Mathematics - 2016-09-17 (page 1 of 2)

Mambo

05:18

hello

ahorn

06:52

How does one create a hyperlink for a tag in meta? e.g. feature-request

Mew

08:32

Hello

Mithlesh Upadhyay

10:46

Please, have a look?

http://math.stackexchange.com/questions/1930141/2-s-complement-of-100-is

Mew

what is compliment

user116211

@ahorn [tag:feature-request]

Alex

12:15

0

Q: Finding cyclic polygon for given values of lengths

A cyclic polygon is a polygon with vertices upon which a circle $C_0$ can be circumscribed. (All vertices lie on circle $C_0$). We are given the lengths of the cyclic polygon $\{L_1, L_2,..., L_n\}$. We need to find the coordinates of the vertices $\{(x_1, y_1), (x_2, y_2),...(x_n, y_n)\}$ such t...

does anyone understand the answer there ?

user147690

@Alex What don't you understand?

user4612744

Hello, please explain like I'm five, any good techniques for simply multiplying a polynomial expression like this (3+i)^4 ? I've forgotten the most basic algebra it seems

Danu

12:31

With complex numbers, the best way to go is to write it in the form $r e^{i\varphi}$

Then multiplication amounts to multiplying radii and adding angles.

user4612744

But if it were just real numbers, an expression like this (3 + 2)^4 for instance...

user147690

@user4612744 If that $i$ is the complex $i$, and that was what was making it tricky see Danu's message, if you mean in general you should try manually doing it step by step, and if that isn't working, see binomial theorem. By manually doing it, I mean $(3+i)^4=(3+i)^2(3+i)^2=(3^2+6i+i^2)(3+i)^2$ you know, the ground work.

Danu

What are the necessary conditions for dividing out the action of a Lie group on a manifold to yield a manifold?

Free + proper is sufficient, but can it be weakened?

user4612744

it's not the complex number that's making it tricky, believe it or not.. I tried doing it manually step by step and got the wrong answer several times (according to my text-book..)

thanks for answering @Danu and @AlexClark

user147690

What answer did they give?

user4612744

12:35

28+96i

user147690

What did you get?

user4612744

for clarification, that particular expression is part of a bigger polynomial equation

37+69i

user147690

Did you convert to $re^{i\theta}$?

Danu

Just do what I said @user4612744

That makes the step-by-step a triviality

user147690

Indeed

Danu

12:40

That's why it's better than the usual method

Geometry is so useful sometimes :D

user147690

And you can answer check immediately, since your radius isn't 100, whereas $3^2+1^2=10$ so the radius initially was $\sqrt{10}$

user4612744

I did not. I may have missed something, but I was not aware this was necessary for a simple equation like this? I'm simply supposed to show that the complex number 3+i is a root in the complex polynomial P(z) = z^4 -8z^3 + 39z^2 + 122z + 170

Danu

What do you mean by "necessary"? This is very easy!

user147690

To take this away from looking scary, note that this is done by grade 11 students in my city, it's just not taught everywhere. Finding complex polynomial roots is usually done later.

Danu

Complex numbers are not part of high school curriculum in the Netherlands

But it's true that this multiplication is really easy.

Also division, etc :)

user4612744

12:51

I meant as in converting the complex number from rectangular to polar form seems like it would be an "extra step"

Danu

I think that, if you timed how much time you spent on doing it your way, versus how much time you'll spend doing it the geometrical way you'll see it's faster :D

Krijn

Although mathematicians may not have the best LinkedIn-profiles, I'm still gonna ask it here: Should you list that you are a blood donor on LinkedIn, and how?

Because it keeps asking me to list my volunteering experience

Danu

No, why?

I think that is mostly personal information

Also I don't think anyone in math gives a crap about linked in profiles

Krijn

A recruiter that checked my CV once told me to put that on there, for some reason I forgot

And no, but then again I might not end up in mathematics.

Danu

Welp, I don't think I'd like it but go for it if a recruiter told you so haha

Krijn

13:00

Then again, most of what recruiters say is rubbish of course.

Balarka Sen

13:23

Hi @Danu, @Krijn

Krijn

Hey!

Danu

@BalarkaSen Hi

Krijn

Im finishing Nayak tonight, because I was interrupted last time.

Balarka Sen

What's up?

@Krijn What do you think of it?

Krijn

So far it's interesting, although not much is happening really

Balarka Sen

13:26

Oh, I misread that "I'm finishing"

thought you said "I've finished"

how far are you?

Alex

@AlexClark I see no easy way to compute this programmatically .

Krijn

@BalarkaSen Almost 1 hour in.

Balarka Sen

Been past that sinking-in-the-sea-of-money sequence then?

Krijn

I stopped after that.

Weird dream

Very symbolic, I suppose

Balarka Sen

heh, you'll see why he dreamt that afterwards

Fredefl

13:36

In an assignment I am asked to prove this equation - I, however, don't understand it in the first case.

(the use of it, that is) X, by the way, is a random variable.

As far as I understand, E is the expectation and argmin is used for getting the x that gives the lowest f(x).

Adeek

13:51

hi @BalarkaSen

Balarka Sen

Hi

Adeek

I learned some cool things in complex analysis from video list I am watching.

I learned the deeper meaning for why cauchy riemann equation hold

Balarka Sen

Mhm?

Adeek

because the derivative mapping in C is conformal so that is why angle is preserved

Balarka Sen

Ah. Yes.

Adeek

13:52

Complex Variables (Lecture 2): Complex Functions as Mappings

►

Very nice play list

Balarka Sen

I think of it in a different way

Adeek

how do you think of it ?

Balarka Sen

Do you know what a differential form is?

Adeek

no not yet

can you maybe explain it quickly ?

Balarka Sen

Then it'll probably be hard to read. C-R equations are so to speak conditions for $df$ to be a closed 1-form, given any holomorphic $f$.

Adeek

13:55

alright

Balarka Sen

@Adeek Maybe not now

Adeek

@BalarkaSen you should this guy youtube.com/playlist?list=PLPH7f_7ZlzxTi6kS4vCmv4ZKm9u8g5yic

this is lectures in differential geometry very nice

he has also lectures in QM if your interested

youtube.com/channel/UC6SaWe7xeOp31Vo8cQG1oXw/playlists

Very nice prof

Balarka Sen

Thanks

Krijn

@Balarka, Have you ever looked at the modularity theorem properly?

Balarka Sen

@Krijn Every elliptic curve is modular?

That one?

Krijn

13:57

Yes

Balarka Sen

I don't actually know what "modular" means.

I'd listen if you want to explain.

Krijn

It means that if you look at the $L$-function associated to $E$, then there is a unique modular form $f$ which has the same associated $L$-function

Balarka Sen

Hmm

Krijn

Which is strange

I don't know enough about modular forms, I think

Balarka Sen

What's the $L$-function associated to a modular form?

Using the coefficients of the Laurent expansion of the form to construct the L-function?

Krijn

14:02

A modular form is periodic, so it has Fourrier coefficients

Take those

Balarka Sen

Aha. Gotcha

Krijn

I hope to learn why anyone would conjecture that

Adeek

there is actually a really interesting talk

I am attending next week

Balarka Sen

Me too.

There's a related version which says any elliptic curve admits a parameterization using a modular curve. That is more intuitive.

parameterization here means a rational map from a modular curve

Secret

Hey guys, I am trying to evaluate integrals of this form here $$\frac{d}{dx}\int_a^{b} f(x,x')dx'$$. However because I knew $f(x,x')$ is in a form that Lebniz rule will fail, I obviously cannot differentiate under the integral sign, therefore the question boils down to the evaluation of $\lim_{h\rightarrow 0}\int_a^{b}\frac{f(x+h,x')-f(x,x')}{h}$. Is the method of evaluating that is basically the same as general limits, or we still have tricks to simplify it and make it easier to evaluate?

Details on next message

0

Q: Exploring the properties of the limit when Lebniz rule fails

Consider the following simple case of differentiation of an integral, where x' is just a dummy variable and $a,b$ are constants $$\frac{d}{dx}\int_a^b f(x',u(x))dx'$$ Now wrote the differentiation in first principles $$\lim_{h\rightarrow 0}\frac{\int_a^b f(x',u(x+h))dx'-\int_a^b f(x',u(x))dx'}{h}...

real-analysis integration functional-analysis indefinite-integrals big-list

Adeek

14:08

@BalarkaSen did you learn elliptic curves?

Balarka Sen

I just know elliptic curves over $\Bbb C$ :P For other fields, not really.

Krijn

I did a course on them, and read a bit about them in Hartshoren

Balarka Sen

Everything related to Euler characteristic and manifolds is such a "circle of ideas" that I sometimes don't know how to teach it without going around in circles. — Tom Goodwillie Mar 27 '12 at 22:18

lol

I can connect to that

Secret

14:26

...Had I started my classical mech self study much earlier, someone would already have provided leads to the above question...

but now, the user had said they will nto visit this chat again, bummer...

Danu

15:14

So I've got the following definition of the Schubert varieties:

$ \Omega(W_\bullet):=\{W\in Gr_k(V)
\mid \dim(W\cap W_j)\geq j,\ j=1,\dots,\ell\} $

So how the heck am I supposed to tell what $k$ is from $\Omega(W_\bullet)$?

I wanna say that I need an extra index somewhere

Balarka Sen

@Danu $k$ is the dimension of $W$, isn't it?

Because only then would it represent a point in $Gr_k(V)$.

Literally by definition of $Gr_k(V)$.

Danu

@BalarkaSen Sorry, I didn't give all the infor

The point is that $(W_j)$ is a flag of subspaces

Balarka Sen

Yes.

Danu

So the bullet will give this information

So do we run over all $k$ or what?

The reason I'm confused is because, right after the definition, he says "so if $\ell=k=1$ then [...]"

So he fixes $k$?

Balarka Sen

No, like, just fix a $k$.

Danu

15:26

Yeah so if I'm fixing $k$

Then I should indicate this in my notation for the space

how will I distinguish $k=1,2,\dots$ if I just have $\Omega(W_\bullet)$?

Balarka Sen

Yeah, that's not a good notation.

Danu

Okay

Good

Balarka Sen

However, the dimension of the largest vector space in the flag is $k$.

So from the flag you'd already know what $k$ is.

Danu

No, why?

Balarka Sen

Sorry, no, that's dimension of $V$ - not $W$. Nevermind.

OK, not a good notation at all.

Danu

15:29

Also here he allows flags that do not "occupy all dimensions" AFAIK

so partial flags, I guess?

Balarka Sen

I confuzzled $k$ with the dimension of $V$. Sorry about that.

Danu

But even the dimension of $V$ plays no role here

Balarka Sen

Well, a flag always ends at $V$.

Danu

Well it plays some role but only in the backgground

No these flags don't

I guess he uses flag for partial flag

Balarka Sen

That's non-standard then. A flag, to me, is a hierarchy of vector spaces $\{0\} \subset W_1 \subset \cdots \subset W_{m-1} \subset V$.

Danu

15:32

Yeah, here it's not :\

(example 2.1.19, Huybrechts)

Balarka Sen

I don't have Huybrechts with me, but in that case, cruds.

Danu

He allows, for instance flags of one subspace

(which already makes it impossible to be in accordance with your definition)

Balarka Sen

Ah, ok, so they need not occupy all dimensions.

Fair enough.

Danu

I'm just going to use $\Omega(W_\bullet)_k$ then

Balarka Sen

Good idea. Weird variety, by the way.

Danu

15:36

Yeah

So for $k=l=1$ we get $\Bbb P(W)$ where $W$ is the single subspace

Balarka Sen

mhm.

SAWblade

Is it possible to assemble a cube from 2000 other cubes?

Balarka Sen

yes; take one of them and throw out the rest of 1999 cubes

Huy

my solution exactly

Danu

lel

SAWblade

15:41

lol fuck i can't believe i didn't think of that trolly answer

absolutely using that

Danu

@SAWblade The word "cube" is suggestive

Is 2000 a cube of anything in any other sense? ;)

SAWblade

I don't catch your drift. xD

Danu

What does the word "cube" mean in mathematics, usually?

SAWblade

A number that is cubed.

Raised to the third power if you wanna be specific.

Danu

Exactly

So if something is a cube

Then what is it, in usual math terminology?

SAWblade

15:43

I know 2000 isn't a perfect cube, but you can make cubes out a number that isn't a perfect cube.

Danu

Wait what? :P

How?

Balarka Sen

arrange the 2000 cubes in a 12.59921049... x 12.59921049... x 12.59921049... cube.

Danu

lol

Balarka Sen

@Huy great minds think alike

SAWblade

$4^{3} = 3^{3} + 1^{3} + 1^{3} + ... 1^{3}$
That's a total number of 28 cubes used to make another cube.

So the question is can I make a cube out of 2000 other cubes? xD

Balarka Sen

15:49

then your cubes are not of the same dimensions?

are you restricting integral dimensional cubes or what?

SAWblade

Integers only!

Balarka Sen

ok. no absolute idea.

SAWblade

I get the feeling it's no, but I don't know why.

Balarka Sen

a number theorist would know. ask in the main site.

SAWblade

Okey doke!

0

Q: Can you assemble a cube out of 2000 other cubes?

In more rigorous terms, does there exist some $x \in \mathbb{N}$ such that: $$x^{3} = \sum_{i=1}^{2000} a_{i}^{3}$$ for $a_{i} \in \mathbb{N}$? Note that the $a_{i}$'s need not be distinct. After tinkering around in Mathematica for a couple of hours I'm beginning to believe the answer is no, b...

combinatorics number-theory

Balarka Sen

16:00

@Danu So, tell me something interesting from complex geometry.

I'm sort of envious that you're learning that stuff. :P

Danu

Not every SES of holomorphic vector bundles splits (?)

I'm tyring to see why they do for smooth and why they don't for holomorphic

Balarka Sen

Interesting. I am guessing that should relate to holomorphic vector bundles not always having a nonzero holomorphic section.

0celo7

16:16

@BalarkaSen Something you probably know but I think is quite cool: $\Bbb R^n-Q$, $Q$ countable, is path connected.

Balarka Sen

That's true.

0celo7

Which part? It being cool or you knowing it?

Danu

I want to say $n\geq 2$?

0celo7

@Danu Right. Forgot that bit.

Balarka Sen

The statement is true, I meant.

0celo7

16:18

@BalarkaSen I know it's true.

Danu

@0celo7 $1\implies 2$

0celo7

In fact, the path can be taken to be two line segments.

Balarka Sen

Yep.

The proof is non-constructive of nature.

0celo7

Yeah.

Danu

It is a very unsurprising result though (you know, take a point out of a disk and it's still connected), isn't it?

Balarka Sen

16:19

I don't think it's cool though. Just a fact worth knowing.

0celo7

@Danu Countable.

Balarka Sen

@Danu What if $Q$ is very messed up?

Like $\Bbb Q$?

0celo7

Not finite.

Danu

@0celo7 Discrete would already reduce to exactly what I said---not finite.

But yeah, I'm not saying it's obvious to me how to prove it

Balarka Sen

Yeah, the point is, $Q$ may not be discrete.

Danu

16:20

I realize that

But it remains unsurprising

Balarka Sen

The key idea is to take a line passing through a given point which doesn't hit $Q$.

Danu

I saw it a while back on MSE

It was a HNQ or something

0celo7

@Danu Really? Is it obvious that you can draw a line without hitting $\Bbb Q^2\subset\Bbb R^2$?

Balarka Sen

I think it was just for $\Bbb Q$ but the same proof pushes through.

@0celo7 Obvious is not synonymous to unsurprising.

Danu

^

Balarka Sen

16:43

Hi @Anubhav

Anubhav

18:04

@BalarkaSen @BalarkaSen hii, how are you? what are you reading now a days?

Balarka Sen

I am good. Learning topology and analysis, mostly.

What about you?

0celo7

18:41

@BalarkaSen Is there any topology on $\Bbb R^\omega$ in which $(1,1,1,\dotsc),(0,2,2,2,\dotsc),(0,0,3,3,3,\dotsc),\dotsc$ will converge?

I don't have an idea what the limit would be, except for perhaps $(0,0,0,0,\dotsc)$

Adeek

18:56

hi @BalarkaSen

I have a quick question

I agree that the minor is open in M(mxn,R)

but why is the original matrix open ?

0celo7

@Adeek The original matrix?

Adeek

we started with matrix A right @0celo7

what I am trying to ask is the following

0celo7

@Adeek Yeah, but what does it mean for a matrix to be open?

Adeek

We would like to show that the set of matrices of rank A = m is open.

I agree that the set of matrix with size mxm having non-zero determinant is open.

because determinant is continous.

so why is the set X now open, i.e the set of matrices of rank m.

Huy

the answer is written in the image you posted

0celo7

19:01

@Huy You're alive!

Huy

no

0celo7

good, I have some questions about Lie groups

Adeek

I don't understand their answer.

0celo7

What part exactly don't you understand?

what line is confusing

Adeek

which implies that A has a neighborhood contained in M_m(mxn,R)

I don't understand this part why

0celo7

19:05

Ok, ok

Huy

intersect

0celo7

Do you know about $GL(n)$?

do you understand why this is open in $\Bbb R^{n^2}$?

Adeek

yes I do

0celo7

Ok, why?

Adeek

because the function $det : \mathbb{R}^{n^2} \rightarrow R$ is continous and so suppose we have the set of points in R which isn't zero, which is open by definition right let us call the set U? $Gl_n = det^{-1}(U)$

0celo7

19:07

Right.

So here we have the same thing

you have that block, which is nonsingular, right?

There's an open set of blocks close to it which are also nonsingular

this is the neighborhood you need for openness

Write the proof formally if you're not convinced

Adeek

I agree that if we cut it to the minors then it is open.

But why is the original matrix open ?

the orginal undeleted matrix

0celo7

Ok, a matrix is not open.

A matrix is a singleton and this is a $T_1$ space, so it's closed.

Adeek

I don't know why I keep doing what I am trying to say

Is that I agree that the set matrices with minors mxm is open and it is a subset of the matrices of rank m right ?

Huy

nobody knows

Adeek

let us just consider a specific case.

suppose m = 2 and n = 3. Then I agree that set of 2x2 matrix who is non-singular is open, but why is true that set of matrices of rank = 2 open ?

Huy

19:17

the definition of a set being open is that for every element of the set, there exists a neighbourhood of it that is still contained in the set...

0celo7

@Huy How is a neighborhood defined?

Huy

please don't do this

I have dealt with enough kids this week

0celo7

What?

Huy

"how is a nbhd defined"

you know exactly what a neighbourhood is

Adeek

ok I understand

0celo7

19:19

@Huy An open set containing a point.

So defining open sets in terms of neighborhoods is ridiculous.

Adeek

I see

0celo7

Of course, we're working in a metric space.

Huy

yes, and you know that

0celo7

@Huy Do I?

I'm an engineer.

Huy

cool

Adeek

19:20

ok I see.

I undestand @Huy thank you

Huy

np

0celo7

@Huy What's the limit of $(1,0,0,\dotsc),(\frac{1}{2},\frac{1}{2},0,0\dotsc),(\frac{1}{3},\frac{1}{3}, \frac{1}{3},0,0,\dotsc),\dotsc$

is $(0,0,0,\dotsc)$ a good guess?

Huy

0, .... ?

would be my first guess

but it's a stupid exercise

it's like "continue the sequence: 3, 5, 9, 12"

0celo7

It's a ridiculous exercise.

I have to compute a bunch of limits in fucked up topologies on $\Bbb R^\omega$

Adeek

@Huy You go to ETH right ?

Huy

19:23

yes

not very often though

Adeek

we have a really cool prof where I am right now originally from ETH

Huy

namely ?

Adeek

sites.ualberta.ca/~favero

He works in algebraic geometry

Huy

never heard of him

probably was at the ETH long before I was

Adeek

he is actually a chair

0celo7

19:30

@Huy do you know about the BCH formula

Huy

ofc

0celo7

In the usual sources (Hall, Michor), the formula is said to hold only if $X,Y$ are close to the identity of $\mathfrak g$.

But the Wiki article does not mention it at all

Reading those proofs makes it seem like the infinite commutator series need not converge unless $X$ and $Y$ are sufficiently small.

Huy

I also only have heard it for close elements of 0

0celo7

Physicists use it in a special form without that consideration

Namely, when $[X,Y]$ is central

$e^Xe^Y=e^{X+Y+\frac{1}{2}[X,Y]}$.

Huy

what do you mean by it being central

0celo7

19:34

Is this physics math or is it actually correct?

@Huy Commutes with $\mathfrak g$.

Huy

if it commutes with g, the other terms all vanish because it comes up within a commutator of higher terms

0celo7

Yes, but that formula doesn't even need to hold for $X,Y$ not close to $0$.

Huy

hm

I've seen it happen in QM too, but I thought that were small X, Y

0celo7

You use it with creation/annihilation operators

Which are probably on an infinite-dim Hilbert space

Don't know what the notion of "small" is there.

Huy

Hall thm 5.1 seems to state it without conditions of X, Y being small

whereas BCH is stated with those conditions

look at the proof

it probably works in the "central" case then, for any X, Y

then again that's finite-dim

0celo7

19:39

Correct, the main ingredient is Picard-Lindelöf.

Huy

ok, then I actually don't know if the special form for any X, Y is "physics maths"

I took QM years before Lie, so I didn't know better

Semiclassical

Hi chat

Huy

@0celo7: what mousepad do you use. please don't tell me none

Semiclassical

And yeah, i love BCH

0celo7

@Huy Not using a mouse rn

no time to play games

Huy

19:48

wow

much adult

Semiclassical

Main thing I associate it with nowadays are coherent states

0celo7

@Semiclassical Did you have a moment to think about my convexity issue?

@Huy How do you show $S^1$ is not homeomorphic to $S^n,n>1$ without homotopy?

Or homology

Huy

connectedness ?

0celo7

Remove two points from $S^1$ to disconnect it?

Huy

yes

0celo7

19:56

Ok, but how do you prove that disconnects $S^1$ and doesn't disconnect $S^n$?

Huy

you could project to R^(n-1)

0celo7

Come to think of it...

I'd have to prove $S^n$ is connected, period.

Transcript for

$Mathematics$ Mathematics