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12:21 AM
I wonder what my next mathematical journey will be...
The division stuff was fun and I learned a lot.
I still have yet to even implement it because it's not a high priority in my project right now, but it's still fresh in my mind on how to do euclidean divides in O(log n) (or O(M(n) log n) depending on what you do).
Modular arithmetic is too OP
Most of what I'm doing in programming right now is more about graph manipulation so it's less algebraic.
There is something of interest to me, however, concerning what I'm doing right now and algebra.
How do I read by rows and columns algebraically as a function of the inputs where the result is equivalent to a geometric transform but of digits instead of points on a plane?
Moving individual bits, I'm limited to a throughput of 4, but if I wanted to interpret four 64-bit integers as a 64 by 4 array instead of as 4 by 64, a numerical solution would be desirable for increasing throughput.
4 hours later…
4:29 AM
@TedShifrin I was today years old when I learned that you were responsible for adding several problems to Spivak's Calculus. So I guess I have you to thank, and loathe, for my journey through his book and the roller coaster of emotions throughout that journey:)
LOL, depends if you did my hundreds of problems. Which edition?
3rd editiion
yea I did almost every problem from chapters 1, 5-24, except for a few triple starred problems here and there
Yup. He also chsnged some stuff in the text according to my notes/suggestions.
I wish his calculus on manifolds had a similar flavor. Perhaps his pedagogy bone hadn't kicked in that early on :)
Use my book instead.
He did that literally right out of grad school. It had to be short — too short.
4:39 AM
oh your online lectures helped me through some parts of the end of his book, but by that time, I had already suffered at the hands of his brevity.
I found your lectures a little too late into my multivariable journey
Keep the book in mind :)
since you're the only person with teaching experience I have access to right now, what do you think about teaching the Lebesgue integral instead of Riemann for students who already have a decent background (say Spivak's calculus). Because after learning both approaches, I found myself wondering why wasn't I taught the Lebesgue integral earlier (and my classmates seem to agree)
I mean instead of Riemann integral again for multivariable calculus, why not go straight into the Lebesgue integral and measures
idk if I'm an odd duckling, but I found the Lebesgue approach much simpler, technically, than Riemann (too many details to keep track of with Riemann integrability, Jordan-measurability etc)
4:59 AM
My combinatorics skills are trash at this point. If anyone up at this hour has any interest in it, I just posted a question to the main site: math.stackexchange.com/questions/4479984/…
It seems like something for which there would already be a solved model for somewhere.
You need both the Riemann & Lebesgue integral.
5:19 AM
@peek-a-boo I don't approve. You can't actually compute any Lebesgue integrals. Learn multivariable calculus rigorously and then deal with your question at the level of more advanced courses.
I could barely fit my course into 112 lectures. Doing Lebesgue integration is a whole other course.
5:40 AM
What's the Union and Intersection symbol called when it's used as part of an aggregation like a $\sum_i^n$ or $\prod_i^n$?
Big Union/Intersection?
Can I repost something I posted on this chat earlier as the discussion was not completed? (or atleast as much as I intended it to be)
At this hour, it's probably not very disruptive. It depends on who you're trying to resume the conversation with.
@Axoren No one in particular
because what was told was really less
Means anyone could pick up
Reposting a message in a live chat window once or twice isn't gonna be very disruptive unless it becomes excessive, but the chat's really slow right now and you've still got more to say about it.
I'd say it's okay.
I posted it long ago
ok thanks
So as it is homogeneous, by putting $a=b=c$, I got $a=\frac{3}{2}$
(This is lagrangian multipliers so $\lambda$ is not important here)
The OP told me that reparametrisation could help and that Lagrangian Multipliers do lead to such calculation
6:00 AM
Wolfram seems to agree with you: https://www.wolframalpha.com/input?i=32a+%2B+%288%2B2l%29bc+%2B+3l%28b%2Bc%29+%3D+0%2C+32b+%2B+%288%2B2l%29ca+%2B+3l%28c%2Ba%29+%3D+0%2C+32c+%2B+%288%2B2l%29ab+%2B+3l%28a%2Bb%29+%3D+0%2C+2abc%2B3%28ab+%2B+bc+%2B+ca%29+-+27+%3D+0%2C+a+%3D+b+%3D+c
What's exactly your question about the above?
Or are you just stating you solved the minimization?
An inequality
@Axoren No I was saying I could find one minimisation
But others too had to be evaluated
imm trying to find the link to that ineq
Q: prove or disprove that:$a^2+b^2+c^2\leq \frac{27}{4}$

AnasI tried to solve the below problem, I spend more than 5h just for prove it but without any result , this is the best attempt I did ,just I need to show that $a^2+b^2+c^2\leq \frac{27}{4}$ if $a,b,c>0$ and $2abc+3(ab+ac+bc)=27$. prove that;$16(a^2+b^2+c^2)+8abc\geq 135$ My attempt: $2abc+3(ab+...

I mean if I do not find all solutions, how could I prove the minimum condition
Ok so there is one solution only
4 hours later…
10:29 AM
Please help undelete & reopen this.
11:22 AM
Hi @Calvin!
11:50 AM
@TedShifrin So i tried what you suggested last night, sadly i did not reach a result.Could you maybe help out if you have interest ?
6 hours later…
5:32 PM
@Mad What I was suggesting was this. Given a multilinear $\phi\colon V\times V\times V\to Y$, define $\phi_v\colon V\times V\to Y$ by $\phi_v = \phi(\cdot,\cdot,v)$. Note this is also linear in $v$. $\phi_v$ induces a linear $\bar\phi_v\colon V\otimes V\to Y$. Now define $$\psi\colon (V\otimes V)\times V\to Y$$ in the obvious way and show it's bilinear, hence ....
1 hour later…
6:34 PM
any one ...who can solve this math.stackexchange.com/questions/4479307/…
6:59 PM
Howdy, demonic @Alessandro.
Was there any further discussion yesterday after my purported counterexample on that homeomorphism stuff?
I don't think so
Things are much nicer in the smooth category with the inverse function theorem to give control. :)
I don't understand the compact-open topology, I only care about it on homeo groups of compact metric spaces, where it is the uniform convergence on compacta topology, or even better on the isometry group of a compact metric space, where it is the pointwise convergence topology
7:09 PM
Well, I think in the case of $\Bbb C\subset S^2$ uniform convergence on compacta should suffice?
I have taught compact-open topology, but long long ago, and it certainly never showed up in my research as far as I know.
As long as the codomain is metrizable the compact open topology on $C(X,Y)$ is uniform convergence on compacta
iirc it works as long as the codomain is a uniform space
Anyhow, I was surprised to find the flaw where I did, since I thought he had an argument for $\subseteq$.
But oh well.
Claiming to have an argument and having one are very different things, as I am reminded every time I get comments on my draft from my advisor
7:13 PM
LOL ... I remember those days on both sides. I also remember some joint work with two colleagues where fake proofs of one particular result kept washing ashore no matter how diligent we were to avoid the fallacy.
In particular, I remember preparing a talk in the airplane going to a conference and realizing it had duped us for a third or fourth time.
Luckily, this is long enough ago that I no longer remember the issue. We did finally resolve it carefully.
I'll give a talk about the draft I mentioned in a conference in July (my first ever conference talk), hopefully I won't have an epiphany like yours on my way there!
Great that you're giving a talk. It's not a bad idea to rehearse in a seminar with your colleagues first.
there are always holes
i would never present something of consequence without a rehearsal
even if on your own
i see message from Ted i go online
Simple man
I will try your suggestion later today, ty
OK, it should work out fine.
7:28 PM
Ted asking a general question.
Does it matter, if a student takes longer than needed to finish his degrees? is it bad in the future?
Will his chances become worse or something in academia.
I cannot speak for European traditions. In the US it's bad grades that are the issue, not an extra year, typically — at the undergraduate level.
What is bad? in your standards
Before continuing I have to say that there should be no presumption that everyone earning an undergraduate degree in math should be considering going on to graduate work. In the US, it's a very small percentage for whom it's the right thing to do. ... Less than Bs in standard advanced undergraduate coursework or beginning graduate coursework are a warning sign.
Makes sense.
maybe 3 out of 40-45 of us went on to grad school, from my graduating class of math majors.
and that seems on the high side.
7:39 PM
what would fall under "hard algebra" and "hard geometry", analogously to "hard analysis"?
huh. maybe computational stuff where you actually care about quantitative bounds and worst case scenarios?
for algebra, anyway. for geometry maybe something similar but with integral estimates.
?? just thinking out loud.
"hard" is a technical term there, not a descriptor.
when i think 'hard analysis' i think problems like "everybody knows |Tf| <= C |f|, but now, for some horrible reason, we actually need to know what C is"
Geometric analysis (as a main subfield of geometry) consists mostly of hard analysis.
C's gonna depend on curvature.
it always does.
7:43 PM
I think hard analysis refers in general to estimates, as opposed to soft analysis (e.g., operator theory) :D
yeah, most operator theory is soft. although you can pose hard problems in it, most people don't work on them.
But not routine estimates like one finds in Baby Rudin or standard complex analysis. It's a leap forward.
yeah @leslietownes thats hard analysis for me too
yes, @TedShifrin it looked like harmonic analysis last I looked at it
Hard analysis, is just pretty mathematics, that you find so good, you get excited. Thus hard analysis
I will see myself out
Good idea.
7:47 PM
thanks for the poem.
some classical operator theory had hard-ish analysis proofs early on but was later boiled down to, choose the right level of abstraction, and take some very cleverly constructed operator, and use the triangle inequality.
come again soon.
just study convex optimisation
Go look at Yau's proof of the Calabi conjecture. Third- (and fourth, maybe) order estimates on solutions of PDEs.
geometric analysis looks really cool
7:49 PM
I sat through many lectures he gave on that when I was in grad school, but I confess I never read the paper.
@MadSpaces BOOOOO!
it was actually a mediocre joke. you gotta hand it to the kid. his mind is in the right place.
I don't find mathematics interesting because the results are pretty. I find it interesting because I don't know it yet
@JoeShmo C- at best.
I'll admit that it's a new take on "thinking about mathematics" but nonetheless
7:51 PM
success is the mark on the forehead of the man who aimed too low
@copper.hat I get the sentiment, but I'm not sure that I concur.
People have different tastes in mathematics. I personally find the stuff most interesting/appealing if it lies at the overlap of different fields of mathematics. Arcane technical things have never interested me.
i have a weakness for arcana.
Given your field and your personality, not surprising.
I like pretty intersects practical of late
practical as in not esoteric theories that nobody needs
7:53 PM
@TedShifrin I enjoy technical and arcane topology, but not in other fields. Working at the intersection of different fields for my research has a big problem though: I don't understand any of them
That gives you goals.
ted: munchkin was vaccinated this morning. after belly aching about it all week, she didn't even feel the shot. she asked when she was going to get it after they had given it to her.
i was like, oops, you missed your opportunity to start crying, maybe get an extra lollipop out of it.
Well, she will get a sore arm, I'm sure.
sore leg. yes, covid.
7:55 PM
@leslietownes Oh, jeez... I need to get my brother in for another booster.
i think the first time i saw the boomtown rats was in the arcadia in cork city, speaking of arcana
Oh ... leg! Is that what they do with munchkins?
yeah, most of her shots have been in the leg. she got ice cream anyway because we spoil her.
I think the latest editions are not as violent as the initial strains
Ugh... but the pharmacy is not open on Saturday afternoons. :/
7:56 PM
@XanderHenderson too many let others determine the meaning of success
@copper.hat Sure, but that seems orthogonal to the quote you gave.
Xander: Is your brother in the same boondocks you are?
@TedShifrin Yes. He lives with me. I take care of him.
it took a while for us to find a pharmacy that had slots. about a 40 minute drive in fact (most of this was traffic).
munchkin's usual health care provider was clueless about when they'd be able to offer shots.
Oh. Is this the attorney brother or a different one?
7:57 PM
@XanderHenderson that quote came from doug cooper of oh pascal fame, i TAed for him for a while. quite the character
i used to have a life
@TedShifrin Not the lawyer brother. This one has Downs Syndrome and is severely mentally disabled.
Oh, now it makes sense.
poor lad
Wow, you are quite the good brother.
7:58 PM
i used to do some volunteering. heart breaking
and uplifting at the same time
you meet many really decent people
i saw u2 in the arc in cork as well. i am sure they learned some of their stagecraft there
@TedShifrin Sometimes. Mostly, it was convenient for everyone. Prior to COVID, my parents were planning on retiring to Tucson, and were going to put him into a group home there (which would probably be a better environment for him).
But then my father died and the world shut down and I got a job out here, so I was able to rent the house from my mother, which kept my brother in the same place he's been in for the last 20 years.
Oh wow. Yeah, a group home would be better for him ... more interaction and activities.
@TedShifrin Maybe. But definitely more structure at home.
As it is, he goes to work five days a week.
Oh wow. Great.
So he gets interaction and activities through that organization.
I think that they work, like, 20 hours per week, and spend another 20 hours per week doing stuff.
8:01 PM
So where has your mom gone?
For example, I went with them to Petrified Forest last week.
@TedShifrin Oh, she's in Tucson.
My parents bought a house down there 15 years ago, about the time that my father first started planning for retirement.
Ah, I see. ... Well, the whole world is roasting, but I don't think I could take the heat where y'all are.
I won't even think of politics ... because it's globally beyond depressing.
(Because he worked for the House of Representatives in Arizona just out of law school, he already had a decade in the Arizona retirement system, and could have retired at 60, which would have been around 2010; instead, he taught until about a week before he died; he was planning lectures for a co-taught anthro/bio class from his death bed. :/)
@TedShifrin It's only 84° here.
Wow. Did covid get him?
Being at 5000 ft helps.
8:04 PM
That's the way to go, though. Staying active until the very end and not suffering.
@TedShifrin No. Lung cancer. The obvious and inevitable result of 45 years of smoking.
Ah, yup.
Good it wasn't prolonged.
He was diagnosed in mid-February, and was dead by the end of March.
That's the way to do it.
He'd been hiking in Petrified Forest at the end of January.
8:05 PM
The last week or two was pretty bad, but he was on so many drugs that I'm not sure that he felt it. I'm pretty sure that he was hallucinating through the entirety of the last phone call I had with him.
Hello, are éléments of $H^1_0$ are convexe?
@Jakobian Sort of agree with this sentiment
Oh gosh darn it, i am a kid : (
I found in a proof that $u\in H^1_0$ then u and u' are convexe
8:13 PM
Annyways, i have question!. Nevermind i actually do not.
I would go one step further to say I like the feeling of total confusion. Things I can't make head nor tails of.
@Vrouvrou if $u$ is in $H^1_0$ then so it $-u$ so it would seem unlikely.
yeah, does H^1_0 have some weird definition? or does 'convex' mean something special here?
it is a subspace last time i went there
Are the zermelo axioms accepted in silicon valley?
8:16 PM
with choice
you need to suspend belief to live in si valley
I lived in silicon valley when I was a younger one
I am very sad to read that story @XanderHenderson Since i lost both my grandfather and father to smoking, and me myself being a smoker, with Asthma even. It is a true curse, my condolences.
i used to break my dad's cigarettes, he used to get mad
8:18 PM
I know what you mean, i used to hide my father cigarete package, but he was abusive and would ocassionaly hit me. I actually stopped doing it since i stopped caring about him lol
Quitting smoking is extremely hard
Smoking is such a cancerious trait, its the perfect human drug, you do not feel the consequences until its too late.
especially when people smoke throughout the day
almost everything is a trigger
Vaping is not a bad alternative.
yes it is
8:19 PM
It just does not look as manly.
smoking is manly?
Its a cliche. men with cigars, you dont ever see that on Tv ?
well maybe cigars
what about long cigarettes?
@Asinomás that's what the tobacco companies tell us
8:20 PM
those also have an associated person
Cigars, suck , bottom. They are tastless, and you cant even inhales them. i tried them once and i was like "bruh,no"
cigarettes ads were targeted mostly towards women in the olden days
When I was in high school one of my friends would take badass cigars to parties
and we would always be surrounded by other 15 year old guys
but no chicks :'(
I am not going to lie, but when a woman talk in that deep sad cigarette hardened voice and those tar tainted vocal cords.. damn thats "hard analysis"
sure, but so is a large percentage of stuff women do
8:23 PM
Yes. Like talking about Kähler manifolds.
like when they speak in a terrible ****** accent
there are worse things than smoking
$-{ 1 \over 12}$ for example.
you mean the math trick thing ppl do?
You know, this is widely misunderstood. If people get the concept of analytical continuation, that form will not like wierd at all.
If -u is in the space then u is convexe ?
8:24 PM
smoking is the leading cause of preventable death, but more hours of human life are probably lost to alcohol.
vrou: we were asking about this up above. what is the domain? what is H^1_0? why would you expect its elements to be convex?
@copper.hat here is a reason for you to hate $\frac5{12}$ as well.
the alcohol situatation is very interesting to me
it's a drug, but it's constantly made sexy by mixing it into drinks
@Vrouvrou i am trying to understand what you are doing, if you are working with Sobolev spaces, surely you have encountered convexity, etc.
even fancy drinks and stuff
I wish people could just take it as shots and wash it down later
I may as well die of a cringe remembering how wildely i used to drink alcohol, going back at home while the sun is rising, and partying like there is no tomorrow. I can never imagine myself being that guy again. jesus. I didnt have a drop of alcohol for 3 years.
8:26 PM
instead of making up elegant stuff around it
just do what you want to man
@robjohn its a slippery slope...
smoke, drink, shoot morphine
asinomas: people do, in fact, do that
just dont endlessly talk about it
8:27 PM
Thats bad advice.
what do you mean endlessly
i think it is best to drink constantly and maintain an equilibrium
@robjohn LOL
now I'm starting to think a 12 in the denominator is the devil in disguise
@copper.hat there is a very funny film about that, it is danish, called "another round" or something
8:28 PM
maybe it'sa good thing we use decimals :)
what do you mean constantly
isnt 7/12 like pie or something
like each day a similar amount?
@Asinomás i am joking
8:29 PM
i believe the best drinking is in the extremes
In the Film, a school teacher tries to implement a method, of increasing alcohol percentage in blood to some certain level, since he read, it makes him more attractive, better at thinking, and more bold.
then i wrap myself in my closed convex hull
notice that the name joe is in joke
@copper.hat "no it's not!!"
8:30 PM
should I try to understand the joke
i would say go with the flow
Today my neighbour asked me to help her get the key out of her door
new neighbour
I managed to do it
but then she asked me to open the door and I failed
@copper.hat BURN IT WITH FIRE!
@Asinomás and thus she disqualified you as a potential partner, life is hard analysis.
And two days earlier I had told my friends on discord that I was going to learn locksmithing because locksmithing was a scam trade
8:32 PM
my brother is an ortho surgeon and studied various injuries. one correlation was for skiing & snowboarding that there was a minimum accident rate after one drink
But well, I havent learned locksmithing yet
@XanderHenderson i never realised you were a $-{ 1\over 12} $ afficionado.
@copper.hat $$\Huge \text{BURN IT WITH FIRE!}$$
@copper.hat i would not take medical studies with much confidence, a very high procentage of them are just proven wrong after a while.Whats the corilation ? what mind part gets hyperactivated in order to reduce the accidents?
well I don't know she apparently isn't my neighbour and she's only coming a week to take care of a cat or something
she told me "hey I'm the friend of *****"
and I was like: "who is *****"
8:35 PM
the other neighbour in my floor of 5 asked me to fix his cable tv and I also failed
I'm such a shit neighbour
to be fair, there is not reason to be able to fix cable tv
even cable tv can't fix cable tv
@Asinomás the key and the lock are just codes
@copper.hat I fixed it for you.
(because the spacing at the end made my head hurt)
8:36 PM
@Asinomás Are you a friend of Dorothy's?
I've got no friends called dorothy
is that something from the wizard of oz?
@Asinomás Not exactly.
@copper.hat but we see the standard definition of $u(tx+ (1-t)y)\leq tu(x)+(1-t)u(y)$
$H^1_0$ is a subspace.
8:37 PM
In gay slang, a "friend of Dorothy" (FOD) is a gay man; and more broadly, any LGBTQ person. Stating that, or asking if someone was a friend of Dorothy was a furtive shibboleth used for discussing sexual orientation while avoiding hostility. == Dorothy from Oz and Judy Garland == === Dorothy from Oz === The precise origin of the term is unknown. Some believe that it is derived from The Road to Oz (1909), a sequel to the first novel, The Wonderful Wizard of Oz (1900). The book introduces readers to Polychrome who, upon meeting Dorothy's travelling companions, exclaims, "You have some queer friends...
@Vrouvrou if $u $ and $-u$ are convex then they are affine on connected components.
that would be a bit limiting for $H^1_0$.
i am wondering if this is candid camera for mse?
No we don't suppose that u and -u are convex
Is the following correct:

If f is an analytic function in a neighbourhood of z_0. Then we know that f can e written as a convergent sum \sum_n a_n (z-z_0)^n on |z-z_0|<ROC where ROC=radius of convergence.
We had a statement telling us that if a sum of the form sum_n a_n (z-z_0)^n converges on some |z-z_0|<r then the derivative exists on |z-z_0|<r. But is it true that r\leq ROC always?
So the largest set where the derivative exists is |z-z_0|<ROC
In general, yes.
8:50 PM
okey perfect thanks!
There are conditions under which the derivative exists at points of the boundary, but in general it is known inside
It is important to note that |z - z0| < ROC is not the largest set where the Taylor expansion of f centered at z0 converges. It is, by definition, the largest such disk, centered at z0
@BalarkaSen right I didn't thought about that thank you!
I recommend studying an example. Say, $f(z) = 1/(1 + z)$. Compute the Taylor expansion centered at $0$, prove that the radius of convergence is $1$. What can you say about boundary behavior? Is $f$ not analytic beyond the unit disk centered at $0$? (Also compute the derivative if you want)
It will always be a subset of $|z-z_0|\le\text{ROC}$
If we're talking about the convergence of the series
That doesn't mean the function cannot be analytically continued to a much larger set
8:55 PM
That's right, I should have rephrased that.
The example should make the situation clearer, in any case.
i recommend studying 1/(1 - z) instead. fewer signs :D
fewer sign changes
Classic pedant.
@BalarkaSen so I mean if I take $1/(1-z)$ then the expansion is $\sum_{n=0}^\infty z^n$ for $|z|<1$ where ROC=1. So we have uniformly convergence on $|z|<1$ and divergence on $|z|>1$ but also on the boundary we have divergence so all in all it diverge on $|z|\geq 1$ right?
Or we can study one of copper hat's favorites: $\sum\limits_{n=1}^\infty\frac1{n^s}$, especially at $s=-1$ ;-)
9:00 PM
@Wave Let us denote $f(z) = 1/(1 - z)$ and $g(z) = \sum_{n = 0}^\infty z^n$. You have proved that the domain of definition of $f, g$ both contain the open unit disk, and $f$ agrees with $g$ there. $g$ diverges everywhere on the boundary of the unit disk, also correct.
But, nonetheless, $f$ has a larger domain of definition than that of $g$. (What is a maximal domain in $\Bbb C$ on which $f$ is analytic?)
right, so f is analytic on $\Bbb{C}\setminus \{1\}$
Precisely. So just because the analytic function, when Taylor expanded at a point a, has some radius of convergence R, does not mean that the domain of the analytic function -- nor its derivative -- is necessarily equal to the disk {|z - a| < R}
It contains it, sure.
The moment you take a Taylor expansion, you should think of the Taylor series as a totally different function than the original analytic function. They agree on the disk of convergence.
but this only means that for all $z_0\in \Bbb{C}\setminus \{1\}$ we can find an $r(z_0)$ such that on $|z-z_0|<r$ we can write $f$ as \sum_{n=1}^\infty a_n (z-z_0)^n. But only on the unit disk we have such a nice function g(z) right?
wave: for a not equal to 1, the radius of convergence of the taylor series of 1/(1-z) about a will be |a - 1|
The Taylor series of $f$ at other points has decidedly uglier coefficients, but certainly not incomputable.
9:06 PM
yeah, you could totally verify this by hand although i'm not sure i would want to
Ah okey so all in all we can say that if a function is analytic at some point z_0 then we can express it on a disk |z-z_0|<ROC as a convergent series, this is independent of the point of z_0 (except when f is not analytic at z_0 which is clear). But this means that in each of this disks |z-z_0|<ROC one can compute the derivative ect.
and on some disks the function has a nice expression with a power series and on some disks it's really ugly but not impossible to compute one
It is usually shown using the Cauchy Integral Formula that the ROC is up to the closest singularity
sorry I don't get this one
$\frac1{1-z}$ has a singularity at $z=1$, is that okay?
9:21 PM
If you expand this function at $z_0=5$, the radius of convergence will be $4$
that is $\frac1{1-z}=\sum\limits_{n=0}^\infty a_n(z-5)^n$
that is because the distance from $5$ to the singularity at $1$ is $4$
right I see
If the ROC were bigger, the function would not blow up at $1$
so that is the largest the ROC could be.
sure. What i don't see is what you mean by the largest. As i thought about it, the ROC is the largest of all radius with this property
It would be the smallest of the distances to a singularity, if there are more than one
One of my favorite homework exercises to assign was to prove that there must be at least one point on the circle of convergence across which the function cannot be analytically continued.
9:26 PM
@robjohn ah okey I see thank you
Beginners confuse convergence of the series with analyticity at the point, too.
@TedShifrin so I mean if it is convergent on |z-z_0|<r for some r then it is also analytic on the same disk. But analyticity on a whole open set \Omega does not mean that it converges on the whole set uniformly right?
That what converges?
You've already seen the easiest example. $f(z)=1/(1-z)$ is analytic everywhere except at $z=1$, but the power series expansion at $0$ converges only on $|z|<1$.
sorry the serie f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n
It will have different power series expansions centered at different points in the domain.
9:33 PM
so then we do not have uniform convergence on the whole domain where it is analytic
You're still not being specific enough. Read what I just finished typing.
what do you want me to specify?
Read the sentence you typed just before this. Uniform convergence of WHAT? Where WHAT is analytic?
So if I take an analytic function f. Assume f is analytic on $\Bbb{C}$ except some points \{z_i\}$ then for all $z_0\in \Bbb{C}\setminus \{z_i\}$ we have a power series expansion as I wrote above. But this means that $f$ converges only on this disk where we have the power series expansion and not out of the disk. So for example taking our example from above then at $z_0=0$ f only converges on the unit this and on $z_0=5$ it only converges on |z-5|<4$ but not necessiarily on the remaining part
Where f is analytic
$f$ doesn't converge. $f$ is a function. You're talking about a particular power series expansion which gives you $f$ on a certain disk. As I've said several times, there are zillions of different power series expansions converging on different disks.
9:43 PM
yes sorry the particular power series converge.
I see thank you!
As I said earlier, the power series might converge everywhere on the circle of convergence but the function might not be analytic at every one of those points. Consider $f(z)=\sum z^n/n^2$, for example. So there are lots of subtle questions.
I think my example would be Taylor series of $\sqrt{1 - z}$ at $z = 0$.
I'd have to work too hard to find the series :P
Easier to argue geometrically, but yeah
And then there are the classics which can't be analytically continued across the circle of convergence at any point.
Branch points are a whole other kettle of fish, a @Balarka.
9:49 PM
The unit disk at the origin intersects the standard branch cut at $z = 0$, where it is still single-valued, but not in any neighborhood of $z = 0$.
True, but it does the trick here, I think.
But how do you know a priori that the series converges everywhere on $|z|=1$?
when i first learned complex analysis i liked how it allows you to understand why the real taylor series for 1/(1+x^2) has a finite radius of convergence even though the function is plainly real analytic on the whole real line.
Yes. That issue had bothered me as a calculus student. Also, as I've brought up here, why the integrals of $\dfrac1{1-x^2}$ and $\dfrac1{1+x^2}$ look so different, when they shouldn't ... Just throw in an $i$ here or there.
The coefficients of that series are quite bizarre, @Balarka. Are you sure you meant that as an example of the phenomenon I had just typed?
If $(M,d)$ is Riemannian connected manifold with induced metric $d$ and $f:(M,d)\rightarrow Y$ is locally K lipschitz where $Y$ is some metric space, why does it follows that $f$ is globally K Lipschitz?
I don't think it does.
Take out all the fancy stuff and just ask about a function on $\Bbb R$ or $\Bbb R^2$. Are you sure there's not a compactness assumption?
9:55 PM
The lipschitz constant is uniform @TedShifrin
Well, try to write a proof on $\Bbb R$, then.
Looks like you'll want to use a lub argument and you can do the same thing with the radius of a geodesic ball in the manifold case, I guess.
@TedShifrin Right, I am not sure anymore.
lub argument?
least upper bound
I was told that the key idea is that for any two points $x,y$ in $M$, $d(x,y)$ is the infimum of the lengths of the rectifiable curves from $x$ to $y$. and this can be generalised to a metric sense
on spaces called length spaces
9:59 PM
What does that have to do with $f$?
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