Mathematics - 2020-11-14 (page 2 of 2)

user2103480

18:03

@MikeMiller Ok cool. It's somewhere in Hatcher I suppose?

@MikeMiller and thanks for the illustration!

Mike Miller

Chapter 4 in Hatcher

user2103480

Ok thanks

user193319

18:25

How do I prove that $e^{\frac{\pi i}{b}}$ is a simple pole of $\frac{1}{1 + z^b}$? I was thinking of using the geometric series expansion, but that won't work because we need $|z| < 1$.

Note $b > 1$

geocalc33

I have a question about metrics. $ds^2=dx^2+dy^2$ is the usual Euclidean metric. What is $ds^2=dxdy?$

Ted Shifrin

What do you mean, what is it? It's a Lorentz metric.

@user193319 Think about zero rather than pole.

geocalc33

oh okay that's what I thought, I just always see it written like this $ds^2=dy^2-dx^2$ lol

user193319

@TedShifrin So show that it is a simple zero of $1+z^b$? Is that easy even if $b$ is not an integer?

Thorgott

how do you go about computing the order of a zero

user193319

18:40

Well, for polynomials that's "easy", but I don't know how to do it in general.

Thorgott

can you relate the order of a singularity to the coefficients of the Laurent expansion at that singularity?

user193319

Yes, I think that's originally what I was going to try, but Ted told me to think about zeros of $1 + z^b$ instead of singularities of $\frac{1}{1+z^b}$ (although they end up being related).

Ted Shifrin

@geocalc Think about how you can rotate coordinates and turn $xy=c$ into $x^2-y^2=c$.

Thorgott

yeah, the order of zero $e^{\pi i/b}$ of $1+z^b$ is the negative of the order of the pole $e^{\pi i/b}$ of $1/(1+z^b)$

it generally holds that $\operatorname{ord}(f/g,a)=\operatorname{ord}(f,a)-\operatorname{ord}(g,a)$ when $g$ isn't zero, this is a good exercise

user193319

Clearly $e^{i \pi/b}$ is a zero.

If $z_0 = e^{i \pi/b}$ were of greater order than $1$, we'd have $1+z^b = (z-z_0)^kh(z)$ for some $k \ge 2$ and some analytic $h$ with $h(z_0) \neq 0$.

Thorgott

18:44

anyhow, if you know how to relate the order of a singularity to the coefficients of the Laurent expansion, you're good, because I claim that you know a very easy way to compute the coefficients of a power series expansion of a holomorphic function (and that's all you need at a zero)

user193319

Oh, so compute the power series expansion of $1+z^b$ at $z_0$?

If $a_1 \neq 0$, does this mean it is of order $1$?

That is, a simple zero?

geocalc33

@TedShifrin yep got it. So I think I know how to generalize this metric to $n$ dimensions

Thorgott

no to the former, yes to the latter

geocalc33

$ds^2=dxdydz\cdot\cdot\cdot dn$

Thorgott

I'm telling you there's an easy way to get these coefficients without computing the power series

well, of course, if you have the coefficients, you have the series, but you only need one of the coefficients, namely $a_1$

geocalc33

18:47

actually that may be wrong...

Ted Shifrin

@geocalc Nope.

Metrics are quadratic forms/bilinear forms.

geocalc33

it's gonna be so complicated.....

user193319

@Thorgott Oh, isn't it a certain integral $\frac{1}{1+z^b}$ around the boundary of a circle enclosing $z_0$?

Thorgott

too complicated

forget the complex stuff

this is a fact about power series

how do you get the coefficients of a power series if you know the function it defines

user193319

Oh, it's $\frac{f^{(n)}(z_0)}{n!}$ for the $n$-th coefficient.

Thorgott

18:51

correct

geocalc33

@TedShifrin okay I can generalize $ds^2=dx^2-dy^2.$ to n dimensions, but I'm having a problem with the other one

Thorgott

now the rest is easy

user193319

Okay, so this shows that $a_1 \neq 0$, and therefore $z_0$ is a simple root of $1 + z^b$ and hence a simple pole of $\frac{1}{1+z^b}$. Is this the correct line of reasoning?

Ted Shifrin

You have to add a positive definite metric in $n-2$ dimensions.

Thorgott

it's a correct line of reasoning

It's possible to compute the Laurent expansion of $1/(1+z^b)$, but I believe the computation would be a lot uglier than this

geocalc33

18:58

so the final metric in say 3 dim. would be of the form $ds^2=(dxdydz)A$ where $A$ is a positive definite matrix?

Ted Shifrin

19:12

What? Reread everything we’ve discussed.

@Thorgott You really need this to do the beginning of the Laurent expansion.

Thorgott

what do you mean?

Ted Shifrin

The only way I know how to compute the Laurent expansion is to use the Taylor expansion of the denominator (around the relevant point, of course).

Thorgott

well yeah, but it's, in general, an ugly computation

Ted Shifrin

19:30

I like doing those Laurent series computations even for residues. I always drilled my students on those (as oppose to various higher-order derivative computations). But the latter aren't good if you want the actual Laurent series for a few terms.

geocalc33

what is the geometry induced by the metric $ds^2=\log(dx)\log(dy)$?

Ted Shifrin

That makes zero sense.

geocalc33

which part?

Ted Shifrin

The expression.

geocalc33

you can't take the log of dx?

Ted Shifrin

19:34

Not if you want to make mathematical sense.

geocalc33

can you explain better I'm confused

Ted Shifrin

$dx$ is a one-form; you can't do random functions with those.

Remember that a metric needs to be a bilinear form on tangent vectors.

geocalc33

@TedShifrin the problem is that I derived that $\log(x)\log(y)=1$ is a preserved metric of a transformation

sorry, I mean that $\log(x)\log(y)=1$ is a preserved metric of $(x,y)\mapsto (x',y') = \left(x^{\exp(\Delta b)}, y^{\exp(-\Delta b)}\right)$

trying to get the metric from this

Filip98

21:06

Hello everyone!

AttractorNotStrangeAtAll

21:25

Hello!

Filip98

Hey there

AttractorNotStrangeAtAll

Guys, should I worry if I hate doing epsilon-delta proofs? It's not fun at all.

Filip98

why do you hate it?

AttractorNotStrangeAtAll

It's much more interesting to prove more general statements, not applying formal definitions to explicit functions.

I mean, I prefer proving that the inverse function of a continuous function that has a compact domain is continuous than proving sqrt(x) is continuous for the domain [0,\infty)

Is this nonsense?

Filip98

Oh Gosh another topological man

AttractorNotStrangeAtAll

21:30

Hahah I'm just starting to get a taste for all of these things, but it does seems more laborious than it should to do epsilon-delta proofs

And I'm not saying that I prefer another definition of continuity, for example, but it's just that when working with specific functions (such as a simple sqrt(x)), the proof is very specific and usually involves intense algebraic manipulation or some trick

But I might be wrong, as I said, I'm really new to all of this

Filip98

You are right, it can be tricky somehow if you do not manipulate correctly

Ted Shifrin

A lot of mathematics depends on estimates (not algebra, but analysis, numerical analysis, differential equations, geometric analysis ... and the list goes on), so that's what you're starting to learn to do.

Nothing is fun until you learn to get better at it.

I never found manipulating symbols in abstract algebra at all fun.

I still don't, but now I understand that there's a lot of beautiful concepts going on in that algebra.

AttractorNotStrangeAtAll

@TedShifrin That is probably true, as these tricks don't come easy at me, so I get lost quickly and that's obviously not fun.

I guess the way to get better is keep trying and doing exercises?

Ted Shifrin

There are some standard methods to learn, and yes, as with everything in mathematics, you have to practice a lot.

I wrote a handout for my Calculus with Theory students some years ago on basic $\delta$-$\epsilon$ proofs. If you want to email me, I'll forward it to you.

AttractorNotStrangeAtAll

Sure! May I email you at your UGA email?

Ted Shifrin

21:39

Yes, absolutely.

AttractorNotStrangeAtAll

@TedShifrin Sent you! Thanks!

Ted Shifrin

Lots more exercises to practice, if nothing else :P

Sent.

AttractorNotStrangeAtAll

Received! Thank you!

Thorgott

learning to use $\varepsilon-\delta$ in actual proofs is important and pops up everywhere in analysis, but proving that a specific function is continuous using $\varepsilon-\delta$ is most often just an extremely ungrateful exercise

Ted Shifrin

Well, it's surprising how many calculus books (and even analysis books) make a hash of it all.

I still remember when I first encountered this stuff reading Thomas's Calculus book in high school, he had ad hoc garbage with cube roots of $\epsilon$. I would never allow my students to do that nonsense.

Thorgott

21:49

I don't believe anyone has ever generated as much as an ounce of insight out of proving that some random degree $3$ polynomial is continuous via $\varepsilon-\delta$, but I used to see a lot of questions in that vein on main

Ted Shifrin

Once you know how to do a quadratic, I agree you're not learning too much, but doing it once is good. The key thing to learn is what I always called the "stipulate" game, and then of course you'll have to know how to divide your cubic by $x-a$.

But if you can do synthetic/long division, it really doesn't matter what the degree of the polynomial is if you do this right.

Balarka Sen

In case of finite measure spaces $L^p$ sits inside $L^q$ if $p > q$. Not true in eg $(1, \infty)$ anymore. What's a unifying way to say these stuff?

Thorgott

it feels very redundant to prove anything polynomial is continuous using $\varepsilon-\delta$ when it's sequentially obvious. Students should learn about both continuity in terms of sequences and in terms of $\varepsilon-\delta$ in a course and why they are equivalent.
Once they have that, forcing them to apply one definition when the other is much more natural just feels very counter-intuitive (there are equivalent ways of defining things everywhere in math and students should learn to actively decide which definition is best suited for their purposes and then apply that). There are still …

Alessandro Codenotti

22:05

@BalarkaSen The general phenomenon is that there is such a nesting iff $X$ does not contain sets of arbitrarily big measure iirc

Thorgott

@Balarka $L^q\not\subset L^p$ iff $X$ contains set of arbitrarily small measure and $L^p\not\subset L^q$ iff $X$ contains sets of arbitrarily large finite measures

wait, I mean not subset

Balarka Sen

Garbage phrasing Thorgott

@AlessandroCodenotti ya

the operator norm of the natural embedding is $\mu(X)^{1/p - 1/q}$

Thorgott

you can fix the phrasing yourself

Balarka Sen

measure spaces which does not contain nonempty sets of arbitrarily small measure -- lol

also known as completely garbage atomic spaces

Can't you say something with norms

Thorgott

I guess you can also say the finite measure space thing is a consequence of convexity of L^p spaces + $L^{\infty}\subset L^1$ iff $\mu(X)<\infty$

Balarka Sen

22:08

its a consequence of direct calculation

Thorgott

it's all just direct calculation

Balarka Sen

$\|f\|_{L^p} \leq \max\{\|f\|_{L^q}, \|f\|_{L^r}\}$ if $q < p < r$ is true maybe

yeah just Holder innit

Thorgott

something better is true

Alessandro Codenotti

math.stackexchange.com/questions/66029/… see here

Thorgott

we just discussed this yesterday in here lol

20 hours ago, by Thorgott

I think I got it. Pick $\alpha$ such that $\frac{\alpha}{p}+\frac{1-\alpha}{r}=\frac{1}{q}$. Then $p/\alpha q$ and $r/(1-\alpha)q$ are conjugate and Hölder gives $\int|f|^q=\int|f|^{\alpha q}|f|^{(1-\alpha)q}\le(\int|f|^p)^{\alpha q/p}(\int|f|^r)^{(1-\alpha)q/r}$. Taking the $q$-th root gives $\lVert f\rVert_q\le\lVert f\rVert_p^{\alpha}\lVert f\rVert_r^{1-\alpha}$. Lastly, observe $x^{\alpha}y^{1-\alpha}\le\max(x,y)$ for $0\le\alpha\le1$.

Balarka Sen

22:09

@Alessandro see what exactly

Alessandro Codenotti

A proof

Balarka Sen

well the proof is direct

i was hoping for something conceptually easier; the inequality seems like something explanatory

@Thorgott yeah

well $p,q,r$'s are switched

but yeah

Thorgott

up to conjugation of variable names

Balarka Sen

@Thorgott what does convexity mean

im making a list of all things banach; give me content

Mike Miller

@BalarkaSen Look up interpolation techniques in L^p theory

Thorgott

22:16

I didn't mean convexity, I meant $L^q\supset L^p\cap L^r$ for $p<q<r$ (consequence of the above inequality)

Balarka Sen

time to terrytao.wordpress.com

Thorgott

But convexity is also a thing: $\lVert f\rVert_p^p$ is log-convex in $p$

Balarka Sen

@Thorgott Yeah OK

Thorgott

wait, inclusion is the wrong way round

Balarka Sen

yeah man its alright

human brain cannot understand inequalities according to Misha Gromov

Thorgott

22:19

give $L^p\cap L^r$ the norm "sum of $p$-norm + sum of $r$ norm", then the inclusion $L^p\cap L^r\rightarrow L^q$ is continuous

Balarka Sen

@MikeMiller Too dank, bookmarked. I'll read

geocalc33

I’m on mobile

Thorgott

you can read the chapter on $L^p$-spaces in Folland

lots of cool stuff in there

Balarka Sen

Too dank, bookmarked

Just tell me the important theorems and I'll prove them by myself

Only way to learn analysis

geocalc33

Yeah!^

Balarka Sen

22:25

time to learn about uniformly rotund Banach spaces

Thorgott

prove L^p is dual to L^q or sth

Balarka Sen

easy

already did

geocalc33

Yes! Banach spaces are what my professor studied

Alessandro Codenotti

@BalarkaSen There are a lot of very cool posts there

I still have to read the one about Gromov's theorem

Balarka Sen

its hard to read because he goes off on so many tangents

analysis brain

Thorgott

22:28

prove Riesz-Thorin or whatever

geocalc33

Geocalc 🧠

Balarka Sen

No Thor

Thorgott

good decision

Alessandro Codenotti

@BalarkaSen But he shows interesting connections with other stuff

Balarka Sen

lol does the proof use holomorphic functional calculus

@Alessandro its too much for my brain man i am a simple man i stick to one picture at a time

ok i have to sleep, be back tomorrow

Thorgott

22:32

gn

geocalc33

Have a good one

Alessandro Codenotti

good night

Astyx

22:45

hi chat

Edward Evans

hey

Astyx

Did you see my message ?

Edward Evans

which?

Astyx

In a private room I made on this chat

Edward Evans

I did not lol

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$Mathematics$ Mathematics