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00:00 - 15:0015:00 - 00:00

00:09
@AkivaWeinberger DogAteMy — I don't believe this is right.
You're not considering the power series as a power series any more.
Heya @Semiclassic
Yeah, but to see if it's constant or not you have to evaluate it
The question is about the function you get from it
Also, hi
But power series are different from convergent power series.
Only in the latter case do you have a function with a domain.
And, hi :P
00:12
But it's always convergent at at least one point (namely, $0$)
Yes, but a power series in general does not define a function.
It defines a function if you set its domain to be where it converges.
This distinction is even important for polynomials. Working over $\Bbb Z_p$, for example, the polynomial $f(x)=x^p-x$ is definitely not the zero polynomial, and yet all its values are just $0$.
Think about what I just said, rather than tangling up with power series.
Ah, right…
so would you not call that "constant"?
Does that polynomial look like a constant to you?
00:14
Right. So, no.
Polynomials are very different from functions once you get out of calculus class.
But then the question is trivial.
It would be by definition.
What would? I'm losted.
Oh, that the zero power series must have only zero coefficients. Yes.
i'm not sure what the question is anymore
LeakyNun's question.
00:15
LOL @Semiclassic ... I'm not sure I ever knew :D
1 hour ago, by Leaky Nun
@Semiclassical Have you ever seen any constant non-trivial power series?
That was what I was answering. ^
The identity principle says that an analytic function cannot have anything but a discrete set of zeroes unless it is identically zero.
Depends on what one means by constant. If it's simply 'constant everywhere' then no.
So we're back to what we mean by "values" of a power series.
I guess the question is ill-defined, DogAteMy.
Hi chat
00:18
I think you can find examples of Laurent series converging to $1$ inside the unit circle and to $0$ outside, but 1) I'm not sure I'm remembering right, 2) it's constant inside and outside of the unit circle, but different constants
Hi Kareem
You can't have one Laurent series that converges both inside and outside, @Semiclassic.
I suppose that, when I'm working in the ring of formal power series, I should write $e\exp(\exp(x)-1)$ rather than $\exp(\exp(x))$, where by $\exp$ I mean the power series corresponding to $e^x$.
hmm. then what am I remembering.
i know i'm remembering something, though. and even better i do know where it's from.
Or just $e^{e^x}$
You can't define a formal power series for $e^{e^x}$ because you end up adding up infinitely many $1$'s.
@Semiclassic: Yes, you are remembering something.
00:20
(I might write $ee^{e^x-1}$, if only to emphasize that all the coefficients are rational multiples of $e$.)
You're remembering defining a function by an integral around the unit circle.
eh, it wasn't that (or at least not explicitly so)
Well, that's what I'm remembering. :P
$e^{e^x-1}$ is… something to do with Bell numbers? I think?
psh
think that's their EGF? not sure, though
00:21
What is EGF? That sounds like a valve in a car engine.
exponential generating function
Hi @TedShifrin
what are you doing akiva ?
complex analysis ?
okay, i found what it was I was thinking of
DogAteMy dabbles in whatever subject the conversation steers him to.
00:24
^
Is that a bad thing?
consider $f_n(z)=\dfrac{-iz^{-1}}{1-z^{-n}}$ with $n\in\mathbb{N}$.
I'm reading something on category theory, actually
I have no complaints. Sometimes you're right; sometimes you're wrong. Sometimes we get confused. :)
@Semiclassic: That converges only on $|z|>1$.
oh
Well, hold on. You're not summing. So what are we doing?
00:27
well, take the domain to be an annulus which includes the unit circle
wait, scratch that. let me make sure i'm thinking of this right.
You just taking the limit?
if $|z|<1$, then $f_n(z)\sim -i z^{n-1}\to 0$ as $n\to\infty$. by contrast, for $|z|>1$ we have $f_n(z)\sim -i z^{-1}$
What does this have to do with a single function/Laurent series?
well, for one, $f_n(z)$ is just a rational function with poles on the unit circle. so if one takes the domain to be, say, the interior of the unit disk, then as $n\to\infty$ one has a rational approximation to $f(z)=0$.
What I was referring to earlier was the function $$f(z) = \frac1{2\pi i} \int_{|\zeta|=1} \frac{d\zeta}{\zeta - z}.$$
Still, @Semiclassic, I claim this has nothing to do with our earlier discussion :P
00:33
whereas if one looks at the exterior, it gives a rational approximation to $-i z^{-1}$
Morning.
G'night, @MikeM.
Missed you earlier; I was in office hours.
think you forgot an $f(\zeta)$, @ted :p
Meanwhile, on FB, one of my old friends and math geeks has discovered the definition "A set M in a metric space R is said to be compact if every sequence of elements in M contains a subsequence which converges to some x in R" in Kolmogorov/Fomin. He's apoplectic (understandably).
00:35
oh, wait. i'm misremembering what you said.
sigh
No, @Semiclassic, I'm defining $f$.
Try $|z|<1$ and $|z|>1$.
yeah, i agree now.
Yippee. :)
winding number etc.
For example. Or any number of other ways.
00:36
this is related to that, though. not that i seem able to speak well today :/
@Semiclassical LOL.
We still don't have a Laurent series converging on $|z|\ne 1$.
eh, no
on that i was misremembering
what you do have, at least, is this
suppose you've got $|z|<1$. then you can expand $f_n(z)$ in a Laurent series as $$f_n(z)=i z^{n-1}+i z^{2n-1}+\cdots$$
Sure, you're going to get two different Laurent expansions on the two different domains. Nothing unusual there.
yeah, that's right.
that's what i was misremembering.
and outside it behaves instead like $f_n(z)=-iz^{-1}-i z^{-n-1}\cdots$. nothing too strange.
I guess what's interesting (I don't know why the $z^{-1}$ has to be there in the numerator) is that $f_n$ has no pointwise limit as $n\to\infty$ when $|z|<1$.
No, that's wrong.
There's nothing interesting :P
00:43
pffff
where this emerges is in the context of approximations of contour integrals
lemme find a nice example (there's a canonical one)
Presumably relates to Runge's Theorem somehow ...
okay, this isn't a canonical example, but this is the one the text i'm referencing cites
@MikeM: Anything good come up in office hours?
suppose you've got a function with a convergent Taylor series inside the unit circle $u(z)=\sum_{j=0}^\infty c_j z^n$
suppose you wanted to compute one of those $c_j$ numerically rather than analytically. one way to approach that is to use contour integration i.e. $\displaystyle c_j = \frac{1}{2\pi i}\oint_C z^{-j-1}u(z)\,dz$
@TedShifrin Well, it's true, for metric spaces.
00:52
Nope, DogAteMy. You fail reading just as I do. Look carefully.
(Right?)
…oh. Oh.
in R.
Their next sentence is that any bounded set in a metric space is compact!! :P
Discrete space with infinitely many points, anyone?
to make things simple in the bit ahead, i'll assume that the radius of convergence is strictly bigger than $1$ so that I can take $C$ to wind just around the unit circle.
I pointed out to my FB friend that most texts (particularly in Europe) would call this precompact.
00:54
I thought you were complaining because it doesn't generalize to Top well.
No, no, I was complaining that it's not correct by our definition of compact ! :P
now, the numerical issue is to approximate that integral. to that end, suppose we replace the factor of $-i z^{-1}$ with the $f_n(z)$ from earlier
Is there a name for sets with compact closure?
yes, precompact :P
00:56
and to that end define $$c_j^{[n]}=\frac{1}{2\pi i}\oint_C f_n(z)u(z)\,dz=\frac{1}{2\pi i}\oint_C \frac{u(z)}{1-z^{-n}}\,dz$$
I have an article saved to my phone that shows how "compact" meant different things throughout its history
BTW, @Semiclassic, this reminds me of something that came up when I taught graduate complex the last time. If $f_n\to f$ pointwise, can you say anything about the convergence of the Taylor coefficients (or vice versa)?
(No issue with uniform convergence.)
Yeah, and most of them are wrong.
this sounds like a Fourier series question in disguise. (which is appropriate, since Fourier series in $e^{i\theta}$ are just Laurent series in $z$)
00:58
yeah, a lot of people assume Hausdorff with compact ... not just Bourbaki.
i haven't done formal complex analysis recently enough to trust my thinking on that question, @ted.
It turns out to be very interesting, @Semiclassic. One of my students made a claim in homework, and I struggled to find a counterexample, but it was a false claim.
which is a fancy way of saying 'i dunno
i think the $f_n(z)$ i gave earlier would be one where it converges pointwise everywhere except the unit circle?
Yes, I think that's clear.
(Despite my thinking to the contrary earlier.)
But that doesn't tell us anything about the $c_n$'s.
right
i know that the analyticity structure of $u(z)$ does translate into claims about how the $c_n$'s behave
for example, if you know that $u(z)$ is analytic in the region $r<|z|<r^{-1}$, then the $c_n$'s have to decay like $r^{n}$.
01:05
I meant, $c_k^{(n)}$ for the $n$th function ... as $n\to\infty$ ... ?
Yeah, totally not what I'm talking about.
Sowwee.
the reason that i bring that up, btw: if $u(z)$ has radius of convergence $R>1$, then the error of those estimates $c_j^{[n]}$ from the true $c_j$ is on the order of $R^{-n}$
and the reason that's convenient is because each of those estimates can be computed by applying the residue theorem to the $n$ poles on the unit circle of $f_n(z)$.
and that turns out to just be a trapezoidal approximation :)
So you're saying you can prove convergence of the coefficients in this case? Is it clear whether the $f_n$ converge uniformly or not?
well, I am assuming something pretty strong about $u(z)$, namely that it's analytic within $|z|<R$.
01:09
Well, I was assuming $f_n\to f$ pointwise, everybody analytic.
might as well link the paper I have in mind: people.maths.ox.ac.uk/trefethen/sirev56-3_385.pdf
it's a review paper, though, so not necessarily that convenient
It's OK. I'm not gonna read it.
lolkay.
here's theorem 2.1, though, which is indicative of what I've got in mind
@Ted Nah, first week of complex analysis.
Oh, it's summer now, @MikeM. I didn't know.
01:13
Let $u(z)$ be some function defined on the unit circle $|z|=1$. Let $I=\frac{1}{2\pi i}\int_{|z|=1} u(z)\frac{dz}{z}$ and $I_n=\frac{1}{n}\sum_{j=1}^n u(e^{2\pi i j/n})$.
Then Theorem 2.1 is: Suppose $u$ is analytic and satisfies $|u(z)|\leq M$ in the disk $|z|<r$ for some $r>1$. Then for any $n\geq 1$, $|I_n - I |\leq \frac{M}{r^n-1}$ and the constant $1$ is as small as possible.
Assuming $u$ is analytic on a larger disk is a strong assumption.
yeah
they've also got a version where it's only required to be analytic on some annulus, and the inequality is similar.
Presumably, analytic on $|z|<1$ and continuous up to the boundary won't get much at all. I've forgotten.
well, that's not entirely right. the choice of $|z|=1$ is a bit arbitrary
01:19
wait, no, i'm mischaracterizing things
that'd work if could take the contour of integration itself to be smaller than $|z|=1$.
but i've already said it is :/
Right.
i mean, that does matter in applications because typically one isn't wedded to $|z|=1$ owing to Cauchy's theorem
Of course $1$ can be anything, but for all our discussions, $1$ is fixed.
So my comment stands.
01:21
key point is that you need to know something about $u(z)$ in the neighborhood of $|z|=1$
and being analytic on $|z|<1$ doesn't give you that.
Yes, but that + continuity up to the boundary gives you Hardy spaces.
(if I remember correctly).
Sounds right.
I won't vouch for that.
And that is related to stuff i was interested in last year, so in principle I should know that stuff. (I don't.)
oh oh, DogAteMy is back.
01:22
The reason it was relevant for me is because a lot of the stuff I was doing involved elliptic integrals
yes, I so recall.
and for that, you're usually happy to pick an annular domain that excludes the branch cuts
so those nice estimates are actually useful.
how does an annulus exclude branch cuts?
i have in mind the scenario where you've got two branch points inside the unit circle and two outside
in that case you can draw one branch cut entirely inside the unit circle and the other entirely outside, so one can pick an annulus that doesn't touch either cut
Whoa. So far the EU referendum votes are verrrrry close.
Oh, OK, @Semiclassic.
Gotcha.
01:26
it's nice stuff, though hardly earthshaking
I don't need more earthquakes for a while.
oof, that is close.
When's the last time 68.5% of eligible Americans voted?
American Idol?
01:28
that's not true either, of course.
Google shows it as 1.29 million to Remain versus 1.44 million to Leave, but it also says only 10.7% reporting...
so i dunno :/
Google is way behind. See this.
I worry that too many important decisions are made by uninformed people.
@JMoravitz: Indeed. You may end up with Trump that way.
We may end up with Trump that way.
Uninformed and ridiculously bigoted and hateful.
01:32
You can be plural @semi
@Ted Yup, we just started.
How far did the geometry course get, @MikeM?
@JMoravitz: Had I not left the South, I would have said "y'all" :)
It may be plural, but it's usually not first-person :p
First??
glares at Semiclassic
You'll end up with him too, I'm afraid to say
01:33
Oh, I see ... I can't exempt myself. Yes.
Can we just have a president whom I don't have to worry about blowing up the world?
2
Maybe my cancer will come back and I'll die quickly, @Semiclassic.
Trump versus cancer.
He would be a global cancer :(
Yeah, Trump being at the top of a chain of command that has a thermonuclear stockpile...
that's what scares the crap out of me
01:36
Everyone is scary, honestly.
elaborate?
Hillary is a hawk war-monger.
ah, that.
yes.
yup, that.
Hence my desire for a president who won't blow up the world.
01:37
Good luck with that.
Well, when it happens I'll bring the nachos
Bring me something heart-healthy, @JMoravitz.
You might be surprised @Ted. My mother would be proud. My roommates and I haven't eaten out for dinner for the past two or three months.
01:40
Cool, @JMoravitz. So I don't even need to offer you cooking lessons? :)
Always cooking at home, trying different recipes. I made a rather tasty Kung Pao the other day which I'm having leftovers of now
Nice :)
Granted, one of them doesn't know much about cooking, so his turns usually consist of grilling sausages or quesadillas
@Ted the last quarter was algebraic topology. We got as far as defining cohomology rings - no Ppincare dualiry!
Not so impressive. Can he tear up lettuce and put on salad dressing?
01:42
I would have liked to tell them about non-oriented PD.
What good is topology w/out Poincaré duality?
For being more than 50 years old, that song feels surprising topical.
(it was either that or "End of the World as We Know It")
Who knows.
see everyone later ... dinnertime here.
 
2 hours later…
03:38
@TedShifrin exp(exp(x)) is holomorphic on all of C so has globally convergent Taylor series
you'd add a buncha 1s if you did 1/(1-exp(x)), but with exp(exp(x)) the coefficients of the outside exp() weigh things down when expanding
03:57
@arctic: I didn't stop to think, I guess. But in general you cannot compose formal power series unless the inner one has 0 constant term.
fun fact: in the p-adic world it can happen that $f\circ g$ evaluated at $x$ does not equal $f$ evaluated at $g(x)$
[where composition is done in formal power series ring]
that sucks
 
2 hours later…
06:29
…wow
So Brexit won
They're leaving
06:54
Yeah I'm surprised.
07:08
if it's what the Almighty Milo Yiannopolous had wanted then it's good that it happened
Britain must build a wall too now, and only admit legal immigrants, enough with the refugees
All these centuries Europe was epicenter of most ideas of the mathematics we use today, now it's in ruins thanks to the Middle East
Huy
Huy
maths is overrated
even if you consider so (although it certainly is far from the truth), 97% of today's science and technology was built by the Western civilization
which is collapsing now thanks to Angela Merkel
Huy
Huy
that's certainly not some random number you just made up
and her leftist likes
well let's be real-give me a single example of a major scientific/technological breakthrough before the 1980's that didn't happen in the West
Huy
Huy
07:23
the clock
alcohol
gunpowder
paper, printing etc.
inoculation
Chinese authors had been familiar with the idea of negative numbers by the Han Dynasty (2nd century AD), as seen in the The Nine Chapters on the Mathematical Art,[27] much earlier than the fifteenth century when they became well established in Europe.
just a note ... please keep discussion civil and "adult" :)
well I wouldn't consider them as 'major' in the sense that they weren't nearly as breakthrough advancements as the invention of computers
Huy
Huy
lol
for example that is
Huy
Huy
that's quite a narrow point of view
but I guess everyone's entitled to their opinion
07:35
@SoumyoB do I actually have to say that computers aren't nearly as new as you think and what you probably think of as computers is just the electrotechnically advanced model of Turing Machines and basic boolean algebra?
07:47
I didn't think of them as new... My point was that all of this technology was from the west
well I should quit this topic now, my views are extremely towards the Right and I could get banned quite easily from here
Huy
Huy
lol
I think you're funny
@SoumyoB which is relevant to Angela Merkel and the "runing of maths by the Middle East" how again?
@Vogel612 that's not what I meant, but I request you to please leave discussing this topic with me for now, I really don't want to get banned
you know as long as you we keep it civil I don't see a problem with discussing this. but okay
08:03
Rightist ideas are hardly seen as civil, I've been kind of ostracized from my social circle since most of my friends identify themselves as 'progressive' leftist liberals
the problem is not the ideas, but the way they are proposed
also the seeming lack of coherence
eh well, I'll be off :)
08:34
@MikeMiller dou you have any suggestion for this doubt mathoverflow.net/questions/242970/…
 
1 hour later…
09:39
If I have a probability question that my c++ program that simulates the system 100000 times to give an approximation of the probability(I'm using the law of large numbers) differs substantially from my mathematical results, do I post the mathematical part here and the program over at stackoverflow?
by "here" I mean create an MSE question.
10:00
Old times are gone here.
The old say that the times are getting worse, the truth is that the old are getting worse.
2
10:59
This weekend I am going to see Andrew Granville - The pretentious approach to analytic number theory, that provide us the Institut des Hautes Études Scientifiques from its official channel in YouTube. I will understand only a few facts , but will be about 200 minutes eating popcorns. I say it if some user want to see these videos (1/3), (2/3) and (3/3). Good weekend.
11:25
@Huy "if you don't agree with my opinion i will punch you in the face"
@BalarkaSen Related; check out this bounty: physics.stackexchange.com/questions/46237/…
haha
lolz y u sighing?
user116211
@Danu something is for sure; he is not having his sense.
11:30
becausing i'm tired of loling at these kind of behavior
Huy
Huy
@BalarkaSen you need more sleep
@MatsGranvik "your old roads are rapidly fadin' / please get out of the new ones if you can't lend your hand / for the times they are a-changin' "
not that i completely believe in that philosophy.
I believe in Bob
lol I remember Lennon's response to that
@Huy i need to bunk school for a week
Huy
Huy
I need to get a response regarding the flat I applied for
I hate waiting
11:42
@BalarkaSen I don't believe in most things I say in chat. I merely say them.
12:17
Hi @JuanFran
 
1 hour later…
13:30
While (d/dx) denotes derivative with respect to x, is there such a prefix for integrals?
I mostly see integrals defined as: $\int{f(x)\cdot dx}$ or $\int{dx\cdot f(x)}$. But is $dx \cdot f(x)$ fine too?
Don't think I've ever seen one
Guess not.
Hm.
No biggie, just wondering.
Without the internet symbol, f(x)dx is a differential form
Integral
Kindle cannot into math
Given that $\frac{d^n}{dx^n}$ is the $n$-th derivative, perhaps $\frac{d^{-1}}{dx^{-1}}$ could signify integration? Not that I'm proposing to actually use this, but I find the idea somewhat funny
In the context of fractional d derivatives maybe
13:36
On the other side, is $f(x)\ \frac{d}{dx}$ accepted?
For derivatives?
@SteamyRoot The "inverse" of the operator $d/dx$ is indeed integration (once you fix some integration constant), so that is a probably valid - although nonstandard - notation.
Vs the d/dx in front
@Owatch Eh.
@Owatch It is really $\frac{dy}{dx}$ which is similar to $\frac{\Delta y}{\Delta x}$
Hi @anon
13:37
Then $\frac{df(x)}{dx}$ becomes $\frac{d}{dx}f(x)$
@Owatch No that would be a differential operator which differentiates first and multiplies by f(x) second
I think some physicists use it tho.
The context I work in doesn't have functions defined by variables.
So I use d/dx (expression)
Hello. Can anyone take a look at my proof, is this valid? math.stackexchange.com/questions/1838152/…
@anon Okay, thank you.
13:41
@IlanAizelmanWS that A is just lambda times the identity no?
hi @BalarkaSen
13:53
morning chat
@Semiclassical hi expert
hah
@SteamyRoot I don't think I've ever seen that usage, but I have seen people write $D=\dfrac{d}{dx}$ and then take $D^{-1}$ as notation for antidifferentiation
I see.
and even more common than that is stuff like $(D-a)y=f\implies y=(D-a)^{-1} f$
that's particularly valuable if you think in terms of Fourier transform pairs
Either way, I do think such notation might be dangerous if people carelessly assume that $D^{-1}D = Id$
13:58
$$(D-a)y(x)=f(x)\implies (ik-a)Y(k)=F(k)\implies Y(k)=\frac{1}{ik-a}F(k)$$
so then $y(x)$ will be a certain convolution product
@Anubhav 1) A 3-manifold that has a Morse function with (1,g,g,1) critical points is the same thing as a 3-manifold with a Heegaard splitting of genus $g$. 2) Yes. Try to do it for surfaces first to see the idea.
14:24
Hi, apologies in advance if I violate protocol; new here.
I am creating a "puzzle" that I don't want to be googlable. If I post it here, will it show up in google?
My issue is that it's a hard enough problem (for me) that I am not confident enough in my solution to publish it.
If you post it as a question on the site, it'll definitely show up in google
This chat room is indexed by Google, yes.
Thanks. Sigh!
@Semiclassical a nice assignment for you
$$PV \int_0^{\infty} \frac{\cos(ax)\cos(bx)}{p^2-x^2}\textrm{dx}$$
14:32
@HectorPefo You can invite people to private chats over irc if you're worried about indexing
hmm hmm hmm
morning
Hello everyone
@Semiclassical I guessed that might be an integral you wanna calculate. Not sure 100% (I have to check that) but I think that the integral can be reduced to one that can also be found in Paul Nahin's book.
@Semiclassical My teacher gave me Griffith's book on quantum mechanics. This stuff seems to me to be mostly mathematics, but rigorous in the sense of a physicist :P
Weird to see the insight/intuition getting pushed below the technicalities in physics books too.
14:44
@Semiclassical Yes, I'm right. I think my memory is still in a good shape, but what I find fantastic is that in the first second I saw it I established the connection with that integral, without thinking over it, all happening instantaneously.
@BalarkaSen have you read Griffith's books on Particle Physics and Electrodynamics?
Nah, I don't read physics.
Shame, he's a brilliant author, and definitely knows his mathematics too.
I can believe that.
As opposed to the professor who taught Quantum Physics at my university. He introduced a "Hilbert Space" as "A vector space such that the sum of two vectors lies in the same vector space again."
14:49
Wolfram has failed me here. While a circle can't have imaginary radius $i$, how does one plot $x^2 + y^2 = -1$?
@SteamyRoot ew
It probably tries to plot in $\mathbb{C}^2$, which is four-dimensional
so it can't plot it at all
14:52
@Axoren That has no real solutions. in C^2 there are of course solutions.
@SteamyRoot @BalarkaSen No doubt the solutions exist in $C^2$, but I settle in my own head if it still "looks" like a circle in that space.
Well, there is a "circle" in the complex space that is a solution
well, take $(X,Y)=(ix,iy)$
then you've just got $X^2+Y^2=1$.
if $x = a+bi$ and $y = c+di$, then a circle in the $db$-plane works.
@Axoren It's a hypersurface inside C^2. The graph is of real dimension 2: a circle doesn't make sense.
Sections of it may look like circles.
14:54
but there are more solutions, of course
so if you're working in $\mathbb{C}^4$, then it's just the same hypersurface as the usual 'circle' would be
It would be easier for me to visualize it if I had a concept of what imaginary distance worked like. But all the foundations for distances are in real units.
I can't fathom that hypersurface.
You want to imagine 4 dimensions simultaneously? Good luck.
Try projection somehow?
@anon What anon?
@anon It's anti hermitian.
14:58
@BalarkaSen The way I see about working with solids in higher dimensions is that if I have a metric in that space, I can deal.
Circles and Cubes are just $l_2$ and $l_\infty$, regardless of dimension.
I don't see how a metric (which C^2 of course has) would help visualize things in higher dimensions in any way. But sure
00:00 - 15:0015:00 - 00:00

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