Mathematics - 2017-02-25 (page 1 of 5)

Hey has someone an idea how ti find a $C$ such that $C||a+b||<=||\lambda/x+(1-\lambda)b/y||$ with $x,y>0$ and $\lambda \in [0,1]$?
Where $a,b \in \mathbb{R}^n$

And the norm is the 2-s norm. Any help highly appreciated!

Maks

12:46 AM

Hi !

Does anyone know how to calculate the basis of the intersection of subspaces?

Zach Hauk

same dimension?

Sophie

what is the definition of the correlation between two variables? In a context like "the correlation between a parent's IQ and their child's is 0.75"

WDUK

okay so i had to cut a wire in to two parts one for a circle and a triangle, i created an expression for the sum of both these shape's areas, and now told to find the minimum value by differentiating, but when i do this the only solution for x is 0 so i think i have gone about this all wrong =/

pretty sure im going to fail this assignment lol

Sir Jony

Hint: Circle will give you more area per perimeter than any other shape, so the larger the circle the better.

Use all the wire for the circle, because using some for the triangle would be "inefficient".

WDUK

1:01 AM

hmm not sure that made it much clear

Sir Jony

Oh, my bad. You're asking for minimum. :)

WDUK

im thinking my area sum is wrong before i even got to the part of differentiating

Sir Jony

Ignore me. :)

WDUK

ah okay lol

i thought i had to differentiate then solve for x

but it only solves for 0

Sir Jony

I'm too busy to solve this, but I have a feeling you might need to take the double-derivative to find out where the minimum "takes place".

WDUK

1:04 AM

oh you mean finding the local minimum/maximum

hmm im not sure on that

i found a youtube video on it so will see how that goes

Forever Mozart

???

HERE

@AkivaWeinberger

Semiclassical

1:31 AM

hi chat

Forever Mozart

@AkivaWeinberger come back

Secret

Hey guys, it is known that every neighbourhood is an open set in a topology. Does the converse hold in general, or it is possible to find a topology such that an open set V of the topology does not contain any open set containing a point x, hence V failed to be a neighbourhood of x?

Ted Shifrin

@Secret: Although I agree that "neighborhood" means "open set," apparently not everyone agrees with that. But remember that no matter what topology you take on $X$, for any $x$, the entire space $X$ is always an open set containing $x$. :)

Rehi @Semiclassic

Is DogAteMy lost?

Zach Hauk

hey @Ted

Maks

Ted ! Could you give me a quick hand with linear algebra?

Ted Shifrin

1:44 AM

Will a foot do, @Maks?

rehi @Zach

Maks

And first of all Hi @TedShifrin how are you doing? Hahaha

Zach Hauk

how was your break from me?

Maks

@TedShifrin a foot will do well too

Ted Shifrin

all 2 hours, @Zach?

Zach Hauk

yep

Ted Shifrin

1:45 AM

insufficient? :)

Zach Hauk

ha

Maks

I'm given two subspaces and asked to find the basis of their intersection

Ted Shifrin

How are you given the subspaces?

Maks

I looked in internet and some people say to make some combination with the null matrix

@TedShifrin how are they expressed you mean?

Ted Shifrin

It depends on how you're given the subspaces. I have exercises like this in my own book :)

Maks

1:46 AM

One is parametric (with the equations) and the other one with the basis

Ted Shifrin

Are you given them as solutions sets of homogenous systems of equations, or are you given bases for them?

Ah ...

Maks

One way was to make a matrix with the equations of the vectors

But it was wondering if there was another way

I thought of creating an equation system

Because if v is in the intersection of V and W

Then it has to be a linear combination of the basis of each subspace

Ted Shifrin

Right. So it's best to get both in terms of bases or both in terms of equations.

Either way.

Maks

But in my case that doesn't look like an easy solution

Ted Shifrin

Why not?

Maks

1:50 AM

What's the best way to get the equations from the basis?

I place the vectors as columns I equal it to coordinates and reduce it

Ted Shifrin

Ah, find vectors orthogonal to all of 'em by doing a nullspace.

No, put them as rows!!

This is the hardest part of linear algebra — deciding when to use rows and when to use columns :P

Maks

Vectors as rows ?

Ted Shifrin

If I want vectors orthogonal to $v_1,\dots,v_k$, I want $v_1\cdot x = \dots = v_k\cdot x = 0$, so that's $Ax=0$ when $A$ is the matrix of rows. :)

Kasmir Khaan

Hi ted !

:D

Zach Hauk

my wrist is really starting to hurt now

Ted Shifrin

1:52 AM

Did you try to be a gymnast, @Zach?

Maks

Should the sum of first element of each vector be equal to the coordinate?

Ted Shifrin

Rehi @Kasmir.

Zach Hauk

@Ted, no, i'm just playing stupid games

that hurt my wrist

Ted Shifrin

Huh? @Maks

Kasmir Khaan

I got 1 question that i just need an answer to it

its for a friend

Maks

1:53 AM

If I have for example $(v_1,v_2,v_3),(w_1,w_2,w_3)$

Ted Shifrin

"a friend," @Kasmir? Uh huh.

Kasmir Khaan

-.-

Ted why -.-

Ted Shifrin

Isn't it middle of the night/morning, @Kasmir?

Kasmir Khaan

3 am soon

Zach Hauk

can someone do "my friend"'s geometry homework?

Ted Shifrin

1:53 AM

Uh huh.

No, @Zach, you have to do it.

Kasmir Khaan

Come on guys

Zach Hauk

I'm kidding

Kasmir Khaan

:D

well now you both discouraged me to post it

Ted Shifrin

We're horrible people, @Kasmir.

Kasmir Khaan

nah just Zack

not you

Zach Hauk

1:54 AM

:P

Kasmir Khaan

=p

Ted Shifrin

I'm horribler than you realize.

Kasmir Khaan

its olympiad question

Zach Hauk

But I'm still the terriblest

Ted Shifrin

Olympiad? Way too hard for me.

Zach Hauk

1:55 AM

interesting

continue

Kasmir Khaan

super hard yes but not hard for you Ted

Zach Hauk

post it, i want to think i know the answer but get it wrong and give up :D

Kasmir Khaan

><

one second and ill post it

well prove that any real solution of x^3+px+q =0

satisfies 4qx <= p^2

from mathematical competion baltic way 2003

Maks

OK I'm back

Something strange happened

Ted Shifrin

You left without permission, @Maks.

Kasmir Khaan

1:58 AM

Iam not sure there is a solution online

Zach Hauk

:P

WDUK

stranger things huh maks?

Maks

Stack exchange refused to acknowledge I was logged in

Ted Shifrin

I paid them off.

Interesting, @Kasmir. I've never seen such a thing before.

I assume $p,q\in\Bbb R$?

Maks

As I was saying if I have $(v_1,v_2,v_3),(w_1,w_2,w_3)$

Then each vector created with those basis

Zach Hauk

1:59 AM

Looks vieta-y to me, but probably cause im stoopid

Ted Shifrin

No, I thought of that, too, @Zach. Leave out such remarks.

Maks

Is a linear combination of them

Kasmir Khaan

@TedShifrin thats all the info they gave

Zach Hauk

the sum of the roots is 0

Maks

So I would equal $x = av_1 + bw_1$

That's why I put them as columns

Ted Shifrin

2:00 AM

Sloppy, @Kasmir, but with an inequality in there, $p^2$ and $q$ have to be real, so I'm assuming $p$ is as well.

Secret

The following is my attempt at understanding why open sets are so named (without referring to open intervals in metric space). Given a topology $\tau$, a set is defined to be open if it does not contain all of its limit points, that is, there exists a net in the set such that it converges to a point not in the set. This thus generalise the intuition that open sets are missing points that sort of "bound" it

Ted Shifrin

I was answering your question about how to give equations for a subspace if you had a basis, @Maks.

Maks

Yay

Ted Shifrin

That is totally NOT right, @Secret.

Maks

Go ahead, I read you

Ted Shifrin

2:01 AM

Give me the exact question (numbers don't matter, but the sense of it matters).

@Maks

Zach Hauk

to make it neater ill just rewrite as $qx \leq \left(p/2\right)^2$

Maks

Given V vectorial space, be $W_1,W_2$ subspaces of V, give the basis and equations of $W_1$ intersection (don't remember the symbol) $W_2$ and $W_1+W_2$

I am given the equations of $W_1$ and basis of $W_2$

Zach Hauk

$\cap$?

Maks

They are $R^6$ and Dim = 4

Ted Shifrin

@Maks: So I told you how to give equations for $W_2$ (that was the rows stuff). Then you have equations for everything and you just find the nullspace of that big matrix to find $W_1\cap W_2$.

For $W_1+W_2$, it's probably easiest to find a basis for $W_1$ (standard algorithm), and then use the vectors and the vectors for $W_2$ as columns and find constraint equations (when is a vector in the column space of that giant matrix?).

Maks

2:04 AM

OK, let's go slow, I find the equations by creating a matrix with vectors as rows and equal it to (x,y,z...) And then I reduced it the null rows are the equations

Ted Shifrin

There are other ways to do this, but that seems easiest to me.

Equal it to ????

Maks

That's what I wanted to know

Ted Shifrin

You know how to find the solutions of $Ax=0$?

Maks

How do I find the null space

Ted Shifrin

You put the matrix in reduced echelon form.

You should watch my lectures on YouTube :)

Maks

2:06 AM

@TedShifrin coordinates x,y,z,u,v etc

To get the equations

Or at least that how we do it

Uhmm I will check them

@TedShifrin I just reduce it using gauss method

Zach Hauk

@KasmirKhaan suppose $a,b,c$ are the roots of that

Ted Shifrin

Right, and once you're in reduced echelon form you can read off the solution of $Ax=0$.

Maks

Reduced echeleon matrix or something like you call if

We call it reduced by row matrix

Ted Shifrin

OK

Zach Hauk

I get using Vieta's that $-abc(a+b)\leq \left(\frac{-ab-bc-ac}{2}\right)^2$

Ted Shifrin

2:07 AM

rref

Maks

I do a matrix like 1 0 2 x

Ted Shifrin

@Zach: You probably need to explain what you mean by "using Vieta's."

No, no, @Maks. That's for consistency. That's what I talked about for the second problem.

Maks

That's what I meant when I said "equal to coordinates"

Kasmir Khaan

Zack I dont really know anything about it

><

it was given to me from a friend

Ted Shifrin

@Kasmir: Tell your friend to come ask, then.

Kasmir Khaan

2:08 AM

I wanted to post it here just that

Ted Shifrin

Or post it on main for your friend.

Kasmir Khaan

she dont speak english

french speaker

Yes yes

thats a good idea

Maks

I will try put them as rows and get the equations

How do you call that process?

Ted Shifrin

If @Zach actually figures it out, he can post a solution.

Maks

Parameterization? Or something like that?

I want to know so I look for it

Kasmir Khaan

2:09 AM

Okay :)

Zach Hauk

hmm

Ted Shifrin

No, @Maks. I'm talking about a basis for the solutions of $Ax=0$.

Maks

And do a bit of research

Zach Hauk

all of the terms go to positive, so looking good so far

Ted Shifrin

Oh, you mean the lectures?

Maks

2:09 AM

@TedShifrin ohhh

Forever Mozart

does anyone here know about homology?

Maks

But that doesn't allow me to solve anything in my case

Ted Shifrin

What's the question, @Forever? You might want PVAL or MikeM.

Maks

Does it?

Ted Shifrin

@Maks. Reread everything I've typed.

Maks

2:10 AM

Sure

Ted Shifrin

I'm not going to do it over and over.

And it's almost my dinnertime.

Forever Mozart

well I know nothing

Ted Shifrin

You didn't take a basic alg top course in grad school?

Forever Mozart

Homology (mathematics)

In mathematics (especially algebraic topology and abstract algebra), homology (in part from Greek ὁμός homos "identical") is a general way of associating a sequence of algebraic objects such as abelian groups or modules to other mathematical objects such as topological spaces. Homology groups were originally defined in algebraic topology. However, similar constructions are available in a wide variety of other contexts, such as groups, Lie algebras, Galois theory, and algebraic geometry. The original motivation for defining homology groups was the observation that two shapes can be distinguished...

Ted Shifrin

That doesn't help.

Forever Mozart

2:13 AM

how the hell does it give path connected components if $H_0(S^1)=\mathbb Z$?

Ted Shifrin

Because you get a generator for each path component.

Daminark

Hey everyone!

Forever Mozart

@TedShifrin no, algebraic topology is the one topology that I avoided

Zach Hauk

Hi @Daminark

Forever Mozart

but I was good at algebra

and topology

Ted Shifrin

2:14 AM

That was pretty dumb, honestly. Two points ($0$-cycles) $a,b$ are homologous if there is a $1$-chain (interval in the space) with boundary $b-a$.

Heya @Daminark.

So two points are homologous iff there is a path joining them. Thus, you get a generator for each path component.

Forever Mozart

oh

Secret

Ok it seems I mixed up boundary points with limit points. There are limit points that are not boundary points

Ted Shifrin

Ponders saying "Hey no one" to Daminark

@Secret: That was just totally messed up. Is $[0,1)$ open in the usual topology on $\Bbb R$? Yet it's missing a limit point (boundary point).

Forever Mozart

so $H_0(X)$ is the set of generators of path components

of $X$

Ted Shifrin

Well, it's the free abelian group generated by them, yes. :)

Forever Mozart

2:17 AM

what if the path components of $X$ are singletons?

Ted Shifrin

Then you get a big group.

Forever Mozart

is $H_0(X)\simeq X$?

Ted Shifrin

You'd better check your algebra. How do you write the group?

Daminark

Lol sorry I was sending an email to the teacher for our Guillemin-Pollack class

Forever Mozart

yeah I have to remember algebra now

hmm

Ted Shifrin

2:19 AM

You want a free abelian group with one generator for each point of $X$, @Forever.

Forever Mozart

ah

but it will have lots of other stuff too?

Ted Shifrin

No?

Forever Mozart

it seems too big

Ted Shifrin

If your space is totally disconnected, then it's totally disconnected.

Forever Mozart

no it's connected

just no paths

Ted Shifrin

2:22 AM

Sorry. My fault. We're talking path components, not components.

I said what I said. What is the issue?

How do you say "it will have lots of other stuff too?" and then complain it seems too big?

Forever Mozart

it will not be a nice group

Ted Shifrin

shrug

Forever Mozart

like something I'm used to seeing

Ted Shifrin

It's not a nice space.

Forever Mozart

yes

Ted Shifrin

2:25 AM

Looking at $\Bbb Q\subset\Bbb R$ ... the 0th homology is a free abelian group with countably many generators.

Secret

@TedShifrin that one is a neither set, since its complement is not open and itself is not labelled as open in the usual topology (where all open sets are open intervals). Hmm, I have no idea if open sets can be described by referencing to only limit points as e.g. an open interval in the reals clearly contains some limit points

Forever Mozart

what if you go to $H_1$

Ted Shifrin

@Secret: I don't think you can do what you're thinking.

$H_1 = 0$ since it's a $0$-dimensional object. There are no $1$-cycles other than trivial ones, and they're all trivially boundaries, too.

Forever Mozart

huh really?

oh

for $\mathbb Q$ you mean

Ted Shifrin

@Forever: $k$-dimensional homology measures $k$-dimensional stuff that is not a boundary of something $(k+1)$-dimensional.

Yeah. That's what I meant. I have no idea what your space is.

Forever Mozart

2:27 AM

it is a subset of the plane

and 1-dimensional I think

Ted Shifrin

If you're doing stuff that is not a CW-complex, homology is going to work weirdly.

But you need $1$-cycles to start with; so you need closed paths. You have none of those other than points. So ... no $1$-dimensional homology.

Forever Mozart

yes certainly 1 dimensional

:(

lol

they say the Cech homology measures quasicomponents

so maybe I need something more like that

Ted Shifrin

Right ... Cech is different. They agree on "reasonable spaces."

So go read up on Cech. It's built on open coverings.

Forever Mozart

I looked but it is not in most Alebraic topology books

Zach Hauk

@Kasmir hasn't posted his question yet, so would I look stupid if I just posted it myself? because i'm nearing a solution

Ted Shifrin

2:30 AM

@Zach: Make sure you have it all right.

Zach Hauk

im checking

:P

Ted Shifrin

It shows up usually when people learn sheaf cohomology, @Forever, but it's in some books. And there should be something on wiki.

@Forever: I found this upon googling.

I need to leave now, but there are sources.

Bye all

Maks

@TedShifrin You mean If I put them as rows, then I reduced it I end up with the equations ?

BAYMAX

2:46 AM

Bye @TedShifrin

Maks

@TedShifrin Me ? No, we werent given algebra of calculus on high school

Secret

One of my major intuition problem I am facing is I don't really understand why continuity of a function f in a topological space is defined to be "the image f(U) of an open set U is open. For example:

math.stackexchange.com/questions/321714/…

BAYMAX

@Secret Mainly the concept is a continuous function pulls back open set to open set

Secret

Is it true that the only continuous function that map from open sets to closed set and vise versa are the constant maps as shown in that MSE? NB I am learning point set topology, I don't think I really understand what a pullback is

Ok sorry wrong definition, let me try again...

BAYMAX

I am also thinking about it! ok, pulls back is nothing but pre-image of $f$ , that is if we take open set in the co-domain then $f^{-1}$ must map it open set...

Secret

2:54 AM

A function is continuous if the premise of any open (resp closed) set is open (resp closed)

BAYMAX

Yes

Secret

3:14 AM

Why I cannot pullback from an open set to a closed set even if f is not injective. As far I know, I struggle to get a counterexample of this but I am not sure how to prove it is impossible for any topology (housedoff or not)

NB autocorrect error : premise=preimage

Yash Farooqui

I have a bit of a soft question -

why are alternating forms the "right" objects to study when dealing with multilinear algebra?

not sure how else to phrase it

it's just that I don't see why alternating forms are natural - unless we already know about determinants

but then the question would become "why are determinants 'natural'?" which really doesn't help

sorry if that's too elmentary

Semiclassical

@Ted would be a good person to ask that. What comes to mind off the top of my head is that you want to have a notion of orientation.

Yash Farooqui

mm. The volume aspect probably follows somehow from that - am just trying to build my intuition

how would I ask @Ted?

Semiclassical

Well, he's been pinged by doing that. So if he's interested he'll respond.

Also, the fact that they're alternating means that if two of the vectors are identical then the volume is zero (as it should be in that scenario)

Yash Farooqui

yeah

There is a natural basis given by the structure of the tensor product I guess so I'm not sure basis independence is necessary

I misspoke

A k-multilinear form or k-tensor is effectively a linear transformation from V^k to your base field, right?

Semiclassical

3:32 AM

Sounds right.

Yash Farooqui

So is there a basis independent way to define an alternating k-tensor?

because when we talk about bases, it makes it seem algebraic but

Semiclassical

hmm.

Yash Farooqui

then it seems to be used as a geometric property?

Semiclassical

I would presume yes, but I'm not actually versed in this stuff.

Yash Farooqui

I'm not either sadly.

Alternating forms should form a subspace.

It's just that they seem out of nowhere to me.

Is there a way to define wedge products without alternating forms?

Secret

3:36 AM

All the proofs I had found so far that continuity implies preimage of open set is open make use of metric arguments. But what about topological space where a metric is not defined?

Yash Farooqui

Isn't the definition of continuity that

the pullback of an open set is open?

Semiclassical

Probably not. I mean, what is a wedge product without the antisymmetry?

Secret

@YashFarooqui it is, but I don't understand why other than I have spent hours trying to find a counter example to understand why

Yash Farooqui

intuitively, open sets are sets of things that are "close" in some sense

that should help intuition

it's a definition though

afaik

unless you're using a different definition of continuity

Intuitively the definition makes sense because it says if f(x) and f(y) are close, then x and y must be close too?

that seems backwards hold on let me rethink

okay. So if you want to get close to a point f(x), you just have to get close to x

Balarka Sen

@YashFarooqui Yeah, the point is that, volume is alternating. Volume of parallelogram spanned by vectors v and w is minus the volume spanned by w and v.

That's why you use alternation, that's all.

Yash Farooqui

3:45 AM

sure... but that particular example doesn't help because you kind of just gave the definition of signed volume

err like

v and w have negative volume of w and v

Balarka Sen

Yes, I did. Signed volume is the right thing to look at when studying forms. So what are you looking for?

Yash Farooqui

Okay. then why is signed volume the right thing to look at?

Balarka Sen

Ah, that's a good question. The point is signed volume carries more geometric data than volume alone. Eg, magnitude of cross product of two vectors v x w in R^2 is precisely the signed volume of the parallelogram spanned by them. And |v x w| = - |w x v|. That's sort of the thing which you are trying to generalize.

On the other hand, in the continuous picture, note that differential forms (continuous generalizations of alternating k-forms) can be integrated, and you want it to satisfy nice properties like $\int_0^1 fdx = -\int_1^0 fdx$. So there's always the story of "oriented volume" hovering around.

Yash Farooqui

okay. That isn't completely satisfying because the cross product itself seemed to be pulled out of nowhere but the latter part seems to suffice for now

I guess I'll work through a few more exercises and maybe move on to the definition of differential forms (am working through Spivak) to understand

thank you

Secret

@YashFarooqui Suppose I take the usual topology in the reals thus all open intervals are open sets say (1,6) and (4,9). By referencing to these open intervals, 2,3,4,5 etc. and 5,6,7,8 etc. are close to each other. Now since the union of any open sets is open, by taking the union I get the open interval (1,9). Now 2,9 are close to each other. So if I have a map that maps (1,3) U (5,9) to (1,9), I have a weird scenario where 2 and 9 are close to each other and far away at the same time?

Balarka Sen

3:53 AM

No problem, you do raise interesting questions. I guess volume with respect to some "right-hand rule"/orientation has become so natural to me that I can't give you an easy intuitive picture :)

But it's certainly more interesting than just raw unsigned volume, I can tell you that.