête pas en France

@TedShifrin I have a new laptop now, it's much better

Ted Shifrin

oooh, a new laptop ... félicitations

no, Thanksgiving celebrates the arrival of the pilgrims to invade the lands of the Native Americans ... :)

Ainz Ooal Goal

When you put it that way it sounds weird

Ted Shifrin

5:47 PM

imperialism often sounds weird

I see bananas is lurking ... haven't seen him in ages

iwriteonbananas

@TedShifrin Ted!!!!!!!!!!!!!!!!!!!!!!!!!

I'm always lurking. Watching you.

Chris's sis the artist

Back to my work. @robjohn, btw I generalized that integral I showed you privately in db last time, and it's very weird.

Ted Shifrin

How boring a life you lead ... :)

How're you doing?

iwriteonbananas

Tracing your every step.

robjohn

@Chris'ssistheartist @r9m I believe the same proof works for the other integral of the Sophomore's Dream: $$\int_0^1\int_0^1(xy)^{-xy}\,\mathrm{d}x\,\mathrm{d}y =\int_0^1x^{-x}\,\mathrm{d}x$$

Numerically, it checks out!!

Chris's sis the artist

5:49 PM

@robjohn sure, I almost missed that.

robjohn

That is just crazy!

iwriteonbananas

@TedShifrin Quite good, wrapping up this exercise sheet. And then moving on to the diffgeo exercise sheet.

Ted Shifrin

Any good diff geo to share?

iwriteonbananas

My life consists mostly of solving exercise sheets.

Ted Shifrin

Well, you could be a philosopher instead and your life would consist of reading tomes and writing long essays.

iwriteonbananas

5:50 PM

@TedShifrin Yeah, we just introduced differential forms, Lie derivates and all that stuff. Give me a minute to write down this solution to the alg top problem.

Ted Shifrin

Good stuff.

iwriteonbananas

@TedShifrin Meh, I prefer this. Though writing essays isn't too bad either.

Chris's sis the artist

Is it possible some people working for some magazine don't work this week due to the Thanksgiving day? Maybe. I wait for a feedback. In general the response is 100%.

robjohn

@Chris'ssistheartist I've just added that as a comment to my answer.

Chris's sis the artist

@robjohn OK. Is it possible some people don't work in US this week due to the thanksgiving day?

robjohn

5:54 PM

@Chris'ssistheartist Most companies only give Thursday and Friday off.

Chris's sis the artist

@robjohn I see.

robjohn

@Chris'ssistheartist The equation for the negative exponent does not extend to three variables either.

Chris's sis the artist

@robjohn Maybe also the 3 variables form can be brought to a nicer form in 2 variables.

robjohn

@Chris'ssistheartist probably, but it is not as interesting if it is not the same :-)

Chris's sis the artist

@robjohn hmmm, what if this happens again for a higher number of variables like ...

2^0, 2^1, 2^2 ,...

Let me check that numerically.

robjohn

5:58 PM

@Chris'ssistheartist doing so now

should use my faster computer...

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$Mathematics$ Mathematics