12:01 AM
@Eran Cayley's theroem
Every group acts transitively on itself by left multiplication

12:23 AM
@orbit-stabilizer I'm not the best at the dihedral group either, and the only practice I know of is to just find a book with good problems and work through it

12:48 AM
Hey there is something stupid that I am stuck on

@Daminark sounds good

I am just trying to understand why is this goes from $C^{2}(X)$ to $C^{0}(X)$?

Each of the derivations knocks down a dimension
So if you compose two it'll knock down two dimensions, an then you take the difference as usual

Claim: If $F \subseteq \Bbb{R}$ is such that $m(\Bbb{R}-F) = c < \infty$, then $F$ is uncountable and unbounded. Proof: If $F$ were countable, then $c = m(\Bbb{R}-F)=m(\Bbb{R}) - m(F) = \Bbb{R}$, which is a contradiction. Again, if $F$ were bounded, then $m(F) < \infty$, then $c = m(\Bbb{R} - F)$ implies $m(\Bbb{R}) = c + m(F) < \infty$, which is another contradiction. Hence, $F$ is uncountable and unbounded. Does this sound right?
Oh, by the way, $m()$ denote the Lebesgue measure.

1:04 AM
yeah but we are almost knowing all dimensions ?
@Daminark can yeah I agree with this but here it goes from $C^2$ to $C^0$

I mean maybe it's that you're dealing with surfaces or something? Like in principle it'd be l to l-2

yeah
I need to stop focusing on such minor details
it is stupid
I think we are picking here a specific l
being l = 2

1:24 AM
hi need help solving (z+i)^5 = 1

@Daminark Let $G$ act transitively on $X$. Let $N\triangleleft G$. Show that any two $N$-orbits in $X$ have the same cardinality.
Ah, but what if G is infinite...

1:39 AM
I have a serious problem trying to prove $\bigcap_{i=1}^kker(f_i)\subset ker(f)\iff f\in Span(f_1,...,f_k)$ Can someone help me :( ?

@Mathematica, $e^{5\log(z+i)}=1$, then use the complex definition of the log function

1:57 AM
@Bvss12 It isnt clear what your notation means. What are $f$ and $f_i$? Im also not sure what $Span(f_1, . . ., f_k)$ means; usually I only see span with elements of a vector space, but your $f$ look like they are maps..

2:09 AM
Hows everyones exams going?

Not goin yet here

@KevinDriscoll math.stackexchange.com/questions/2546630/… Thanks Kevin for answering.
I asked the question and found a pretty good proof in this link
But that's part is not too clear for me...

@orbit-stabilizer never think about infinite groups, you lose cardinality arguments

@Daminark lol

2:41 AM
How do I represent a function in in the form of a/(1-bx) as a power series?
nvm, I think I got it... Would it be a/(1-bx) = Sum[n=0,inf]: a(bx)^n ?
What have I done wrong?
nvm
I miscalculated the interval of convergence
I hope you guys are ok with me using this chat like a rubber ducky
nobody actually online here to stop me ;)
Actually legit need help here: The hell am i doing wrong?
plz halp?

3:38 AM
pls?

4:01 AM
@Daminark hey

4:28 AM
Hey

4:44 AM

he has left us
when's your group theory exam @Daminark

Friday

Next week?

Yeah, and probably
I was okay on the first midterm, better on the second, so hopefully this will go nicely

That's great!
Mine's Tuesday and I'm not ready at all

4:57 AM
Damn, well good luck

Currently learning nilpotence and solvability. Sighhh
Wow that was fast entrance and post @Secret

Hmm, I knew algebra can be visual, but I never thought associative algebra can be done similarly

Wait
I think I was just reading her blog earlier today!

@orbit-stabilizer I used to that. Sometimes, I drop a link to seed a discussion and then disappear, other times I just sit back and watch the discussion unfolds

4:59 AM
10 hours ago, by Silent
Hey! Let $F$ be an archimedean ordered field. Suppose $\bar F$ be Dedekind completion of $F$. Does $\bar F$ has to be archimedean?
Please have a look at this

@Silent sorry, I don't know Dedekind completion

@Silent I think the answer is yes, as $F \subset \bar F$ is dense (right?)

@orbit-stabilizer That one is a subset of the cartesian product between two shapes. In general, if you do a cartesian product between two 2D objects, you will get a 4D one, so to make a torus, it is a subset of that cartesian product. I forgot the details but I will revise later
Dedekind cuts cannot really create infintesimals from a set that does not have any, do they?

Oh no visual stuff I'm outta here bye everyone

lol

5:05 AM
Okay not really but lol

When I finally get back to algebra, I should explore both symbolic and visual ways to represent them. By that time, I will be ok with infinities enough to visualise most dense structures and infinite groups I hope

I think topology is pointless.

Not really if you know what kinds to focus on
On a more joking note, there exists pointless topology

Dammit Secret

Twas a joke

5:08 AM
I can't have you sniping me now
Also hey @Mathein!

Hey @Daminark

The cohomology discussion in recent weeks confuse me like hell, though,which is why I am kinda less active on main and mostly in my own rooms

But yeah I dunno, I've been less than interested in the very fancy stuff in point-set, like blah only works in $T_{screw\ you}$ spaces

well actually, cohomology is still ok because I have been to some lecture. It is those fundamental groups stuff that really confuse me

Fundamental groups are an easier concept than cohomology, I'd say
The idea is that you say two functions $f,g:X\to Y$ are homotopic if there is some $h:X\times I \to Y$ such that $h(x,0) = f(x)$ and $h(x,1)= g(x)$
(Everything here being continuous ofc)
So, one thing you have in topology is called the compact-open topology
And these things called compactly generated spaces

5:13 AM
so all open sets in those topologies are compact sets?

Nope
I don't know much about it
But it's something like
Choose $K\subset X$ compact and $U\subset Y$ open, and let $V(K,U) = \{f:X\to Y |\ f(K)\subset U\}$
And let that be a subbasis for the topology
Or something vaguely to that effect

Aren't those called $K$-spaces or something @Daminark
Also morning everyone!

The point is, when you let your spaces be compactly generated, it turns out that currying becomes an isomorphism
$C(X\times Y, Z) \cong C(X,C(Y,Z))$

"currying becomes an isomorphism"

That's a homeomorphism of topological spaces

5:16 AM
Yeah, that's basically the main reason why one should care about compactly generated spaces

Oh shit currying is actually a legit term

Currying means you let $f(x,y) = f(x)(y)$

I thought it was a pun on curries or something

Currying is used when you look at group actions. You have a homomorphism from G to the symmetric group and you have a map from G x X to X that satisfies a couple properties. The relationship between the two is from currying

Ahhh I see

5:20 AM
@Mathein and @Perturbative what's going on with you?

@Daminark doing fine, thanks. Could use some more sleep I guess. And yourself?

@Daminark It seems that I've lived up to my stereotype, by being Indian and hungry and mistaking "currying" for a pun on actual curry

Doing aight. My finals aren't until Friday so I've got some time to rest
@Perturbative OH LOL

Apart from that, I'm doing aight

I mean the food curry is the fake one and currying as above is real ofc

5:24 AM
@MatheinBoulomenos I searched for that, and I got this question, saying that no non-archimedean field is dedekind complete.

@Silent interesting

Well, in that question, we are starting with non-archimedean field, while I was asking if we can end up with non-archimedean field

@Daminark @Perturbative obviously, Haskell Curry was not only an important mathematician and theoretical computer scientist who inspired multiple programming languages, but he also made an important contribution to Asian cuisine

Lmao

@Silent As I wrote above, I think this should follow more or less directly from the fact that any poset is dense in its Dedekind completion

5:31 AM
i just ate curry 10 hours ago, and am hungry again. I am from Asia. I don't know if curry is only food in Asia.

@CaptainBohemian Curry is a global object.

@MatheinBoulomenos wow, can a mathematician and programmer be interested in cuisine? I thought scientists are generally not interested in cuisine.

I was just joking

no wonder

Hi chat!
If we have any set $A \subset \Bbb{R}$, then any method by which we can say that $A$ has an open interval?
Like if we are given this set----> $A = \{\sum_{i=1}^{\infty}\frac{a_{i}}{5^i}, i =0,1,2,3,4\}$

5:43 AM
Oh my lord

@BAYMAX $A$ has an open interval or $A$ is an open interval?

I wish there's a way to separate the geometric series portion from the arbitrary series $a_i$. If we can do so, then if $a_i$ does not dominate the geometric series, then the fact the partial sums of the geometric series converges to zero means the end where the limit is will become open. My guess that in such condition, A can only be a half open interval

@Perturbative We need to check if $A$ contains an open interval.

what are the contents of $a_i$?

sorry
A typo $a_{i} = 0,1,2,3,4$

6:00 AM
I think $A$ will be half open of the form $[0,\frac{1}{1-\frac{1}{5}})$. This is because $5^i > a_i$ for all i, thus the asymototic behaviour of the summand, hence the series will be dictated by the geometric series $\sum_{i=1}^{\infty}\frac{1}{5^i}$
Actually, the open end will be something slightly larger than $\frac{5}{4}$. But as you can see, it closely follows the behaviour of the geometric series

[Integral Project] Hmm, I might be able to use methods like these to compute Waiting's many series
$$\sum_{i=1}^{\infty} \frac{b_i}{c_i}$$

In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. Put more plainly, the nth term of an arithmetico–geometric sequence is the product of the nth term of an arithmetic sequence and the nth term of a geometric one. Arithmetico–geometric sequences arise in various applications, such as the computation of expected values in probability theory. For instance, the sequence ...

hmm

The series you provide has a closed form. Thus yes, A should be half open with the form $[0,a)$ where $a$ is the infinite sum of the arithmetico geometric series $\frac{i}{5^i}$

6:52 AM
in The h Bar, 1 min ago, by DanielSank
$$\int_0^T dt_1 \int_{t_1}^T dt_2 \int_{t_2}^T dt_3 \cdots \int_{t_{n-1}}^T dt_n$$
Anyone have any idea how to do integrals like that?

Have you tried induction, or something of that sort?

@BalarkaSen Yeah but I'm not sure what I expect the result to be so induction is still not solid.

@BalarkaSen Maybe I can figure it out by evaluating the $t_n$ integral...
If the original expression is $I_n$, I get $$I_n = T I_{n-1} - \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-2}}^T dt_{n-1} t_{n-1} \, .$$
I now suspect the result should be $T^n/n!$.

Hmm

7:06 AM
...which perhaps we can prove by showing that this nested integral is related to a term in the Taylor series of the exponential function.

It seems you get a bad recurrence relation. $I_n = TI_{n-1} - T^2/2 I_{n-2} - \cdots$

Yeah but I bet if we take derivatives w.r.t. $T$ or some nonsense we get something good.

2 hours later…
9:05 AM
Suppose we have a set X which is not a linear subspace. Can the span of X then have a basis?

The span of $X$ will necessarily have a basis, since the span of any set is a vector space

@MaryStar define basis

9:23 AM
But if it is not a subspace then it does not pass through the origin, so it cannot be a vector space

@Secret Mary was talking about the span of $X$, not $X$ itself

@Daminark oh lol, I also missed that part

must span X pass through the origin if X is a plane that does not cross the origin?

@Secret 0 is in the span of anything
for it is the empty linear combination

Wait so in R^3, given an off centred plane, it's span will be all of R^3?

9:30 AM
@LeakyNun A basis of S contains linear independent vectors and every element of S can be written as a linear combination of the elments of the basis, right?
We have the subsets of $\mathbb{R}^2$: $X_1 := \{(x,y) \in \mathbb{R}^2 : x + y = 1\}, X2 := \{(x,y) \in \mathbb{R}^2 : x^2- y^2 = 0\}$. I have shown that these are not a linear subspace of the $\mathbb{R}$-vector space $\mathbb{R}^2$.

The span $\langle X_i\rangle$ contains every linear combination of the elements of $X_i$, right?

We have that $X_1 := \{(x,y) \in \mathbb{R}^2 : x + y = 1\}=\{(x,y) \in \mathbb{R}^2 : y = 1-x\}=\{(x,1-x) \mid x\in \mathbb{R}\}=\{(0,1)+x(1,-1)\mid x\in \mathbb{R}\}$

well, the span is $\Bbb R^2$ if you ask me
@Secret that should be correct
yes, that is correct, because you have an easy orthogonal basis: take the vector that points (from the origin) to the closest point on the plane, and then take two mutually perpendicular vectors from the plane starting from that point

I see
Hmm, I guess the reason why the span of sub spaces can never be the full space then might be because some of the degrees of freedom becomes zero vectors and hence cannot contribute to the span

if it's a space, then the span is itself, by definition

10:04 AM
Hey I have a doubt . Just now I got 200 reputation . So I earned a badge. Now what would happen if I get more up votes . Will they be added tomorrow

10:21 AM
1 message moved from Mathworks (Not the main chat!)

@AtharvaShetty No. Extra reputation is discarded.

11:03 AM
@AtharvaShetty The only way to earn over the threshold is to be awarded a bounty, users who downvoted you to be removed, questions that you create a bounty on are deleted, accepting answers to your questions, or getting your answers accepted (after the threshold is reached).

11:32 AM
I am reading Grang's Linear Algebra (Introduction to linear algebra), but the exercise there (OK, upto the second chapter which I read) looks kinda meh and almost each one can be done in <3 minutes, none requiring thinking. Is there a good book for supplementing the exercise of this ?

Perhaps the exercises are nontrivial in the later chapters?

Linear algebra only gets progressively interesting
Not many exercises that can be made out of matrix multiplication...

OK, so what should be done as an exercise supplementing the first few chapers ?

What is covered?

Well basic things. Just usual matrices, algorithm for inverse etc, transporse and algorithm for multiplication and I don't find something particularly interesting/nontrivial here.

11:48 AM
Let $A$ be a $2\times 2$ matrix and $A^{10}=0$. What can you say about $A^5$? (Take whatever field you wish)

@Narcissusjewel quite a lot
e.g. $A^5$ commutes with every matrix

I meant that to be an exercise for Alex K Chen

oops I was out for a while.
Yeah, nice exercise.
How do you type matrice in latex ?

e.g. $A^5$ isn't invertible
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$ \begin{bmatrix}1&0\\0&1\end{bmatrix}
e.g. $A^5$ is diagonalizable

Leaky >:(, don't annihilate the exercise

11:53 AM
@Narcissusjewel did I?

If $A^5 = \begin{bmatrix} p & q\\ r& s\end{bmatrix}$, then then $(p, q)$ is a multiple of $(r,s)$.

If he knows what those statements imply then he would know the answer beforehand

It's necessary and sufficein condition.

@AlexKChen you can say more
and I would ignore my remarks

Well I still don't know what "diagonalizable" means, but yeah it's obvious it's not invertible.
What ohter properties ?

11:55 AM

I can't find any other stuff except obvious one (eg. determinant is zero/commutes)

think about it until you can
can you give an example of $A^5$ other than the obvious one?

Like $\begin{bmatrix}1&2\\4&8\end{bmatrix}$ ?

what is $A$?
what is $A^{10}$?
Like say $A^5 = \begin{bmatrix} 4a &2a\\ 2a& -4a \end{bmatrix} 12:03 PM then$A^{10} = \begin{bmatrix} 20a^2 & 16a^2 \\ 16a^2 & 20a^2 \end{bmatrix}$so you're still wrong The exercise may be hard without some heavier tools. @Narcissusjewel I agree so it's ultimately unfair but @AlexKChen shouldn't post examples without verifying them first It's fun to think about. Maybe he can't get it today. OK, I have checked it 3 times:$ \begin{bmatrix}2 &1\\-4&-2\end{bmatrix}A^{10} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$indeed, but what is$A$? 12:11 PM I'm stuck and I will think about it later. BTW, obviously$A^{-1}A = AA^{-1} = I$, then why it's written in Grang's book that for square matrix (doesn't it holds for all matrix ?), if$BA = I$, then$AB = I$, in this case$B = A^{-1}$(and he tells the proof is hard)? because you don't get to pretend$A^{-1}$exists before you prove it Have you read of$\text{GL}_n(\Bbb F)$yet? (Other notations$\text{GL}(n,\Bbb F)$,$\text{GL}(n)$or over an explicit field$\text{GL}(n,\Bbb R)$, etc) @Narcissusjewel Nope... I'm just in the second chapter :D (but I have seen the symbol in PftB) (BTW is it in later chapters of Grang ?) @LeakyNun What do you mean ? I don't understand let$A$be a matrix I haven't used Grang sorry. 12:15 PM theorem: if$AB=I$for some$B$, then$BA=I$@Narcissusjewel Grang is a short for Gilbert Strang your proof: clearly$B=A^{-1}$, done Yeah I haven't used Gilbert Strange sorry. Exactly. 12:16 PM my objection: what is$A^{-1}$? how do you even know that it exists? let's say$A^{-1}$is the matrix$B$such that$AB=I$then how do you know$BA=I$?$A^{-1}$is just a symbol you need to use the theorem @LeakyNun Given$AB = I$, suppose$BA = S$. Then$ABA = AS \Rightarrow IA = AI = AS \Rightarrow A(I-S) = 0$... this is where I got stuck Good song. Oh, hi Mark who is mark ? @AlexKChen he did say that the proof is not easy in particular there's no way you can do it purely algebraically only onsidering matrices because it is false for infinite-dimensional matrices, but they obey the same rules which you will ever use 12:23 PM wow infinite dimesional matrices means something like an infinite 2d array of numbers ? for now you can consider them so if you know about 1. vector spaces and 2. the connection between matrix and linear transformation then I can tell you more zero for #1, and very very less for #2. Probably I'll know something about both the next saturday, so then. you should really know about #2 even if you don't know #1 1 hour later… 1:42 PM @TedShifrin why is (n choose 2) the dimension of SO(n)? 2:16 PM @LeakyNun do you know how to prove that SO(n) is a manifold? the dimension falls out from the proof they say you need to consider it as a subspace of$\Bbb R^(n^2)$and solve for the rank dim of preimage of a regular point = dim of domain manifold - dim of range manifold just a linear algebra theorem, locally rank-nullity In the typical proof that O(n) is a manifold you consider the map from$M_n(\Bbb R) = \Bbb R^2$to the space of all symmetric matrices given by$M \to M^T M$O(n) is the fibre of the identity matrix 2:20 PM yee SO(n) is then just an open subset of O(n) 2:50 PM Hey does anybody know a good way of measuring confidence for measurements when the timestamps don't overlap? 1 hour later… 4:00 PM Grüß gott! @ÍgjøgnumMeg Ja 4:16 PM Let$\Bbb H=[0,1]^{\Bbb N}$, equipped with the product topology, is there a "natural" way to embed$\ell^p$and$L^p$spaces in$\Bbb H$? 4:31 PM Hi @Ted @LeakyNun The best way to see this easily is to think about the Lie algebra, which is the space of$n\times n$skew-symmetric matrices. This is the vector space$\Lambda^2(\Bbb R^{n*})$. :) Heya demonic Alessandro. :) Hi @Ted For the Hilbert cube, don't you want$\prod [0,1/n]$? Heya Balarka. @TedShifrin I think they're homeomorphic I don't think so, actually. 4:34 PM I don't see why they are not homeomorphic Component-wise checks in the product topology work for the target (range), but not the domain. I mean if$X_i$is homeomorphic to$Y_i$for all$i$, how can$\prod X_i$not be homeomorphic to$\prod Y_i$? I haven't thought about this before, but why would everyone define it the other way? Why isn't the obvious map$h(\cdots, a_i, \cdots) = h(\cdots, ia_i, \cdots)$continuous? It's continuous on the factors, so continuous on the whole thing. That's what product topology means Be careful. When you're mapping into a product, not out of. 4:36 PM I am mapping into a product. It's \prod [0, 1] What about the inverse, then? Something's fishy. That's also mapping into a product :P @TedShifrin interesting. How to see that the Lie group and the algebra have the same dimension? And is continuous on the components @Leaky: Because the Lie algebra is the tangent space at the identity. 4:37 PM afterall, the dimension is a topological one for the former and a linear-algebraic one for the latter Right, that's the power of manifolds. ah, the beauty of maths Wiki says that they are homemorphic and that one usually uses$[0,1/n]$so that it inherits the metric from$\ell^2$Aha. so$L^2$is self-dual! 4:39 PM So? just a remark ah, it's just the inner product! are inner products defined for other$L^p$then No. They aren't inner product spaces. Google Hölder. I know that inequality Do you know how to test if a normed vector space is an inner product space? this I don't 4:42 PM It's geometric :P Polarize! Let no inner product evade your eyes! Remember why the good lord made your eyes, so don't shade your eyes... how? and Polarize, Polarize, Polarize! Balarka wants to be Tom Lehrer. Hey everyone! 4:42 PM hi Perturb Ugh functional analysis is frying my brain Hey @TedShifrin :) @TedShifrin It's hard to be a god > Then the underlying theorem, attributed to Fréchet, von Neumann and Jordan, is stated as: In a normed space$(V, \|\cdot \|)$, if the parallelogram law holds, then there is an inner product on V such that$\|x\|^{2}=\langle x,\ x\rangle$for all$x\in V$. @Alessandro: Fritto misto? 4:44 PM @TedShifrin Nope, just my brain :P Diff geom notation is frying my brain Stop your whining! I was just joking @Ted :p @Perturbative$\nabla\partial\wedge\Bbb P$@TedShifrin which software did you use for your avatar? Mathematica, about 20 years ago. 4:48 PM$\mathfrak{o}(n) = \mathfrak{so}(n)$Indeed. so "Lie algebra" is some kind of a cross product kind --> generalization Cross product only works in 3 dimensions. Otherwise you need$n-1$vectors in an$n$-dimensional space. The right viewpoint, however, is exterior algebra, whence the wedge in my reply to you earlier. :) but is wedge an algebra? Yup. 4:50 PM I mean, it isn't closed The whole exterior algebra is$\bigoplus \Lambda^k$. oh, fair enough$\operatorname{Aut}(\Delta) = S_3$:P A physicist would tell you that the cross product of two vectors is a pseudo-vector, not a vector. how so? Because it doesn't transform the way a vector should if you change basis. You're using an isomorphism$\Lambda^2\Bbb R^3 \cong \Bbb R^3$(star operator). 4:52 PM$\operatorname{Aut}(\Bbb R^n) = GL_n(\Bbb R)$Can I safely say that the ANTI-DERIVATIVE gives you a function whereas the Define Integral gives you a number?$\operatorname{Aut}(\operatorname{Aut}(\Delta)) = \operatorname{Aut}(\Delta)$@Trey: More or less. @Trey no need to shout; we are civilized people well there's the constant of integration... 4:53 PM @Trey a parametered function, maybe Antiderivative if you fix a value gives a function. Similarly, definite integral needs an interval. But, yes. dem, sniped Right, what Ted says thanks Did anyone get my joke... too obscure rip 2 mins ago, by Leaky Nun$\operatorname{Aut}(\Delta) = S_3$:P 4:54 PM "Suppose$M$is a smooth manifold with our without and$p \in M$,$v \in T_pM$and$f, g \in C^{\infty}(M)$, if$f$is a constant function then$vf = 0$." - Why does pointwise multiplication of$v$and$f$feel like taking the derivative of it? It is not pointwise multiplication at all!!!!!! @Perturbative what is$v$? afterall derivatives and tangents are closely related A tangent vector at$p$acts on a smooth function by taking its directional derivative at$p$in the given direction. It is derivative :) <-- resigns and leaves the room to Balarka. 4:56 PM lol @TedShifrin$C^{\infty}(M)$is a commutative ring, I thought$vf$was implicity$v \cdot f$where$\cdot$was the pointwise multiplication operation in the ring No, it's an action NOOOO lol all the screams in this room How does one see that the$L^p\$ spaces are not locally compact?