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00:00
eigenvectors
Yes. This is the key to the problem .
So if the matrices $A$ and $B$ are unitarily equivalent then they would have the same eigenvectors?
No. Why same?
That's true they are different matrices.
But there is a unitary matrix carrying …. To ….. .
00:07
Ohhh...if it clicked correctly in my mind. The unitary matrix $P$ would be carrying the basis of eigenvectors of $A$ to the basis of eignevectors of $B$.
Right ..
If that didn't come out right $P$ would be the "change of basis" matrix using the basis of $A$ and $B$ in it.
so one basis is a rotation of the other.
why couldn't they just explain it in such a simpler way instead of using these fancy words......THat also actually clarifies what unitarily equivalent actually means to me too...
You should've done a series of videos on linear algebra using Insel and Friedberg......
It was supposed to :)
Well, have you watched my videos on chapter 9? 🤷‍♂️🤷‍♂️
Nothing on unitary, though.
00:11
@TedShifrin not there in your book yet.....so maybe all this struggling will come full circle when I get to those chapter...
I do understand the basic rotation matrix ideas of Ch.2 of Insel and co, it is this inner product stuff.
haha, full circle! if that's a pun on some 2x2 unitaries, well done.
So, what’s the resolution of the question ?
great now I have to find a counter example using this idea. Interesting, but not really, this whole "solving" session still resorts down to a specific way of thinking about things. When I don't deviate form that thinking I don't run into problems..
In particular I shouldn't be able to get a matrix going from the basis of eigenvectors of $A$ to $B$.
@leslietownes it wasn't intended to, but seeing as rotation is an orthogonal operator I could see why you said it....vaguely....
Review rotation matrices from section 4 of chapter1 ;)
00:29
there's a ted exercise for everything.
That was text, not exercise …
But we established earlier that sometimes Ted is insufficiently careful/pedantic.
If Leslie would only read the text like he is supposed to so he can be fluent in its contents and answer questions with knowledge of reference.
But then again the fancy PhD thinking skill set is put on display when you are able to pluck all the ideas out of thin air without much reference
Has Munchkin broken all her new year’s resolutions?
I aspire to the greatness of you too..................perhaps by the time I reach 60
@D.C.theIII Don’t worry. I’ll be dead soon.
00:34
Well......that was rather morbid.............
That’s life sometimes.
Dark humour....I like it....
01:11
@TedShifrin My grandmother said the same thing.
For more than 25 years.
I won’t be that slow. :)
So I know that the derivative of $\|Ax - b\|^2$ is $2(Ax - b)^tA$, but how would I derive that expression? I remember how we got the derivative for $\|x\|$, but that was with a hint. Also the "linear approximation" part doesn't jump out when I try using the difference quotient.
How about chain rule?
That’s actually done in chapter 5 in treating inconsistent systems and the normal equations.
oh I should've been more specific what I meant.
nevermind I see what you mean and I saw it in the video too............I was just trying to use a first principles approach....cause of my masochism.
01:27
You can do that. Just show the error is small.
That's where I was having problems: $f(a +h) - f(a) = \|Aa + Ah - b\|^2 - \|Aa - b\|^2$
Go on. Expand and simplify.
How do I expand it? that's actually my hiccup
Grr.
What is $\|v\|^2$?
ah.... $\|v\|^2 = v \cdot v$
01:36
Yup.
Ahhh....now it makes sense.......use all the tools in the toolkit.......
You can’t afford to forget basics.
😉😉😉
well, you absolutely can, but wait until you're no longer working in school or working in math.
then, you can even forget some things that aren't basics.
Yeah, just work for Rethugnicans.
Each stooopider than the preceding.
01:41
Or you could wing it like George Santos.......fail upwards
the only thing i'm fairly sure about george santos is that his name isn't george santos.
🤣🤣🤣
We know that. And of everything he denies we find proof.
Jorge would make more sense him being Brazilian and all
Devolder …
01:45
I just looked that up....wow......just wow............I have no idea what to make of the US Gov't. I always thought the minimum was a university degree needed.
as I said...thought...
Absolutely not. Plenty in the House have barely a high school diploma, if that.
harry truman was president with hs diploma.
@TedShifrin this explains a lot....
but a degree doesn't guarantee competency....then again those educated Repubs that are pushing the nonesense also have very insidious personal agendas so they benefit by being duplicitous.
finally get to use that word in a sentence....
Boebert barely got a GED. I’ll look up Gaetz and MTG.
dc you make a good point, a lot of effective legislators do not have great credentials and some of the least competent ones have "prestige" credentials.
01:57
Older I get those old jokes such as "A politician is a failed (choose all the fields you want)" have hidden truth to them.
02:25
Greene got a worthless business degree from UGA (yup). Gaetz has an. Infernal law degree.
Yes. Some of the stupidest people I know have PhDs.
A lot of politicians are convicted felons.
I question “a lot.”
Two Wikipedia pages worth.
Out of many thousands …
Those are just for the US.
02:39
I meant in the US
Sure it's a global issue.
I have no idea of the extent of corruption in Russia and China and the “third world.”
I expect it’s worse there than the US.
The world cup showcased the extent with no compensation for the migrant workers.
03:36
The US has the best politicians that money can buy...
Sep 12, 2021 at 4:07, by PM 2Ring
The Mona Lisa Twins (originally from Austria) love 60s music, and have been doing great Beatles covers since their early teens. Their latest original song is a hilarious sarcastic piece with a music hall flavour. I Bought Myself A Politician.
Some geometry trivia. In the right triangle with sides (85, 132, 157), the angle opposite the 132 side is almost 1 radian.
04:02
R.I.P. David Crosby
I hadn’t pondered it before. But since $\cos 1$ is irrational, there can be no triangle with integer (rational) sides.
As immortalized by contradiction.
04:20
Minimizing the function $f(x) = \|x-b\|^2$ where $b = (3,7,1)$ subject to the constraint equations:

$$g(x) = \begin{pmatrix} x_1 - x_2 + 3x_3 \\ 2x_1 + x_2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$I get the following Lagrangian setup:

$$2 \begin{bmatrix} x_1 - 3 \\ x_2 - 7 \\ x_3 - 1 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix} + \mu \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} $$

After all the simplifying and manipulation I will have the following equations and two constraint equations:
04:50
Is this from my book or elsewhere? You want to eliminate the multipliers. Rewrite the second thing (dump the 2) in matrix notation. Same for first.
your book of course
When I did it I got the two lagrangian multipliers and then solved for my variables
well I should say tried to solve for my variables
Look at section 5 of chapter 5, right before subsection 5.1.
I didn't want to use the projection matrix approach yet
I just wanted to some raw lagrangian practice
Forget projection. It shows you how to eliminate the multipliers by linear algebra.
I don’t want to type it all now.
04:56
Yeah I which part you're referring to
see
You have one key matrix here.
rows being transposed
So call it $A$ and write both equations.
Yup I have a good picture of what you're saying. I'm going to write it out hopefully I will report back tomorrow with good news
05:06
Beautiful stuff now seeing what is going to play out. Ok it is bed time. A bientot
Bonne nuit.
 
1 hour later…
06:12
How do we show that the dual direct l_p sum of Banach spaces is isometric to l_p sum of the dual of Banach spaces?
One problem is that I don't understand how to define a norm on 'l_p sum of the duals of Banach spaces'.
Suppose that $X_n$'s are Banach spaces. Then is the following correct: $f\in \oplus E_n'\implies f=(f_1,f_2,...), f_i\in E_i'$ and $\|f\|=(\sum_n \|f_n\|^p)^{\frac 1p}$?
$\|,\|$ on RHS is operator norm.
Corrected version of the first sentence: How do we show that the dual of direct l_p sum of Banach spaces is isometric to l_q sum of the dual of Banach spaces?, where q=p/(1-p).
06:30
@Koro is that not $q=\frac{p}{p-1}$?
so that $\frac1p+\frac1q=1$
indeed. sorry for the typo.
07:12
I recently found out about a site called AoPS, is it similar to MSE?
Nope, AoPS is more like a tutoring site.
Is tutoring like a lecture?
there's some segment of that site that is also a little more oriented toward contest math than MSE is
a tutor - generally a teacher of one-to-one or one-to-few pupils who advises, coaches or supports a student's learning, for example a dissertation/thesis/project tutor. a lecturer - a teacher who lectures, give lectures or informative classes to whole groups of students
07:21
the meaning of these terms depends on the institution. some places may not have formal roles by either name
in the US it's pretty common for "tutor" to not have any formal meaning whatsoever, just, to be anyone who helps with a subject. maybe in connection with a school, maybe not
i think its not unusual for "tutor" to mean something more formal in the UK
"lecturer" in the US is usually a paid position associated with a school, but where it ranks in the academic hierarchy can vary depending on the place
at many schools it implies a contract worker who might work on a term-by-term basis with no guarantee of continual employment, but at some schools it is not like that
@onepotatotwopotato MSE is a Q&A site that does not offer lectures or tutoring.
AoPS does offer some kind of tutoring and i think also classes? it began as a kind of problem solving forum i think
i think specific to contest math, originally
Yeah if one needs to prepare competiton math, tutoring would be helpful.
AoPS also publishes books.
2
Q: The dual of p-sum of Banach spaces is isometrically isomorphic to the p'-sum of their duals

FilburtThe problem: Let $p\in(0,\infty)$ and $(E_n)_{n\in\Bbb N}$ a sequence of Banach Spaces. If we define $$\Bigl(\sum E_n\Bigr)_p:=\biggl\{ (x_n)_{n\in\Bbb N}~|~x_n\in E_n, \forall n\in\Bbb N, ~\text{and}~\|(x_n)_{n\in\Bbb N}\|_p:=\biggl(\sum_{n\in\Bbb N}\|x_n\|_{E_n}^p\biggr)^{\frac{1}{p}}<\inft...

not sure how to show the reverse inequality.
Are some more tools needed like Hahn Banach theorem etc. to get the reverse inequality?
 
2 hours later…
09:12
no, it should just be some epsilonology
ok thanks.
I'm stuck at one step here:
5
Q: Help proving that a Banach space is reflexive

Willian HansI'm having trouble in proving that the following space is reflexive: $$E = \{ x= (x_n) : x_n \in \mathbb{R}^n \text{ and } \sum \|x_n\|^2_\infty < \infty\}$$ with the norm $$ \|x\| = (\sum \|x_n\|^2_\infty)^\frac{1}{2}$$ I already tried an analogous of the proof that the $l_p$ spaces are refl...

In answer, under the **proposition**, the answer says that 'For $\alpha \in A_1$, choose $e_\alpha \in E_\alpha$ with
$\lVert e_\alpha \rVert_{E_\alpha} = \lVert f_\alpha\rVert_{F_\alpha}^{q-1}$ and
$\langle \delta_\alpha(f_\alpha),e_\alpha\rangle > (1-\varepsilon) \lVert f_\alpha\rVert_{F_\alpha}^q$
what I don't understand is what guarantees existence of such $e_\alpha$'s.
i'm having difficulty interpreting this notation, the setup is way more general than OP is asking about. if you have a textbook that just does the duality between ell^p and ell^q for sequence spaces (i.e. E_n = F_n = R for all n and A is the positive integers) it will contain the essential idea.
i wouldn't wade through a more general treatment unless you find a higher level of abstraction more helpful than harmful.
I have proven that dual of ell^p is ell^q. But I am having difficulty in showing the general duality in the answer.
OK. have you specialized the general proof to those very concrete ingredients? does the mystery go away?
Should I just keep spamming L'Hopital's rule whenever I see an indeterminate form, or try to be creative instead?
09:29
Leslie: here is the simplified notation(s): $\alpha$'s are elements of index set A. We can for simplicity consider $A=\mathbb N$, then $\alpha$'s become just the natural numbers. And $\langle \delta_\alpha(f_\alpha), e_\alpha\rangle := f_n (e_n), e_n\in E_n, f_n\in E_n'$
mm, experience has to be your guide there. i wouldn't randomly spam a rule over and over if it didn't seem to be leading anywhere nice. l'hopital's rule also (somewhat annoyingly) has hypotheses that can become difficult to check after one or two or an arbitrary number of applications.
even after zero applications, i guess.
Yeah, I guess I'll just try to be more creative. The limit has a quartic polynomial in the numerator and a cubic in the denominator. They're probably factorable but its gonna be lengthy. Especially since the cubic has a double root that I checked by graphing it
koro, oh, i meant down to the level of numbers. it looks like you still have a layer of notation there. "E_n" is a funny way of writing R.
l'hopital is certainly overkill for rational functions, imvho.
brb.
@leslietownes In case of ell^p spaces: Showing that some T:ell^p--->ell^q preserves the norm worked out well (we have to show two inequalities) but in the present case, one can show the inequality form one side using Holder's but the other side is creating problem.
Probably is, I'm just being really lazy right now
09:33
(does my last comment make sense?)
it seems that the answerer to the post does not use MSE anymore.
Oh I understood the step now. :-)
10:25
"I think it's rarely about what you actually learn in class . . . it's mostly about things that you stay motivated to go and continue to do on your own." - Maryam Mirzakhani
11:14
@copper.hat: If $\zeta(-1)=-\frac1{12}$ gets on your nerves, consider the number of derangements of $-1$ item ;-)
$-0.69717488323506606877-1.15572734979092171791i$
$-\frac{\operatorname{Ei}(1)+\pi i}e$
How's the recovery coming along? @robjohn
it is. I wish it were going faster.
11:31
we wish you speedy recovery @robjohn.
Thanks
Don't try to rush it.
Jan 8 at 23:30, by user2236
You have to be extremely careful afterwards to take it slow.
11:53
2
A: How do I prove that the following map is onto?

Yiorgos S. SmyrlisLet $\varphi\in \ell'_p$ and $\{e_n\}$ the standard Schauder basis of $\ell_p$, and $y_n=\varphi(e_n)$. We shall show that $y=(y_1,\ldots,y_n,\ldots)\in \ell_q$ and $\varphi(x)=\langle x,y\rangle$, for all $x\in\ell_p$. Let first $\varphi_n(x)=\sum_{k=1}^n x_ky_k$, where $x=(x_1,\ldots,x_n,\ldot...

I don't see why: $\|\phi_n\|\le \|\phi\|$.
nvm, I understood it now.
12:11
Why the $n$th coefficient of $\left(\sum_{n = 0}^\infty z^n\right)^k$ equals to $\binom{n+k-1}{k-1}$?
@onepotatotwopotato have you looked at negative binomial coefficients?
$(-1)^n\binom{-k}{n}=\binom{n+k-1}{n}=\binom{n+k-1}{k-1}$
I see. Thanks
That is because $\sum\limits_{j=0}^\infty z^j=\frac1{1-z}$, so you're looking at $(1-z)^{-k}$
13:12
Hey
I can't seem to understand the "norm" in Behrend's popular construction (from 1946)
Here's a question where I have elaborated: math.stackexchange.com/questions/4622963/…
Could someone please help? Thank you
13:28
@TedShifrin Of course, but I like it when I find nice rational approximations. :)
@user726941 Here's Croz singing Joni's For Free in a duet with Sarah Jarosz:
 
2 hours later…
15:17
Suppose that I have $x_k^{(n)}\to x_k$ as $n\to \infty$. Then can something be said about $\sup_k\|x_k^{(n)}-x_k\|$?
Given any $\epsilon>0$, $(\forall k\in \mathbb N) \exists N_k\in \mathbb N, n\ge N_k\implies \|x_k^{(n)}-x_k\|<\epsilon/2$.
I can't do sup_k both sides as that would change $N_k$ as well.
Is there any way by which I can conclude that the supremum is $0$ for large $n$?
16:12
Given $f(1) = 2$, $f'(1) = 1$ and $f''(1) = -2$, $P(1, 2)$ can't be an extremum that lies in the domain, but can it be an edge extremum (an extremum that lies on the edge of the domain (for example for $D_f = (-\infty, 1]$))?

In that case, shouldn't $f'(1)$ be undefined?
 
1 hour later…
17:22
0
Q: Showing that a sequence converges.

KoroAssume $x_n^k, x_n$ to be real for all $k$ and for all $n$. Suppose that I have $x_k^{(n)}\to x_k$ as $n\to \infty$ for every $k\in \mathbb N$. Then can something be said about $\sup_k|x_k^{(n)}-x_k|$? Given any $\epsilon>0$, $(\forall k\in \mathbb N) \exists N_k\in \mathbb N, n\ge N_k\implies |x...

18:00
For what complex numbers $\alpha \in \Bbb{C}$ does $\log z^{\alpha} = \alpha \log z$ hold for every $z \in \Bbb{C} \setminus \{0\}$? Note, I don't believe we are performing any branch cuts.
18:13
@AlessandroCodenotti do you know anything about z-filters?
18:39
I found this math.stackexchange.com/questions/4020961/… but this asks whether we have equality for a branch cut.
in the realm of "is this a pointless use of L'hopital": math.stackexchange.com/a/3414145/137524
the second use of it is definitely pointless, b/c that limit's just the definition of $(f')'(z)$
but i'm not remembering if it's silly vs useful for the first
19:02
Okay guys, I'm going to boggle your mind for a bit with Knuth's TAoCP's exercise.
I am trying to prove this by induction on n. This looks suspiciously similar to Lagrange polynomials.
@user193319 but then the 'equality' sign is not the usual = sign, it means equal as sets.
isn't that right?
This is how I managed to transform the sum
And then the book gives the hint. And I can't understand what it points me at.
Which 'values'? Why do they turn out? How to argument that? No idea.
2
Q: Exercise 1.2.3-33 from TAOCP -- Prove a formula by induction.

Nick AllenExercise: One evening Dr. Matrix discovered some formulas that might even be classed as more remarkable than those of exercise 20: $$\frac{1}{(a-b)(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)}=0$$ $$\frac{a}{(a-b)(a-c)} + \frac{b}{(b-a)(b-c)} + \frac{c}{(c-a)(c-b)}=0$$ $$\frac{a^2}{(a...

I've seen this question, but it only clarifies what I already figured out by the time.
@Jakobian I've heard the name before, but not really
19:25
@Koro Yes, I believe so...unless you take a branch cut(?)
@Semiclassical You need $C^2$ to use L'Hôpital. Far better is to use a second-degree Taylor polynomial with remainder. Then you need only $C^1$ and existence of the second derivative at the point.
@TedShifrin makes sense, and follows the theme of "just use Taylor"
Yeah, AP teachers turn calculus students into L'Hôpital robots.
20:12
For $\alpha \in A_1$, choose $e_\alpha \in E_\alpha$ with
$\lVert e_\alpha \rVert_{E_\alpha} = \lVert f_\alpha\rVert_{F_\alpha}^{q-1}$ and
$\langle \delta_\alpha(f_\alpha),e_\alpha\rangle > (1-\varepsilon) \lVert
f_\alpha\rVert_{F_\alpha}^q$
Can anyone please explain the above part in this answer: math.stackexchange.com/a/530452/266435?
this is in proposition in the answer.
I think that there should have been mod sign on LHS
But then the proof won't work?
Taylor is all I know and all I ever need
everything else is just a corollary
and Maclaurin?
on mse there is probably only one answer to the above question.
in fact I looked up and that answer is the only answer where all links related to the question on this site direct to.
But I don't see why the quantity <$\delta_a(f_a),e_a$> is positive.
@Koro Just a special case. I never use that name.
I forgot what that even is
just Taylor at $x=0$?
Yup.
I actually never looked up the history to see the chronology.
20:28
I think Maclaurin came first
then came Taylor's.
(otherwise anyone could have put a=0 in Taylor's series)
just as anyone could have put $f(a)=f(b)$ in Lagrange's theorem to get Rolle's theorem.
Not so fast!
From C.H. Edwards's The Historical Development of the Calculus:
If a million grains of rice are called rice
Then is a single grain called rouse?
In the Acta Eruditorum of 1694 John Bernoulli published a series that was sufficiently similar to Taylor's for Bernoulli to accuse Taylor of plagiarism when the Methodus incrementorum appeared 20 years later. ... The case $x_0=0$ of Taylor is often called Maclaurin's series. Colin Maclaurin (1698-1746), perhaps the most successful of Newton's disciples, employed Taylor's series as a fundamental tool in his Treatise of Fluxions (1742).
So Taylor appeared 1714, Maclaurin 1742.
Hahaha. I just recalled a comment somewhere on mse that said to a post that: 'it's like claiming heavier objects fall faster.'
Often nomenclature is NOT faithful to history. There are zillions of examples, although right now I can't summon them.
20:33
ohh
Just shows to go: You should not jump to conclusions!
yeah.
I just realized I received a bronze badge for the tag geometry. And soon for the tag euclidean-geometry too. Progress!
You’re cleverer with synthetic geometry than I am, for sure.
@TedShifrin I also own more hice than you
20:45
Indeed. But it’s not hard to beat 0.
(And owe more property tax too)
@TedShifrin jump to the other side of a stream.
Jump behind?
You back to your usual walks?
All of Math.SE on a cruiseship, what could go wrong?
No. Not for a few weeks. I see the doctor on Tuesday, and we’ll figure things out.
20:48
Cool. Keep recovering :)
We’re going to go get food in a bit, and spend some time just watching movies.
21:01
Enjoy!
21:35
I ate too much
I've become a convert to Linear Algebra for questions with vast computations....
Hi @D.C.theIII!!
Hi @Gokuカカロット!!
@Koro how's it going
Hey Koro....late night on a Saturday for you..
but then again as a mathematician this is the life we chose as our "weekends"...😪
6:40AM Sunday here in Japan. Freezing cold outside
21:40
I just finished an assigment.
Why are you up at 6:40 AM on a Sunday? DOing math?
@Koro Lol...ah the never ending assignment grind
@D.C.theIII yes, I've been studying since 4:00AM
:-)
And I need to go to the manga store at 7:00AM quickly
You prefer to do the studyinbg in the early part of the day?
21:41
@D.C.theIII and very late at night as well
Interesting......I might have to consider this...I'm still trying to find an "optimum schedule for me that would allow a decent social life....precisley being able to go to the gym all the time and practice Muay Thai...tht's all I ask
DC: How's your linear algebra going?
I went to gym only probably 3 to 4 times in my life so far.
@D.C.theIII since I have practically no social life, I can dedicate more time to studying and other activities
or may be 5 to 6 times.
I just do taekwondo
21:45
@Koro Of course I should've been done long ago, but you know taking that long lay off, which was probably for the best, has delayed things. I'm finishing up Ch. 6: with Unitary operators and then Orthogonal projections and Spectral theorem. Then CH.7 will be Canonical forms.
@Gokuカカロット other activities count as a social life, well at least a life away from math
@D.C.theIII by "other activities" I mean coding, gaming, watching anime and such
exactly so you're more than active. What programming language?
@D.C.theIII Currently? Learning JavaScript
I know a bit of C++ too but I need to improve at it even more
Nice. I've been working on C++, but "math" always squeezes into the schecule
@D.C.theIII I see.
21:47
cause I'm learnign R too...
I'm still not so comfortable with Canonical forms.
@D.C.theIII that's nice. I forgot the cpp that I knew.
And you're in a masters program, which tells me I'm going to be in for spending more time on the chapters than I had originally anticipated....
@Koro probably just the advanced stuff you forgot, cause you can probably write all the simple recursive loops which is the foundation, everythign else builds off of that
I mean the syntax etc.
like in C, should I put ; after for(...,...,...) or not, these type of things.
chatGPT can fix that for you.
it asks for mobile no. to register.
I won't give them that.
21:51
@leslietownes curse the bots and A.I. making life too easy...
Waiting for the store to open
nobody appreciates a lil elbow grease anymore..
@D.C.theIII exactly. I wanna go back to the days of hunting your own food. Current life is too easy
hot take: in the very near future we'll start obtaining definite answers for open questions . but since they'll be given by advanced AI, we'll have to study years and years to reverse engineer the answers. university departments will start to become specialized in reverse engineering determinate answers, an entreprise taking decades or more
give me money, i'll invest in stocks related to this
Is there even an IPO related to this?
21:55
can you invest in Nasdaq directly from outside USA?
I can not do that directly from here, I think.
@Koro ETFs many kinds
you heard it here first lads, you'll be millionaires for this idea. thank me later
hmm, I do it through mutual funds. I'll try ETFs. These are relatively new instruments here.
and an international bank
@Koro They don't have stock trading platforms in India?
By new, I mean I see that the buyers, sellers seem very less in numbers in ETFs.
@D.C.theIII There are stock trading platforms but the instrument ETF is relatively new.
21:59
ah, but you should have access to US markets and if you can buy stocks in US markets then you can buy ETFs in US markets because it is traded like a stock
So say I put money in ETFs, when I want to withdraw, there may not be any buyer.
ETFs trade just like stocks, and the ETFs are traded in the same markets
@D.C.theIII that's why I asked above if one could directly invest in US markets from outside US.
you should. If your bank is an international bank they should have a trading platform available
koro there is no US legal obstacle to doing so. trading platforms may have their own hoops.
there could be weird tax consequences too, most likely on your end than from US law, but i dunno.
22:02
using the stock trading apps that I use, one can invest only in Indian markets (or may be I don't yet know the full functionality of the app, and I have not yet tried commodity exchange) and in US markets via mutual funds.
@leslietownes I think the tax part is reduced if the investment is for long term.
@leslietownes of course the lawyer's first frame of reference is legal ramifications...lol
trading is cool though :-).
@Koro yeah that is pretty common in taxation (i have no idea how it works in india!). and depending on your ties to the US, maybe no US taxation on gains.
centralize investment
Ah, leslie is here for the FA question koro had.
22:05
overtake the means of investment
i mean
take them, not overtake them
to overtake them is the polar opposite of what i mean
i don't see it in scrollback but maybe it will resurface. i saw something on completeness of ell^infty on MSE.
Class equation, Sylow theorem all covered in just 2 to 3 hours.
depends on what you mean by 'covered', but yeah, i believe it.
Leslie: my confusion here is clear now-
2 hours ago, by Koro
Can anyone please explain the above part in this answer: https://math.stackexchange.com/a/530452/266435?
ah, cool.
one drawback of a lot of newer texts is L^p theory is either a short special case of some general excursions into integration, or you get some really abstract banach space treatment with little connection to the classical roots.
koro if you have never looked at riesz/nagy functional analysis (available in cheap dover reprint in US) it is a good resource for less abstract proofs of things. and usually with no sacrifice in generality (just sometimes weird old terminology that people don't use anymore).
22:11
I also realized that to prove a space X to be Banach: I was only finding the limit candidate x for a Cauchy sequence (x_n) without checking whether x is in X or not.
@leslietownes Thanks. I'll take a look at that.
Leslie: In functional analysis class, if some introduction to categories has been given, then what is it for exactly?
i'm sure pdf copies are also available gratis on hotbooks dot biblioteka dot ebiz dot ru.
maybe just to give you the language of morphisms and such, so you don't have to write the same sets of hypotheses over and over. maybe to develop analogies with other areas (e.g. when does this 'direct sum' construction behave like a categorical sum). it isn't necessary, at least not for the classical stuff.
Well, after two and a half years of teaching here, I have just learned that Verizon will give me a discount on my phone bill for being a member of the faculty.
for banach space theory (which is almost its own thing separate from functional analysis if you really dive down the rabbit hole) you probably do "actually" need it, it isn't just a convenient language.
at least eventually. banach space theory can get very woo-woo, but the galaxy brain stuff tends not to be the stuff you need for functional analysis.
@leslietownes oh I see. I thought it was to prepare a background for Uniform boundedness theorem .
@robjohn my first time hearing the term derangement! i have neen described as deranged, but i am sure my mother did not mean it
22:23
Hi @copper.hat!!
Hi @Koro! Beautiful day here
Oh nice!! It's cold here (but not as cold as it is in Delhi).
went for a nice cycle this morning imgur.com/FxCFGrz
@Koro possibly, but, i think most functional analysis books manage to introduce that stuff without any intro to categories. the theorem certainly predates categories. :D
22:38
oh I see.
22:53
Got the manga, somehow
Breakfast time now
life isn't worth living without making statements like "the forgetful functor from Banach spaces and linear contractions to Sets that sends a space to its closed unit ball preserves and reflects limits"
who's the french guy that did topos theory. lewviere, leviere, lewfviere???
google is not helping me out here
since when is it so picky with its french orthography
found him. it's Lawvere
i wanted to say, he has a paper citing mao
this needs to be public knowledge. category theory is inherently attempting to overthrow western society
huh, he's not french
23:09
yeah
he was also big on Hegel
when i heard the hegel part i decided i would learn topos theory, at some point in my life
23:40
hi
i like quandles for breakfast :)
@copper.hat yeah, I’d never thought about analytically continued derangements before.

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