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5:00 PM
(pretty basic matrices stuff)

the (no sum on j) line?

Yes
also does anybody know how to get latex on iOS?

it can be done---use the "latex in chat" link, and see the instructions for mobile at the bottom

Okay I’ll keep an eye on this chat whilst I do that
what I don’t understand is a) what he’s doing and b) how you get essentially Ax when it should only be one row acting on each Xi

the relevant step is this one: $A_{k\ell} (x^{(j)})_\ell = \lambda_j (x^{(j)})_k$
with implicit summation on $\ell$

5:02 PM
Can I use the uniqueness of Haar measure to conclude that the only measurable solutions of $f(x+y)=f(x)+f(y)$ are $f(x)=xf(1)$?

To make life easier, let me drop the $j$ subscripts, i.e. let's just deal with a particular $x$ such that $Ax=\lambda x$

@Semiclassical

?
so that the problem becomes just $A_{kl} x_l=\lambda x_k$
But the first is just $(Ax)_k$ by the definition of matrix multiplication
so $(Ax)_k=(\lambda x)_k=\lambda x_k$
So this is just using the fact that $x$ is an eigenvector of $A$ with eigenvalue $\lambda$
The point, as I see it, is that you can think of matrix multiplication as such. Suppose X has column vectors x1,x2,...xn i.e. X=(x1,x2,...,xn)
then AX=A(x1,x2,...,xn)=(Ax1,Ax2,...,Axn)
so the action of A on X is to map the column vectors {x_k} to {Ax_k}
@JakeRose does that help?

@Semiclassical what do you mean by this line?
could you justify how you can think of matrix multiplication as that way?
did that tell you what line I replied to btw it’s not showing it to me

$A_{kl}x_l=(Ax)_k$
(subject to implied summation on $l$)
I mean, what is the definition of matrix multiplication?

5:14 PM
$(AB)_{ij}=A_{ik}B_{kj}$

need some {}'s
The point is that a column vector can be thought of as a matrix with one column
e.g. $x_k=X_{k1}$

Im just having trouble seeing how that implies you can do A acting individually on each vector

in which case you have $A_{ik}x_k = A_{ik}X_{k1}=(AX)_{i1}$
trying to figure a better way to say this
Let's try a different approach. What are the components of $Ax$?
i.e. what is $(Ax)_k$?

You know how math papers will refer to certain properties as "good"? E.g. a good pair

sure

5:25 PM
The paper I'm looking at defines an "excellent arc"

haaah

$A_{ki}x_k$
?

Not quite. For one, $(Ax)_k$ has $k$ as a free index and therefore not a dummy index

(It's when the intersection of two surfaces is isotopic to something of a certain standard form, or something)

@AkivaWeinberger I saw an Amer. Math. Monthly problem lately that was formulated in terms of "nice pairs"

5:26 PM
@Semiclassical "Good" is a dummy term

Oops yeah sorry, habit of having k as the dummy variable

yeah, i sorta made things annoying

$A_{ki}x_i$
bugger I did it again

that's what you already wrote :/
okay, sure. Or, to match what they had, $(Ax)_k=A_{kl}x_l$
But we've chosen $x$ to be an eigenvector of $A$
So what is $Ax$?

Oh are we assuming x is the full matrix? Then shouldn’t we have another freeindices?

5:29 PM
no, but the notation is a bit irritating

but the X matrix is a matrix no?

sure. But if you look down the jth column of X, you'll get the column vector $(X_{1j},X_{2j},\cdots)^\top$
i.e. the jth column of X is the column vector with components $X_{kj}$

Okay sure
so when we multiply these two together we have

to distinguish that from $X$ itself, we write that column vector as $x^{(j)}$
with components $(x^{(j)})_k=X_{kj}$

$(AX)_{ij}=A_{ik}x_k$

5:35 PM
almost. the only issue now is that our earlier choice to suppress $j$ doesn't make sens anymore
so it should be $(AX)_{ij}=A_{ik}x_k^{(j)}=(A x^{(j)})_i$

Ahh I see
so that now it’s the same as A acting on each x

In more detail: $(AX)_{ij}=A_{ik}X_{kj}=A_{ik}x_{k}^{(j)}=(Ax^{(j)})_i$

pis the matrix now 3x1 (counting the vectors as 1)

@JakeRose yeah
This is also rather natural if you think about how matrix multiplication works. Suppose you imagine multiplying out AX and you want to compute a certain column

@Semiclassical was that yeah to my most recent comment? I couldn’t tell if it came before I sent it or not

5:38 PM
well, yeah to both
When you're computing that column, you don't care about any column of X but that one
and you read off the dot products of that column with the rows of A

sure

But that's the same as what you do when you multiply that column vector by A on the left
So you naturally get AX=A(x1,x2,...)=(Ax1,Ax2,...)
(try a case like A being 2-by-2 and X being 2-by-3 if you want to convince yourself of that)

Didn’t we just show that the matrix gets ‘rotated’ though?

What?

sorry that’s a poor term

5:42 PM
are you worried about the order of indices?

we showed above that the ith row is A acting on xj right?

well then

the point is that the ith row of A, acting on the jth column of X, gives (AX)_ij

I’m like 90% there in understanding

5:45 PM
@JakeRose which equation did you take as indicating this?

:47615713
Did that work?

nope

wtf

reply with an actual message

@Semiclassical this one

5:46 PM
I find something humorous in the line "$p\in\mathbb{R}\cup\{ p\}$"

So $(AX)_{ij}=(Ax^{(j)})_i$

Yeah

so that's saying: the element in the ith row and jth column of AX = the ith row of [A*(jth column of X)]
you multiply the entire matrix A on the jth column vector, giving a new column vector which is the jth column of AX

oh and Axj is just another vector and so it’s still a column?

5:49 PM
yeah agreed
i hate how complicated I find this simple thing
@Semiclassical how much linear algebra did you have at undergrad?

i had a few sweeps of it
but my famliarity with index notation is more from physics
we love implied summation, lol

I’ve been comparing that Natsci course to the mathematics course and the disparity is crazy
they get 21 lectures on it

we have 4

21 lectures seems like a bit much
oh, 21 lectures on linear algebra?

5:52 PM
They’re mathmos
to be expected that they go a bit too far

thought you meant 21 lectures on index notation

Oh god no
just linear algebra

lol
even 4 lectures on index notation seems like too much

Oh no I mean linear algebra

yeah
by contrast, just four lectures on linear algebra seems entirely insufficient
unless you're assuming that it's just review

5:54 PM
Agreed
I believe we do more Jin the folllowing terms
this term is just sort of giving us all the skills we need quickly to do everything else in future

sure

it’s a very quick run through of everything
next few terms are more pure

and the main point of the above really is that matrix multiplication can be viewed as one matrix acting on the columns from the left
(or it can be viewed as the rows of the first matrix being multiplied on the right by the second matrix---same difference)

Tbh we don’t do much more matrices, but do topics like storm Louisville theorem and stuff
if that’s even linear algebra
our course is weird

ugh, sturm-liouville yuck
definitely sounds like a mathematical physics course
Sturm-Liouville is a subject I can respect but not love

5:58 PM
We spend 5 lectures doing greens functions solutions to Laplace and poisons equations which sounds fun

we haven’t really used that method in our emag course
but greens functions are just nice
4 lecutures on Cartesian tensors
setting us up for GR I suppose

well, you sorta do use them. you just dont' ncessarily realize it
for instance, suppose someone gives you a charge distribution in free space and asks you to compute the electric potential

yeah

@Semi, do you have time?

6:00 PM
what you do for that is work out what the electric potential of a single charge, then use that to express the electric potential of the entire charge configuration as an integral

Ah yes I’ve seen this
convolution right?

nah, simpler than that

@Semiclassical when did you do GR ?

I didn't

Oh how come?

6:01 PM
I went down the quantum route instead

I suppose it’s less useful for a quantum guy
Didn’t you do it in undergrad?

the potential of a point charge $q$ at $\vec{R}$ is $\Phi(\vec{r})=\frac{kq}{|\vec{r}-\vec{R}|}$

or was it an option?

nah, it wasn't an option at my (relatively small) liberal arts college

ahh that sucks
I’m not sure how much detail we do it in in 3rd year

6:03 PM
then the potential of a differential charge is $d\Phi= k\dfrac{dq}{|\vec{r}-\vec{R}|}=k\dfrac{\rho(\vec{R})dV}{|\vec{r}-\vec{R}|}$
so the potential of a charge configuration is $$\Phi(\vec{R})=k \int_V \rho(\vec{R})\frac{1}{|\vec{r}-\vec{R}|}\,dV$$

Hmmm, what's a space that the one-point compactification of $\mathbb{R}$ under the discrete topology would be homeomorphic to?

@Rithaniel cofinite
no
well cofinite is part of the picture
it wouldn't be homeomorphic to any familiar space
any open set containing the special point would have to be cofinite

But that second factor is (up to a normalization factor I forget ) just the Greens function $G(\vec{r},\vec{R}))$ for Laplace’s equation

any set not containing the special point is open

Yeah, co-finite if p is contained. Alright. If there isn't a commonly used topology I can just explain it the long way.

6:07 PM
Ah I see

just say the one-point compactification lol

Greens functions hidden away I suppose

So you have $\Phi(\vec{r}) = \int G(\vec{r}-\vec{R})\rho(\vec{R})\,dV$
Yeah
It’s just that things are sufficiently simple that it’s not evident what you gain in thinking about Green’s functions
The place where they come into play is when you impose boundary conditions other than just “the potential goes to zero at infinity”

Yeah, I'm supposed to be describing the one-point compactification. Though, if I were using this in an actual proof, I would just refer to it, instead of explaining what it is.

6:23 PM
This problem screams “use inclusion-exclusion” to me: math.stackexchange.com/q/2988227/137524

Hello!!

Let $\gamma\in \mathbb{R}$ and $A=\begin{pmatrix}1 & \gamma \\ 0 & 1\end{pmatrix}$. I want to calculate the norm $\|A\|_2:=\sup_{x\ne 0} \frac{\|Ax\|_2}{\|x\|_2}$.

We have that $$Ax=\begin{pmatrix}1 & \gamma \\ 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}x_1+\gamma \ x_2 \\ x_2\end{pmatrix}$$
Then $$\|Ax\|_2=\sqrt{\left (x_1+\gamma\right )^2+x_2^2} \ \text{ and } \ \||x\|_2=\sqrt{x_1^2+x_2^2}$$

So \begin{equation*}\|A\|_2:=\sup_{x\ne 0} \frac{\|Ax\|_2}{\|x\|_2}=\sup_{x\ne 0} \frac{\sqrt{\left (x_1+\gamma\right )^2+x_2^2}}{\sqrt{x_1^2+x_2^2}}\end{equati

@MaryStar typo: should be $\|Ax\|_2=\sqrt{(x_1+\gamma x_2)^2+x_2^2}$
that seems important to me, since it ensures that $\|Ax\|_2\to 0$ as $x\to 0$

Oh yes! But how can we calculate the supremum? @Semiclassical

My first impulse is to note that $$\frac{\sqrt{(x_1+\gamma x_2)^2+x_2^2}}{\sqrt{x_1^2+x_2^2}}=\sqrt{\frac{(x_1+\gamma x_2)^2+x_2^2}{x_1^2+x_2^2}}$$
ugh, where's my typo
and since the square root function is monotonic, it suffices to find the supreumum of that rational function
which can be further written as $$1+\frac{2\gamma x_1 x_2+\gamma^2x_2^2}{x_1^2+x_2^2}$$
not sure what one does from there tho

6:42 PM
good evening :)

Hiya lush

6:57 PM
Regarding the cesàro sum, discussed also yesterday, I need some suggestions to rigorously write the proof. Let me reintroduce the problem. We have a sequence $(a_n)$ that converges to $a$. We want to prove that $\frac{1}{n}\sum_{k=1}^{n} a_n$ also converges to $a$. To prove it we state the implication of the supposition by definition. But when showing what we want to show, do we chose the same $N$'s of the first implication for each $\epsilon$?

I was thinking about that a little. The proof we attempted was to note that $$|s_n-a|=\left|\sum_{k=1}^n \frac{a_n-a}{n}\right|\leq \sum_{k=1}^n \frac{|a_n-a|}{n}$$

Yes, exactly.
We said that: if we bound the left most one, that means we also bound the middle one.
Which is what we want to do, eventually.
I hope you tolerate my grammatical mistakes :)

I think you mean: If we manage to get good enough bounds on the $|a_n-a|$, we'll get a good enough bound on the $|s_n-a|$

Why you exclude the $n$'s in the denominators?

because it's the same value in each of them
so it seems superfluous to include it
The trouble being that, for any $\epsilon>0$, the fact that $a_n\to a$ only implies that there's some $N$ such that $|a_n-a|<\epsilon$ for any $n\geq N$
it says nothing about $|a_n-a|$ for $n<N$

7:04 PM
Yeah, indeed but I want to make a little pause here.
Let's say, we managed to bound the terms for $n<N$, that means what?
I mean, we pose the same $N$'s for the new proposition, right?

well, what we want to do is bound them so much that we get $\sum_{k=1}^n \frac{|a_n-a|}{n}<\epsilon$ despite the fact that convergence only tells us about $|a_n-a|$ for $n\geq N$

Hello !
I'm looking for a 3rd year project, I've been searching for a long time and I can't find any exotic interesting questions, do you have any ideas in mind ?
It can be fractals, chaos, graph, NP-Algorithms, Math applied to real life.. anything exotic I enjoy it!

My impulse is to say: We're under no compunction to say that it's the same $\epsilon$ for $a_n\to a$ as $s_n\to a$

So, we use the same range $N$ for the bound of the $s_n$.
We don't pose another range for the new sequence.
We utilise the same range.

What I have in mind is this: Let $\epsilon'>0$. Then there exists some $N$ such that $|a_n-a|<\epsilon'$ for any $n\geq N$
That bounds the terms with $n\geq N$, but for $n<N$ we can't say that $|a_n-a|<\epsilon'$

7:09 PM
By range I mean the $N$ we find every time for $\epsilon$ by the way.
@Semiclassical I agree.

Hey everyone. I vaguely remember some identity, but I don't remember what branch of mathematics it comes from.
Something along the lines of $e \wedge e = 0$?
And I think it had something to do with forms.

Lol, so I was going to try and ask for the Principia Mathematica for Christmas, but...it's $400 on Amazon. However, it is nevertheless true that there is some max value for$|a_n-a|$for$1\leq n\leq N$Does that ring a bell for anyone? Scratch that$500

7:11 PM
@Semiclassical, yes I followed that also.
But things become super sloppy rapidly.

call this max value $\mu$, so that we have $|a_n-a|\leq \mu$ for n<N

Ok, I'm following

Hmm. This does indeed start to feel sloppy
oh, blah. typo earlier: we want to bound $\sum_{k=1}^n \frac{|a_k-a|}{n}$

I think I found it. Wolfram MathWorld says that if $a$ is a differential k-form of degree 1, then $a \wedge a = 0$, where the $\wedge$ symbol denotes the wedge product.

(the cesaro sum is for $s_n=\frac1n \sum_{k=1}^n a_k$, not $\sum_{k=1}^n a_n$)

7:16 PM
I see
I also read some of proofs written in m.se but I don't think I'm convinced.

yeah
I'm not much of a fan of this tbh

It feels indeed really true, explicative, but in text, I think better can be made.
:)

lol, I know what you mean
Let's try again. Pick some $\epsilon'>0$; there exists $N'$ such that $|a_n-a|<\epsilon'$ for $n\geq N$. Additionally, we have $|a_n-a|\leq \mu$ for $1\leq n<N$
So if $n\geq N'$, we have $$\sum_{k=1}^n \frac{|a_k-a|}{n}=\sum_{k=1}^{N'}\frac{|a_k-a|}{n}+\sum_{k=N'}^n \frac{|a_k-a|}{n}$$

is there a website to simplify complicated algebraic expressions in say 6 variables?

(If $n<N'$, then the second sum disappears)
From our prior statements, we can bound these sums as $$<\sum_{k=1}^{N'}\frac{\mu}{n}+\sum_{k=N'}^n \frac{\epsilon'}{n}$$
Hmm, I think I may see it. Take $\epsilon$ to be the larger of $\mu$ or $\epsilon'$
In that case, we have $\mu\leq \epsilon$ and $\epsilon'\leq \epsilon$, so we can bound this by $\sum_{k=1}^{N'}\frac{\epsilon}{n}+\sum_{k=N'}^n \frac{\epsilon}{n}=\sum_{k=1}^n \frac{\epsilon}{n}=\epsilon$
I think this may net us a proof
no, no

7:26 PM
I've been afk, let me read it.

it doesn't work. if $|a_1-a|=1$ is the largest one in the entire sequence, then you've got $\mu=1$ and all we get is that the sum is bounded above by $1$---not useful
we need to be able to get $\epsilon$ arbitrarily small

@Semiclassical In this line, we have or we have to show? I think it's the latter.
Oh, pardon me.

no, it's "we have". but I wrote it wrong regardless
$$\sum_{k=1}^n \frac{|a_k-a|}{n}=\sum_{k=1}^{N'-1}\frac{|a_k-a|}{n}+\sum_{k=N'}^n \frac{|a_k-a|}{n}$$

about the indices, I suppose.
Yeah, that looks right.

All I’m doing is splitting it into those terms with k<N’ and k>=N’
But this isn’t enough as far as I can tell

7:32 PM
math.stackexchange.com/questions/2995244/… How to prove that map $ϕ$ is homomorphic

Ok, I've read it.

in the accepted answer

@Semiclassical yes, seems so.

What we have right now is $\sum_{k=1}^{N’}\frac{|a_k-a|}{n}\leq N’(\mu/n)$
Hmm. The only saving grace is that, if n>>N’, then that term will be small
So if I require n to be sufficiently large, then I can make that bound arbitrarily strong

That is the more precise expression of our initial motivation

7:37 PM
Right

is $\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$ a isometry matrix? if so, is it rotation matrix or symmetry matrix?

@Semiclassical The upper bound of the sum is $N'-1$, right?
not $N'$.

Yeah, oops

Ok, np.

I’ll probacly keep doing that by accident
I think we’re on the edge of having an answer
But not quite there

7:42 PM
Yeah, it'll boil down I think to choose the bigger one of the two ranges of which one will come appropriately from the sufficiently large $n$.

My thinking being: if we require $n \geq N$, then $\mu(N’-1)/n\leq \mu(N’-1)/N$
So if N is large enough then we can make that bound tight
The problem, as you say, is to balance this with the second sum so as to get any arbitrary epsilon
But I can’t stare at this more
hmm. maybe it's worth playing with a specific example
take $a_n=1/n$. Then for any $\epsilon'>0$ we can pick an integer $N\geq 1/\epsilon'$ and guarantee that $|a_n-a|=1/n<\epsilon'$, so $1/n\to 0$
For instance, take $\epsilon'=1/10$. Then $n\geq N'=11\implies |a_n-a|=1/n<1/10=\epsilon'$
To what extent does this allow us to bound $s_n=\frac{1}{n}\sum_{k=1}^n \frac{1}{k}$?

8:22 PM
@user330477 wolframalpha

8:35 PM
@Semiclassical There are two things that worth noting.
First, I strongly think that I got the learning outcome that I should get, from this exercise. What I can write as a proof (or justification) is enough, I suppose.
The second is that it can be useful to remark that we have a proposition states that if we have a null sequence $(c_n)$ and a bounded sequence $(b_n)$, then the sequence $(c_nb_n)$ is also null.
We have $1/n$ being null as a sequence.

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