Mathematics - 2018-01-19 (page 3 of 4)

Silent

16:01

I think this should follow from uniqueness of exponent but really can't see it.

Semiclassical

which step are you struggling with exactly? problem 7 outlines a proof

Silent

@Semiclassical, no no! I don't need proof of those problems! I need to apply problem 7 to show $\left(b^{\frac 1p}\right)^p=b$

Semiclassical

Well, you've also got part (a) of 6 there

and since $r=p/p=1/1$, that implies $(b^{1/p})^p = (b^1)^1=b$

Silent

@Semiclassical but that part is for rationals, ie if p was an integer, i could use that

Semiclassical

hmm

point

Silent

16:08

thanks :)

Semiclassical

well, it should also work if $p$ is rational

Silent

Yes, but what about irrationals? Do I have to use all the $\sup$ arguments, or is there an easier way?

Semiclassical

not sure.

Akiva Weinberger

@BalarkaSen So, final verdict is, $SM$ is not a manifold and is not homeomorphic to $S^4$, but $SM\times\Bbb R\simeq S^4\times\Bbb R$?

Semiclassical

This question puzzles me: math.stackexchange.com/q/2612170/137524

I thought the convolution theorem for F.T. was $(f\star g)(t)\leftrightarrow F(\omega)G(\omega)$

not $F(i\omega)G(i\omega)$

Akiva Weinberger

16:21

I'm still hoping for a simpler example.

Is there a pair of manifolds that works?

Hm, that would require a pair of manifolds of the same dimension that are homotopy equivalent but not homeomorphic. Does such a thing exist?

(Possibly with boundary?)

Abhas Kumar Sinha

Is there any way to get the length of chord, by knowing arc length and radius? (without trignometery)

Semiclassical

Given that the result involves sin(theta)=sin(arc length / radius), no

(well, more properly it's sin(theta/2). but same point holds)

Abhas Kumar Sinha

@Semiclassical what?

Akiva Weinberger

The final answer is $2\sin(\theta/2)=\sqrt{2-2\cos(\theta)}$ I think

Semiclassical

@AkivaWeinberger for a unit circle

Abhas Kumar Sinha

16:29

I said without trigonometry

Semiclassical

And I said no, it can't be done without trigonometry

Abhas Kumar Sinha

no other way without that?

Akiva Weinberger

The final answer contains trigonometric functions, so to get there you'd have to use trigonometry

Semiclassical

^

Abhas Kumar Sinha

okay

Okay, let me think of something else

If there's a alternative of doing that

Akiva Weinberger

16:31

@AbhasKumarSinha For specific angles, like 45 degrees or 45/2^n, there might be a way

Abhas Kumar Sinha

doesn't work :(

Semiclassical

I mean, if you have an arc length of 2 and a radius of 1, then the chord length will be 2sin(1).

Akiva Weinberger

In fact, for 45/2^n degrees, the answer ends up using nested roots and stuff and looks quite nice

I don't remember what it is, but it's something like n nested square roots

Semiclassical

and you're not going to be able to express sin(1) in simpler terms.

For a certain definition of 'nice.' :P

Akiva Weinberger

In fact, that's how you derive Viète's formula for pi.

Hm, yeah, it's less nice than I remember

Semiclassical

16:34

the idea is cute, to be sure

Abhas Kumar Sinha

That's what I'm trying to do

Akiva Weinberger

Oh, wait it does look nice

Abhas Kumar Sinha

for ex- to get value of sin 1

and sin43.343

let angle bac = 1

Semiclassical

I should note that i mean sin(1 radian)

not 1 degree

Abhas Kumar Sinha

I mean sin 1 degree

not radian

Akiva Weinberger

16:36

For 90/16 degrees, it's $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}$

Abhas Kumar Sinha

without trignomtery tables

Semiclassical

Good luck with that.

Akiva Weinberger

(Cont'd) Calculate 32 times that value, and it'll approximate pi

Abhas Kumar Sinha

let AC be 1 unit and angle BAC = 90 deg then, BOC = 2 degree, then I just need to fin how to get BC without Trigonometry, so that I can get trig ratios without sine tables

for triangles having less than 90 deg

Also, AC is diameter

Semiclassical

Once again: Good luck.

Abhas Kumar Sinha

16:38

@Semiclassical thanks :D

Don't you think that its possible with few constructions and calculations?

Semiclassical

No.

Abhas Kumar Sinha

hmmmmm

Balarka Sen

@AkivaWeinberger That is right.

@AkivaWeinberger I wondered that. Closed examples are certain Lens spaces

L(5, 1) and L(5, 2)

Akiva Weinberger

@BalarkaSen What dimension are these, again?

Abhas Kumar Sinha

@Semiclassical So, should I start with something else? from scratch?

Balarka Sen

16:42

3

Abhas Kumar Sinha

for that?

Balarka Sen

those are quotients of S^3

Akiva Weinberger

Mhm

Semiclassical

I think you should use trig and accept that that's how these problems work.

Akiva Weinberger

Weird

That'd be the smallest dimension in which that's possible though I guess

Wait, actually

Balarka Sen

16:43

Closed 2-manifolds are determined by their homotopy type, yeah

Akiva Weinberger

Mobius strip and cylinder ($[0,1]\times S^1$)

Well, with boundary

Balarka Sen

Closed is the right condition here

Akiva Weinberger

They have different products with $\Bbb R$ anyway (orientability)

Balarka Sen

Mhm

Hmm

How about the double annulus (twice punctured D^2) and a neighborhood of the 1-skeleton of the torus?

Akiva Weinberger

Oh, right

(AKA punctured Tory)

Semiclassical

16:46

@AbhasKumarSinha the trouble is that, while you know sin(30 degrees)=1/2

Akiva Weinberger

(torus)

Balarka Sen

Yeah

Akiva Weinberger

(Not punctured UKIP)

Semiclassical

you need to divide that angle by 2, 3, and 5

first isn't too bad. second also isn't horrible, though it's hardly trivial.

Akiva Weinberger

Wait, does that actually work??

Oh, wait

Semiclassical

16:48

but the last one is baaaad

Akiva Weinberger

It might if we were to multiply by $[0,1]$… but if we multiply by $\Bbb R$ then they have a different number of boundary components

Semiclassical

(for comparison, note that it's provable that trisection of angles can in general not be done with compass and straightedge. 5-section of angles is worse.)

Akiva Weinberger

@Semiclassical sine(72 degrees) has a closed form, so that brings you down to 3 degrees

Semiclassical

hmm

3 degrees not nine?

Akiva Weinberger

Oh whoops

9, yeah

cos(36 degrees) is half the golden ratio

Semiclassical

16:51

Oh hey: math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf

Balarka Sen

@AkivaWeinberger Boundary components?

Semiclassical

That's a nice table

Balarka Sen

They look the same to me. Am I missing something?

Akiva Weinberger

@Semiclassical Oh, wait:

36 is doable, hence 72 is doable. 60 is also doable; use the difference formula to get 72-60=12

and then 6 and 3.

Balarka Sen

Hey @MikeMiller

Semiclassical

16:52

@akiva yeah, see the last two pages of what I just linked

Akiva Weinberger

Meh, you looked it up faster than I could figure it out from memory

Semiclassical

lol

Akiva Weinberger

@BalarkaSen Double annulus has three boundary components; punctured torus has one.

Multiplying by $\Bbb R$ shouldn't change that

($\partial\Bbb R=\emptyset$)

Mike Miller

hi

Semiclassical

@AbhasKumarSinha check out what I just linked for the details of how you get exact values of sin(n degrees) for various values of n

it isn't nice.

Adeek

16:54

@MikeMiller hey

Balarka Sen

@AkivaWeinberger Oh, right, I was taking the topological boundary to account

So x [0, 1]

Adeek

right @BalarkaSen

user8469759

is there a reason why a differential solid angle is usually represented by a cone?

Balarka Sen

Was thinking of an $\epsilon$-thickening

Semiclassical

for instance, $$\sin(3^\circ)=\frac{1}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}$$

Balarka Sen

16:56

@MikeMiller Is there a way to prove $M \times \Bbb R \cong S^n \times \Bbb R$ implies $M \cong S^n$ (topologically) without invoking the Poincare conjecture?

Akiva Weinberger

For closed manifolds, mind

Balarka Sen

Thanks, Akiva

Akiva Weinberger

And maybe even closed manifolds with boundary, we haven't gotten a proper counterexample yet

Balarka Sen

S^n x R has no boundary, so M cannot have boundary

$\partial(M \times \Bbb R) \cong \partial M \times \Bbb R$

Akiva Weinberger

Oh, $S^n$, whoops. I thought you had two arbitrary manifolds there.

Balarka Sen

16:59

Ah, no. But I do want examples of compact (without boundary, ideally) manifolds $M, N$ such that $M \times \Bbb R \cong N \times \Bbb R$ but $M, N$ are not homeomorphic

Semiclassical

by comparison, if you note that that's 3pi/180=pi/60 radians, which is small, you get the estimate sin(3 deg) ~ pi/60 = 0.05235987... when the exact value is 0.052335956

Alessandro Codenotti

@Eric do you also happen to be familiar with the usage of the axiom of choice in functional analysis? Specifically is it possible to prove that the weak topology is Hausdorff without any form of $\sf AC$?

Semiclassical

so the seemingly 'obscure' sine representation yields a very useful approximation

Akiva Weinberger

I suppose if we had an example with boundary, we could cap off the boundary components and get an example without boundary

Mike Miller

@BalarkaSen No, I don't think so

Alessandro Codenotti

17:02

weak topology on a Banach space*

Balarka Sen

@MikeMiller Hm, weird

Akiva Weinberger

@Semiclassical There are much nicer approximations, such as the aforementioned $\pi\approx32\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}$

Semiclassical

ehhh

again, definition of 'nice'...

Akiva Weinberger

You prefer $15\sqrt{8-\sqrt3-\sqrt{15}-\sqrt{10-2\sqrt5}}$?

Semiclassical

I prefer to accept that $\pi$ is a useful constant absent approximations :P

Akiva Weinberger

17:08

which is a (slightly) worse approximation

Semiclassical

i.e. sin(3 deg) is to me a far more useful answer than the above mess of square roots

Akiva Weinberger

It's sine 45/8 I think

(The bit after the 32)

Semiclassical

sure

Studentmath

Cute question:
Consider a set of triples, each member appears in 6 triples exactly. Show you can two-colour the triples, such that each member appears in both colours

Semiclassical

I mostly just think it's silly to prefer the exact forms of something like sin(3 deg). there's a reason why we use trig functions---a closed-form expression for a sine value is by and large pretty useless.

to the extent that you get anything useful out of it, it's by crunching out the resulting value. but in that case there's easier approximations which you might as well do instead

(blah blah soapbox blah)

Eric Silva

17:12

@AlessandroCodenotti i think you need hahn banach which can be proven without full choice but im not certain about that

Akiva Weinberger

@Studentmath We don't know how many members in total there are?

Alessandro Codenotti

@EricSilva I see, thanks

Studentmath

@Akiva nope. You can think of it as a 3-uniform 6-regular hypergraph if you want (I think), and show that there are two disjoint hyper-edge covers of it

@Akiva you can assume they are finite, though

Eric Silva

i think all you need is the ultrafilter lemma? idk

Semiclassical

Now I find myself wanting to make up a story for this setting :P

Eric Silva

17:16

something weaker than choice in any event

Alessandro Codenotti

@EricSilva That's enough for Hahn-Banach, yes

Eric Silva

cool

Akiva Weinberger

@Studentmath And the triplets don't care about order?

MatheinBoulomenos

@AlessandroCodenotti @EricSilva I don't think you need any form of choice. The weak topology on $V^*$ is induced by the familiy of seminorms $\|f\|_v := |f(v)|$ for $v \in V$. Since we trivially have $(\forall v \in V \|f\|_v=0 )\Rightarrow (f=0)$, the topology is Hausdorff

Studentmath

@Akiva precisely, that's why 3-uniform 6-regular hypergraph is a good representation (well, for me at least it helps)

Semiclassical

17:17

Maybe something like

Akiva Weinberger

I don't quite know what 3-uniform 6-regular means

Hypergraph is like edges have three vertices instead of two?

Studentmath

Yeah precisley

That's 3-unform hypegraph

And 6-regular just means each vertex is in 6 hyperedges

Akiva Weinberger

Ah

Balarka Sen

I want to think of it as a 6-dimensional simplicial complex.

Alessandro Codenotti

@MatheinBoulomenos What do you mean with "is induced by a family of seminorms"?

Semiclassical

17:19

You've got a bunch of people practicing group presentations. Each presentation must involve three people, and each person must participate in six presentations.

Akiva Weinberger

The other type of "group presentation"

Semiclassical

lol

hadn't even thought of that

Mike Miller

@BalarkaSen Closely related but not precisely the same, I think

Akiva Weinberger

And then everyone participates in at least one morning presentation and one afternoon presentation. @Semiclassical

Semiclassical

hmm

that may work, yeah

Studentmath

17:20

It's really cute though. Haven't yet thought of a complete proof

Akiva Weinberger

Or, rather, it can be scheduled like that.

Mike Miller

I think to each hypergraph you can associate a simplicial complex, by saying that {v_i} form a simplex if they're connected by some hyperedge (which may connect more things)

Akiva Weinberger

Is what we need to prove.

Balarka Sen

@MikeMiller Maybe it won't be a simplicial complex. I was thinking, like, let $T_i$ be the triples. If $T_i$ and $T_j$ intersects, put an edge. If $T_i \cap T_j \cap T_k$ is non-null put a 2-simplex.

MatheinBoulomenos

@AlessandroCodenotti if you have a family of seminorms $\| - \|_i$ on a vector space, then they induce a topology. You define your open balls by $B_{\varepsilon}(x)=\{y \in V \mid \forall i \|y-x\|_i < \varepsilon \}$

Studentmath

17:21

Hypergraphs are cool, since they can represent more systems than regular graphs

Semiclassical

Another way to do it would be to have two different observers

Balarka Sen

And so on

Alessandro Codenotti

@MatheinBoulomenos I'm thinking about the weak topology on $V$ rather than $V^*$ as well. But I have to be away for a while now, I'll ask you more about this later probably

Akiva Weinberger

And anything with "hyper" in its name gets a lot more exciting

Semiclassical

And you want to schedule the talks so that each observer sees each student at least once

Akiva Weinberger

17:21

Except for maybe hyperplanes

though that sounds like a really good airplane

Studentmath

Haha

Balarka Sen

I'll think about this for a bit.

Semiclassical

anyhow

MatheinBoulomenos

@AlessandroCodenotti oh yeah I think you need Hahn-Banach for that

Akiva Weinberger

Concord 2.0: Hyperplane

Eric Silva

17:23

@MatheinBoulomenos yeah it shows up somewhere at some point

Semiclassical

Can you draw hypergraphs?

Studentmath

Sure

Hypergraph

In mathematics, a hypergraph is a generalization of a graph in which an edge can join any number of vertices. Formally, a hypergraph H {\displaystyle H} is a pair H = ( X , E ) {\displaystyle H=(X,E)} where X {\displaystyle X} is a set of elements called nodes or vertices, and E {\displaystyle E} is a set of non-empty subsets of X {\displaystyle...

Semiclassical

right, i see it now

yeah

Studentmath

Considering 3-uniform hypergraphs, you can try to imagine it as some triangles-sort-of-thing. I never really found it too helpful, but you can then throw some theorems about triangles in Graphs

Semiclassical

right

so it's sorta like a vertex in a polyhedra being the point of contact for multiple faces.

Studentmath

17:26

The point of the proof, is just like v2 in the above picture being in two different colours - for triples such that each member appears in 6 triples, you can 2-colour the triples (hyperedges) and cover all members s.t. they have two colours on them

Semiclassical

it's not a perfect analogy, though

Studentmath

Similar in a way, I'd say

Balarka Sen

I wonder if thinking of it as a bipartite graph with one subset of vertices corresponding to the triplets and the other subset corresponding to members help. Each member is connected to 6 triplets and each triplet is connected to 3 members

You want a 2-coloring on the subset of triplet vertices

Such that each member makes it to two colors

It seems more efficient than thinking about hypergraphs

Semiclassical

I can see a particularly simple example: Form a triangular lattice, with each vertex corresponding to a member and the faces corresponding to groups

If you want finitely many people you'll have to take a lattice with periodic boundary conditions and the right proportion of people, but that's not too bad

Studentmath

@Balarka That's nice

Semiclassical

17:32

and then it's rather plausible that you can color the faces so that adjacent faces have opposite colors

in which case you're done.

Buuuuut that's an easy version.

Studentmath

Maybe turning it the other way around would make it easier.

I mean, consider vertex colouring and not edge colouring

But then the structure won't be that neat.. nevermind

Akiva Weinberger

The complete thingy with five people satisfies this, right?

Each object would be in $\binom42=6$ triplets

Studentmath

@Akiva not sure I follow

Balarka Sen

@Studentmath We want a bipartite subgraph of the thing I said such that it looks like 2n "triplet" vertices on one side and n "memeber" vertices on the other and each member vertices is joined by two hands to one black "triplet" vertex and one white "triplet" vertex

Akiva Weinberger

@Studentmath Say there are five people, and all possible groups of three of them gives a presentation

Then each person will be in 6 presentations, if you work it out

Person A has six ways to complete his triplet out by choosing two people out of {B,C,D,E}.

And so for everyone.

Studentmath

17:48

@Balarka Sounds like that, preciesley
@Akiva right

@Akiva you can see the "theorem" is true in that case

Balarka Sen

@Studentmath It cannot happen that two distinct member vertex do not have a pair of edge coming out of them ending at different triplet vertex, yes? By counting

Studentmath

Sure. Each member vertex has 6 edges going to the other side

Balarka Sen

Well what if it looks like the complete bipartite graph K_{2, 6}?

I guess it can happen and I am wrong

Studentmath

Actually, there is a cute short embedding lemma for this. For such a bipartite graph, we can embed every 6-vertex 3-degenerate graph

(Not that it helps)

Balarka Sen

I can embed K_{3, 6} as a bipartite subgraph but not K_{4, 6}. OK, that's the limit.

Studentmath

17:54

I don't think it can look like the complete bipartite graph K_{2,6}

Balarka Sen

Why not?

Studentmath

You mean the subgraph?

Or the entire structure?

Balarka Sen

A bipartite subgraph

Studentmath

Ah, sure then

K_{3,6} is the "certain" subgraph

Balarka Sen

Yeah

Studentmath

17:55

I think

Balarka Sen

But you can't get K_{4, 6}, is the point

Okay okay okay

Anderson Felipe Viveiros

Anybody?

https://math.stackexchange.com/questions/2612101/conditions-on-a-lipschitz-function-fu-subset-bbb-rm-to-bbb-rn-which-guara

Studentmath

I think I found out where my friend took the question from
http://math.mit.edu/classes/18.095/lect6/notes.pdf

First exercise, think it's pretty much the case (maybe a coincidence, will ask him)

Balarka Sen

Aha

mercio

is it possible to ignore some users so you don't see their questions anymore ?

Balarka Sen

18:02

Put black tape on your monitor over the questions of that users that show up in your feeds

Simple

Akiva Weinberger

does

Oh no too much black tape

Studentmath

Let's say, for contradiction, no such colouring is possible.

Akiva Weinberger

Hrm

Studentmath

It means there is at least one member..

Akiva Weinberger

More Kepl than ever

Studentmath

18:06

@Akiva that's cool

Balarka Sen

Coolio

@Studentmath I considered that but it gets complicated quickly

Studentmath

It means there is at least one member, such that in every two colouring of the triples, it will be in one colour only (stating the obvious here)

@Balarka Well yeah..

Balarka Sen

Like, the condition for not being able to change the color of the triple

Studentmath

But I think a constructive proof will be even uglier

Balarka Sen

That means it's connected to something which is connected to everything of opposite color

(except one)

And ugh

Studentmath

18:10

It could be even worse, could be a chain of connections that ends up ruining all up

Balarka Sen

Yeah

Studentmath

And it could be more than one member... maybe contradiction isn't the best idea here, yeah you are right

Balarka Sen

I'll think about it. It doesn't look like a super easy problem to me

But I do think the bipartite formalism helps

Studentmath

It's definitely tougher than it seemed to me at first

Semiclassical

Ideally you could write down an actual algorithm

Studentmath

18:25

@Semi That was where I started. Somehow it just got to too many options to cover, but ideally I'd guess that's the idea

Akiva Weinberger

Hey, the intersection of compact sets is compact, right?

Wait

Dammit

In Hausdorff spaces, the intersection of compact sets is compact, right?

Studentmath

@Akiva yep

Iirc in non-Hausdorff spaces that's not always true

Silent

If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.

*Proof.* Take $n>(1/\varepsilon)^{1/p}$. (Note that the archimedean property of the real number system is used here.)

Where is archimedean property used in this proof?

Is it used to assure that there is an $n$ with $n>(1/\varepsilon)^{1/p}$?

Akiva Weinberger

@Studentmath Yeah, line with two origins. Take the two versions of $[-1,1]$ in it and intersect them

Semiclassical

My reaction to this problem: math.stackexchange.com/q/2612497/137524

be praying

Eric Auld

19:02

Suppose I take the singular value decomposition of a matrix like $\mathbf{I} + \mathbf{u}\mathbf{v}^*$.

My temptation is to take my right singular vectors to be $\mathbf{v}$ along with a basis for $\mathbf{v}^\perp$.

I know in general I can't choose my right singular vectors, but here what doesn't work? If $\mathbf{A} + \mathbf{I} + \mathbf{u}\mathbf{v}^*$, then either $\|Av\|\geq 1$ or $\|Av\|<1$.

anonra

Heya. I've been asked this question by a friend, but as a poor CS student my knowledge of probability isn't good enough to crack it. Could someone point me in the right direction?
A band whose concerts are usually 80% booked wants to book a concert hall for their next concert. How many seats should the place offer so that there's a 90% chance of only 100 seats remaining empty but also not more than 100 people not being able to get a seat.
I tried playing with the CDF of the binomial distribution and confidence intervals, but that got me nowhere and I haven't been able to find any similar qu…

Balarka Sen

Hey @EricAuld. Haven't seen you around in this chat before.

Eric Auld

In the former case $v$ is a good first singular vector. In the latter it is not, but why can't I choose a bunch of singular vectors corresponding to $\sigma = 1$ in $v^\perp$?

Hi @BalarkaSen ! Thanks for all your good answers.

Balarka Sen

Haha, the proportion of good answers I posted is not very high, but I'll take it :) I did remember enjoying answering/learning from the answers to a few of the questions you posted.

Eric Auld

Oh, I'm starting to think I can just do this the way I want. I was confused. The matrix is singular iff $\langle v, u \rangle \neq -1$, and that jives with $v + \|v\|^2 u = 0$.

Alessandro Codenotti

19:12

@Balarka remind me where your avatar comes from

Balarka Sen

Oh yeah that's from "Tale of Tales" by Yuri Norshteyn

Alessandro Codenotti

thanks

Studentmath

@Anonra I would indeed go for $p=0.8$ for a seat being taken, and play with that

@Balarka I think there are some theorems about matchings/1-factorisations in bi-regular graphs. Maybe something there could help us, with your formalisation

Balarka Sen

Mm, I don't know much graph theory. I'd like to see those theorems

Daminark

19:42

@Balarka we should retitle my analysis class "Ergodic theory" at this point

Balarka Sen

Sounds fun

Daminark

Last class we did Krein-Milman, generalized it slightly today from normed spaces to locally convex spaces

Balarka Sen

You're going to end up doing it for the space of probability measures on a compact space eventually, I suppose

Daminark

Then we said alright, so if $K$ is a compact metric space and $\phi:K\to K$ is a homeomorphism, then if you look at the space of $\phi$-invariant probability measures on $K$ (which sits in $C(K)^*$), call that $M(\phi)$, then $\mu$ is ergodic iff it's an extremal point of $M(\phi)$

Balarka Sen

Mhm

Daminark

19:46

We did this modulo the statement that

$\frac{1}{n}\sum_{k=0}^{n-1} f\circ \phi^k \to \int_K fd\mu$ in $L^2$ if $\mu$ is an ergodic measure

Balarka Sen

That's right

Daminark

This we will prove next class using an even stronger ergodic theorem

philmcole

Hello. Does anybody have some recommendations for reading material on ODEs? I'm searching for a comprehensive and well structured book which doesn't lack of detail and explains the concepts well (not for some loose Wikipedia articles).

Balarka Sen

That's the same as saying $\mu_n \to \mu$ in the weak topology, I think, where $\mu_n$ is the average delta measure along a length $n$ orbit of $\phi$

Daminark

Yeah that should be right

Balarka Sen

19:52

I need to get some chemistry done

5

Faust

$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y)}{x^2+y^2}$

Balarka Sen

Oh come on

Daminark

l m a o

Faust

L seems to equal 0 yes?

Dattier

Hi,

A sequence of real functions $f_n \in C^2(\mathbb R)$ with $\exists M>0, \forall n\in \mathbb N, f_n '' \leq M$, and the sequence simply converge to $g$.
Is-it true that $g$ is continuous ?

Dattier

20:14

The answer is true, but why ?

Balarka Sen

Year of No Light - Perséphone

►

This is very good

Daminark

20:36

Hey @Mathein!

MatheinBoulomenos

Hi @Daminark

Daminark

How's it going?

MatheinBoulomenos

Pretty good, thanks. I've got a bachelor thesis advisor I wanted, though we don't have a topic yet

And for you?

Alessandro Codenotti

Nice, do you have some ideas for the topic already?

Daminark

Hey @Alessandro! And I'm just waiting on one of my classes to upload the pset yet

Alessandro Codenotti

20:45

Hi @Dami

Daminark

In the meantime I have no purpose in life

Really it's just like, our prof in functional is already a couple days late in uploading the pset and it's prob gonna be heavy so I want to get started on it as soon as it comes up

Alessandro Codenotti

What are you doing now in functional analysis?

Akiva Weinberger

Hhey

MatheinBoulomenos

I'm probably going to do something related to number theory. Probably something that involves both algebraic and analytic techniques (I'm taking a course on modular forms before I'll start on my thesis and a second course on algebraic number theory.) I'd like to do something with elliptic curves as well

Balarka Sen

hey hey hey

Daminark

20:46

@Alessandro right now we're doing ergodic theory

MatheinBoulomenos

@Daminark your functional analysis course sounds pretty advanced

Balarka Sen

I want to look at number theory again at some point

Alessandro Codenotti

I see, sounds cool @Mathei

I'll probably do some descriptive set theory for my thesis

Daminark

@Mathein yeah it's doing good stuff. I found that one of the students in the class is TeXing notes for it

MatheinBoulomenos

I know next to nothing on descriptive set theory, but what I've seen was pretty cool

Balarka Sen

20:47

I have my copy of Iwaniec-Kowalski lying around gathering dust on my bookshelf

I feel bad about that book

Daminark

(I have classmates doing it but like, I know a grad student who has a webpage and is uploading notes to it)

math.uchicago.edu/~dsli/notes/math313_w18_notes.pdf

Tobias Kildetoft

@MatheinBoulomenos Have you read mathoverflow.net/questions/290776/… ?

MatheinBoulomenos

@Tobias no, thanks for the link

Vrouvrou

Hello, can someone explaine me why we have this: Let $X$ be a Banach space and $f,g\in X'$ then for every $\lambda\in\mathbb{R}$:$$ \sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle=\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f-\lambda g,y\rangle\leq \sup_{\|y\|=1}\langle f-\lambda g,y\rangle,$$

Daminark

Hasn't uploaded today's notes but today we briefly talked about a generalization of Klein-Milman from normed spaces to locally convex spaces, defined ergodic measures, and showed that a measure is ergodic iff it's extremal in the space of $\phi$-invariant probability measures. Except for one lemma, that $\frac{1}{n}\sum_{k=0}^{n-1} f\circ \phi^k$ converges in $L^2$ to $\int_K fd\mu$ for any $f\in L^2$ (e.g. any continuous function) if $\mu$ is an ergodic measure

@Vrouvrou so if $y\in \ker(g)$, you know that $\langle f,y\rangle = \langle f-\lambda g, y\rangle$, right?

Vrouvrou

20:52

yes my problem is with $\leq$

and this: that is, $$\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle\leq \|f-\lambda g\|,~\forall\lambda\in\mathbb{R},$$

@Daminark

Daminark

So in one case we're trying to take a sup over the set of points in the kernel of $g$ such that $\|y\| = 1$

In the second case we're also allowing a larger set, namely anything with norm 1, without restricting to the kernel of $g$

So the sup is at least as large

Vrouvrou

i don't understand you

Daminark

Okay so if $A\subset B$, then $\sup A \le \sup B$, right?

Vrouvrou

we have for $y\in\ker g$, $\langle f-\lambda g,y\rangle=\langle f,y\rangle$

@Daminark yes it is right

Daminark

Okay, so take $\{\langle f-\lambda g, y\rangle : y\in \ker(g), \|y\| = 1\} \subset \{\langle f-\lambda g, y\rangle : \|y\| = 1\}$

This is true because every $y$ in the first set satisfies $\|y\| = 1$, so it is in the second set. We're removing conditions that we require so we get a larger set

Vrouvrou

21:01

i understand

and how we get this $$\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle\leq \|f-\lambda g\|,~\forall\lambda\in\mathbb{R},$$

Daminark

$\sup \{\langle f-\lambda g,y\rangle : \|y\| = 1\} = \|f-\lambda g\|$

So by the above inequality we're done

Vrouvrou

why $\sup \{\langle f-\lambda g,y\rangle : \|y\| = 1\} = \|f-\lambda g\|$

Daminark

That's like, by definition

I mean like maybe you used a marginally different definition, but show the two are equivalent, it's a good exericse

Vrouvrou

ooo there is 4 equivalent norm on dual space right

Daminark

Well, equivalent is a technical term, but there are 4 easily equivalent definitions of the same norm

Vrouvrou

21:12

i see ok thank you very much

Leaky Nun

21:39

the concept universal property is really useful

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$Mathematics$ Mathematics