@leslietownes from "them"

yeah, you can put the emphasis either way. the point is, weight on getting info via the thing and not via stuff around the thing that might tell you about the thing.

You could also get that by adding the word 'willingly' to the definition, eg "take steps to willingly obtain knowledge from other beings"

maybe even "willingly and consensually". maybe that is part of communicate. knocking a being out and reading its neural matter for signals is "communicating" with it in some physical sense but not what we would think of as questioning. more like "interrogating" broadly defined.

at least, my aliens don't seem to like it when i do that.

Yeah, we'd like you to stop-- I've said too much

"Take steps to cooperatively obtain knowledge from other beings"?

But perhaps defining cooperation is thorny.

i like asking questions to undermine authority

like a child making a sequence of why questions to infinity

then you're not looking for an answer

I guess in that case, it's not out of an understanding that other people know things you don't

right, it's that they don't know what you don't know either

"why is the sky blue, father, you absolute buffoon, contemplate now that you know nothing"

but son, it is probably due to some gases, or some mixture of it

"father, are you a gas scientist? a scientist of the skies? and what is even, making colour out of mixing gases? once again, i see that your authority rests on nothing but lies and deceit"

(replies the 4 year-old)

I see you have interacted with children before.

Wait, let me rephrase that

Have you ever met a child??

He's in the same boat as I

But in that case I feel like there's some level of irony - using words to mean something other than their literal intention

This isn't the literal intention of the words of the question.

"May the fight begin."

A suggestion I got from someone on Facebook: "Communicate (1) one’s own ignorance and (2) one’s dissatisfaction therewith."

Hm

"Mother, what is this? Only 5 chicken nuggets? Have I not been forever reiterating how utterly unsatisfactory this is?", said the 4-year old

"I would like to know how you would define questions." vs "I don't know what questions are and I would like to know."

(my definition vs the one above)

Hm, maybe that isn't the best realization of my definition... "Please tell me how to define questions."

maybe you want a subset of the set of questions

Akiva, I'd send the alien to watch Bill Clinton's "it depends on what the meaning of the word is, is"

what is $U$ here?

U_1

U should be U_1

"Where is the ball?" => [Subject: ball location] [response requested]

or

[Subject: ball location] [help needed]

Maybe an even more fundamental question would be, what does "asking for help" mean? Once we know that, asking a question would be asking for help obtaining knowledge.

It’s wrong without an additional hypothesis.

Ted, stupid question - is it possible to long divide $1 + z + \frac {z^2} {2} + \frac {z^3} {3!} + \ldots$ by $1 + \frac 1 z + \frac 1 {2 z^2} + \frac 1 {3! z^3} + \ldots$

@JoeShmo Sure... but (1) it isn't going to be particularly nice, and (2) why?

there are general methods for that. they basically reduce to computing the coefficients of the power series for 1/g(z) when a power series for g is known (the coefficients of the power series of f(z)/g(z) = f(z) * 1/g(z) can be computed from the coefficients of the factors using the usual formulas from multiplication of polynomials). the formulas get ugly really quick and are the kind of thing people use computers to do. or if you're a human, you often find another way of looking at the problem.

when i say 'basically reduce to,' i mean 'are as difficult as,' not that you'd necessarily want to compute a given power series for f/g by first computing 1/g and then multiplying.

it's the same general level of difficulty (in terms of hand calculation) when the numerator is 1, is all im saying there.

@leslietownes This. But, in this case, why?!

i dunno, if the quotient is a generating function for something that counts cool stuff, you might care?

more common is just to need a small number of coefficients, not the whole sequence.

@leslietownes Right, but isn't the given quotient just $e^{z-1/z}$? Why not just expand that out...

this kinda arises just with power series. for many purposes, it is not much use to have a 'general formula' for the nth term. maybe you want to know something about the nth term, or about the function having that nth term.

xander: i wasn't even looking at the specific example.

@leslietownes I figured.

I mean, in general, there are ways of working out the $n$-th term of a quotient of series (as you say, consider the Cauchy product of $f$ and $1/g$, where there are techniques for working out the terms of $1/g$).

if you look at people using power series before "modern" complex analysis (say mid 1800s) they sometimes do quite a lot in terms of that, and as clever as it all is, the modern reader is often asking xander's question.

@leslietownes Them people was quite clever.

@XanderHenderson yes, its $\exp(z - \frac 1 z)$, which gives you $\sum_{n=0}^\infty \sum_{k=0}^n \frac {z^{n-2k}} {k! (n-k)!}$

now I want to know if I can get there with long division (for the usefulness of the technique in general)

because, for example, like I said earlier, I don't know how wolfram computed the coefficients here: wolframcloud.com/obj/a1e1363d-2667-445c-b882-25cf69806fc5

my comments above sketch this out at a high level. if you want general formulas, the answers in math.stackexchange.com/questions/2050216/… address this a little bit.

i have no knowledge of what WA is doing, but i do note the formulas/procedures in those answers are not very human friendly.

yes, I have seen (some of) them. not anything pretty or particularly interesting

but note that e.g. the recursion in marty cohen's answer would be easy for a human to to implement on a computer.

btw, what I have up top looks to give an essential singularity of $\exp(z - \frac 1 z)$ at $z=0$, which doesn't track because at 0, you have $\exp(0 - \frac 1 0) = \exp(- \infty) = 0$

there's an error there, I also need a $(-1)^k$ factor inside, but I doubt the terms will compensate away to remove the essential singularity at $z=0$

@JoeShmo have you looked around $0$? The function goes wild.

oh duh

for example approaching from the negative real axis sends the function haywire to $\exp(\infty)$ this time.. 🤦‍♂️

then wolfram does in fact get it wrong

Hi! I'm reading the Radon Nikodym Theorem in papa Rudin. Can you please tell me where $w<1$ is used?

I think only $w>0$ and $w\in L^1(\mu)$ are used in the proof.

Something I just learned: if you multiply 123456789 by any number smaller than and coprime to 9, it shuffles the digits

This seems to generalize to any base

I think I see why this works but I'm not sure I can put it into words yet

Example:

123456789 * 7 = 864197523

In base 5, 1234 * 3 = 4312

Probably similar to what happens when you multiply the decimal for $1/7$ by any reasonable integer.

Similar results for other numbers (nest it n-1 times)

I found this while solving a puzzle, but I have no idea how I'd prove it if I were given it outside of that context

Perhaps that's a better way to put it

Oh, well, to the extent that linear fractional manipulation is the same as matrix manipulation…

Why is f bounded? Is every linear functional in $(l^1)'$ bounded?

Ahh no no..

Somehow I didn't think about this:

Is this question correct? I think (AB)^T is not even matrix mul compatible.

@Semiclassical Yes there is. If there is an upper bound u, then let k∈ℕ+ such that S(k) = 1/1+1/2+1/3+...+1/k > u−1/2, and then observe that S(2k) ≥ 1/2+(1/2+1/2+1/4+1/4+1/6+1/6+...+1/2k+1/2k) = 1/2+2·S(k)/2 = 1/2+S(k) > u.

@darkexodus My -guess- would be that they meant that B is 3-by-2, so that AB is well-defined as a 1-by-2 matrix and thus (AB)^T is 2-by1 as written

as written, though, you're right that it doesn't make sense

oh, I misread: they're saying that the second column of B is 2-by-1. so that indeed would have to go with B being 2-by-3 which doesn't make sense with AB

so yeah, something's nonsense here

I want to find the dual of a particular space.

the space is {the sequences in $l^\infty$ which are conditionally convergent. } with the norm being: $\|(x_n)\|= \sup_k |\sum_{n=1}^k x_n|$.

It seems to me that the dual would be $l^\infty$.

but I am having difficulty proving this.

@JoeShmo what did you feed Wolfram?

@MartinSleziak: good morning!

Good evening!

Let's take any T in the dual of the space. {e_i} is a Schauder basis of the space.

whatever...

Hi @robjohn @MartinSleziak !!

@Koro Does this actually give a norm on this space? In the other words, is $c\subseteq\ell_1$?

If you take $x_n=\frac1n$, you'll get $\|x\|=\infty$.

Let's call the space X whose dual we have to find. So for every $x=(x_n)\in X$, we have $x=\sum_i x_ie_i$. $|Te_i|\le \|T\| $ whence $\sup |Te_i|\le \|T\|$. This gives me a reason to believe that the dual is $l^\infty$ (I want to map $T$ to the sequence $(Te_i)$).

@MartinSleziak this sequence is not allowed for we want a sequence $x_n$ such that $\sum x_n$ converges.

Oh, I see, I probably misunderstood what space you mean. You are taking exactly the sequences such that the corresponding series is convergent.

I managed to confuse myself by thinking - what that actually means that a sequence is conditionally convergent.

Let me write it in more detail: $X=\{(y_n)\in l^\infty: \sum y_n \text{ converges.}\}$

Sorry, I erroneously wrote it in a confusing way earlier.

This X is complete.

But the problem I have is in showing that the dual of X is actually $l^\infty$.

By $e_i$ do you denote the sequence $(0,...,0,1,0,...)$?

yes

I guess we should be able to show that, for every $x\in X$, the sequence $(x_1,\dots,x_n,0,0,\dots)$ converges to $(x_1,\dots,x_n,x_{n+1},x_{n+2},\dots)$ in this norm.

Consequently, for every $T\in X^*$ we have $T(x)=\lim_{n\to\infty} T(x_1,\dots,x_n,0,0,\dots)$.

yes, for the first assertion.

The equality $T(x)=\lim_{n\to\infty} T(x_1,\dots,x_n,0,0,\dots)$ is a consequence continuity of $T$.

I see as T is bounded (hence continuous).

We also have $T(x_1,\dots,x_n,0,0,\dots)=\sum_{i=1}^n c_ix_i$ where $c_i=T(e_i)$.

True.

So $\sum_{i=1}^n x_i Te_i\to x$

And thus $T(x_1,\dots,x_n,0,0,\dots)=\lim_{n\to\infty}\sum_{i=1}^n c_ix_i=\sum_{i=1}^\infty c_ix_i$.

indeed.

But the question is whether the two norms are the same.

yeah.

I.e., whether $\|T\|_{X^*}=\|c\|_\infty$.

yes. One side is clear viz. $\|c\|_\infty \le \|T\|$.

This is a bit unusual norm, I'd say.

haha

To show $\|c\|_\infty\le\|T\|$ you're using the fact that $e_i$ belongs to the unit ball of $X$, right?

the fact that $\|e_i\|=1+0+...=1$ by the norm definition.

You mean $\|e_i\|=1$ right?

yes ofcourse.

(fixed my last comment.)

@MartinSleziak I think one typo here. LHS should have $T(x_1,x_2,...)$ the whole sequence $(x_n)$.

Sorry, I'll have to go.

np

see you later. :)

There was a phone call.

Let us denote $y_n=(1,-1,1,-1,....,1,-1,0,0,0,0,...)$. We have $\|y_n\|=1$ w.r.t. this norm.

Maybe I should use upper index - so that I distinguish between terms of a sequence and elelemnts of $X$.

Let us denote $y^{(n)}=(1,-1,1,-1,....,1,-1,0,0,0,0,...)$. We have $\|y^{(n)}\|=1$ w.r.t. this norm.

And now if we define $T(x)=\sum (-1)^n x_n$ then $|T(y^{(n)})|=2n$.

@Koro I might have missed something, but it seems that from the sequence $c_n=(-1)^n$ (which belongs to $\ell_\infty$ we get a linear functional which is unbounded.

BTW this question is about the same space: Proving that the space of sequences with bounded partial sums is complete. (But about completeness - not about the dual.)

And here is a question whether this is equivalent to the $\ell_1$-norm: Are the norms $\|\cdot\|_1$ and $\|\cdot\|$ equivalent?

@MartinSleziak $(c_n)\notin X$

as $\sum c_n$ is not convergent.

@MartinSleziak Their space seems to be different (does not seem to require conditionally convergent.)

@darkexodus You're correct, of course. $B$ needs to be $3\times m$ for some $m$.

This is badly written. You're starting with $x_n\in S^\perp$?

Ah sorry sure

Yes, it's right.

$x_n\in S^\perp$ otherwise it doesn't make sense

Thanks! Do you mean with badly written that it is a bad "proofstyle"?

I meant that you didn't tell me where the $x_n$ lived.

ah okey sorry

If you know that the inner product gives you a continuous function of one variable (which is what your proof shows), you can just say $\langle x,s\rangle = \langle \lim x_n,s\rangle = \lim\langle x_n,s\rangle$.

But why can I show like sequential continuity

you're in a metric space

ah in a metric space it is equivalent?

You're doing Hilbert spaces and you don't know that, huh?

No we have not seen that in a Hilbert space sequentially continuity is equivalent to the epsilon delta continuity

But does this follows from the underlying topological space?

it's the same thing in any metric space

this is usually one of the first things one proves when introducing metric spaces

We haven't proved it explicitly but what I know from topology that continuity is equivalent to say that the preimage of an open/closed set is open/closed. Is this the point since every metric induces open sets?

that's another characterization of continuity, but it's different from this one

ah okey is there somewhere a good online proof about the equivalence?

Oh my gawd... the first day of the semester could not possibly be going worse. The heat is out in the building, the LMS is fighting me, no one can tell me what I am supposed to be teaching, I am sleep deprived from scrambling to get things up and running all weekend. I kind of just want to cry. :'(

On the bright side, there is a "take a book, leave a book" shelf by the library, down the hall from my office. There was a copy of Apostol's Mathematical Analysis on it. So I have now filled a hole in my collection of texts.

That's actually a decent book.

Well, a quarter of mile from my apartment the road is still deep in water — they had pictures all over the country of the drowning cars near us.

@TedShifrin Yeah, I've had access to it in the past, via the math library at UNR, and via a friend's library at UCR, but I've never had my own copy.

So I am very happy to have it now.

It is not one of the books I saved when I retired.

@Koro But $c$ belongs to $\ell_\infty$. You're trying to show that for every $c\in\ell_\infty$ you get a linear continuous functional (which has the same norm as $c$.) In any case, I would suggest to continue in another chatroom so that it is not mixed with the other conversations here. (But not today - it is already late in my timezone.)

The question is, what book should I leave in its place?

Without seeing your library, I cannot answer :P

What is LMS?

@TedShifrin "Learning Management System". Moodle, in this case.

Oh. If technology can go wrong, it always will. Axiom #1.

@TedShifrin In this case, it is not the technology. It is the people who were supposed to transfer classes over to me. Because four of my five classes were canceled on Thursday for low enrollment, and I had to get other classes to replace them.

I think that this is what I get for asking to teach classes first thing in the morning. I want to be up then, but the students do not.

I taught only one 8 AM class in my entire career. It was a precalc winter quarter that I took over from someone who was fired. I actually had a couple of good students in that class.

I did teach a few afternoon classes, but hated those. Students were either starving or falling asleep because they'd just eaten.

@TedShifrin My preference is to get all of my teaching done by noon. I tend to get into the office around 5am, am super productive until my first class, and rarely get much of anything useful done after 1 or 2 in the afternoon.

jesus, office at 5

But you said you have 5 classes? How do you fit those all in the morning?

he teaches them all at once, in a farcical series of pretend bathroom breaks.

@TedShifrin I don't.

It is my preference, but it never quite works out that way.

Ah yes, the good ol' sitcom ploy.

morning people are scary

I confess that most of my career I taught two classes a term. A few years (the reward for the highest college teaching award) I taught 3 a quarter.

do you drink the same amount of coffee everyday @XanderHenderson

Most semesters, I have a calc class from 8:30--9:30 MTWR, then a MW section and a TR section of precalc from 9:30--10:45 am. I often get another calculus class from 11--1 two days a week, and then some godawful Thursday evening thing from 6-9.

@shintuku Nope. When I work from home, I tend to either have just one cup early on, or continue to make coffee all day. When I go into the office, I tend to brew up a triple or quadruple macchiato, and polish that off during my first couple of hours of work.

hm i see

i had hypothesized that a morning person would have constant coffee intake, apparently not

No.

@TedShifrin I'll bet you had a real research career, too. :P

Yes, I was supposed to. But I also had 6-10 office hours, well attended, every week and graded homeworks for all my non-calculus classes. So not the typical research faculty.

I wrote some good papers, but I would say I'm prouder of my textbooks that leslie loves to ridicule than of my research productivity. I think they have had a positive effect on more people :D

In the last two years, I have sat down on at least four occasions to write one of the papers that I need to write. Every single time, I pull out my thesis, spend a week or two trying to remember what the hell I was doing, and then have to deal with teaching duties.

This was a bit interesting.

I know you'll find this hard to believe, but my research productivity was highest when I was busier with teaching. It kept me energized. But most of my research was joint work; I enjoyed that a lot more.

@TedShifrin Yeah, I find it very hard to go back and forth from a "teaching" mindset to a "research" mindset. It is a hard gearshift for me.

Also, I totally get the desire to work with others. The one paper that I have managed to get out has three coauthors, and was the result of a very pleasant evening at a pub in Coventry.

My father, too. Never published a sole-author paper, to the best of my awareness.

Collaborative work is easier now that we're all in Zoom/LaTeX mode. You don't need someone in the next office.

When I have collaborators, I feel accountable to them, so work gets done. With no collaborators, other things can feel more urgent / important.

Yes, there's accountability. But it's also more fun and more productive bouncing ideas off one another.

@TedShifrin Oh, absolutely.

All the best work is done on a bar napkin, right?

I remember one of Kirby's best students at the time teaching me Kirby calculus (3-manifold stuff) over beer and pretzels.

Lovely!

@TedShifrin What is a TA? 🤷🏿‍♂️

The Apple

@D.C.theIII teaching assistant

can help with correcting, can sometimes teach class when teacher isn't available, or can be in charge of non-theoretical classes

is available as someone to ask things about the class content

How is the unit vector for $(-1+i, 1)$, $\frac{1}{\sqrt{3}}(-1+i, 1)$? Something very simple yet I'm confused

@shintuku Oh I was being facetious to Ted because he said he had 6-10 office hours, but more importantly graded homework assignments for all his non-calculus classes........something that you would assign to a TA as a menial task below a prof's time

or rather how to get $\sqrt{3}$ as my "normalizer"

@D.C.theIII What is the length of $(-1 + i, 1)$?

dc: staffing/funding is often an issue.

when i was an undergrad TA grading for non calculus classes was kind of sporadic. some semesters it would be available, and some it wouldn't (and then it was up to the prof). i thought it was a worse look for the department than if it were never available, because it gave the impression of a department teetering on the edge of financial insolvency.

@leslietownes Less cynically, TAs are usually graduate students, and receive tuition reduction for TAing. Also, graduate student TAs are receiving valuable teaching / instructional training!

Though I think that the cynical interpretation is somewhat closer to the truth.

there's certainly a sense in which grading non calculus classes isn't "menial" work. you can't outsource it to just anybody who got an A in the class, for example. which you might, for a calculus class. and it helps a prof to keep up on how their students are doing. you'd want to look at the assignments anyway, even if you're not grading them.

@XanderHenderson my complex numbering game sucks, but I essentially thought it was $\sqrt{(-1+i)^2 + 1^2)}$ if I'm being pedantic

@D.C.theIII Close. You need to take the moduli.

it is the length of the vector in complex space @D.C.theIII

$$ \| (-1+i, 1) \|^2 = |{-}1+i|^2 + 1^2 $$

What is $|{-}1+i|$?

nein, what is the vector corresponding to $-1+i$

@XanderHenderson $\sqrt{1^2 + 1^2}$

@D.C.theIII Which is...?

which will give me after all the other calculations my 3

@D.C.theIII Молодец!

@XanderHenderson sent me to DeepL to translate what I think is Greek?...... glad I chose to do the complex number question to brush up on things....going to have to do a brush up on concepts

@D.C.theIII Russian.

It means "Good job!"

@leslietownes I was under the assumption that at least in the Math Dept all TAs would be grad students. I'm also cynical with the "menial" work comment because of how U of T treats undergrad grading...

I had my best students from the honors multivariable math class grade the proofs in the next year’s class. I could never count on a beginning grad student to know the material or grade competently. Truth.

Ditto for diff geo, diff top, or even algebra.

well seeing how rigourous you are I could see why you would trust them.

@TedShifrin I might trust a beginning grad student with grading multivariable, depending on how "deep" the class is. I would have difficulty trusting them with the grading of much of anything upper div.

Of course I would've been the head TA for you at the time.....🙃🙃🙃

On the other hand, I don't trust beginning grad students to teach precalculus or calculus.

I expect them to know the material, but not the pedagogy / androgogy.

And those classes require a little finesse, as they are primarily for non-majors.

It was worth the time to me to do it right and know where the students were. For probability I used a grad stident who knew it better than I.

where i went, much of the grading was done by undergrads. thats how i paid for food most summers. :D

@XanderHenderson In my endeavours I've come to realize they really are two different things. I appreciate it too

In any event, it has been a long (and ultimately rather frustratingly unproductive day), so I am going to go home.

Xander, my multivariable math class was my blue book: proof-based plus computation -— linear alg + mult analysis, msnifolds, forms.

@TedShifrin Yeah, I likely wouldn't trust a first year grad student with that.

Hell no, not at UGA.

That sounds like a better multivariable class than what is taught at any institution I've ever been affiliated with.

I'm used to multivariable calculus being Stewart (or Thomas), part III.

So, a cookbook class, primarily aimed at engineering and physics majors, light on proof or reasoning.

@XanderHenderson that's cal 2 no?

Taken only by a handful of best and motivated students. Maybe 25-30 max.

@shintuku Typically, Calc I is differential, Calc II is integral, and Calc III is multivariable.

oh i see

Though I suppose it varies from institution to institution.

Anyway, I really am going home now. I'm beat. Later, all.

thanks for help, enjoy your evening

In Japan we had differential and integrals in a single course called "calculus". This is back in High school

Some ODEs too

@Gokuカカロット Sure, but my impression (from watching too much trash anime) is that Japanese high schools are far more specialized than American high schools, and that a university-bound high school student will usually complete the equivalent of the first year or two of American college while in high school, but won't take as many courses outside of their "major" as a typical American.

Well, more and more places in the US students are finishing AP BC calculus junior year and taking "multivariable" calculus and "linear algebra" either joint enrollment in community college or in their own high schools. This is just a mess.

Not good mathematics preparation at all, except for a few exceptional cases.

Sakigake!! Otokojuku ...proper anime that.