Mathematics - 2017-03-25 (page 2 of 4)

Alessandro Codenotti

4:00 PM

No, all and only those satisfying the parallelogram law do (one direction is easy, one is not)

@trilolil is this in the context of Lagrangian mechanics or something related?

trilolil

@AlessandroCodenotti yes I am reading about orbital mechanics.

lagrange differential equation

Alessandro Codenotti

It's just a double derivative, the $\dot r$ is treated as a coordinate

Eric

Wait @Alesssandro, which direction of the parallelogram law thing is hard

trilolil

@AlessandroCodenotti strange that they write a dot on top of it, usually in mechanics a dot denotes a derivative...

Alessandro Codenotti

$T$ (kinetic energy by any chance?) is probably a function of $T(t,r_1,...,r_n,\dot{r_1},...,\dot{r_n})$

trilolil

4:03 PM

@AlessandroCodenotti absolutely kinetic energy!

Eric

Ah nvm I guess showing that what should be the inner product is actually an inner product isn't easy

Balarka Sen

what's new

Alessandro Codenotti

@Eric showing that a norm satisfying it comes from an inner product it's not trivial, see the first answer here for example

Meow Mix

Hi @Balarka

Balarka Sen

Hi

trilolil

4:05 PM

@AlessandroCodenotti actually the equation I am looking at describes the motion of a spacecraft using polar coordinates when orbitting around a celestial body.

Alessandro Codenotti

You probably want to look at generalized coordinates to better understand that $\dot r$

trilolil

@AlessandroCodenotti what should I read about on wikipedia to understand T(t,r1,...,rn)?

@AlessandroCodenotti Thanks!

Alessandro Codenotti

@BalarkaSen I started reading my professor's notes about homology yesterday instead of studying analysis and probability as I'm supposed to

Balarka Sen

@Alessandro Excellent! Or probably not.

Meow Mix

@Balarka I have this linear algebra thing to prove

Mahmoud

4:18 PM

Hi.

mr eyeglasses

Hello Balarka Sen, how are you doing

Meow Mix

Prove that, for linear subspace $V$ (not necessarily closed), its orthogonal complement $V^\perp$ is closed

(Don't answer yet, I'm still thinking about it!)

Semiclassical

hi chat

Astyx

Hi @Semi

And chat more generally

Meow Mix

Hi @Astyx

Balarka Sen

4:26 PM

Hi @mreyeglasses. I'm pretty good.

Astyx

@Meow Closed under which norm ?

Meow Mix

The norm defined by its inner product

Sha Vuklia

He guys. If $(ax)^p=(bx)^p$ for some $p$, can we say then that $a=b$?

Astyx

He @ShaVuklia

Balarka Sen

@MeowMix Are you working infinite dimensions already?

Sha Vuklia

4:28 PM

Hi @Astyx

Assume that $x\neq 0$ I guess

Astyx

What's the context in this question @ShaVuklia ?

Alessandro Codenotti

@ShaVuklia context?

Sha Vuklia

I am not given the context XD

haha

Meow Mix

@Balarka Not really, Ted just gave me this to think about. He showed me an example of one which isn't closed, but whose orthogonal complement is.

Sha Vuklia

someone asked me this, I think they're working with group theory?

Secret

4:29 PM

so, closed in the topological sense, or closed as in $a+b \in S$

Balarka Sen

@MeowMix Every subspace is closed in finite dimensions.

Meow Mix

Closed as in, contains all it's limit points.

Astyx

@Balarka In finite dimension, this exercise is quite trivial

Alessandro Codenotti

(And every finite dimensional subspace is closed in general)

Balarka Sen

@Astyx I am aware. I just didn't know Meow is doing stuff beyond finite dimensions already.

Astyx

4:30 PM

Of course you are aware :)

Meow Mix

For example, consider our vector space to be the set of infinite sequences with finite sums. Our inner product will be the usual inner product, which is defined because, well, it is.

Secret

@ShaVuklia if this is an abelian group, then yes you can act the inverse of $x$ both sides $p$ times and rearrange to get $a=b$, if this is not an abelian group, it does not necessary holds unless there are some more relations between the elements

Meow Mix

Then consider the subspace $V$ of infinite sequences with finitely many nonzero components

Balarka Sen

Ya, and that's what I meant by infinite dimensions

Sha Vuklia

@Secret Ah okay, I've asked for more context!

Meow Mix

4:32 PM

Then, we can show it doesn't contain all it's limit points. In fact, the complement of $V$ is all limit points, because for any vector in the complement of $V$, you can make the distance arbitrarily small, by adding more and more components of the infinite sequence, and it will always be finite.

Ted told me that it's called "dense"

Astyx

@ShaVuklia We don't have enough information to answer really. Is it for all $x$ ? Is it commutative ? Can $p=0$ ?

Sha Vuklia

@Astyx Sorry, I have no idea either. I've asked the one who asked this question to elaborate more on the context:d

Meow Mix

By the way, a metric space naturally defines a topological space, right?

Balarka Sen

@MeowMix Another example: consider the space of all continuous functions $[0, 1] \to \Bbb R$. This is a vector space - given two functions $f, g$ you can just add $f + g$ and multiply by a constant $cg$ and whatever. Call this $V$. Let $W$ be the subspace of all polynomials functions. Then $W$ is dense in $V$ (This is nontrivial and known as the Weierstrass approximation theorem) in an appropriate norm.

Astyx

@Meow Yup

Sha Vuklia

4:35 PM

But how is it true for an abelian group @Secret? I can reduce the problem to $a^p=b^p$, but assume the order of $a$ and $b$ is $p$, then they don't have to be the same right?

Meow Mix

@BalarkaSen Oh, approximation because you can approximate functions to an arbitrary degree using polynomials?

Balarka Sen

(In layman's term it means any continuous function can be approximated by polynomials)

That's it, @Meow

Meow Mix

Here's my idea of it.

So, given $n$ points, we can construct a polynomial through those $n$ points, right?

So, just take a lot of points

And you'll approximate it as much as you want?

Astyx

@MeowMix Not really

Alessandro Codenotti

The same space from @Balarka's example also has the nowhere differentiable functions as a dense subset @meow

Balarka Sen

4:37 PM

It's not that easy. Try approximating $f(x) = |x|$ by polynomials as a baby case.

Secret

@ShaVuklia yeah sorry, I overlooked there are $p$ $a$s multiplied together, then yes you need something more to show $a=b$, e.g. $a^pb^{p-1}=a$

Astyx

The theorem is for the infinite norm, so it's not about having a lot of points in common, but rather about being close

Balarka Sen

It also means totally bad functions like Alessandro said about having that property.

@MeowMix Any "reasonable" topological space is a metric space, even. By reasonable I mean Hausdorff (any two points admit disjoint neighborhoods) and a technical thing known as second countability.

Sha Vuklia

@Secret ok cool thanks!

Alessandro Codenotti

@MeowMix call $P_n$ the $n$ degree polynomial through $(1/k,f(1/k))$ where $f$ is the function you wish to approximate (so you're using interpolating polynomial on equally spaced nodes), then it's possible that $\lim_{n\to\infty} ||f-P_n||_\infty=\infty$, this is known as Runge's phenomenon in numerical analysis

So just taking a lot of points won't work

Secret

4:44 PM

Is it provable that there does not exist an explicit form (infinite series and integral expressions are also ok) for the indicator function of transcendental numbers?

Alessandro Codenotti

My points above are all wrong, derp

Just take equidistant points in [0,1]

Semiclassical

I wouldn't be surprised if it's not provable, but I'd be shocked if one existed.

trilolil

@AlessandroCodenotti $\frac{d}{dt}(\frac{\partial T}{\partial \dot r}) - \frac{\partial T}{\partial r} = ....$
Does this make any sense to you? the first part AFAIK shows the change in momentum while the second part shows smth else. if $\dot r$ is a (generalized) coordinate Shouldn't they have been consistent in their notation and write a dot on the second part as well?

those equations are pretty hard to intuitively understand...

Alessandro Codenotti

No, the second part is derived wrt to $r$, which is another coordinate

trilolil

... = $-\frac{\partial W_r}{\partial r}$

any suggestions about how I should interprete this formula?

Alessandro Codenotti

4:54 PM

That's the Euler-Lagrange diff.eq. for your system, its solutions are the equations of motion

Secret

One reason I suspect that is that polynomials alone don't really distinguish between the set of all algebraics and transcendentals, because for the former at least one polynomial will give zero as a result, and for the latter, all polynomials will give nonzero values, but there seemed to be no way to "average" all the possible values of all polynomials (with rational coeffcients) which there are $\aleph_1$ many of them when acting on the reals, which there are also $\aleph_1$ many of them, without false positive results like cancellations such as 1-1=0

Alessandro Codenotti

If you want to know why they work you should look how they're obtained from the Newton's laws

Secret

That is, attempt to write some function that group all the results when polynomials were plugged with algebraic numbers and transcendental numbers and you always end up losing information somewhere along the way, at least for my attempts so far

Alessandro Codenotti

Why are you limiting yourself to rational polynomials? I'd call a combination of real polynomials a closed form

Meow Mix

Sorry the internet stopped working for a second

trilolil

4:58 PM

@MeowMix we forgive you.

Meow Mix

Had to turn the router off and on again

Secret

because transcendentals are defined to be not roots of any rational polynomials. If we start allowing transcendentals as coefficients, then it might have a transcendental root I guess, unless you mean a coefficient defined by linear combination of reals that give algebraic numbers?

Fawad

@Secret $aleph_1$ is what?

Secret

cardinality of power set of the naturals, which by continuum hypothesis (CH), is the same as the cardinality of the reals

Meow Mix

Hi @skillpatrol

Fawad

5:04 PM

Hi pal @skillpatrol

skill patrol

Hello friends :-)

Fawad

@Secret that cardinality of $\aleph_1$ is total number of subsets,1 null set and one set itself or cardinality of total elements of all sets ?

In this cardinality of $\aleph_1= 3$?

Secret

$\aleph_1$ is the cardinality of the set containing the empty set, the whole set of countable numbers, and all subsets of said countable set

Cardinality of a set $S$ is denoted by $card(S)$ or $|S|$

Thus for reals under CH, $|\Bbb{R}|=\aleph_1$

and for $|[x,y,z\}|=3$

For $\mathscr{P}(\{x,y,z\})$, the cardinality is $2^3=8$ as shown in your diagram

where $\mathscr{P}$ means power set, or the set of all subsets of a given set

Fawad

@Secret this makes easy to understand.

Alessandro Codenotti

Actually I think that function might be in some high Baire class, even the indicator function of, say, the cantor set is and it's quite a mess

Fawad

5:12 PM

@Secret then what is $\aleph_0$ and $\aleph_1$ ?

Secret

$\aleph_0$ is the cardinality of countable sets, the typical example is the natural numbers and rationals.

$\aleph$ are cardinal numbers. They denote how large a set is in terms of a bijective function

Meow Mix

They don't change, either

Secret

For example, since for integers we can count from 0 1 -1 2 -2 3 -3 and so on, it has a 1 to 1 mapping with counting numbers 1 2 3 4 ... Thus we can use the bijective function to "count" the number of elements in a set. We then say this set has a cardinality of $\aleph_0$

Meow Mix

Each set doesn't have its own $\aleph_1$, as you seem to think

Daminark

Hey everyone!

Meow Mix

5:15 PM

Hi @Dami

Fawad

Hi

Meow Mix

Ted gave me a question about infinite dimensional vector spaces

Daminark

Oh that's my thing (at least right now), what is it?

Balarka Sen

i promise i had an extensive plan for stuff after the exams. i dunno what to do.

Meow Mix

Don't answer it yet (like @Akiva does /s ), but prove that for any subspace, it's orthogonal complement is closed

Balarka Sen

5:18 PM

i can, like, procrastinate on nothing

Fawad

@Secret I didn't remembered. But I will ask you,which is bigger, $\aleph_0$ or $\aleph_1$ ?

Meow Mix

The latter

Fawad

And why/how?

Daminark

Nice, this is a good problem

Meow Mix

To see this, take a set with cardinality $\aleph_0$ and one with $\aleph_1$

Our set with cardinality $\aleph_0$ will be $\Bbb Z$

Semiclassical

5:19 PM

@BalarkaSen Give me a pencil and a pad of paper, and I'll procrastinate.

5

Meow Mix

And our set with cardinality $\aleph_1$ will be $\mathcal{P}(\Bbb Z)$. Okay?

Balarka Sen

@MeowMix Technically speaking, no. You need CH for that

$\aleph_1$ is by definition larger than $\aleph_0$

Semiclassical

Power set and all that.

Daminark

Assume GCH! :P

Meow Mix

Obviously, the power set contains $\Bbb Z$ itself

Secret

5:20 PM

@AlessandroCodenotti I see. I wonder if there are upper bounds on the Baire class this indicator function might be, that might help further investigation, but as you implied, the cantor indicator is already quite messy. If this function is in a Baire class > 2, then it might be intractable to even write it down as some recurrence relation or series approximation

Meow Mix

But, it also contains all other subsets of it. Therefore, there are more elements

Daminark

@Meow Careful

Infinite sets can have the same cardinality as proper subsets

Meow Mix

Yeah, good point.

Semiclassical

As an example, the even positive integers are a proper subset of the positive integers.

Daminark

$\mathbb{Q}$ contains the prime numbers along with infinitely many other elements, and yet they have the same cardinality

Meow Mix

5:22 PM

I haven't studied any set theory. slowly walks away

Daminark

The way you would show something like this is to attempt to construct a bijection between $S$ and $\mathcal{P}(S)$ and show that things break

Fawad

@Semiclassical so there are proper/improper subsets? What proper sub set is?

Semiclassical

What?

A proper subset of a set A is, by a definition, a subset of A that isn't equal to A.

The even positive integers are a subset of positive integers which is not equal to the positive integers, ergo it's a proper subset.

Fawad

I see

Meow Mix

Are there vector spaces over a field of finite elements?

Krijn

5:25 PM

@MeowMix Yeah

Secret

@MeowMix any finite fields will do e.g. $\Bbb{Z}/7\Bbb{Z}$

Fawad

@Semiclassical null set can be proper sub set?

Krijn

And it's just $\mathbb{F}_q^n$ so not really super hard

SteamyRoot

You can even have the vector space over the field with one element

Semiclassical

It always is, so long as the set A of which it's a subset isn't itself the empty set.

Krijn

5:27 PM

@SteamyRoot Some men just want to watch the world burn

Semiclassical

And some just want to fiddle.

Secret

Field with one element

In mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist. This object is denoted F1, or, in a French–English pun, Fun. The name "field with one element" and the notation F1 are only suggestive, as there is no field with one element in classical abstract algebra. Instead, F1 refers to the idea that there should be a way to replace sets and operations, the traditional building blocks for abstract algebra, with other, more flexible objects. While there is still no field with a single...

Balarka Sen

And some are like okay with that

Secret

Well, I still don't fully comprehend that, thus will be dealt with later

Alessandro Codenotti

Vector spaces over finite fields are very popular in oddtown @meow ;)

Balarka Sen

5:28 PM

Nobody knows F_1

Krijn

@BalarkaSen Although if we did, number theory would be a lot easier

Balarka Sen

So I hear

Meow Mix

@AlessandroCodenotti is that ahint?

Sha Vuklia

Let $I\subset\mathbb R$ be an open interval and let $f\colon I\to\mathbb R$ be a differentiable function in $a\in I$, with $f’(a)=c$. Now consider $r(h)=f(a+h)-f(a)-ch$. We know that $\lim_{h\to0}\vert h\vert^{-1}\vert r(h)\vert=0$.
My book says that differentiability of $f$ in $a$, implies that $r(h)$ is of smaller order in $h$ than a linear function in $h$.
We can rewrite the equation as follows: $f(x)=f(a)+c(x-a)+r(x-a)$.
I just don’t understand where this linear function comes from. I know that $\vert r(h)\vert$ is of smaller order than $\vert h\vert$ for $h$ close to 0, but is it becau…

Alessandro Codenotti

Yes @meow

Semiclassical

5:30 PM

That's rather confusing notation. You might want to write (x-a)c instead of c(x-a).

Balarka Sen

@ShaVuklia $\lim_{x \to 0} r(x)/x \to 0$ means that $r(x)$ grows slower than any linear function on $x$.

Meow Mix

Neither my parents nor my math teacher (yes, let that sink in) think I should pursue something pure math-related

Sha Vuklia

@BalarkaSen oh of course,

we can simply apply limit theorems then

for $\lim_{x\to0}r(h)/cx$

Semiclassical

@MeowMix What do your parents think you should do?

Balarka Sen

yeah, sure.

Sha Vuklia

5:33 PM

@MeowMix lol, what are their suggestions for you then?:P

Meow Mix

Well my brother is getting a medical degree and that makes a lot of money.

Semiclassical

ugh.

Sha Vuklia

^

so your parents want you to make a lot of money, how about your teacher?

Semiclassical

I mean, there is an argument for being at least somewhat practical about what you do. (A degree in art history, however satisfying it might be at the time, has a fairly low return-on-investment.)

But math is hardly an impractical degree.

Sha Vuklia

true

Semiclassical

5:36 PM

There's also an argument for not committing to a specific major/career too soon. But, again, starting in math makes it a lot easier to move into other majors/careers.

Balarka Sen

Whoever says education is about making lots of money, know that they are ideas about education are all garbage.

Mike Miller

They're right, it's about making lots of money for admin

4

skill patrol

^true dat pal

Secret

Many of my friends from professors to classmates to teachers to students all said we are all buried by the admin and not having enough time to do actual studying nor research

and that paperwork just get thicker

Fargle

@BalarkaSen Or "schools should compete like businesses to improve the quality of the product", because if there's one thing we want to be zero-sum, it's our damn future

Semiclassical

5:39 PM

I'm not going to argue for/against pure math research as a career. I'm cynical enough about academia at this point that I can't do the former, but I don't actually know enough about pure math research to do the latter.

Meow Mix

Hi @Fargle

Fargle

Hiya @Meow.

Semiclassical

But pursuing math at the undergrad level? Anyone who considers that to be impractical just strikes me as silly.

Daminark

Back! Hey @Fargle, @Sha, @Mike, and @skill!

Fargle

Heya @Dami

Meow Mix

5:40 PM

@Dami turning a cold shoulder to me. I thought we were friends! /s

Secret

Anyway, I think I need to go to sleep, that excel file full of ugly numbers have bored me to exhaustion by having me to do copy and paste 10 times

skill patrol

Welcome back pal @Daminark

Meow Mix

And you forgot @Balarka too, and @Secret

Daminark

I mean I'm not in the North so my shoulders are reasonably warm

:P

Balarka Sen

Oh, and hi @MikeMiller

Daminark

5:41 PM

Weren't they here before as well?

Lmao, well, how's everything going with you guys?

Balarka Sen

My shoulders are burning man.

Mike Miller

Hi

Daminark

(And Zach what progress have you made on proving that orthogonal complements are closed?)

skill patrol

Chillin

Semiclassical

@MikeMiller There was a nice talk here yesterday. (It was part of the history of science/technology colloquium series, but I think you'd have enjoyed it too.)

Mike Miller

5:42 PM

what on?

Semiclassical

@MikeMiller physics.umn.edu/events/calendar/spa.all/2017/spring/…

Sha Vuklia

Hi @Daminark ! Welcome back:P

Mike Miller

i would have for sure

Sha Vuklia

My book says that $f$ is differentiable in $a$ with derivative $c$, is equivalent to saying that $f$ is differentiable with derivative the linear map $A:h\mapsto ch$. However, there is a different between $c$ and $ch$, isn’t there? Or do they actually mean with linear map the corresponding matrix?

Daminark

Pretty much, be careful to make the distinction between a matrix and a linear map though

Semiclassical

5:45 PM

(I wonder to what extent being a TA can be understood under the rubric of 'maintenance', come to think of it.)

Daminark

But let's say you already chose a basis

Sha Vuklia

$\{1\}$ I'd say

Daminark

Then yeah, you have a matrix $c$, then you get the linear map $h\mapsto ch$

Semiclassical

@MikeMiller But yeah, it was quite enjoyable.

And it ended a little early so there was time for a good bit of discussion, which was nice.

LegionMammal978

Small question: For any classes $a,b,c$, does $a\subseteq b\cap c$ follow from $a\subseteq b\wedge a\subseteq c$?

Meow Mix

5:48 PM

yes

wait, what?

do you mean $\subset$?

LegionMammal978

@MeowMix whoops

Daminark

@Sha Are you currently dealing with mappings between arbitrary $\mathbb{R}^n$ and $\mathbb{R}^m$?

Meow Mix

If you do, then yes

LegionMammal978

And it works with $\subseteq$ as well?

Meow Mix

mhm

Actually, it only works with $\subseteq$ I think

Sha Vuklia

5:50 PM

@Daminark Well, I'm about to read the definition of differentiability between such mappings!

but in this case it was still $\mathbb R$

Astyx

Those two symbol kind of mean the same I beleive

Semiclassical

Depends on the author.

Daminark

Well, what is the definition you gave me right now? Because the use of lowercase sounds a bit like it's still one dimension

Ah, yeah

So a linear mapping from $\mathbb{R}$ to itself is a function $f(x) = ax$ for some real number $a$

Sha Vuklia

Yep

Meow Mix

$\{2\} \subset \{1,2\}$ and $\{2\} \subset \{2,3\}$ but $\{2\} \not\subset \{2\}$

Daminark

5:55 PM

When you define the derivative more generally it'll start to make sense

Spoiler alert

Sha Vuklia

@Daminark hahahah, yea I reckoned :P

Daminark

$f$ is differentiable if there exists a bounded linear operator $T$ such that $\lim_{h\to 0} \frac{\|f(x+h)-f(x)-Th\|}{\|h\|} = 0$

I mean boundedness is automatic if you're dealing with $\mathbb{R}^n$ anyway

But yeah, if you swap things around slightly in the definition for the case of $\mathbb{R}$, you note that $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ is the same as $\lim_{h\to 0} \frac{f(x+h) - f(x) - f'(x)h}{h} = 0$

In general you just have to stack on a norm because you can't divide by vectors

Sha Vuklia

yea, and then they just added these norms, so we can work with the limits

haha yea:P

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$Mathematics$ Mathematics