Mathematics - 2015-07-01 (page 1 of 4)

3:29 AM

evening @ted

Quiet night?

evidently. i only just arrived myself

I saw

Well, I am here now

So now the party can begin.

Oh the party starts?

3:32 AM

I brought the butter beer.

The what?

You haven't read harry potter?

The primary alcoholic drink is butter beer

i never understood what that was supposed to be

Nope, never read it.

Me neither

Uh well that's where my name comes from :P

3:36 AM

Your pseudonym?

Yes :P

according to google, JK Rowling said at one point: "I made it up. I imagine it to taste a little bit like less-sickly butterscotch."

which, uh

I keep thinking of the Mass Turnpike :D

Ugh, I hate butterscotch.

doesn't sound all that great :/

poors butter beer down toilet

3:39 AM

what's a bit weirder to me is that google also gives made-up recipes for it

Including this whole list of them

That's even weirder.

But people have a tendency to try to make fantasy things real.

@TedShifrin do you read fiction ever?

3:56 AM

math.stackexchange.com/questions/1345369/… I got an answer in first!

lol @Semiclassical, I went to a party a while back where "butter beer" was served. It actually tasted very close to what I imagined from the books.

was that a good thing or a bad thing? :P

Oh no it was quite good

ah

Hey guys, I've got a linear algebra question which has been bugging me for a while.

Is there a linear map T: L^2 -> L^2 such that the null space of T is precisely its range?

4:00 AM

Yeah $T:x\mapsto 0_{L^2}$

So, like, everything goes to 0?

What @AlecTeal said

That doesn't quite work, since the null space of T would be the whole of L^2 while its range would be the singleton {0}

Oh of course, what was I thinking

I'm looking for the null space of T to be precisely equal to its range.

Ohhh, I see

How on Earth did we both fall for that @Alec ?

4:01 AM

@KajHansen real butter beer?!? :D

It's a slightly trickier condition. Of course, on a finite-dimensional space with even dimension, it's not too hard to write one down

It's 5am my time, and I've not slept, what excuse are you going with @KajHansen? :P

haha, @StanShunpike, who knows what you mean by "real"

Lol

@StanShunpike certainly not finite dimensional. Not sure about in general!

4:03 AM

Hmm. Do you think it's worth posting on Math.SE then?

I just wanted to be sure I wasn't missing an easy example

I think it's worth it

math.stackexchange.com/questions/1329495/… today this question and the 1 answer to it got a downvote. WTF

Sometimes I get randomly downvoted too @AlecTeal. Personally I think one should have the common courtesy to explain downvotes in non-obvious cases (like no-effort homework questions).

Yeah that's why I linked it. I wanted a second opinion

It looked fine to me, but I only looked for a quick second.

skill patrol

4:41 AM

They're called "drive-by" down votes.

Oh

5:04 AM

@AlecTeal finite dimensional butter beer?!?

Kartik Watwani

5:32 AM

here is the question:Prove that no GP can have three of its terms(not necessarily consequtive) as three consequtive non zero integers.

@skillpatrol math.stackexchange.com/questions/1329495/… it got another downvote!?

WTF

Kartik Watwani

what are your views on this question?

anybody there to talk on this question?

skill patrol

I up-voted it @AlecTeal

Don't worry about it pal, if it is serial-down voting the system will reverse it.

JiK

5:47 AM

math.stackexchange.com/questions/1345401/… was edited which completely changed the question (and the new version is a duplicate) after I answered it. Should I delete my answer or ask that the question is reverted back to the original form (which is not a duplicate AFAIK) because valid questions (after the first edit) shouldn't be changed? I couldn't find any relevant meta topics about this.

thanks @skillpatrol

6:10 AM

@AlecTeal I love that book!! +1

those were unrelated lol

?

the book on your question

Tensor Analysis on Manifolds

I love that book!

Oh!

Yeah I love Dover Books in general TBH!

exactly. cheap and rich in content

Hold on a sec @StanShunpike I want to show you something but I need to alter htacess first

6:11 AM

sure

@StanShunpike nailed it

I use that a lot

@AlecTeal wow, yeah what are these diagrams? OHHHHH and I LOVE that book too!

It's something I am inviting others to test actually!

Oooo

I want to know what I have and have not read.

6:20 AM

It looks like a tetris barcode

So now I write stuff like [11- then [11-15] when I study pages 11 to 15 and mark them off on this.

Wow, brilliant

Blue is unread, green is read, orange is read twice to max-1 times and red is most read (I need to alter this to do a gradient)

How do you make them?

It's important because.... well suppose a book has a really good bit of info, but I don't want to read chapter 1 because of being bored at the time. I can cherry pick from books without knowing what I haven't read.

For example:

What do you mean?

You're welcome to try this BTW, I do want "testers"

6:23 AM

Well, they are all the same format? Do you use a program?

I wrote the program

Yeah that was what i was gathering

Wow, thats really smart

It does help.

I think one of the most interesting aspects of that is how you can identify if you read in a particular pattern

what might be even more cool is that that format could be used to connect to like your Kindle

since the Kindle is monitoring what pages you read

that would be a fantastic graphical way of illustrating it

rather like a standardized format

I mean if people cared lol

Like i dont know how many would

I don't like the kindle for things you don't read in order

6:26 AM

Hmmm

Why?

@BalarkaSen good morning

I dont really like it except for books i dont care about

feeling any better?

It's very easy to flick through real pages, not so much on a kindle. Also you can "remember (ish) where you saw something" and flick back. It becomes an NP search with a page-by-page kindle!

@iwriteonbananas yes, started on antibiotics.

6:27 AM

cool

have you heard of the ham sandwich theorem?

wondering what the guy means by "one linked circle" in the question.

@AlecTeal hahahaha best description ever of my kindle reading

@iwriteonbananas er. that's calculus.

lots of knots out there

it can by anything

true

im guessing he meant a hopf link

then what is "two linked copies of circles"

a circle winding the other circle twice?

6:30 AM

@BalarkaSen it's a corollary of the borsuk ulam theorem

oh, aha.

i recall.

yes, yes.

@BalarkaSen gosh, i dont know :P

@BalarkaSen im trying to prove something very similar to the ham sandwich theorem:

if $A_1,...A_n$ are measurable subsets of $S^n$, then there is a great $S^{n-1}$ cutting each $A_i$ exactly in half

the statement of the ham sandwich theorem was that if $A_1,...,A_n$ are measurable bounded subsets of $\Bbb{R}^n$, then there exists a hyplerplane cutting each $A_i$ exactly in half.

right

you can one-pt-cptify to get the desired, right?

im thinking there must be an easy way to do this using that $\Bbb{R}^n \cup \{\infty\} \cong S^n$

lol

6:33 AM

lol

but havent formalized it yet

if $A_1, ..., A_n$ are measurable sets in $S^n$, then you can find a point not lying in any of them

why?

you're the one who knows the definition of measure.

:D

if you can find a point, then you're done

6:36 AM

no, that's not true. $S^n$ is a measurable subset of $S^n$

ok.

yikes.

but the lebesgue measure of a point is 0

then you'd have to do more work than just applying ham sandwich blindly as you go

@BalarkaSen why are we done then?

homeomorph to $\Bbb R^n$

your sets are measurable subsets of this

bisect by hyperplane

6:40 AM

yeah

consider the point back in it's place

the hyperplane is now a great circle

done

and does it still divide each $A_i$ exactly in half?

um, probably it won't.

oh well

ok

actually i think it will

7:40 AM

i take that back, it wont

risdel

could anyone help me out with a (seemingly) simple conditional probability proof?

but i think we can do this:

It was quite amusing. I took a look at the profile of one of my friends on MathSciNet to see which of his papers he had gotten published so far. And it turned out that he has a publication in the same issue of Journal of Algebra as me.

(though his already has a review there, which mine is still missing)

take measurable subsets $A_1,...,A_n$ of $S^n$ not containing the north pole. map them to $\Bbb{R}^n$ via stereographic projection and then homeomorphically to the interior of $D^n$. by ham sandwich theorem, choose a hyperplane cutting each $A_i$ exactly in half. this hyperplane mapped back onto $S^n$ via the same homeomorphisms should be a great circle cutting each $A_i$ in half

well, that's precisely what i meant to do above, but i am not convinced that it'd work

7:47 AM

why not?

the stereographic projection is horrible at preserving isometry

@iwriteonbananas you have to prove that one, at the very least

that's why we compose w/ a homeomorphism from $\Bbb{R}^n$ to the open unit disk

@BalarkaSen yeah, there's a lot to be justified lol. im having trouble making it rigorous

oh, wait, i missed that bit

eh. i am still not sure that composing witht that homeomorphism would do any good

gah this is frustrating

glad i am not doing this problem.

:P

7:54 AM

and i think it's pointless, i can't make it rigorous. it's not a shortcut anymore, i think im gonna do to borsuk ulam appraoch

yes, and you have to consider cases where you can't find a point not lying the complement of the intersection

yeah, yikes

user183297

8:07 AM

Hi, everybody!
Can someone please comment my question: http://math.stackexchange.com/questions/1344723/multivariable-calculus-an-algorithm-for-this-kind-of-questions

I dont know if it is permitted to write such asks here, so my apology if its not.

8:36 AM

@user183297 Your question is terribly unclear. How can the person move ? along the mountain's borders ? Along the mountain's surface ?

@BalarkaSen hmm what about this:

@iwriteonbananas o/

@iwriteonbananas wanna bet on that you'll have RB's answer here in 12 hours, self-promoting each and every work he has done on van Kampen theorem for higher homotopy groups?

view $S^n$ as $D^n/S^{n-1}$

ok

8:37 AM

@BalarkaSen lolol i was thinking the exact same when i saw that question

let's do it for n=2. take measureable subsets $A,B$ of $D^2/S^1$. let's assume they don't contain the $S^1$ that is collapsed to a point

sure, ok

so $A,B$ are subsets of $D^2\subset \Bbb{R}^2$. choose a hyperplane that cuts $A,B$ in half.

go on

that hyperplane included in $D^2/S^1$ is a great $S^1$ cutting $A,B$ in half

no, i don't think so.

8:46 AM

why not?

you have to homeomorph to $S^2$ to do that

and that homeomorphism might not preserve the halving thing

well, it might cut it in half, but you haven't really proved anything

:P

true

but i think this works lol

shurgs

I am not gonna try out this problem.

I have forgotten how you used to prove ham sandwich theorem using borsuk-ulam anyway.

alright

user183297

9:04 AM

@Hippalectryon along the mountain surface. But i need only the direction, from the point where he is.

@user183297 Hmm sorry I can't help you a lot on this one :/

user183297

If you remember, if he could go any direction, and not only above the surface, it was just the negative gradient of the temprature function. But Here we have a restriction on the moving subject, it can move nly above the surface, so we need to take the negative gradient (of the temprature formula) and find its projection on the surface.
And that what I want to be corrected, im not sure about that.

ok thanks )

Anybody, gradients, multivariable-calculus, i qould glad to get some help, plz: math.stackexchange.com/questions/1345534/…

user20997

has anyone got Vakil's Foundations of Algebraic Geometry? math.stanford.edu is down for the second day in a row today

@user20997 here dropbox.com/s/qb495k68l1f4pzq/…

user20997

thanks Tobias!

@TobiasKildetoft thanks :)

9:17 AM

your welcome @user20997

9:58 AM

Hello @robjohn @DanielFischer @TobiasKildetoft
How are you?

@evinda Good, thank

Are you an undergraduate student? @TobiasKildetoft

Hi @evinda ! How are you ?

@evinda No, postdoc

@Hippalectryon Fine, thanks :) And you?

@TobiasKildetoft In which field?

10:00 AM

@evinda representation theory

@TobiasKildetoft So do you do research in this field? In which semester are you?

@evinda I am not a student

but yes, I do research in that field

@TobiasKildetoft Oh, sorry... I misunderstood and thought that you are a phd student.
So you do teach too at the university?

@evinda Not currently, but I will be teaching in the fall

@TobiasKildetoft I see...

@TobiasKildetoft How is the weather in Denmark?

10:07 AM

@evinda Hot, as far as I know (I live in Sweden, where it is certainly hot)

@TobiasKildetoft Ah I see.. Have you also studied in Sweden?

@evinda No, I studied in Denmark

10:18 AM

Hello@TobiasKildetoft Did you see my diagonal argument?

@Rememberme No

I did prove it using diagonal is closed when space is Hausdorff

@Rememberme Ahh, great. I felt that was the most natural way to do it, as any other way I could think of essentially ended up reproducing the arguments for that anyway

Really, the main thing one shows is that if $f,g: X\to Y$ are continuous and $Y$ is Hausdorff, then the set of element of $X$ where $f$ and $g$ agree is closed

@Tobias lets say $A_\alpha $ be a subspace of $X_\alpha$ . Now if I equip one with box topology and other with product topology .. Will they still be subspaces...

The actual question was what if I equip both them with the same topology that is either box or product

Hi @MatsGranvik

10:26 AM

@Rememberme subspace just means subset where we fix a certain topology. So you are asking if the box/product topology on the product of the subspaces gives the subspace topology on the product in the box/product topology?

@Rememberme if you give one box and the other product it clearly fails in general (take $A_a = X_a$ for all $a$)

Mats Granvik

@evinda hi

What's up? @MatsGranvik

Mats Granvik

i am in horzontl position because of back pain.

@MatsGranvik Get well soon!

Mats Granvik

@evinda thanks

10:41 AM

Hello!! We can create in 3 steps individuals mixes of coffee, in the following same order.

In the first step the mix can be chosen from 10 coffees. all the 10 coffees could be used.
In the secod step the degree of grid can be chosen from 8 degrees. Only the choice of one degree is possible.
In the third step can the etiquette ban chosen from 50 motives. The client can chose only one.
How many possibilities of mixes are there?

I thought that it would be 10*8*50 but I am not sure because at the first step it is possible to choose all 10 but at the other steps only one.

Which is the number of possibilies at the first step?

@MatsGranvik @TobiasKildetoft are you familiar with combinatorics?

@MaryStar In the first step, it seems like you can choose any number of coffees, not just one of them

how can we count the number of possibilies at the first step?

@MaryStar So you want to count the number of ways to select any number of objects from $10$ objects

Is it maybe 2^{10} for the first step? So, in total $2^{10} \cdot 8 \cdot 50$ ? @TobiasKildetoft

@MaryStar Well, how did you get $2^{10}$?

10:48 AM

for each of the 10 coffees there are two possibilies, we take it or not. @TobiasKildetoft

is this wrong?

@MaryStar No, that is completely correct

Q: The quadratic and cubic versions of a tough intregral

so the answer is $2^{10} \cdot 8 \cdot 50$, right? @TobiasKildetoft

Chris's sis the artist

@Hippalectryon @r9m see below

0

In this post, Proving that $\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$, it's proved that $$I_1=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$ but then some natural questions arise. Might we possible h...

:D

when the answer is $2^{10} \cdot 8 \cdot 50$ are we sure that we take one at the first step?

@TobiasKildetoft

or is it possible that we don't take anyone? @TobiasKildetoft

10:54 AM

@Tobias Oh.. Thats the case..

Huy

@evinda: How's it going in Greece these days? Doesn't seem to be a very good time according to newspapers. :(

@MaryStar Ahh, yeah, that is if we are able to select none

Yes some kind of Bankruptcy @Huy

@TobiasKildetoft How do you construct a matrix from a characteristic polynomial ? i.e. given a charact poly, how do you get a matrix that has that charact poly ?

So the answer is $2^9 \cdot 8 \cdot 50$, right? @TobiasKildetoft

10:56 AM

@Hippalectryon Over the complex numbers, you just put the roots on the diagonal

@TobiasKildetoft Not if we don't know the roots

@Huy Yes, it isn't a very good time. There can't be a new agreement before the referendum on Sunday.

@Hippalectryon in general, I am not sure if it is possible

@TobiasKildetoft (see math.stackexchange.com/questions/1345647/…)

@evinda its dictatorship in Greece??

10:58 AM

@Rememberme No, democracy.

Okay but referendum??

Huy

@evinda: I read the Greece basically voted out the previous politicians who made the previous bailout deal with the EU and the new politicians stopped the austerity policies, which is why they now need another bailout (which the EU understandably doesn't want to offer). Is that correct?