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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?
@AlessandroCodenotti yes I am reading about orbital mechanics.
lagrange differential equation
It's just a double derivative, the $\dot r$ is treated as a coordinate
Wait @Alesssandro, which direction of the parallelogram law thing is hard
@AlessandroCodenotti strange that they write a dot on top of it, usually in mechanics a dot denotes a derivative...
$T$ (kinetic energy by any chance?) is probably a function of $T(t,r_1,...,r_n,\dot{r_1},...,\dot{r_n})$
4:03 PM
@AlessandroCodenotti absolutely kinetic energy!
Ah nvm I guess showing that what should be the inner product is actually an inner product isn't easy
what's new
@Eric showing that a norm satisfying it comes from an inner product it's not trivial, see the first answer here for example
Hi @Balarka
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.
You probably want to look at generalized coordinates to better understand that $\dot r$
@AlessandroCodenotti what should I read about on wikipedia to understand T(t,r1,...,rn)?
@AlessandroCodenotti Thanks!
@BalarkaSen I started reading my professor's notes about homology yesterday instead of studying analysis and probability as I'm supposed to
@Alessandro Excellent! Or probably not.
@Balarka I have this linear algebra thing to prove
4:18 PM
Hello Balarka Sen, how are you doing
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!)
hi chat
Hi @Semi
And chat more generally
Hi @Astyx
4:26 PM
Hi @mreyeglasses. I'm pretty good.
@Meow Closed under which norm ?
The norm defined by its inner product
He guys. If $(ax)^p=(bx)^p$ for some $p$, can we say then that $a=b$?
He @ShaVuklia
@MeowMix Are you working infinite dimensions already?
4:28 PM
Hi @Astyx
Assume that $x\neq 0$ I guess
What's the context in this question @ShaVuklia ?
@ShaVuklia context?
I am not given the context XD
@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.
someone asked me this, I think they're working with group theory?
4:29 PM
so, closed in the topological sense, or closed as in $a+b \in S$
@MeowMix Every subspace is closed in finite dimensions.
Closed as in, contains all it's limit points.
@Balarka In finite dimension, this exercise is quite trivial
(And every finite dimensional subspace is closed in general)
@Astyx I am aware. I just didn't know Meow is doing stuff beyond finite dimensions already.
4:30 PM
Of course you are aware :)
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.
@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
Then consider the subspace $V$ of infinite sequences with finitely many nonzero components
Ya, and that's what I meant by infinite dimensions
@Secret Ah okay, I've asked for more context!
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"
@ShaVuklia We don't have enough information to answer really. Is it for all $x$ ? Is it commutative ? Can $p=0$ ?
@Astyx Sorry, I have no idea either. I've asked the one who asked this question to elaborate more on the context:d
By the way, a metric space naturally defines a topological space, right?
@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.
@Meow Yup
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?
@BalarkaSen Oh, approximation because you can approximate functions to an arbitrary degree using polynomials?
(In layman's term it means any continuous function can be approximated by polynomials)
That's it, @Meow
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?
@MeowMix Not really
The same space from @Balarka's example also has the nowhere differentiable functions as a dense subset @meow
4:37 PM
It's not that easy. Try approximating $f(x) = |x|$ by polynomials as a baby case.
@ShaVuklia yeah sorry, I overlooked there are $p$ $a$s multiplied together, then yes you need something more to show $a=b$, e.g. $a^pb^{p-1}=a$
The theorem is for the infinite norm, so it's not about having a lot of points in common, but rather about being close
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.
@Secret ok cool thanks!
@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
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?
My points above are all wrong, derp
Just take equidistant points in [0,1]
I wouldn't be surprised if it's not provable, but I'd be shocked if one existed.
@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...
No, the second part is derived wrt to $r$, which is another coordinate
... = $-\frac{\partial W_r}{\partial r}$
any suggestions about how I should interprete this formula?
4:54 PM
That's the Euler-Lagrange diff.eq. for your system, its solutions are the equations of motion
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
If you want to know why they work you should look how they're obtained from the Newton's laws
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
Why are you limiting yourself to rational polynomials? I'd call a combination of real polynomials a closed form
Sorry the internet stopped working for a second
4:58 PM
@MeowMix we forgive you.
Had to turn the router off and on again
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?
@Secret $aleph_1$ is what?
cardinality of power set of the naturals, which by continuum hypothesis (CH), is the same as the cardinality of the reals
Hi @skillpatrol
5:04 PM
Hi pal @skillpatrol
Hello friends :-)
@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 $\aleph_1= 3$?
$\aleph_1$ 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
@Secret this makes easy to understand.
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
5:12 PM
@Secret then what is $\aleph_0$ and $\aleph_1$ ?
$\aleph_0$ 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
They don't change, either
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 $\aleph_0$
Each set doesn't have its own $\aleph_1$, as you seem to think
Hey everyone!
5:15 PM
Hi @Dami
Ted gave me a question about infinite dimensional vector spaces
Oh that's my thing (at least right now), what is it?
i promise i had an extensive plan for stuff after the exams. i dunno what to do.
Don't answer it yet (like @Akiva does /s ), but prove that for any subspace, it's orthogonal complement is closed
5:18 PM
i can, like, procrastinate on nothing
@Secret I didn't remembered. But I will ask you,which is bigger, $\aleph_0$ or $\aleph_1$ ?
The latter
And why/how?
Nice, this is a good problem
To see this, take a set with cardinality $\aleph_0$ and one with $\aleph_1$
Our set with cardinality $\aleph_0$ will be $\Bbb Z$
5:19 PM
@BalarkaSen Give me a pencil and a pad of paper, and I'll procrastinate.
And our set with cardinality $\aleph_1$ will be $\mathcal{P}(\Bbb Z)$. Okay?
@MeowMix Technically speaking, no. You need CH for that
$\aleph_1$ is by definition larger than $\aleph_0$
Power set and all that.
Assume GCH! :P
Obviously, the power set contains $\Bbb Z$ itself
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
But, it also contains all other subsets of it. Therefore, there are more elements
@Meow Careful
Infinite sets can have the same cardinality as proper subsets
Yeah, good point.
As an example, the even positive integers are a proper subset of the positive integers.
$\mathbb{Q}$ contains the prime numbers along with infinitely many other elements, and yet they have the same cardinality
5:22 PM
I haven't studied any set theory. slowly walks away
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
@Semiclassical so there are proper/improper subsets? What proper sub set is?
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.
I see
Are there vector spaces over a field of finite elements?
5:25 PM
@MeowMix Yeah
@MeowMix any finite fields will do e.g. $\Bbb{Z}/7\Bbb{Z}$
@Semiclassical null set can be proper sub set?
And it's just $\mathbb{F}_q^n$ so not really super hard
You can even have the vector space over the field with one element
It always is, so long as the set A of which it's a subset isn't itself the empty set.
5:27 PM
@SteamyRoot Some men just want to watch the world burn
And some just want to fiddle.
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...
And some are like okay with that
Well, I still don't fully comprehend that, thus will be dealt with later
Vector spaces over finite fields are very popular in oddtown @meow ;)
5:28 PM
Nobody knows F_1
@BalarkaSen Although if we did, number theory would be a lot easier
So I hear
@AlessandroCodenotti is that ahint?
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
Yes @meow
5:30 PM
That's rather confusing notation. You might want to write (x-a)c instead of c(x-a).
@ShaVuklia $\lim_{x \to 0} r(x)/x \to 0$ means that $r(x)$ grows slower than any linear function on $x$.
Neither my parents nor my math teacher (yes, let that sink in) think I should pursue something pure math-related
@BalarkaSen oh of course,
we can simply apply limit theorems then
for $\lim_{x\to0}r(h)/cx$
@MeowMix What do your parents think you should do?
yeah, sure.
5:33 PM
@MeowMix lol, what are their suggestions for you then?:P
Well my brother is getting a medical degree and that makes a lot of money.
so your parents want you to make a lot of money, how about your teacher?
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.
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.
Whoever says education is about making lots of money, know that they are ideas about education are all garbage.
They're right, it's about making lots of money for admin
^true dat pal
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
@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
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.
Hi @Fargle
Hiya @Meow.
But pursuing math at the undergrad level? Anyone who considers that to be impractical just strikes me as silly.
Back! Hey @Fargle, @Sha, @Mike, and @skill!
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