Mathematics - 2018-02-05 (page 5 of 6)

Kasmir Khaan

19:00

Do we declare that 2 and 7 are the same for example, in Z/5Z

Ted Shifrin

There are no such things as 2 and 7.

But now you're back on the beginnings of algebra. This has nothing to do with vector spaces.

Kasmir Khaan

hmm that Point is very comfusing

Mambo

@Ted

Ted Shifrin

Aw come on. You've spent months that elements of $\Bbb Z/5\Bbb Z$ are equivalence classes.

Kasmir Khaan

I know Ted

Ted Shifrin

19:01

hi @Mambo. Any progress to report?

Kasmir Khaan

But am asking, are matrices with entries in Z

a vector space over the field Z/5Z for example

it makes no sense to me

Ted Shifrin

No, you cannot multiply integers by elements of $\Bbb Z/p\Bbb Z$.

Mambo

i guess there is no such constructive function. Maybe pull Baire Category or something

Kasmir Khaan

okay makes sense

Ted Shifrin

OR maybe there actually is none, @Mambo?

Kasmir Khaan

19:02

so we need the entries in Z/5Z

neat thanks Ted :D

Ted Shifrin

Right, Kasmir.

Alessandro Codenotti

@TedShifrin pfff maybe I'll think about it after dinner

Ted Shifrin

Or some field extension of $\Bbb Z/5\Bbb Z$ will work.

Kasmir Khaan

Did not read that field extension yet ><

I really so eager to learn about it

sounds fun :D

Ted Shifrin

Those are, in fact, vector spaces over $\Bbb Z/5\Bbb Z$.

Kasmir Khaan

19:03

how we add one root to a poly , open up lots of things D

Ted Shifrin

And you could use matrices with entries in those ...

Kasmir Khaan

Yes yes :D did not sound right to me at start, but wanted to make sure :D

Tuki

Our lecturer said that $\nabla f(x,y)$ is always the direction of largest increase in $f(x,y)$ values and $ - \nabla f(x,y)$ is the largest decrease but is this true ?

Kasmir Khaan

@Tuki where r u from Tuki if you dotn mind me asking

Tuki

Finland @KasmirKhaan

Kasmir Khaan

19:05

Nice :D

Tuki

How about you @Kasmir ?

Ted Shifrin

Yes, @Tuki, it's true. Look at the formula for directional derivative in terms of $\nabla f$.

This is one of the fundamental, important ideas in multivariable calculus.

Kasmir Khaan

@Tuki living in Sweden atm but not from here =p

Tuki

I think i got confused by a which had almost saddle point but it wasn't point such $\nabla f(x,y)=0$ satisfies this condition.

Mithlesh Upadhyay

2

Q: GATE $2018$ - Probability question : occurrence of green and red faces when die is rolled $7$ times

A die is colored green on $4$ sides and red on $2$ sides. if the die is rolled $7$ times. Then which of following option is most likely regarding occurrence of green and red faces? (A) $2$ green and $5$ red (B) $3$ green and $4$ red (C) $4$ green and $3$ red (D) $5$ green a...

Tuki

19:08

$\nabla f(x,y)=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ it was

Ted Shifrin

Saddle points can only occur at points where the gradient vanishes, @Tuki.

MatheinBoulomenos

@KasmirKhaan I failed it intentionally

Hi @KasmirKhaan @Ted @Alessandro

Tuki

you mean points where $\nabla f(x,y)$ does not exist ?

@ted

Ted Shifrin

No, I mean it's the zero vector.

Tuki

oh yes exactly

Ted Shifrin

19:10

On the saddle surface $z=x^2-y^2$ only the origin is an actual saddle point.

Kasmir Khaan

@MatheinBoulomenos No problem, I did that once too, I knew I could get better grade and I like the subject so =p but I Think for your level, you either get an A or dont do it =p

MatheinBoulomenos

The thing is I could have easily gotten an A-, but there was one formula I didn't know, so I couldn't get an A

Tuki

well critical points can exist in point where gradient is not defined but saddles points cant right ?

Ted Shifrin

@Tuki: For me a critical point is a place where the function is differentiable and the derivative (or gradient) is zero.

Your textbook may have a different definition.

Tuki

yes textbook i have would include points where gradient is not defined as well

Ted Shifrin

19:12

I don't like that, but I accept the fact that some books do it.

MatheinBoulomenos

I've already got three As and one A- in that part of the bachelor and at most 4 courses from that area count for the bachelor, so getting an A- would be completely useless for me

Daminark

'hello nerds

MatheinBoulomenos

Since I failed I can write the 2. exam

Hey @Daminark

Mithlesh Upadhyay

please check someone my question?

Daminark

How's it going?

Mithlesh Upadhyay

19:15

2

Q: GATE $2018$ - Probability question : occurrence of green and red faces when die is rolled $7$ times

A die is colored green on $4$ sides and red on $2$ sides. if the die is rolled $7$ times. Then which of following option is most likely regarding occurrence of green and red faces? (A) $2$ green and $5$ red (B) $3$ green and $4$ red (C) $4$ green and $3$ red (D) $5$ green a...

probability expectation

Ted Shifrin

Maybe Demonark failed his exam on purpose, too.

MatheinBoulomenos

I think that's different in the US, isn't it?

Kasmir Khaan

@MatheinBoulomenos mathein you are a beast :D i only got an A once , but lots of B's =p

MatheinBoulomenos

@Daminark Well, my advanced complex analysis didn't go so well, you can read the details above

How's it going for you?

Daminark

Functional has killed me but not intentionally

Now luckily it seems like it killed everyone, even the grad students

But self-confidence is permanently shot

MatheinBoulomenos

19:17

Yeah, what you told about the course sounded really advanced

more advanced than the functional course I took

Ted Shifrin

Your self-confidence is immune to being shot, Demonark. We've established that.

More of Chicago craziness :P

MatheinBoulomenos

Did you have your other exams go well?

At least I've got a ring theory badge now, so there's that

Daminark

Algebra was alright. I mean even functional probably won't literally fail but like... I'm starting to wonder if taking it was a good idea

@TedShifrin have we now? That's a good sign I guess

MatheinBoulomenos

my Master application will be like "I failed my bachelor, but look at all those tag badges"

(that I wil hopefully have then)

Ted Shifrin

@MithleshUpadhyay: My intuition certainly is that D is most likely. I just computed C and D and it's right.

Mithlesh Upadhyay

19:22

@TedShifrin , thanks I answered D, but need a mathematical explanation.

MatheinBoulomenos

@MithleshUpadhyay you didn't specify that all sides of the die are equally likely and the multiple rolls are independet. It's impossible to solve

Ted Shifrin

You just have to calculate $\binom{7}{5}(2/3)^5(1/3)^2$ and compare it to similar other things.

Daminark

@MatheinBoulomenos I've never heard of those

Ted Shifrin

You should be able to do it without brute force. Compare $\binom{7}{4}$ and $\binom{7}{5}$, and compare $(2/3)^5(1/3)^2$ and $(2/3)^4(1/3)^3$ without computing them.

Mithlesh Upadhyay

@MatheinBoulomenos , this was there, I'm editing there.

MatheinBoulomenos

19:24

@MithleshUpadhyay I'm just joking

Mithlesh Upadhyay

@MatheinBoulomenos , in responce, I can troll you, as I do on Twitter :)

0celo7

@XanderHenderson i.gyazo.com/8990b0b8f7ce59336590748d9671f645.png

Xander Henderson

GAH!

Mithlesh Upadhyay

@TedShifrin , how is this relevent,sir ?

Ted Shifrin

Huh?

How is it NOT relevant?

Mithlesh Upadhyay

19:34

@TedShifrin , we have to calculate expectation, right?

Ted Shifrin

Not expectation, but probability of given events.

I'm saying that you can show D is largest by just simple reasoning, without using calculators.

Kasmir Khaan

@MatheinBoulomenos if your not busy , Id like to ask you about soemthing, not ( math perse ) =p

MatheinBoulomenos

@KasmirKhaan okay, sure

Kasmir Khaan

@MatheinBoulomenos invited you to our room =p

Ted Shifrin

@MithleshUpadhyay: I'm saying that when you compare the case of 5 greens with the case of 4 greens, the probability changes by a certain factor. $\binom 75 = 21$ and $\binom 74 = 35$, so that factor increases by $35/21$, but the product $(2/3)^5(1/3)^2$ goes to $(2/3)^4(1/3)^3$, which is a net factor of $1/2$. $\frac{35}{21}\cdot\frac12<1$.

Mithlesh Upadhyay

19:42

, I agreed, could you please explain/answer there, if possible.
Thanks in advance.

it's 1:15 AM in New Delhi, I need to sleep.

MatheinBoulomenos

You're in good company as an unsleeping Indian

Mithlesh Upadhyay

@MatheinBoulomenos , I will see you tomorrow :)

MatheinBoulomenos

Oh I didn't mean myself, I meant Balarka

Adeek

HI @MatheinBoulomenos @TedShifrin

I am gonna apply some yoga on you guys and turn you into invariants

hahahaha

MatheinBoulomenos

Hi @Adeek

Adeek

19:50

@MatheinBoulomenos hahaha

@MatheinBoulomenos in regards to tags lolz

I am gonna learn complex analysis today

@TedShifrin @MatheinBoulomenos In my talk I will mention how there is no magic in our world, because we are governed by laws of science. But, magic does occur in complex analysis

hahaha

I am in a good mood today

MatheinBoulomenos

It's true that complex analysis feels magical

Adeek

yeah

Mambo

Think about a closed rectifiable curve that has every possible winding number

Abcd

20:09

$\arcsec x + \arcsin x = \dfrac{\pi}{2}$

Only possible values of x is $1,-1$.

Because of domain constraints.

But how does it make $x= \pm 1$ the solution?

Just because x is in the domain of both, doesn't imply that it will necessary satisfy the equation.

But answer is $\pm 1$. Why is my reasoning wrong?

values of x are*

Sorry please ignore all the above messages.

I am sleepy and being silly!

0celo7

20:27

@EricSilva go say hi to TB tomorrow. You wrote those notes for her

Eric Silva

20:40

Who dat

0celo7

@EricSilva she's giving a talk at Chiraq tomorrow

or was it today

Slereah

lol Chirac

0celo7

oh no it's ML

@EricSilva call him a hipster

TB is next week

Eric Silva

21:06

Oh at the geom analysis seminar I see

O rip I can't go to those they're during my labs

Dattier

21:21

@AkivaWeinberger ok

MatheinBoulomenos

21:35

Is every real $n$-dimensional $C^\omega$ manifold a $C^\omega$-submanifold of a $n$-dimensional complex manifold? I think this should be true, because locally every convergent real power series also converges with the same radius if you consider it as a complex power series, but I'm not sure how to patch that together to get a complex manifold

Hi @AlessandroCodenotti

Alessandro Codenotti

Random fact of the day: today is the 10316th day since the fall of Berlin's wall, which is also the number of days it stood

Hi @Mathei is every archimedean ordered field an extension of $\Bbb Q$ contained in $\Bbb R$?

MatheinBoulomenos

@AlessandroCodenotti yes

there's even a unique order-preserving embedding into $\Bbb R$ for every such field

Alessandro Codenotti

Even stronger if $f:A\to B$ is a morphism of archimedean ordered fields it's also the only one

MatheinBoulomenos

I don't know the proof off the top of my head, it's in Milne's notes on field theory iirc

Alessandro Codenotti

Well not actually stronger, there could be some bigger than $\Bbb R$ as I stated it

@MatheinBoulomenos thanks for the reference, I'll check it

PVAL-inactive

21:41

What's an archimedean field bigger than R?

MatheinBoulomenos

@PVAL-inactive there is none

Alessandro Codenotti

There is none, but I wasn't sure whether that's the case or not

MatheinBoulomenos

@AlessandroCodenotti you need order-preserving, because else $A=B=\Bbb Q(\sqrt{2})$ is a counterexample

Alessandro Codenotti

I agree, a morphism of ordered fields is order preserving with the terminology we used in the foundations of maths course

MatheinBoulomenos

If you have an ordered field that is closed under taking square roots of positive elements (e.g. constructible (in the geometric sense) real numbers, real closed fields), then any automorphism is order-preserving

Alessandro Codenotti

21:45

(Which makes sense if you think about ordered fields as first order structures and morphisms as morphisms of structures)

@MatheinBoulomenos sure, because then the order relation is definable in terms of the field operations

That's the trick used in showing that $\text{Aut}(\Bbb R)$ (as a field) is trivial

Dom

Hi fellas, if a derived equation contains a -b, where the original equation contained an unsigned b, does that mean flip the sign on b, or flip to negative only if positive?

does that make sense?

PVAL-inactive

@MatheinBoulomenos There are maps like x \mapsto x^3, which is a local diffeo of the Reals around zero but not of \Bbb C

I bet there's some obstruction to this.

MatheinBoulomenos

@PVAL-inactive hmm, true

Dom

Sorry, I didn't know whether to ask if I could ask first or just ask straight out

I think it looks a bit rude

sorry if I offended anyone

PVAL-inactive

oh i guess that stuff doesn't have real-analytic inverse.

What a strange category.

MatheinBoulomenos

21:52

$x \mapsto x^3$ doesn't have a smooth inverse either

it's not a local diffeo

PVAL-inactive

There's still power series with critical points off the real line.

so you cant replace x mapsto S(x) with z mapsto S(z) willy nilly

MatheinBoulomenos

yes, but they're outside the radius of convergence

Hi @MikeMiller any idea on this: chat.stackexchange.com/transcript/message/42693458#42693458 ?

Mike Miller

You can embed them all in $\Bbb R^{2n}$ I believe

PVAL-inactive

yeah but the construction @MatheinBoulmenos suggest they're real submanifolds of a complex n-fold.

MatheinBoulomenos

I'm quite sure what I wrote works locally, but I don't know how to patch it together

Mike Miller

22:02

What I gave is complex

But that comes from the Morrey-Grauert theorem which lets you embed in Euclidean space

PVAL-inactive

Yeah but I don't think your submanifolds are real.

Mike Miller

Googling that suggests they prove that as a starting point by finding a complex manifold which is a tubular nbhd of the real analytic one

So perhaps it is carried out precisely that way

I think Mathei's construction works, building an atlas out of U x U for sufficiently small U

Akiva Weinberger

$$\frac{98765432}{12345679}=\frac{987654312}{123456789}=8$$

MatheinBoulomenos

@MikeMiller oh, right, we can just take the product of the manifold with itself, that was the step I was missing, thanks

Mike Miller

M x iM would be a reasonable name

MatheinBoulomenos

22:10

I agree

Akiva Weinberger

M x M is your name, isn't it? @MikeMiller

Either that or Mi x Mi

Mike Miller

Hehe

PVAL-inactive

sciencedirect.com/science/article/pii/016686419090034Y

I thought that sounded suspicious.

MatheinBoulomenos

So much for that idea

I can't see what goes wrong, though

PVAL-inactive

Your construction might work, but I don't think there's any reason for the thing you get to be the product.

I feel like it's probably open sometimes even if your original manifold is closed.

MatheinBoulomenos

22:17

Oh right, the contruction only shows how to get charts for points in a neighborhood of the diagonal in MxM

so we get an open submanifold of MxM

PVAL-inactive

I don't think that either.

The map on imaginary coordinates interacts weirdly with the map on real coordinates away from your original manifold.

TM is always almost complex (since it's symplectic), but I don't think that structure is usually integrable.

Akiva Weinberger

@BalarkaSen Oh by the way I found the minimal finite approximation of the projective plane in a PDF

PVAL-inactive

IIRC you should be able to obstruct it from having any Kahler str. for instance (in certain examples of course).

Akiva Weinberger

(Remember how a sphere was approximated by that finite topological space with six points?)

Take a cube, and identify opposite sides, so that you're left with 3 faces, 6 edges and 4 vertices

Make it so the faces are open and the edges open-ish (without their endpoints) like with the sphere, so that they partition the space

and quotient them all down to a point

so you're left with a space with 3+6+4=13 points

You could also do this with an octahedron instead, you get something with 4+6+3=13 points, which turns out to be the other 13-point one "upside-down"

PVAL-inactive

A product M\times M has charts of the form (\phi,\phi) where \phi is a chart on M. This isn't what a complexified real map looks like.

Akiva Weinberger

22:28

I guess you can define a convex set's "boundary" regardless of a surrounding space

and similarly there should be a way to define that a convex set is "open" intrinsically

like the affine image of $(0,1)$ in $\Bbb R^n$ should be "open" in some sense

Perhaps the right definition would be, if you delete any point, it is no longer convex. (This is false for $[0,1]$.)

MatheinBoulomenos

@PVAL-inactive we don't need the complex structure to be compatible with the standard structure on MxM as long as it's a complex manifold. We should still get the same structure on M if we embed diagonally

PVAL-inactive

I don't understand what you mean.

MatheinBoulomenos

Probably because it's nonsense. I wanted to say something like it doesn't matter if we get a chart for MxM as long as we somehow get a complex manifold

Mike Miller

I think Mathei is right. If you build a manifold out of charts $U \times U$ for $U$ very small inside $M$, defining it as a coproduct with identifications along intersections of open sets, you get something identifiable with an open subset of $M \times M$ that includes the diagonal, but as long as the $U$ are very small, you can't get pairs of points that are far away from each other.

eh, that's not quite true, because the transition maps should be the complexified transition maps as opposed to the products (like you said above). but you should get an open complex manifold that is not obviously related to a subset of $M \times M$

PVAL-inactive

22:44

I agree with the latter statement.

Mike Miller

the train of thought (which came from Mathei's) was wrong but still led to the right idea

so idk, not the worst

Alessandro Codenotti

@Akiva Extremal points of a convex sets are a concept very similar to what you're talking about I think

Akiva Weinberger

Ahh, you know what, I think I've heard that term before

Right, that's exactly the thing I was talking about @AlessandroCodenotti

So what do you call a set that doesn't have any extreme points?

MatheinBoulomenos

Thank you @PVAL-inactive and @MikeMiller

Alessandro Codenotti

Dunno, I've heard them mentioned only because of the Krein-Milman theorem, but it deals with compact convex sets in topological vector spaces and those have extreme points

Daminark

22:53

You prove it in that context by a Zorn argument

The theorem Alessandro stated says that a compact convex set in a locally convex TVS is the closed convex hull of its extremal points

Akiva Weinberger

In any case, the point is that there is a finite topological space with 13 points that is weakly equivalent to the projective plane

(two, actually, related to each other)

in contrast to the 6 points required to approximate a sphere

(There are none with fewer than 13 points, so they're minimal)

@Daminark TVS?

Oh, topological vector space?

MatheinBoulomenos

Hi @Ted

Ted Shifrin

Hi, @Mathein. I was just looking for a reference for you.

MatheinBoulomenos

on the complex manifold thing?

Transcript for

$Mathematics$ Mathematics