Mathematics - 2019-06-04 (page 2 of 2)

Ted Shifrin

18:00

No, I don't remember why I started.

Maybe rebellion against your apostrophes.

skullpatrol

rebel :P

Balarka Sen

lol

Ted Shifrin

@Mathphile: Have you considered something like unique factorization? Any prime factor of $x!$ must divide $x$.

@nbro I don't know what the derivative of a function with respect to a function is. If $f$ is a function of $x$, and $\Delta x$ is an independent variable, how in the world is $f$ a function of $x+\Delta x$?

Rithaniel

Heya Ted

Ted Shifrin

hi @Rithaniel

MatheinBoulomenos

18:10

@Alessandro si dice p.es. "un gelato della fragola" oppure "un gelato con la fragola"?

Ted Shifrin

oh, hey @Mathein

MatheinBoulomenos

hey @Ted

buonasera! @Alessandro

Ajay Mishra

@Tanvir You there?

FuzzyPixelz

Hello, I'm trying to show that the direct sum of the line spanned by $u=(1, 0, \dots , 0, \dots)$ and its orthogonal complement (the set of sequences with $u_1=0), is not equal to $l^2(\mathbb N)$

Ted Shifrin

Are you sure?

All sequences with $u_1=0$ or all $\ell^2$ sequences?

In other words, orthogonal complement in what Hilbert space?

FuzzyPixelz

18:17

Oh in $\ell^2(\mathbb N)$

Ted Shifrin

So what makes you think the whole space isn't the direct sum?

FuzzyPixelz

Well, as far as I'm concerned it is the direct sum, but I stumbled upon a problem that claims this, the only thing I changed was that instead of the first element being $1$ they chose it for some $n$

Ted Shifrin

I don't believe it.

Balarka Sen

Can you clearly state what the vector space in the other problem is?

If you have a closed subspace of a Hilbert space it always decomposes as that and it's orthogonal complement

Tanvir

@AjayMishra yes

FuzzyPixelz

18:24

It's just $\ell^2(\mathbb N)$, isn't it infinite-dimensional ?

Balarka Sen

No, I mean, what is the subspace?

Ted Shifrin

growls at Balarka's apostrophe

FuzzyPixelz

Oh the line spanned by the vector $u=(1, 0, \dots, 0, \dots)$

Ajay Mishra

@Tanvir How you figure out azimuthal part? or you are getting in that part?

Balarka Sen

@FuzzyPixelz $\ell^2(\Bbb N)$ is $\langle (1, 0, \cdots ) \rangle \oplus \langle (1, 0, \cdots ) \rangle^\perp$. I was referring to what the precise problem in the other thing you saw that made you think otherwise.

Ted Shifrin

18:26

@FuzzyPixelz: Somewhere recently I saw a post about sequences that only had had finitely many nonzero terms as a subspace of $\ell^2(\Bbb N)$ or $\ell^\infty(\Bbb N)$. Are you sure you're not confusing things?

Balarka Sen

Sorry @Ted

Ill try be more careful next time

Ted Shifrin

LOL

Tanvir

@AjayMishra for the hemisphere I get that but for a spherical cap did not get that

FuzzyPixelz

My text precisely says that they consider the set of all sequences $(u_n)$ in $\mathbb R ^ {\mathbb N}$ such that $\sum_n u_{n}^2$ converges, with the typical inner product

Alessandro Codenotti

@MatheinBoulomenos "un gelato della fragola" is wrong, "un gelato con la fragola" means an ice cream with a strawberry added on top, "un gelato alla fragola" means a strawberry flavoured ice cream

MatheinBoulomenos

18:30

@AlessandroCodenotti ah of course!

grazie!

Alessandro Codenotti

Di niente!

Are you studying Italian in your break semester?

Ajay Mishra

@Tanvir It seem to require another parameter.

MatheinBoulomenos

I'll try at least

FuzzyPixelz

I could post the problem here but it's in french.

Balarka Sen

@Ted knows French

Abdullah UYU

18:32

:)

MatheinBoulomenos

je suis une baguette

Tanvir

@AjayMishra Yeah I also think so. Either height of the cap or polar distance from center of sphere

Alessandro Codenotti

I approve of that choice @Mathei :P

Ted Shifrin

Balarka, Ted is trying to unravel the nonsense in this, but Fuzzy's will certainly be easier.

Balarka Sen

I bet what you have in mind is the subspace of $\ell^2(\Bbb N)$ spanned by $a_1, a_2, \cdots$ where $a_n = (0, \cdots, 1, \cdots, 0)$ with $1$ at the $n$'th plane.

That is not closed: it's what Ted said, the subspace of finite sequences with trailing zeroes

Ajay Mishra

18:33

@Tanvir I took height as the parameter, but that is not helping at all.

FuzzyPixelz

Yes

Tanvir

@AjayMishra Did you try with angle?

Balarka Sen

@FuzzyPixelz Yes, OK. So your scenario and the scenario in the problem are different.

Ajay Mishra

which angle, specify it please!

Balarka Sen

The line spanned by $a_1$ is a closed subspace, the subspace spanned by $a_1, a_2, \cdots$ is not.

Ted Shifrin

18:34

Yeah, @Fuzzy, you're thinking $n$ is fixed, and they're taking the span for all $n$, as @Balarka suggests.

MatheinBoulomenos

overkill approach: a Banach space is never countably-infinite dimensional by Baire

Tanvir

@AjayMishra check here theta angle. en.wikipedia.org/wiki/Spherical_cap

FuzzyPixelz

I'm embarrassed now, thank you.

Ted Shifrin

I think it's confusing, actually. But they never said $n$ was fixed.

Tanvir

@AjayMishra that also defines how large will be the cap. if it is 90 means its a hemisphere

Balarka Sen

18:36

@TedShifrin Looks scary

FuzzyPixelz

I asked the person who gave me this whether it's fixed or not and they didn't give me a clear answer

Ted Shifrin

Some day you should learn about Chern-Simons, Balarka.

MatheinBoulomenos

@Alessandro è difficle parlare italiano nella vita quotidiana, volo fare forse un "tandem"

Balarka Sen

@TedShifrin Roughly, what's it about?

I don't have a clue

Ted Shifrin

You know what transgression is, right?

Balarka Sen

18:38

Yup, those arrows in the spectral sequences I never really understood, right?

Ted Shifrin

Doesn't need to be coming from a spectral sequence.

Balarka Sen

Eg one which comes in Gysin sequences but that's easier to interpret as cup product with the Euler class

Huh, I see, I haven't heard of them in other contexts

MatheinBoulomenos

posso parlare solo un piccolo, piccolo po' dell'italiano, probabilmente troppo poco per fare un "tandem" (non so come si dice "tandem" nell'italiano)

Ted Shifrin

Basically, when a cohomology class vanishes (say a 4-dimensional class on a 3-manifold), you can get a canonical "antiderivative" (in this case a 3-form) that is interesting. That's the transgression.

Balarka Sen

Oh interesting

Ted Shifrin

18:42

Gauss-Bonnet has something similar, working on the frame bundle. That's how Chern got the invariant (generalizing geodesic curvature for surfaces) to integrate over the boundary in the case of a Riemannian manifold with boundary.

Tanvir

@everyone, Can you please have a look in this problem? stuck for a quite long time.

0

Q: Area of a spherical cap crossing an intersection

I have a sphere that equally divided in to two hemisphere P and S. There is a plane that separate two different zone. Upper zone called A and lower zone called B. The angle $\alpha$ defined where the sphere is touching the intersection plane.So $\alpha$ larger means sphere is more in zone A and v...

Balarka Sen

I always did the moving frames proof locally but never tried to see what happens if I do it globally on the frame bundle. I should try that out once

Ted Shifrin

So Chern's genius was to intuit expressions in connection and curvature forms to get a combination whose derivative would be the Pfaffian of curvature (namely, the integrand in Gauss-Bonnet).

For surfaces, it's all trivial, of course.

Balarka Sen

Ah.

ÍgjøgnumMeg

Really struggle with people who pronounce "Weierstraß" as "Weierstrauß"

MatheinBoulomenos

18:45

lol

is that really a thing?

Ted Shifrin

I concur, @ÍgjøgnumMeg.

ÍgjøgnumMeg

yeah it happens all the time lol

Ted Shifrin

I even had a brilliant professor who misspelled it.

ÍgjøgnumMeg

Silverman is doing it...

MatheinBoulomenos

Weierbouquet

Ryan Unger

18:46

Leebniz

ÍgjøgnumMeg

Reimann

Balarka Sen

That ^

is the best

Ryan Unger

Youler

ÍgjøgnumMeg

hahaha

GFauxPas

Nuth

Ted Shifrin

18:46

I think this is a good time for a lunch break for Ted.

ÍgjøgnumMeg

Teed

Balarka Sen

Shiffrin

ÍgjøgnumMeg

Teed Sheefreen

@Ted forgive me

MatheinBoulomenos

that reminds me, my mother is an English teacher and she had a student translate "Vogel Strauß" as "bird bouquet"

ÍgjøgnumMeg

hahaha wat

MatheinBoulomenos

18:48

not to mention the translation of "Der Menschenauflauf löste sich auf" as "The human gratin dissolved"

Ted Shifrin

So what is a bouquet of birds?

ÍgjøgnumMeg

the fuuuh

Ryan Unger

does anyone know what the set of "critical values" of the devil's staircase is? It's not all of [0,1]...is it dense. Full measure?

critical values meaning the values of the intervals on which it's constant

MatheinBoulomenos

"Vogel Strauß" means ostrich

Ted Shifrin

oh, how in the world did that come to be?

Balarka Sen

18:50

@RyanUnger Complement of the Cantor set, right?

Ted Shifrin

because of all the colors?

Ryan Unger

@BalarkaSen I mean the values on the y axis

ÍgjøgnumMeg

Apparently "Weier" translates to "fish tank"

Weierstraß was just an aisle in a pet shop

Ryan Unger

complement of the cantor set has connected components which are more than points

Balarka Sen

Oh, what it maps to

Ted Shifrin

18:52

VALUES.

MatheinBoulomenos

Strauß comes from the Greek στρουθίον (strouthion), from where the Latin avis struthio is derived, which turned into the english ostrich

I think there are no actual bouquets involved, seems like a homonym

ÍgjøgnumMeg

That's a very roundabout etymology

Balarka Sen

@RyanUnger Take a Cantor set and mark all the midpoints of each of the intervals during the one-third removal in the process of building the Cantor set

That's the thing, I believe

Ryan Unger

so uncountable but not dense and not full measure

it has measure zero I mean

Balarka Sen

Ya

MatheinBoulomenos

18:54

yes, I checked "Vogel Strauß" and "Blumenstrauß" have completely unrelated etymologies

Ryan Unger

but then maybe you can do this with a fat cantor set

scary thought

Alessandro Codenotti

@MatheinBoulomenos voglio*

But it sounds like a good idea

MatheinBoulomenos

@AlessandroCodenotti grazie

I'm just worried my Italian is too bad for even basic conversations

Balarka Sen

More precisely, I mean the midpoints of each of the intervals in $[0, 1] \setminus C$

skullpetrol

Why isn't Italian a language of modern math?

Balarka Sen

18:58

@RyanUnger It's countable, isn't it.

FuzzyPixelz

Alright, so with this orthogonal complement of $F$ must reduce to $\{0\}$, but then $F$ alone contains the non-square-summable sequences of $\ell^2(\mathbb N)$. I call it a day.

MatheinBoulomenos

penso che il mio italiano sia troppo male per fare un tandem

Balarka Sen

Right, @FuzzyPixelz, the subspace of eventually zero sequences in $\ell^2(\Bbb N)$ is dense.

FuzzyPixelz

Smells like topology, I don't know any topology, but thanks nonetheless.

skullpetrol

\o @Daminark

Daminark

19:00

Hey there ya lot of nerds

Ryan Unger

you dont know topology but are trying to work with Hilbert spaces?

ÍgjøgnumMeg

Hey @Daminark

MatheinBoulomenos

you just added one to the counter @Daminark

FuzzyPixelz

Actually, my course focuses on Euclidean spaces and occasionally inner product spaces, I was just trying to get a conter-example.

Rithaniel

Hey, I am much more of a dork than a nerd, thank you very much.

skullpetrol

19:03

nerd vs dork

Daminark

:0

skullpetrol

vs bookworm

Rithaniel

Join the Dork Side

3

Balarka Sen

Daminork

skullpetrol

Dork Vader

Ryan Unger

19:05

@BalarkaSen basically I am wondering how bad the critical values of a graph can be

can a map $I\to I$ have dense critical values?

Ajay Mishra

@Tanvir Do you know the answer?

Ryan Unger

I know it's possible for $R\to R$

Balarka Sen

the rational bumps thing

Ryan Unger

yeah

Balarka Sen

i dont want to think about that it seems awful

Alessandro Codenotti

19:07

@MatheinBoulomenos just go ahead and try, practice is the best regardless of your level

Ryan Unger

does that work for $I\to I$? I guess it does if you make the bumps go like $1/2^k$ width

Balarka Sen

should

Ryan Unger

"seems awful" yeah welcome to this project

Tanvir

@AjayMishra Nope. Thats why I post it. I have edited the question. It might be helpful

@AjayMishra So far nobody can answer this problem. Not even a hint.

Ajay Mishra

I will.

Balarka Sen

19:09

its like you have to have the height of the bumps converge to whatever the values at $0$ and $1$ are

which seems complicated and ugh

Ajay Mishra

I've written down everything, atlast i realize that it is 3 D, I have everything prepared.

Balarka Sen

i am leaving the chat out of pure disgust

Tanvir

@AjayMishra Lets see. Can you upvote it? so that it draw attention.

Ryan Unger

hmm

Ajay Mishra

My rep = 7

Tanvir

19:11

ohh ok..no worry then

UserX

20:53

@BalarkaSen order of the quotient grp is 6 cause the lcm(ord[2]_4, ord[4]_6) is 6 and Z4 X Z6 has order 24, so S3 and Z6 are the two isomorphism classes

How do I check if the quotient grp is isomorphic to any of those without computing its elements?

Avantgarde

21:29

Hey guys. I have a question.

Ultradark

23:15

Ask away avant le garde

Pour vous avez veux ca

Un bonne phrase c'est bien connaitre: Juste demander; ne pas demander a demander.

Ryan Unger

Parlay voo

Ultradark

Oui je suis courrament en francaise

Ryan Unger

wee wee

Ultradark

Question: What is a cotangent space? It says it's the dual space of the tangent space but I don't understand still.

Okay after reading a bit more I think I understand better

Is there a way to formalize how malleable a symplectic manifold is compared to a quite rigid Riemannian manifold

GFauxPas

23:33

you can think of them as the span of the $1$-forms

the cotangent spaces

Ultradark

okay

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$Mathematics$ Mathematics