Mathematics - 2017-07-19 (page 2 of 3)

lyme

16:23

hello . how can I show if $||x_k-x_l||\leq 1/k+1/l$ then x_n converges?

Astyx

What space does $(x_n)$ live in ?

lyme

R_n

Astyx

Do you know about Cauchy sequences ?

lyme

R^n

I forgot after graduated :( i tried to show like in this by using triangle inequallity but came up with harmonic series which diverges math.stackexchange.com/questions/490107/…

Astyx

Well first you can show it's bounded since $||x_k-x_1|| \le {1\over k} + 1 \le 2$

Therefore by Bolzano-Weierstrass, you get that its set of limit points is nonempty

You then simply need to show it's a singleton

Proceed by absurd : suppose it has two limit points $a\lt b$

Take, $\epsilon = {||b-a||\over 2}$

And $N\in\Bbb N$ such that ${1\over N}\le \epsilon$

lyme

16:31

thank you very much. Ill give it a try

Astyx

You'll take it from here ?

lyme

yes

Astyx

Right, have fun :)

Hi Akiva

Akiva Weinberger

Hi

Semiclassical

So, weird math of the day. For reasons, I'm interested in the following integral: $Z(\mu,g)=\int_{-\infty}^\infty e^{-\mu x^2/2-g x^4/4!}\,dx$

This can be written in closed-form using either a modified Bessel function (and some ornamentation thereof) or a confluent hypergeometric function.

There's at least one funny thing about it. If $\mu>0$, then $Z(\mu,g)\to 1$ as $g\to 0^+$.

But if $\mu=0$, then it instead goes to $+\infty$.

Astyx

16:36

Wut ?

Why does it go to 1 ?

Semiclassical

oh, wait. I'm being silly.

first off, forgot to include a factor of $1/\sqrt{2\pi}$ overall.

Astyx

And $\mu$ also plays a role

Semiclassical

Second off, I misread the paper I'm looking at and forgot that it's $Z(1,g)\to 1$ as $g\to 0^+$.

Astyx

Since you can dominate the integrand you have that the limit of the integrals is the integrals of the limit

Semiclassical

Right.

Astyx

16:38

Right that seems more like it

Semiclassical

Not sure why I thought it was weirder than that.

Nah.

Astyx

It shouldn't :p

Semiclassical

lol

Anyways. In the limit $g\to 0^+$, one has $Z(\mu,g)\to 1/\sqrt{\mu}$ when $\mu>0$.

So that's consistent.

Astyx

Hi Ted

Ted Shifrin

hi @Semiclassic and @Astyx

Semiclassical

16:40

hi @ted

On the other hand, Mathematica fails to get a limit at all when I have it assume $\mu<0$.

Astyx

Weird

You mean limit as $g\to 0^+$ ?

Semiclassical

Right.

Astyx

That should be $+\infty$

Alessandro Codenotti

Hi @Ted

Why amoeba?

Astyx

By restraining to a (sufficiently large) compact

Semiclassical

16:42

Yeah, that's what it gives.

The $\mu<0$ case is where things are weird.

Astyx

Or even by comparing it to the case $\mu = 0$

Ted Shifrin

hi @Alessandro ... Because that's what your shape looks like :)

Astyx

I apparently did better on my analysis written exam than on my algebra one for Polytechnique

I am confused

Semiclassical

The weirdness of it, I suppose, comes from the fact that when $g>0$ the integral converges regardless of $\mu$.

Alessandro Codenotti

Oh, it's supposed to be seven people looking into a well, seen from the bottom

Ted Shifrin

16:44

So what does that all mean, @Astyx?

Semiclassical

But when $g=0$, the integral is only finite when $\mu>0$.

Ted Shifrin

LOL, OK, Alessandro.

Astyx

It's all meaningless anyway

Alessandro Codenotti

I might change it sooner or later though

Ted Shifrin

Is it?

Astyx

16:45

Isn't it ?

Socratic me isn't the best me

(luckily)

Ted Shifrin

Yes, luckily for us all.

Astyx

@Semiclassical Yeah, you could add some degree $6$ term in there and make things even more weird

Semiclassical

Yeah, and the paper I'm looking at even suggests doing that :P

Astyx

I guess if you want information on degree $4$ you could look at degree $6$ to crush the divergence ?

Not too sure where that would lead you though

Semiclassical

$\mu<0$ vs $\mu>0$ is intentionally a bit weird: If $g>0$, the integrand has a single global maximum at $x=0$ when $\mu>0$ but has two symmetric global maxima when $\mu<0$.

Astyx

16:49

I gotta go now

Semiclassical

It's therefore valid to do a saddle-point approximation using the single maximum at $x=0$ so long as $\mu>0$.

Later.

Astyx

I'd be interested in anything interstig you find, keep me updated !

Semiclassical

But if you trust that approximation for $\mu<0$, you'll run into a nasty surprise :)

Astyx

Seeya

Semiclassical

byyye

Adeek

16:58

hi @TedShifrin

GFauxPas

oh no, Ted is here

Semiclassical

oh noooes

Adeek

I got really sick past 2 days. Maybe I overworked myself. I was studying for 13 hours for 2 weeks straight. I have been having migraine for past 2 days :S

GFauxPas

it's tough to find a balance of work/play

Adeek

yeah

GFauxPas

17:00

but stress headaches indicate you're not at equilibrium

Adeek

Yeah I will try find a better equilibrium

Annelise Toft

17:14

studying for 13 hours?? what?

Semiclassical

The latest PHD Comics is pretty good: phdcomics.com/comics/archive.php?comicid=1963

Secret

so tempted to make a closed curve of this

Akiva Weinberger

A what?

Secret

but then it is impossible to turn a tenured professor into an grad student

Well... one unlikely way that will happen is the foundation of a research field need to be revised from the ground up, sending all experts back to the drawing board I guess...

Semiclassical

maybe a dashed arrow from tenured prof to grad student signifying "convinces undergrads to go to grad school"

Secret

17:19

hmm, that might work

Secret

Meanwhile, a new [random] sounds quite physical:

Consider the scenario where you are at a platform and the train is preparing to leave

Suppose the train accelerates at 1 unit s-2 (because maths people don't worry too much about units when going abstract)

Semiclassical

Is this the whole "how fast do you have to run to catch the train" business?

GFauxPas

nice, I landed an internship with my gf's father with an operations research thing

Semiclassical

oh, cool.

GFauxPas

optimizing network connections

Semiclassical

17:23

paid or unpaid?

Secret

nope. it's more subtle than that, it's about the geometry of the space abstracted form this scenario, we'll get to that

GFauxPas

paid :)

Semiclassical

noice.

GFauxPas

find the best way to connect computers to all other computers

sounds like graph theory stuff

spanning trees

Secret

Now you are prepared to hop onto the train. Assuming you only move horizontally from platform to train with no vertical motion, you can choose to hop at any time t onto the train

Semiclassical

17:24

I imagine you also want to do so in a way that's somewhat flexible i.e. you don't have to change the network a lot if you just add/remove one computer.

a stable solution, so to speak.

Secret

Now the modelling part is this. Devise a time dependent eucledian metric tha describes the increase in distance between train and platform at different time t

Another way to put this question is as follows: Suppose you are on the train and the train is preparing to leave. If you hop off the train at t=0, then the horizontal distance between the point you want to land on the platform and the starting point should be effectively zero (or some small number)

Semiclassical

Do you mean "between a given point on the train and the platform"?

Because if you just want to reach the train it'll always be the same distance (supposing you don't miss the train).

Secret

Yeah, a fixed point on the train (where you jump off) and the point that you expect to land when the train just prepared to left the station at t=0. Inspired by missing a station, it makes me curious on what the effective geometry of the space look like the later you hop off the train

GFauxPas

@Semiclassical good point

Secret

Observing the fact that the vertical distance increases in an accelerated manner as the train left the platform, I am trying to convert this scenario into a metric so that the vertical distance with time becomes a geometric structure like a grid

Semiclassical

17:32

@GFauxPas It's a trade-off, of course. If you pick a network configuration which is truly optimal for the given set of computers, it may require a lot of 'fine-tuning'. That'll be a problem if you add/remove a computer, since that fine-tuning will be different for each configuration. But one has to be willing to make some adaptations if the circumstances change.

GFauxPas

right

Semiclassical

@Secret Could also work in the train's reference frame.

(one could even ask what this scenario would look like if you assume Lorentz invariance rather than Galilean invariance.)

Secret

I am guessing any coordinate grid of spacetime in the lorentz case will be contracting along the direction of motion

Semiclassical

Another variation: Suppose that the train is passing by a wall and you throw a ball out the window.

If you throw it horizontally, I'd think it'd bounce back to you (assuming that the collision is elastic so that none of the kinetic energy is lost in the bounce)

Secret

The lorentz case I suspect will be very complicated, the Galilean case will have the ball flying horizontally in the train's frame but forming a v shape trajectory in the platform's frame (not sure how the grid of spacetime will look like for this scenario, though, which is one of the thing I am interested in)

The ultimate reason why I want to get a metric to describe the whole scenario in a "timeless" manner is because any metric will induce a topology, and inspired from me forget to alight at a station and observing the increase in distance from my destination as the train left with me still on the train, I am interested in how the topology between the pair of blue points changes with time as the train leave the station

It will be interesting if I can represent the motion from all reference frames of these two relatively moving objects as a "static" geometric object, so that the metric give me information on how distance between any pairs of points of interest will vary with time

Semiclassical

17:44

Well, one should keep in mind that what will be the same in either the train or the platform's reference frame is the horizontal velocity of the jumper.

nor will the horizontal distance to reach the platform change. so the time $T$ spent in air isn't a function of the train's velocity.

But in the platform frame, the person is moving forward with the same velocity $u$ of the train. Hence they'll travel forward a distance $uT$ during that jump.

If I take the speed of the jumper to be $v$ and the distance to the platform to be $l$, then $T=l/v$. So they'll land a distance $ul/v$ forward from where they jumped.

If you ask for the total distance they travel in the platform's frame of reference, then that'll be $\sqrt{l^2+(l u/v)^2}=(l/v)\sqrt{v^2+u^2}=T\sqrt{v^2+u^2}$, which makes sense as that square-root is the magnitude of their velocity.

Secret

That expression looks familar, it so reminds of the lorentian metric in special relativity

So I am guessing we already have done something like that when in SR courses the minkowski metric is first introduced, usually motivated by consideration of light travel

Secret

18:03

Let vectors be $(x,y)$ and the Galilean metric for this scenario has signature $(+,+)$. Under the standard basis, the metric has coordinates:
$$\begin{pmatrix}Tv & 0\\0 & Tu\end{pmatrix}$$

where $T=\frac{l}{v}$

and in the case of acceleration of the train, we have in general:

$$\begin{pmatrix}Tv & 0\\0 & T\dot{u}\end{pmatrix}$$

Nice. suddenly metric become a lot more visual. Looks like accepting that the more accurate analogy of the expansion of the universe as distance between things literally increases without moving away from each other does help to visualise motion in terms of an array of time varying distances

Secret

18:24

Now I am starting to wonder, whether a metric of this form represents some kind of generalised comoving frame so that everything is rendered stationary relative to some box we draw in the scene, and whether it can always be found for any arbitrary motion

Tobias Kildetoft

@Secret Are you using metric in some other meaning than the mathematical one? Because none of that looks like the metrics I know.

Secret

uh, semi and I have ran through the calculations above on determining how distances are measured between two points and from trigonometry we have found the expression of the distance and thus we canconstruct a metric on that by sqauring that expression and writing it in the standard basis as shown by the 2x2 matrix?

This guy should have d (x,y) > 0 d(x,y) =0 <=> v=0 and triangle inequality. But let me check that real quick...

Tobias Kildetoft

@Secret I don't see how it gives you a number rather than a vector

Secret

Well, metric tensors are symmetric bilinear maps, thus two vectors can be fed into it and it spits out a scalar, which is the squared distance in this case

Let the vector be a=(x,y)=b under the standard basis) and let the metric tensor $\begin{pmatrix}Tv & 0\\0 & Tu\end{pmatrix}$ be $g(-,-)$
1. d(x,y) > 0 and $d(x,y)=0 \implies a=0=b$ holds as long u,v > 0 since $g(a,b) = Tvx^2+Tuy^2$ means (x,y) has to be positive or zero
Triangle inequality holds trivially since the outcome is the eucledian distance

Semiclassical

18:44

I'm not paying too much attention right now, so don't expect me to weigh in :P

Secret

but there is indeed somehing very strange about this symmetric bilinear map: If u and v have values in other quadrants, it can potentially become indefinite

I suppose, however that it should be independent of basis transformation, though so it has the requirement of a metric tensor, which is a metric function

In conclusion, if u and v > 0 (which we can choose it to be so by picking a basis, which is the same as choosing a reference frame) then we have a metric tensor which give us a metric

pregunton

19:13

Hello. In a rooted tree, is there any name for the rooted tree formed by the descendants of a node?

Daminark

I think any subtree would have this property, no?

If you had more than one ancestor, it'd be disconnected, and thus a forest

pregunton

You are right, thank you.

Ajna Sarut

19:38

Hi, i need help about something;
i need to get an additional bank commission from my customers purchase but always im losing some money. if i sell $1000 item, i need to pay $10 to the bank(as a commission of credit card payment system. rate is x0.01 ) so if i add commission like (itemprice+itemprice*0.01), still bank gets more commission because last price is changin to (itemprice+itemprice*0.01)*0.01 and it continues forever. so are there any formulas to solve this issue easliy? I am very happy if someone can help. thanks.

GFauxPas

19:51

@AjnaSarut what is the issue youre trying to solve?

user243301

@reuns I like say to you a thing about a previous comment where you are saying that you've no access to a paper of Michael Rubinstein (A Simple Heuristic...). But notice that if you search the article in Google you find it in JSTOR. An account MyJSTOR is for free, you can read up to 3 articles every 2 weeks (if you need any detail about it tell me or contact with the site). Maybe Michael's paper isn't interesting for you question, but this service of MyJSTOR provide you a large database.

Is not required a response of this message, good day.

Ajna Sarut

:38883281 im selling some items. and im using a banks service to get money from customer in my website, as an example; im selling $1000 carpet, bank says i will get 0.01 of the price as a commission. so when i sell the carpet i get $990, and banks get $10. so i dont want to lose money. i need to get $1000. i tried to sell the carpet to $1010 but then bank gets $10.1. there are $100 and $100.000 products. so this lose is heavy with a $100.000 product. so i need a formula or an aproach to lose less money. but i cant figure it out.

GFauxPas

maybe you want to know, "how much should I charge, to keep 100 dollars?

don't use dollar signs here, it messes up the TeX

Ajna Sarut

okey, so how?

GFauxPas

20:18

I'm saying you need to decide and write exactly what your question is

Astyx

So you want ${99\over100}p= 1000$ ? (Where $p$ is the price)

Daminark

One of the golden quotes from atop: "We can do calculus on R^n, but we want more. Now, we should know why we want more, for funding reasons. For instance, some people care about something called physics"

5

Akiva Weinberger

20:49

@Secret I don't know the context but I like this picture. I wonder if there's a notion of continuously deforming a metric space (in such a way that its homeomorphism class never changes).

Probably a special type of homotopy, of maps $X\times X/{\sim_{\rm cyc}}\to\Bbb R_+$ ($\sim_{\rm cyc}$ quotienting out by permutation)

(and no, $\sim_{\rm cyc}$ isn't a real notation, but it should be)

(That's probably not the best choice. I dunno)

Bubaya

Dumb question, but its late: Is a free and graded module a free graded module?

(asked differently: given a free and graded module, can I always find a basis of homogeneous elements?)

Tobias Kildetoft

@Bubaya I don't think so, but my intuition is not very good here

@Bubaya Ahh, as usual the answer is already on the site math.stackexchange.com/questions/557402/…

Kirill

@TedShifrin Mr. Shifrin, I will write the solution tomorrow. Best regards, Kirill

Bubaya

@TobiasKildetoft: Ah, thank you a lot. The statement of the reply indeed astonishes me.

Akiva Weinberger

How would you write the set of unordered pairs of elements of $X$? I'm thinking $(X\times X)/S_2$

Or like $X^2/S_2$

I mean, I'm sure there's no standard, but if you had to choose a notation

Above I suggested $X\times X/{\sim_{\rm cyc}}$ but that doesn't generalize to higher powers

Oh, there's a similar idea for vector spaces, right?

Balarka Sen

21:04

I think the notation for that is that of the symmetric space. $S^2 X$, maybe?

Is that even called the symmetric space? I think it has a name\

Akiva Weinberger

For vector spaces, $S^2(V)$ is like $V\otimes V/(v\otimes w-w\otimes v)$

Balarka Sen

Ok, it's "symmetric product"

Akiva Weinberger

(I think)

Balarka Sen

$SP^n(X) = X \times \cdots \times X/S_n$ is the standard notation

@Akiva Yeah, the "symmetric algebra"

Akiva Weinberger

Hm, this gives the name $SP(X)$ to the "infinite symmetric product"

Balarka Sen

21:07

It's missing the index

I think the infinite symmetric product is the direct limit of all the SP^n(X)'s

Akiva Weinberger

It's kinda $X^\infty$ but all but finitely many entries have to be the basepoint, and you quotient by permutations

Balarka Sen

There's a theorem which says $\pi_k(SP(X)) \cong H_k(X)$

for a CW complex $X$

Akiva Weinberger

I'm not finding the notation $SP^n(X)$ online, but this 1931 paper by Borsuk and Stanislaw calls it $X(n)$

Thanks, by the way

Balarka Sen

$\text{Sym}^n(X)$ also seems popular

Akiva Weinberger

Oh I found an ncatlab page with it

Yeah thanks @BalarkaSen

Balarka Sen

21:11

No problem

Akiva Weinberger

In any case, I was wondering, @BalarkaSen

We could define a notion of continuously deforming metric spaces

in such a way that the *homeomorphism type never changes

Balarka Sen

homotopy type, you mean?

Akiva Weinberger

Like, a homotopy of maps $X\times X\to\Bbb R_+$ such that, at every instant, it's a metric space, and any two instants game the same homeomorphism type

@BalarkaSen Whoops. Homeomorphism.

Balarka Sen

Ok. We could, probably, yeah.

Akiva Weinberger

And then that gives us an equivalence of metric spaces

where you ask if you can deform one into (something isometric to) the other

and I guess the question is, is this a trivial equivalence?

Balarka Sen

21:16

Sounds like an interesting question. I doubt it.

Akiva Weinberger

Like, can you always do that for any two homeomorphic metric spaces?

I kinda feel like a circle and a knot, given the metric inherited from $\Bbb R^3$, might be a counterexample

Oh, wait, they're not.

Unknot them in $\Bbb R^4$ and give them the metrics inherited from there.

Balarka Sen

I can isotope them in R^3 x R right?

Dang, sniped.

Akiva Weinberger

I don't know what I should do first, try to prove it or look for counterexamples.

(I feel like there's a decent chance it's either trivially true or trivially false)

Also, I wonder if this is, like, what Ricci flow is

Or at least if this is related

Balarka Sen

Ya Ricci flow is a 1-parameter deformation of the Riemannian metric, but with some differential equation it's gotta satisfy

I think, at least

Akiva Weinberger

@AkivaWeinberger (*metric, not metric space)

*have, not game

Hm. Would each such deformation define a metric on $X\times I$?

Where, for two points with the same $I$-coordinate, you define it to be like the metric of the space at that instant

anf for two points with a different $I$-coordinate, you do… something, I dunno

Probably easiest to do a Manhattan-type thing

Balarka Sen

21:26

I want to extend your knot idea somehow. Are any two knots in a 3-manifold $M$ isotopic to each other in $M \times I$?

I don't think so.

Like, take the longitudinal circle in $S^2 \times S^1$ and a circle in a chart :P

Akiva Weinberger

But you could embed that manifold in some $\Bbb R^n$

and then unknot them there

Balarka Sen

Yeah fair.

Akiva Weinberger

(we can assume $n\ge4$)

PVAL-inactive

@Balarka The only invariant should be the image in $\pi_1$

Barring that they are.

Balarka Sen

@PVAL-inactive That sounds super believable.

Akiva Weinberger

21:28

Is that a response to this discussion or something else @PVAL

Balarka Sen

it's a response to when two knots in $M$ are isotopic in $M \times I$

Akiva Weinberger

Oh, yeah, OK

PVAL-inactive

Take an immersed cyllinder bounding them.

Balarka Sen

So I want something that embeds in $\Bbb R^n$ super-messily.

PVAL-inactive

By transversality you can assume the intersections are isolated

Balarka Sen

21:30

Surely a very bad space is needed

Akiva Weinberger

So I think this is equivalent to asking if there's always a bigger metric space we can embed these in in which they are isotopic

PVAL-inactive

now shorten the cyllinder a little and push off those intersection off the edge of the boundary.

Akiva Weinberger

The converse part of that is given by the stuff I said before, with the metric on $X\times I$ defined by the deformation

(Weird, "in in" ^)

Balarka Sen

Do you really get a metric on X x I

Akiva Weinberger

I believe so

though I need to elaborate on what "Manhattan-type thing" means probably

Wait

Can't we just average the metrics

$\lambda d_1(x,y)+(1-\lambda)d_2(x,y)$

We need to check that it satisfies the metric axioms

Astyx

21:35

Isn't any positive linear combination of metrics a metric ?

Akiva Weinberger

Is it?

Does the triangle inequality work?

It should, shouldn't it

Balarka Sen

it does

Akiva Weinberger

Yeah OK

So it's true and kinda trivial

Wait

We also want to check that it preserves the homeomorphism type at every instant

Balarka Sen

should be true, because d1 and d2 give same homeo type

and we're just taking linear combination

Akiva Weinberger

We probably specifically want the identity map from $(X,d_1)\to(X,d_2)$ to be a homeomorphism

but I think we can assume that

I need to walk home

but it's probably some routine-ish calculation on open balls

Astyx

21:40

What are we doing ?

Balarka Sen

Now I want to know the 1-homotopy type

aka does every loop of deformations starting and ending at $(X, d)$ "contract" to the zero loop?

whatever that means

I gotta run now. Have to travel 1843 kilometers.

Astyx

Running ?

Good luck

I have to sleep 14 hours in 8 hours

Akiva Weinberger

Yeah, it does follow routinely

Hm, I have a new question

Say $(X,d_1)$ and $(X,d_2)$ both define the same metric space

and they both can be embedded in $\Bbb R^n$

Then can $(X,d_1+d_2)$ be embedded in $\Bbb R^m$ for some $m$?

Hm. $(X,\sqrt{{d_1}^2+{d_2}^2})$ can.

Speaking of which, $\sqrt{(\lambda d_1)^2+((1-\lambda)d_2)^2}$ would be another answer to the last question

Akiva Weinberger

22:10

Oh. No. Duh. Take $\Bbb R$ and $\Bbb R$ themselves.

You end up with the Manhattan metric, which can't be embedded in $\Bbb R^n$ for any $n$.

(Trying to remember the proof of the above fact…)

@BalarkaSen bill wurtz voice Time to fly through all of India

Most of India

Daminark

22:32

Hey everyone!

Alessandro Codenotti

Hi @Dami

Ted Shifrin

hi Demonark, @Alessandro, DogAteMy

wonders how Astyx will do sleeping 14 hours in 8

Alessandro Codenotti

Hi @Ted

Daminark

How's it going @Ted and @Alessandro?

Ted Shifrin

I've just been shopping and starting to cook. A former UGA colleague is here for a month and he and his family are coming for dinner tomorrow.

Where are you in complex analysis now, Demonark?

Alessandro Codenotti

22:39

I'm on holiday, so quite well

Ted Shifrin

Where on holiday, Alessandro?

Daminark

I screwed up and slept through the lectures today

Woke up at like, 1:45

But in complex, today was Cauchy's inequality, Liouville, which I know, followed by Morera and some stuff on zeroes of an analytic function

Alessandro Codenotti

At home, I just meant holiday as in "no exams nor lectures"

Dodsy

I'm sure you'll still do fine, Amin.

You're a smart lad.

Daminark

@Dodsy I mean it's not like I'll get hit by a lightning bolt, more like, today was a day I was looking for to for the lectures since the two people were really good

(At least I don't think I'll get hit by a lightning bolt...)

Ted Shifrin

22:43

Time to un-Balarka your sleep schedule, Demonark.

Hi, Nate!

Oh, ok, @Alessandro.

Alessandro Codenotti

You should stay up a whole night to reset your schedule @Dami

I'll go to Tuscany in August probably @Ted

Dodsy

Hey Ted, how are you?

Ted Shifrin

I get confused about "home" versus "school" for you, @Alessandro.

Daminark

Good idea @Ted, might try @Alessandro's shtick

Dodsy

:(

I'm sorry about your misfortune, Daminark.

Alessandro Codenotti

22:45

Yeah, I'm not using "holiday" properly, but after 2 months packed with exams that's good enough :P

Dodsy

Glad to hear that lightning will not strike you, any time soon.

My mom went to a native pow-wow and this shaman did a ritual on her and said that she would be safe for a single day.

Which I found funny.

Daminark

Oh Morera is just that for continuous functions, Cauchy is iff

Alessandro Codenotti

What's Cauchy's inequality?

Ted Shifrin

Morera is actually very powerful.

But it's just the FTC in disguise.

Alessandro Codenotti

In the proof I've seen it wasn't even a very good disguise

Dodsy

22:48

I like math.

Daminark

@Alessandro Cauchy's inequality says that if $f(z) = \sum a_nz^n$, then $|a_n|r^n \le \max_{rS^1} |f(z)|$

Alessandro Codenotti

Ah, ok, I didn't know it has a name

Ted Shifrin

No, that's not right, Demonark.

Cauchy's inequalities are the bounds on the derivatives.

Daminark

I mean, the $a_n$ would be derivatives, no?

PVAL-inactive

Hi folks

Dodsy

22:51

Hey PVAL.

Daminark

Well, multiples of them

But yeah I mean, that's what we had referred to as Cauchy's inequality

Alessandro Codenotti

Can we construct a function on a not simply connected domain that has zero integral on every null-homologous chain yet it's not holomorphic?

Ted Shifrin

Oh, I see. I've never seen it that way ...

Dodsy

I am always disturbed at seeing endless chains of theorems, most of them of no interest, and without any stress on the main points.

Ted Shifrin

Good complaint, Nate. Some books don't motivate/explain what's important and what's meh.

Hi @PVAL.

@Alessandro: This is a local question, not a global one.

Alessandro Codenotti

22:54

What do you mean?

Daminark

@Ted Lol, I've checked Narasimhan and it mentions both ways of looking at it

PVAL-inactive

As long as $f$ integrates to zero on every curve in a neighborhood of a point p, f is already holomorphic at p.

Ted Shifrin

Topology introduces non-zero integrals that are zero locally.

If that makes sense to you.

PVAL-inactive

I think I learned the stronger version of Morera.

where the assumption was only for triangles.

Ted Shifrin

Yes, I've taught that version. Rectangles. Whatever.

Daminark

22:57

I also found, especially in analysis books, that the way proofs of theorems are phrased can be, not in the way that you'd approach it

Dodsy

What books are you currently reading, Ted? For leisure or otherwise.

Ted Shifrin

Nothing at the moment, although I have a small pile I should be.

Daminark

As an example for which this is slightly less of a problem but still demonstrates it, proving that if $a_n \to L \ne 0$, that $\frac{1}{a_n} \to \frac{1}{L}$. The way you'd ideally present it is say okay, let's look at $|\frac{1}{a_n} - \frac{1}{L}| = |\frac{L - a_n}{La_n}|$. The numerator converges to $0$ and of course the denominator is bounded away from $0$!

Transcript for

$Mathematics$ Mathematics