Mathematics - 2017-06-28 (page 6 of 7)

@AkivaWeinberger If you let $X = (x,\dots,x)$ then $\det(C_1 + X,\dots,C_n+X) = \det(C_1 + X,C_2-C_1,\dots,C_n-C_1)$ is an affine function of $x$ by developping along the first collumn

Semiclassical

@Liad Let me try again.

If you take the Fourier series based on the behavior of $f(x)$ over $[0,1]$, then does that Fourier series know anything about $f(x)$ outside this interval?

Liad

no so we can think of $f$as a 1-period function

Akiva Weinberger

6:08 PM

@Astyx Oh, cool. So that does it for the special case of $J_{m,n}=1$.

Semiclassical

@AkivaWeinberger That's the only case we need, though.

Liad

so what is wrong with finding the coefficients of $f(xt)$ ?

it is also defined on $[0,1]$

Semiclassical

Well, what I'd say is that the Fourier series will converge to the 1-periodic continuation of $f(x)$ over $[0,1]$.

Akiva Weinberger

@Semiclassical This was part of a larger application?

Or do you mean that the general case follows from the special case?

Semiclassical

@AkivaWeinberger Yes, that of finding $\det(M)$.

Akiva Weinberger

6:10 PM

Ah, OK.

Semiclassical

@Liad Here's the problem. Suppose $t=2$.

Then what is $g(x)=f(2x)$ at $x=1$?

Liad

f(2)

Semiclassical

Hmmmmmm. Hang on, I feel like I've been making this more complicated than it needs to be.

I guess here's the point. What's the natural periodicity of $g(x)=f(xt)$?

Taking as given $f(x+1)=f(x)$.

Akiva Weinberger

$g$ is $f$ shmooshed, right?

(For $t>1$)

Semiclassical

Right. Dilated for $t>1$, contracted for $t<1$.

Liad

6:15 PM

but we can still find $g$'s coefficients , so what's wrong :/

Semiclassical

I guess my point is that there's no reason to expect these coefficients to be at all nice, or even expressible in terms of the coefficients of $f$.

Liad

huh

this is an exercise :P

in the case $f(x+t)$ it was easy

Semiclassical

You can write down the integrals, of course, but $g(x)$ as such is a $1/t$-periodic function.

You can of course take the portion over $[0,1]$ and periodize that, but it doesn't strike me as an operation that'll play well with the Fourier coefficients.

However, what they might want is something a bit different than this.

Astyx

@AkivaWeinberger Kinda, but not quite

Semiclassical

Namely, suppose I consider the Fourier series over the interval $[0,1/t]$.

Liad

6:18 PM

im getting $\dfrac{1}{t}\int_0^t f(u) e \ ^ {-2 \pi i u/t} du$ ,after using $u = tx$

Astyx

You need two values at which you can compute it

Akiva Weinberger

@Liad @Semiclassical Is $t$ an integer?

Semiclassical

Sure, but now the complex exponential isn't 1-periodic

Liad

@AkivaWeinberger it is not written in the ex. i can ask

$t\ne 0$ , that's given :P

Semiclassical

So you won't be able to interpret those as these as Fourier coeffs of f(u) over [0,1]

Typhon

6:22 PM

i dont get the joke

:p

Akiva Weinberger

That's today's comic

The title is "Existence Proof"

Faust7

anyone working on anything intresting?

thot id take a break from my vacaction today

Liad

you can try my question :P

SteamyRoot

Interesting depends on what you find interesting, I guess.

I'm currently wondering if "characteristic subgroup" is closed under direct products

Akiva Weinberger

Given the (complex) Fourier coefficients of $f(x)$, find those of $f(tx)$

is Liad's question, I think

user91500

6:28 PM

25

Q: Multiple Integral $\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\frac1{1-xyzw}\,dw\,dz\,dy\,dx$

In page 122 of a book by William J. LeVeque, namely Topics in Number Theory (1956), there is an exercise for evaluating the following integral in two ways. $$\int_0^1\!\!\!\int_0^1\frac1{1-xy}\,dy\,dx$$ First way is to write the integrand as a geometric series, $$\int_0^1\!\!\!\int_0^1\frac1{1...

Faust7

@Liad not sure ur being serious but i dont see a link :P

Liad

@Faust7 akiva wrote it

Faust7

lol ok looks less intimidating that 91500

i can barely rember the subject trying to find my book

Axoren

6.282017 is an odd estimate of tau.

Faust7

is there a proof that no perfect number can be odd?

Axoren

6:39 PM

Makes me feel as dirty as those who round $\pi$ to $3$

Faust7

lol

Axoren

Hey @MikeMiller, Do you happen to know the term for curvature when generalized to tangent spaces of higher dimension?

Mike Miller

curvature

Axoren

Still curvature?

Thanks.

Mike Miller

the way it works is you don't actually do it for higher-dimensional tangent spaces. you pick 2-dimensional tangent planes inside the big tangent space to take the curvature of

(I was being a bit snarky; this is called sectional curvature)

Axoren

6:43 PM

Oh, I see. If you do it that way, you're not bothered by all the other dimensions.

I've been trying to do it the hard way.

That's exactly the insight I needed, thanks Mike.

Liad

@AkivaWeinberger @SteamyRoot @Semiclassical $t = -1$

Mike Miller

There is another way to package this in terms of fancy linear algebra / differential geometry, called the Riemann curvature tensor. If you know what tensors are, this is a (3,1) tensor on your manifold. But it's really the exact same data as the above

Akiva Weinberger

@Faust7 It's a famous open problem

Faust7

was curious

Zee

@MikeMiller hey mike

Akiva Weinberger

6:47 PM

So there is no known proof or disproof

Faust7

seems logical that they are all even suprised theres no proof

Zee

Known proof*

Axoren

@MikeMiller Assuming I pick an appropriate Normal vector field, I should be able to do the usual stuff with the Shape operator, right? That's the route I was trying to take, but it was becoming very difficult for me to conceptualize what a normal field for higher dimension tangent spaces would be.

Instead, by using sectional curvature, I still have to make the same decision of which normal I'm picking, right?

Zee

I think mike has me on ignore :/

SteamyRoot

Sectional curvature is intrinsic.

Axoren

6:50 PM

@SteamyRoot It's intrinsic to a chosen plane, isn't it?

Or is it independent of the chosen plane (this doesn't make sense to me just yet, if that's the case)

SteamyRoot

Intrinsic meaning it doesn't depend any space the manifold/surface is embedded/immersed in.

You don't need anything "normal" to talk about the sectional curvature of a manifold.

Axoren

Let's say that I have $M = \mathbb R^n$, at each point $p \in \mathbb R^n$, I have $f(p)$. $X_p = \nabla f(p)$, so there's my vector field. Everything's really general so far.

Now, if I want the curvature at $p$. What decisions do I need to make for that to even make sense.

Mike Miller

@Zee I just wasn't looking at this. But I also haven't paid much attention to most people in the chat.

Akiva Weinberger

Was it Secret with whom I was discussing 4D geometry a while ago?

@Secret I need a "new" set of four mutually orthogonal vectors in $\Bbb Z^4$ that's fundamentally different from the basis vectors

"New" meaning it's not obviously from replacing $(1,0)$ and $(0,1)$ with $(1,1)$ and $(-1,1)$ in subsets of the coordinates

Axoren

@AkivaWeinberger What do you mean by fundamentally different?

SteamyRoot

6:59 PM

@Axoren Assuming we're talking about the sectional curvature, you pick two tangent vectors at $p$ that aren't a multiple of eachother.

Mike Miller

@Axoren I don't think about this in terms of the shape operator in higher dimensions or anything like that. Once you pass beyond surfaces I think it's often easier to just work in terms of coordinate-independent Riemannian geometry.

Akiva Weinberger

@Axoren I'm not 100% sure

Axoren

@SteamyRoot Right, so we need to make a selection of a plane with respect to which we calculate curvature.

SteamyRoot

Yes

Akiva Weinberger

But preferably something weird that can't work in 3D or 2D

Zee

7:01 PM

Quaterionians?

Quaterionins*

I can never spell this word

SteamyRoot

Quaternions?

Axoren

@AkivaWeinberger Use the four-horsemen of the apocalypse. It won't work in 3D because there's 4 of them. That's your new basis.

Zee

Yup

Mike Miller

If you want to think about things in the same way you do surfaces, there's a way of calculating sectional curvature as the curvature of a "canonical" surface sitting inside your larger manifold, but that isn't a surface in 3D space

SteamyRoot

The choice of plane can really change the curvature, too. An easy way to realise this, is to consider like, a generalise cylinder $\mathbb{R}^n \times S^k$

Axoren

7:04 PM

@MikeMiller So, in the case of calculating it with coordinate-independent Riemann geometry, I'm going to have to finish reading this book.

I'm not at that level in my studies yet that I understand the Riemann manifold way of doing things just yet.

SteamyRoot

If you have two vectors pointing in the planar directions, or one in the planar and one in the spherical, you'll have sectional curvature $0$.

But if the two vectors you take lie in the spherical part, you'll have the curvature of a sphere

Liad

$2 \int_0^1 f(u) cos(2 \pi n u) du$ , is it equals zero?

$f$ is 1-period function

Akiva Weinberger

I GOT IT i think

$(1,1,1,1)$, $(-1,-1,1,1)$, $(-1,1,-1,1)$, $(1,-1,-1,1)$

Axoren

One of the main reasons I even want to calculate the curvature in the first place is that I want to try and apply Newton's method, but by restricting it to the geodesic and correcting it by the curvature.

But I'm not optimizing over a sphere, so it's going to take some time before I get to that point in the literature where I understand how to do that.

Akiva Weinberger

That is, the set of four vectors whose last coordinate is $1$, and where an even number of $-1$s appear

Does that work?

Axoren

7:09 PM

@AkivaWeinberger wolframalpha.com/input/…

Akiva Weinberger

Special orthogonal

Woo, that's what I want

Oh, wait, that's about the row-reduced version, which is just $I$.

SteamyRoot

wait, what? (Special) orthogonal?

That thing definitely doesn't have determinant $1$ :O

Akiva Weinberger

Yeah, sorry, it has determinant $16$.

Axoren linked to a computation of the row-reduction. Not sure what that adds.

Axoren

That it is indeed a basis. I thought that's what you asked earlier.

By "does it work"

Akiva Weinberger

The question was about mutual orthogonality

SteamyRoot

7:14 PM

Normalise the stuffs

Axoren

wolframalpha.com/input/…

Akiva Weinberger

@user91500 It was all in response to this question.

25

Q: Multiple Integral $\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\frac1{1-xyzw}\,dw\,dz\,dy\,dx$

In page 122 of a book by William J. LeVeque, namely Topics in Number Theory (1956), there is an exercise for evaluating the following integral in two ways. $$\int_0^1\!\!\!\int_0^1\frac1{1-xy}\,dy\,dx$$ First way is to write the integrand as a geometric series, $$\int_0^1\!\!\!\int_0^1\frac1{1...

real-analysis complex-analysis multivariable-calculus closed-form multiple-integral

Axoren

This is the evaluation of the matrix itself, rather than it's rref

Akiva Weinberger

Don't know if it helps the question, but hopefully it's useful

Axoren

Says it's a Normal matrix.

Akiva Weinberger

7:16 PM

Normal means rows (and columns) are normal to each other?

Axoren

Yeah.

Akiva Weinberger

Wait, does rows orthogonal mean columns orthogonal?

Got to go now-ish

Axoren

I think so, but I'm not sure.

SteamyRoot

Yes, that is the case

A normal matrix is defined as a matrix that commutes with its (conjugate) transpose, afaik.

Balarka Sen

I am dead right now.

Axoren

7:18 PM

Guys, I think I'm a medium.

I hear dead people.

@BalarkaSen What happened? Exam?

Balarka Sen

Only good M Night Shyamalan movie that has ever existed

@Axoren Nah, just had to write a really long report

Axoren

It bothers me that there are some people that saw Avatar twice on purpose.

Balarka Sen

It's a nice movie, but certainly not anywhere on my favorites list.

Or anywhere near

Zee

Avatar is so coool

Axoren

The live action movie was a mistake.

Zee

7:21 PM

The visuals are the best I ever seen

SteamyRoot

@BalarkaSen Anything he made between Sixth Sense and Signs was at least okay, though.

Balarka Sen

@SteamyRoot Unbreakable was ok. I don't remember the rest.

SteamyRoot

It's after Signs it all went to hell

Well, unbreakable is the only one he directed in between :P

He did help write Stuart Little :P

Balarka Sen

lol

SevenSidedDie

For a minute I assumed everyone was talking blue-people Avatar, not whitewashed Avatar. Suddenly this makes much more sense. :)

Zee

7:22 PM

@SevenSidedDie I was also confused

SteamyRoot

I've heard his more recent work is actually okay again

Axoren

Unbreakable was confusing. It was like a superhero origin story but then it just kind of ended.

Like, what was the whole point?

SevenSidedDie

@Zee I started typing the same thing you did, that I liked the mechs and sweeping landscape visuals. ;D

Balarka Sen

No, I meant the blue people Avatar.

Zee

How can you not love blue people avatar?

Aren't you a geometer?

Axoren

7:24 PM

Okay, what?

SteamyRoot

Blue people avatar was pocahontas in space

Axoren

How are those connected?

Balarka Sen

I like it. I don't think I said I do not like it.

@SteamyRoot lololol

@Axoren Yeah it's an unconventional superhero movie. Might be the most realistic one I have seen.

Zee

@Axoren visual cortex

Seems like avatar 2 is on the way

SevenSidedDie

@BalarkaSen But Axoren meant The Last Airbender one, by Shyamalan.

Axoren

7:25 PM

@Zee Most things are connected at the visual cortex.

Balarka Sen

OH.

SteamyRoot

Avatar 2 is constantly being postponed

I think it's now planned for 2020

Balarka Sen

That's a shit movie. I was 30 minutes in.

Axoren

@BalarkaSen Wait, did you think I hated the James Cameron masterpiece?

Balarka Sen

And then I turned my TV off.

SevenSidedDie

7:26 PM

@BalarkaSen (I am relieved that Zee and I aren't alone in our confusion! :D )

SteamyRoot

(but, on the other hand, they're also planning a 3, 4, and 5 - and they should all be done by 2025)

Axoren

That one was NOT M. Night Shama$\dots$

Zee

@Axoren I was implying that both of these things are highly visual

Axoren

I forget how to spell the rest of his name

SevenSidedDie

@Axoren …amalamalayamalan

Balarka Sen

7:27 PM

@Axoren vOv man

SteamyRoot

I think people that never saw TLA animation possibly almost didn't hate the live action movie.

Axoren

I know the rest of his name, but there is not enough room in this margin.

Zee

For some reason when I was walking with my painting board in the math dept, only the geometry students and professors took strong notice

Balarka Sen

I am not a big fan of James Cameron. I like some Hollywood movies, but I put most of them away from my favorites list. It's like, a different level of liking.

SteamyRoot

To anyone else, it feels like, well, trash :P

Axoren

7:28 PM

It was kind of upsetting to me because I did see the animation first, as well as the behind the scenes of the animation

Where they actually got martial arts experts to plan out the movements and how it would translate to manipulating the elements.

And then you had like 50 dudes doing a song-and-dance number to throw like 5 rocks at some guards with Earth bending.

I personally find James Cameron to be an interesting person, regardless of my perceptions of him as a director.

SevenSidedDie

I liked the visuals of blue-people!Avatar, but the plot didn't do it for me, and the attempt to gracefully provoke thoughts of race-relations and colonialism were hamhanded and backfired in a bunch of ways. But it was visually stunning and I don't regret seeing it in 3D in the theatre. It's just not his best work, when there's stuff like The Terminator and Aliens in contention for that honour.

Axoren

He's the kind of guy keep high up on the list for that "fly to the meteor and blow it up" mission, just because he's the kind of guy to go do it.

SteamyRoot

youtube.com/watch?v=67MRiMT9cnE mandatory

Zee

Oh god

Balarka Sen

Avatar is an interesting attempt at critiquing colonial relations. There exists better attempts.

SevenSidedDie

7:32 PM

@BalarkaSen That's a good way to put it.

Axoren

You know. I never thought about this.

SteamyRoot

Axoren

But were there other continents they could have colonized that didn't have the Navi (sp)?

SteamyRoot

I think the point was that they could find unobtanium in a lot of places

But the home tree was on top of the largest concentration

Fargle

Even if they could have gone elsewhere, it was much more convenient for them to keep screwing over the native population.

Balarka Sen

7:39 PM

I think there was a quote on directors by directors where at some point David Cronenberg was asked about M Night Shyamalan and he literally said "I hate that guy!! Next question". I can't seem to find it.

Fargle

That was more or less the point--those engaged in exploitation don't care, they merely want maximum material utility for themselves and their own.

Balarka Sen

Ah, there

Fargle

Kevin Smith got his revenge quite nicely. I had forgotten about that exchange.

Balarka Sen

I laughed a lot on that.

There's another where Herzon says Godard is intellectual counterfeit money compared to a good kung fu movie

Fargle

good Lord

Sha Vuklia

7:50 PM

@Steamy als je $x$ links vermenigvuldigd met $a$, dan doe je toch $ax$? of juist $xa$?

SteamyRoot

$ax$, zou ik zeggen... maar het klint toch een beetje dubbelzinnig :P

Balarka Sen

Godard on Tarantino: He named his production company after one of my films. He'd have done better to give me some money.

Sha Vuklia

@Steady ja vond ik ook, thanks!

Balarka Sen

Heh, this math major called Shane Carruth has made two completely mucked up films which has apparently gained a cult following. Gotta watch these next time.

user193319

8:06 PM

I realize this may be overkill, but I just want to make sure my reasoning is correct. Consider the sequence 0,1,0,0,1,0,0,0,1,... I just recently proved that if a sequence $x_n$ in some topological space converges to $x$, then $x$ must be a limit point of $S = \{x_n ~|~ n \in \Bbb{N} \}$. In our case, $S = \{0,1\}$, which means that $S$ is closed (in $\Bbb{R}$ with the standard topology) and therefore contains its limit points.

Therefore, if the sequence does converge, it must converge either to $0$ or $1$, which clearly cannot happen, since given $\epsilon = \frac{1}{2}$, we can always find an $n \in \Bbb{N}$ such that $|x_n - x| = 1$.

Fargle

Your reasoning seems spot-on. Overkill, sure, but sometimes it's good to do overkill just to see that the more powerful method does in fact work.

user193319

@Fargle Yes. Definitely overkill! Thanks!

Mary Star

8:25 PM

Hello!! We have a rectangle triangle ABC. The angle BCA is 30 degree and length of the side AB is equal to $\sqrt{3}$. The bisection of the angle CBA intersects D.
We want to calculate the length of the side CD.

I have done the following:

Since the angle A is 90 degrees and C is 30 degrees, we get that B is 60 degrees.
From the law of sines at the triangle ABC we have the following: $$\frac{\sin C}{AB}=\frac{\sin B}{AC} \Rightarrow \frac{\sin 30^{\circ}}{AB}=\frac{\sin 60^{\circ}}{AC} \Rightarrow \frac{\frac{1}{2}}{\sqrt{3}}=\frac{\frac{\sqrt{3}}{2}}{AC} \Rightarrow \frac{1}{2\sqrt{3}}=\f…

Hippalectryon

8:55 PM

@MaryStar Why don't you just work in BAD, since you know all of its angles ? The definition of sin/cos/tan gives you DA right away, and in ABC it gives you CA right away, just subtract the two

Mary Star

Do you mean the following?

From the law of sines at ABC we get the following: $$\frac{\sin C}{AB}=\frac{\sin B}{AC} \Rightarrow \frac{\sin 30^{\circ}}{AB}=\frac{\sin 60^{\circ}}{AC} \Rightarrow \frac{\frac{1}{2}}{\sqrt{3}}=\frac{\frac{\sqrt{3}}{2}}{AC} \Rightarrow \frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{2\cdot AC} \Rightarrow AC=3$$
From the law of sine at BDC we get the following: $$\frac{\sin \frac{B}{2}}{DC}=\frac{\sin C}{BD} \Rightarrow \frac{\sin 30^{\circ}}{DC}=\frac{\sin 30^{\circ}}{BD} \Rightarrow BD=DC$$

Hippalectryon

@MaryStar Not really. Simply: $\tan(ABC)=\frac1{\sqrt3}=\frac{CD}{\sqrt3}\Rightarrow CD=1$. Likewise, $\tan(ABD)=\sqrt3=\frac{DA}{\sqrt3}\Rightarrow DA=3$. Therefore $CD=CA-DA=2$

Mary Star

9:10 PM

@Hippalectryon Ahh ok!! I see!! Thank you so much!! :-)

Hippalectryon

:-)

Danu

9:31 PM

@ShaVuklia Wait... you're Dutch? I didn't realize...

Dodsy

Netherlands

also Danu

my school (university) offers enriched physics and normal physics

do you think I should take the enriched physics?

physics.uwo.ca/undergraduate/current_students/files/…

Danu

What is that supposed to mean? :P Enriched?

Dodsy

hmm

that's a good question.

I guess in this case, they try to apply it to more things.

@Semiclassical

Danu

I don't like applications, myself. Depends on what you're into

Dodsy

hm.

Fargle

9:39 PM

It pains me to know that anything I ever do might one day be useful, personally.

I'm a hard purist.

Astyx

Such a burden to feel actually useful

Balarka Sen

I like that.

Astyx

So do I. A lot.

Balarka Sen

We three might be homeboys, you know

Dodsy

?

Astyx

9:42 PM

"God damnit, I've actually done something meaningful."

Balarka Sen

you two should def read my personal bible

Astyx

Send me a djvu, I'll read it

Balarka Sen

The Underground is available right at your doorsteps.

Hole up anytime!

Astyx

I take it your real name is Dostoevsky

Hippalectryon

Fyodor Dostoyevsky

Fyodor Mikhailovich Dostoyevsky (; Russian: Фёдор Миха́йлович Достое́вский; IPA: [ˈfʲɵdər mʲɪˈxajləvʲɪtɕ dəstɐˈjɛfskʲɪj]; 11 November 1821 – 9 February 1881), sometimes transliterated Dostoevsky, was a Russian novelist, short story writer, essayist, journalist and philosopher. Dostoyevsky's literary works explore human psychology in the troubled political, social, and spiritual atmosphere of 19th-century Russia, and engage with a variety of philosophical and religious themes. He began writing in his 20s, and his first novel, Poor Folk, was published in 1846 when he was 25. His most acclaimed works...

Astyx

9:50 PM

I know, I know :p

Balarka Sen

It ends with "vsky", so close enough.

I don't like the translation in the pdf I linked. Surely there are better ones out there.

Astyx

$\det(e^A) = e^{Tr(A)}$ right ?

nigel

Let V_\lambda be a highest weight rep of a lie algebra g, and let W be the weyl group of g. let v_0 \in V_\lambda be a highest weight vector. let w \in W. why is it the case that w.v_0 has weight w.\lambda?

Balarka Sen

@Astyx True for diagonal matrices, for one. I guess you can prove from there.

Astyx

density

Balarka Sen

9:52 PM

Prove for diagonalizable matrices, and then note any matrix can be approximated by diagonalizable matrices?

C-H argument

Astyx

continuity

Balarka Sen

The golden words.