so it's somewhat of an accident that so many nice functions have very slowly growing taylor coefficients. i mean it's not, they tend to solve various differential equations etc. from which the growth limitations can be deduced.

ha, yes, that.

he should have just proved the theorem. i hate it when people refer to books, even in response to something that is a reference request.

i didn't know it was called borel's theorem. thank you for finding that.

Ye, thanks for mentioning this. I think it's cool, and I wasn't aware of this result

i first saw it in hormander's PDE book. he gives enough hints to make it easy. the upvoted answer doesn't give any hints.

oh i see someone gives what is described as "peano's proof" in one of the responses. that is a mess, you can do it more simply than that.

but i upvoted because he gave a proof!

<xD

i've even met him. it's such a small world.

hello, someone have an idea about this condition :math.stackexchange.com/questions/4124022/…

i don't have any immediate thoughts, it looks tricky. i'd ask about the intended application but i'm not sure that i want to know. :)

that is a very specific condition without any context

bingo

you sometimes see estimates of that type in providing that operators between various function spaces are bounded. sometimes there is general theory you can use that doesn't quantify what the value of something like tau would have to be, you can just use it like a black box.

those are my senile thoughts.

Monotonic functions have no discontinuities of second kind

Is it true only when the function is defined on an interval

?

if a function has a 'complicated' domain there's a split in opinion about what continuity and discontinuity even mean. the bourbaki books go one way and patriot americans go another.

Or in any case when the monotonic function $f$ is defined on any subset $E$ of $\mathbb R$

the main issue is the potential absence of stuff immediately to the left or right of something that happens to be in the domain of the function.

Let’s go the patriot American way that is I’ll assume that isolated points are points of continuity

And that discontinuity is defined only for points of domain of definition of $f$

i don't know what left and right hand limits at c can mean, if c is not at least an accumulation point of the domain of f, which it might not be for an arbitrary E.

if the domain of f sucks, there's no use to that result anyway. again, i'd ask about the intended application but i'm not sure that i want to know.

I ask for my knowledge. Whenever I see this theorem, it’s always on an open interval.

awfully late in life to realize i'm fundamentally an engineer. "what is this FOR?"

i will say broadly most results of calculus are not optimal, but seeking optimal results is often in the nature of digging rabbit holes and not finding more useful theorems. the best example of this is finding general conditions under which the FTC holds. complete waste of time for almost all purposes except that one.

i'm not an authority, others may disagree.

it's definitely useful to know that the hypotheses in calculus books can usually be weakened. it's not as useful to know how weakened they can be. there's often no best, most general result, that subsumes all of the others, as there sometimes is in algebra.

you just have curiosities.

If the domain of $f$ sucks 😅

that taylor series thing from earlier is a good example of this. calculus books give conditions on the growth of the size of derivatives of f under which the series will converge to the function on some interval. are those optimal? i don't know. do you need to assume something? yes, by what sha kindly identified as borel's theorem.

@leslietownes: I hope I don’t recall this during a serious meeting at my office. 😅

i'm a big fan of being frank about the domain of f.

if it sucks, say it sucks. we all know what it means. :)

if i had to define a key difference between 'analysis' and 'algebra' broadly interpreted, it's that in algebra, 'what are both necessary and sufficient conditions for [something interesting]' is a good question with a useful answer, and analysis it is not.

some may disagree, i certainly was never the world's greatest analyst. that was the vibe that i got

What does it mean to say that discontinuities of a monotonic function need not be isolated?

@leslietownes

it is to say that any neighbo(u)rhood of one discontinuity may potentially contain other discontinuities. a discontinuity would be 'isolated' if you could choose a neighbo(u)rhood of one discontinuity that did not contain other discontinuities.

added those u's in for copper.hat

u are fantastic :-)

Ahh I understand. Thanks @leslietownes

Why? I think they write neighbourhood in UK?

Programme instead of program

petrol. lorries. tyres.

Colour and not color

there's no end to it

copper hat was born on something that as far as i am concerned is not the UK but infiltrated the US.

@leslietownes you are very helpful and fantastic. Thank you.

maths and not math

one thing you can do to rile up brits if you are american is to misunderstand the phrase "taking the piss," which means making light of something in perhaps a mocking fashion. deliberately refuse to understand the term and repeatedly ask "what? you're taking a piss? why are you doing that? where are you doing that?" taking a piss is american for urinating.

No. I think it’s actually Maths and not math. Right?

people go along with it for quite a while because they expect americans to be dumb.

it's math in the USA.

You takin' the piss?

yeah, in this phone booth.

lol

in the last 20 years my wife has seen me do this joke in front of five or six brits. she sees it coming and looks at me, like, "no, don't do it."

highly offensive

it's basically just a joke between me and my wife. the brits are NPCs.

I wonder whether or not you know that I'm a Brit

well, you do now

lol

i certainly was not born in the uk...

You were born in the pretend country floating off the side of Britain

i knew you were.

heh

i have often had difficulty convincing that the republic was actually a separate country :-)

I have that difficulty in Germany too

you've infiltrated another country.

Yeah we've been here before

i got a lottery visa

it's weird that with my propensity for bothering people i have not infiltrated a country.

i think i would be far more annoying in another country

when i came it was very difficult for irish to infilitrate the usa

It's strange that upon arriving in Germany I didn't need a visa and now I do

ahh, freedom. finally free of all the EU strangulations.

Finally we can fish our own waters without any of the pesky Europeans getting in our way! Except not.

i think it's not an overstatement to say, i don't think some of those people thought everything through to the end.

good ole nigel

I can't help but feel it's arrogant to say that the working class in the UK didn't know what they were voting for because of constant media bombardment, but I feel like it's also quite close to the truth.

modern politics is binary, all or nothing.

there's something similar in the US. the media ecosystem has gone off of the rails.

cameron bears a lot of the responsibility for his stupid gambit

Yup. Ergh it's a shame, I can't talk about it and I need to go and do some work :(

i do not work on sunday. i'm not religious, i just find that people get off my back if i say i don't work on sunday.

Alas, I'm a chronic procrastinateur and have a deadline tomorrow

one time driving in rural iowa i pulled up to a gas station and the guy said he couldn't sell me gas on sunday but he would loan me some until monday. i wasn't coming back, so i told him to look for his payment in his mailbox, which was right in front of both of us. it was weird theater but i respect commitment to a bit.

odd

there are lots of unorthodox religious people in the middle of iowa.

various splinter sects from protestant churches, and then one city completely populated by devotees of the maharishi mahesh yogi.

sometimes people would come in to protest in iowa city. i wish i had a picture, a guy had a sign that read "SODOM. GOMORRAH. IOWA CITY."

if i were mayor i'd make it the city motto

@EdwardEvans "Some work"? You?

accurate

Have I missed any interesting math in here?

I just got a weird email from a British person who's a grad student in Genoa, basically saying he/she has a geometry question and am I open to helping? Can you say "vague"?

there was some algebra from love_sodam and some algebraic geometry from sha.

someone from brazil once emailed me with a generic request about functional analysis.

i don't know who these people are but we should find them and stop them.

If I get such emails I reply you can ask and if I have time & can help I will reply, but do not rely on receiving a response.

Ah, I discussed a little bit of that stuff with Sha yesterday — a trivial observation (trivial for me, not for her). I miss that stuff. ... Yeah, I actually gave a bit of advice in my answer and told him/her that the nature of the question should be included in the original contact.

I also said that my research days were long gone and I may not have anything to say.

that's a very ted like response, going above and beyond.

@copper: Did that image of the ball response get somewhere with that OP?

Well, usually people either say they know me from here or they make some reference to some published work I've done. This was unusually vague.

I added an answer but haven't heard anything since.

i did not respond to the random person from brazil. i think she had gotten my information from a former officemate who had postdoc'ed at her university. i didn't see any upside in it.

The professor wrote the mapping in a very confusing way, as far as I'm concerned. Why include the $2^{-m}$ in the $n$-summation?

I was trying to think of a good intuitive way to look at that problem.

Yes, it took me a moment to unravel to see its just a one dimensional range. the whole $\tau$ thing is just noise.

there was a turkish guy who asked me for answers from one of my advisor's books and i responded because he had a hilarious myspace profile where he was shredding on an electric guitar as if it was the 1980s and i respected that.

@leslietownes btw $\tau$ is pronounced taff

no, it's tee.

the au in greek is aff, so Europe 'should' be pronounced evropi

alfie, beetie, gammie, deltee, etc. untill pee and not pi.

Yes, the i sound is ee, as it is in Latin.

i'm trying to fill out some stupid class action thing for health insurance. i think its beyond even my record keeping detail

But I'm not sure you have the vowel diphthong right for tau.

there is no such thing as a stupid class action thing.

all of those are meritorious.

Every class action suit I've ever filled out crap for has gotten me nothing.

As a Greek person to pronounce $\tau$.

Well, our chat resident Greek hasn't been around in months.

class actions are a nice throwback to law as it existed before individuals had any rights.

I have got some, I suspect the return on this will be negligible. I am just surprised that I did not keep more detailed records.

you were wronged, here you go, it's a coupon for the same thing you bought when you were wronged.

I just got something in the mail recently telling me I was eligible and would be included if I didn't withdraw. I don't even remember what it was.

And I used to have a good memory.

that's weird to have an opt-in default.

I'm pretty sure that's right.

there's often a ton of negotiation over what to do with the money left over when people don't opt in. you ideally want that pot to be as large as possible and for the judge to approve a recipient that makes you look socially responsible.

you'd prefer people to opt out and then get a headline about X going to the Coalition to Defend Puppies and Kittens, or whatever.

i'm unusually cynical today but that's how it works.

its a rip. the effort involved is not worth the return

No, you're your usual level of cynical. Copper seems slightly less grumpy than usual.

i wonder who makes money out of class action :-)

i can volunteer a guess.

Guaranteed some attorneys do.

juts a normal headache today :-) and no mse threats this morning

i remember phoning the CEO of a nationally known nonprofit about being the potential recipient of the leftovers of a class action settlement. i thought, there's no way this guy will even acknowledge my existence. he called back in five minutes.

You were stunned.

i am glad to be past the various tricks to get people to call back

the judge approved it, too. he got his money.

@copper.hat a long long time ago it was pronounced 'tau'

I'm still failing at the getting response thing. One of my doctors' office emailed and texted me saying I was past due for an appointment. I phoned and after 10 minutes on hold, hung up. I then responded to the text. Nary a word, 4 days later.

not in Greece

tell him you'll give him money in a class action settlement

ancient greek. BCE

:-) from the recordings?

sooo

matrix inversion and adjugation are continuous

but not pseudoinversion. that's neat

If we know that $4x^3+b_2x^2+2b_4x+b_6$ does not divide $x^4-b_4x^2-2b_6x-b_8$ (where the coefficients lie in some field $K$), then why does that mean there exists an $x_0\in\overline K$ such that the first polynomial vanishes to higher order at $x_0$ than does the second? What does it even mean to vanish at higher order? Does that mean that the first so derivatives of the polynomial vanish too?

(Actually, I'm even given that the two polynomials are relative prime)

@Corellian it is a bit like defining $f(0) = 0$ and $f(x) = {1 \over x}$ otherwise.

that's a good analogy.

also loving the oldschool use of \over. no frac for you.

@Sha: Yes, multiple roots are about simultaneous roots of the polynomial and its derivative(s).

Not dividing is a lot weaker than saying relatively prime.

Copper is the only person here who uses \over in TeX.

that's how we can tell his age.

I never once have used it, and I started learning TeX/AMSTeX in the late 80's.

test ${1+X^2}\over{1-2X+X^3}$

$\frac{1+X^2}{1-2X+X^3}$

i remember seeing a lot of very goofy stuff when working with my oldest colleagues. stuff that would have made don knuth scream.

That is nice.

it really is for frak's sake

Fric and Frac?

:-)

fewer { } to type

ok, i managed to find my health insurance records.

crap i forgot to send rent last nite

paypal?

all good now

I remember the days of yore when I had to send monthly checks for rent.

helps to recall april only has 30 days

yesterday was may day

today is mother's day in spain, portugal, few others

@copper.hat In fact, for $\frac12$ you can even type no braces and no spaces!!

I'm a bit lost with an assignment in Functional Analysis :(

can you say more about that? can you think of the pairing that might associate an element of $\ell^1$ with a linear functional on $c_0$?

if i'm within epsilon of a sequence converging to zero in the sup norm, i can't be too far from zero myself. i'd emphasize that it's the l^infty norm in the first question.

math.stackexchange.com/questions/678911/… has most of the relevant details. not so much in the question as in the dialogue represented by the answers.

@TedShifrin Thanks, that will save me a few strokes!

@LogarithmicDerivative Suppose $x_n \to x$ with $x_n \in c_0$ for all $n$, then show $x \in c_0$.

Thanks, Leslie!

thank you for asking a functional analysis question. a lot of goofballs around here ask stuff with more algebra in it. we need to do our part.

@copper.hat That feels very obvious, but I don't know how to prove it.

Excuse me for my noob analysis question: say I have a function $f : \Bbb R \to [0, 1]$ and I define intervals $I_{n, m} := \left[ \frac{m-1}{2^n}, \frac{m}{2^n}\right]$ for $m = 1, \dots 2^n - 1$ and $[1 - \frac{1}{2^n}, 1]$ for $m = 2^n$. Then I can define simple functions $f_n = \sum_{m=1}^{2^n} \frac{m - 1}{2^n} \Bbb 1_{f^{-1}(I_{n, m})}$. These should converge pointwise to $f$, does that sound reasonable? I want to integrate $f(x) = x\Bbb 1_{[0, 1]}(x)$ wrt the Lebesgue measure

and I think one does this by defining simple functions in this way and then just hitting it with an $\int$

@leslietownes I'm going through the solution you linked now. I feel like this stuff is exactly what I need: Not being given the answer, but something that dances around it enough to calm down and think straight. :)

@EdwardEvans You use the dominated or monotone convergence theorem along with $0 \le f_n \le f$ and $f(x)-f_n(x) \le {1 \over 2^n}$.

I just realised I defined my intervals stupidly, they should be half open

I'll pretend this is a typo

@copper.hat thanks :)

if it is Riemann integrable it will be Lebesgue integrable with the same value.

I'm more worried about my partition of $[0, 1]$ tbh

it doesn't look right

(with the exception of the missing half-openness)

when $n \to \infty$ I'll end up with $\varnothing$ and not $[0, 1]$

Just use $[{ k \over n}, {k+1 \over n})$ for $k=0,...n-2$ and the remainder for the last.

durr thanks

No, for each $n$ (modulo the half open bit) your intervals are a partition so they will work.

but if you want to compute and get the integral of $x$ then you are complicating your life if you use the $2^n$ intervals

I was trying to get something that doesn't give me a $\frac10$

how are you getting a ${1 \over 0}$???

with $n = 0$ your partition gets me a $\frac10$ lol

Why would you take $n=0$??? you want $n \to \infty$

Basically Reimann integration.

idk I'm used to having a $0 \in \Bbb N$

I'll just pretend there is no 0

i'm still lost. you are trying to find $f_n \to f$, why do you care about $n=0$?

Because you're indexing $f_n$ by natural numbers, so $f_0$ would be weird, that's what I'm getting at

maybe i am missing something. it really seems irrelevant to me.

Alright I defer to you, I thought it would be relevant and that's why I have these weird $2^n$s

don't defer.

Well I don't understand why it's relevant lol, if I'm defining a partition by $[\frac{k}{n}, \frac{k+1}{n})$, how is $n=0$ not relevant?

if you are looking at the limiting behaviour, you don't care about any particular finite number of elements of the sequence.

You are creating partitions of $I$. You can start with $n=1$ or $n=10$, it does not matter.

Okay yeah, I was just worried about one of the sets being undefined

This highlights by analysis weakness lol

my*

You can start your sequence anywhere you want in this instance.

if you were summing the values it would matter.

Thank you :P

Grr … once again, $0\notin \Bbb N$.

Damn European idiocy.

$\Bbb N$ is a monoid

additive

in this situation it is irrelevant...

@EdwardEvans just think of it like the fact that the inclusion of a cofinal sequence is a final functor

ROFL