Mathematics - 2023-05-01

00:53

@TedShifrin If $f:M\to N$ is a branched covering map with $p\in M$ a branched point of order $n$, then I can find local coordinates on $p$ and $f(p)$ such that $f:\Bbb D\to\Bbb D$, $f(z) = z^n$ that vanishes on $p$ and $f(p)$ respectively. Then I consider $0$ as a fixed point.

Maybe the expression fixed point is not a very good idea

Ted Shifrin

01:06

When it’s a map between two different spaces, no, fixed point isn’t cook.

It’s the branch-switching map that has a fixed point.

one potato two potato

01:17

Ah I see. Branch-switching map. Then how can I visualize a branched point? Several sheets with points identified?

Ted Shifrin

Right. Think about your $z^n$ local model. With hyperelliptic curves, just do $n=2$.

Lemon

04:49

Quick question, for $S^1$ the circle, why are the two sets $U_x = \{ (x,y) \in S^1: x \ neq 0 \}$ and $U_y = \{ (x,y) \in S^1: y\ neq 0 \}$ not charts but cutting them up into two pieces suddenly is? Are the sets $U_x $ and $U_y$ not homemoprhic to an open set in $R^n$?

Ted Shifrin

05:12

$\Bbb R^1$

You have to give the mappings, not just the sets.

Lemon

It would be $f(x,y) = \sqrt{1 - x^2 - y^2}$ where $(x,y) \in S^1$

Ted Shifrin

What you wrote makes no sense.

Moreover, what is the definition of a chart?

Lemon

05:39

It's a pair of an open set and a homemophism to R. But i think i m seeing what is wrong. If I try to define a chart on my $U_x$, I am trying to write a map $\phi(x = \pm \sqrt{1 - y^2},y) = y$. I think I am violating some part of the implicit function theorem and that's why my $(U_x, \phi)$ can't be a chart.

Fdst Zfsy

0

Q: Find the solution of the differential equation $\frac{xdy-ydx}{x^2+y^2}=0.$

Find the solution of the differential equation $\frac{xdy-ydx}{x^2+y^2}=0.$ I though this was quite an obvious differential equation. Hence, I proceeded to solve like this: Given, $\frac{xdy-ydx}{x^2+y^2}=0.$ We can write this as, $$\frac{xdy}{x^2+y^2}=\frac{ydx}{x^2+y^2}=0\implies xdy=ydx\impli...

Hi all! Can anyone please help me to understand where I went wrong ?

Ted Shifrin

05:55

So the usual chart map will be $\phi(x,y) = y$. This won’t be one-to-one on your $U_x&.

Koro

06:06

2

Q: Can this given $f: S^1\to \mathbb C$ be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C, F$ is holomorphic on $\mathbb D$?

Suppose that $f: \mathbb S^1\to \mathbb C$ is continuous such that $f(z)=f(\bar z)$ for all $z\in \mathbb S^1$. Can it be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C$ such that $ F$ is holomorphic on $\mathbb D$? If $f$ is constant, then the result is obviously true. So suppose...

complex-analysis continuity power-series cauchy-integral-formula

I realized my mistake here: $\int_{S^1} F(z) z^{k-1}dz, k\ge 1$ need not be zero because F need not be holomorphic on $S^1$. We only have continuity of $F$.

Lemon

oh right, one-to-one. Thanks.

Rithaniel

06:59

Late night question: In combinatoric game theory, are there any "fuzzy" games whose sum is also a fuzzy game?

Koro

07:29

In an inner product space, is it true that $(a_n,b_n)\to (x,y)$ iff $a_n\to x, b_n\to y$?

$\Leftarrow$ direction is true.

But what about the $\Rightarrow$ implication?

I think it's not true.

Say $a_n$'s are in some S, and $b_n$'s are in S perp, then the limit is 0 but x and y can be chosen to be anything from S and S perp respectively.

another observation: $(x,y)= (x/c, \bar c y)$ where $c\ne 0$.

so no.

To get rid of this last bit, suppose that I have $(a, b_n)\to (a,b)$. Can I conclude that $b_n\to b$?

Alessandro Codenotti

@Koro Your example with $b_n$ picked in a $a^\perp$ still works

Koro

thanks.

Now, let's put an extra condition: $b_n$ has a convergent subsequence.

suppose that the subsequence converges to some $c$.

Can I say that b=c?

I think we still can't.

But we can say so if I put an another condition, so the revised conjecture is: In an inner product normed space (or let's take Hilbert space) if $(a,b_n)\to (a,b)$ for every $a$ in the space and $b_n$ has a cgt. subsequence with limit c, then c=b.

Proof: $(a,b_{n_k})\to (a,c)=(a,b)$ (because $\Leftarrow$ above is true.). So we have $(a,b-c)=0$ for every $a$, in particular for a=b-c. Q.E.D.

robjohn

08:35

@FdstZfsy Your questions seem to have been answered, no?

leslie townes

koro: i get lost in those last few sentences. what is b? (are you assuming or proving that the whole sequence b_n is norm convergent?)

robjohn

@Koro the same problem arises

leslie townes

note for example that if you interleave an orthonormal basis of an infinite dimensional hilbert space with a sequence of zeros, you get a sequence b_n satisfying (1) (a,b_n) converges to 0 = (a,0) for all a in the space, (2) (b_n) has a lot of subsequences that are norm convergent to 0, (3) (b_n) itself is not norm convergent to anything

robjohn

There can be two or more convergent subsequences.

leslie townes

but so many hypotheses were being swapped in and out in the discussion above that i'm not sure if this bears any relation to your problem

robjohn

08:43

Let $a=\langle 1,0\rangle$ and $b_n=\left\langle0,(-1)^n\right\rangle$

leslie townes

you might search around MSE for questions about various circumstances under which weak convergence can imply convergence in norm. such circumstances do exist but they are not everywhere

a lot of the 'pass to a subsequence' things that people do in functional analysis are sometimes necessary, in that you can't always go all the way back to your original sequence to get the same limiting behavior. this is maybe a big part of the general vibe of why people introduce topologies other than the norm topology

robjohn gives a very good example, my example shows that you can't get around the issue by orthogonality conditions (e.g. "what if nothing is orthogonal to {b_n} except the zero vector", maybe already mentioned above in connection with "S" and "S perp", but not clearly a part of the same question)

Secret

10:00

Trying to define some integral composition of functions along some nonassociative convergent net
$$\int_{y\to\ell_{2}}\int_{x\to \ell_{1}} \mathcal{G}\circ \mathcal{H} d T = \lim_{y \to\ell_{2}}\int_y^{id} G\lim_{x\to \ell_1} \int_x^{\circ} H d[U]d[V]$$
where
$$\int_{x\to \ell_{1}} \mathcal{G}dT =\lim_{x\to\ell_1} \int_x^{id} G d[U] = \left(\mathscr{R} \circ \mathscr{M}\right)_{L_1}\circ G= \left(\lim_{I \to L_1}\prod_{i \in I} F_i\right)\circ G = \left(\lim_{I \to L_1} A\circ B \circ C \circ D \cdots \right) \circ G$$

where > < is some function that conveys something like "< a but > b"

shintuku

10:53

what's a good topological space to study to understand how a cauchy sequence might not converge?

Thorgott

$\mathbb{Q}$

shintuku

oh

i get it

hm. is there a topological space where a cauchy sequence doesn't converge to a point outside the space either?

Thorgott

"point outside the space" doesn't make a whole lot of sense

but to answer the spirit of this question, any metric space (which is the setting in which Cauchy sequences make sense) embeds as a dense subset into another metric space, which is complete

shintuku

i see, thanks for pointing out the proper way to think about it

one potato two potato

11:55

If $U_n$:open such that $U_n\subset U_{n+1}$ for each $n\geq 1$ such that there is a closed set $F$ contained in $\bigcup_{n=1}^\infty U_n$, then $F\subset U_n$ for some large $n$? Ambient space is $\Bbb R^m$ (for simplicity).

PM 2Ring

12:05

@geocalc33 Sure. You don't need to worry about the sum inside the zeta functions, Sage knows how to handle that. Eg, if you ask for zeta(2) it returns pi^2/6. It can also do sums, including infinite sums, symbolically.

OTOH, for this task we really don't want Sage to compute stuff symbolically. If we feed it the Gram series with k going to infinity it will create a symbolic expression that it can't evaluate numerically. If you set an upper limit on k it produces a huge symbolic expression which it can evaluate, but it's rather slow. And you then need to figure out what's an appropriate upper k for a given x, and the numeric precision that you want.

Fortunately, there's an alternative strategy: mpmath, which is included with Sage. mpmath has a variety of tools for handling infinite sums, and it has an excellent built-in zeta function. mpmath.org/doc/current/index.html

one potato two potato

Is it possible that Fatou set is empty?

PM 2Ring

Here's a demo comparing Sage & mpmath computations of G(x). The expression g(xx).n(digits=digits) does the numeric conversion. print g(xx) to see the horrendous symbolic expression. ;)

sagecell.sagemath.org/…

shintuku

@onepotatotwopotato don't we have $\bigcup U_n = \mathbb R^m$?

one potato two potato

Why?

I don't assume that

shintuku

hm right

PM 2Ring

12:15

sagecell.sagemath.org/…

shintuku

so $\bigcup U_n \neq \mathbb R^m$, so the closure of $U_n$ in $\mathbb R^m$ converges to the closure of$\bigcup U_n$ in $\mathbb R^m$, so we can't have $F$ sharing the same closure, therefore $F$'s closure is closer to the center of $U_1$ than $\bigcup U_n$'s @onepotatotwopotato

so there is a distance between the closure of $F$ and the closure of $\bigcup U_n$, so there must be an $U_n$ who's closure is there

if you suppose $\bigcup U_n = \mathbb R^m$ then we trivially have $F$ in there

one potato two potato

12:47

@shintuku The union is taken to be infinitely many

shintuku

yeah, and the union is either equal to R^m or not. if it is, then clearly F in some U_i. if it isn't then the closure of U_i must converge to the closure of the union

and F cannot have the same closure as the union because it is closed

F's closure must be strictly within the closure of the union, since the union is open

therefore there is a distance separating their closures, and there is an U_i whose closure is inbetween

one potato two potato

If it's infinitely many, the inclusion maybe proper

shintuku

proper i.e. strict?

if F is a strict subset of the union then even more clearly we have an U_n who's closure is between the closure of F and the union

one potato two potato

I mean closure

shintuku

the union of open sets is open

wait i don't know the infinite case

confirmed, it is still open

Thorgott

13:17

@onepotatotwopotato $U_n=(-n,n)$ and $F=\mathbb{R}$?

one potato two potato

Three people answer the same answer at once

The note I'm reading use that fact and I wondered why and while I'm typing the question, I realized the ambient space is compact.

geocalc33

13:54

sagecell.sagemath.org/…

@PM2Ring I figured it out last night, but you're right this way is slow. Thanks!

PM 2Ring

14:27

@geocalc33 No worries. Sage has a built-in prime_pi function, but it's a bit slow when you want to calculate it for lots of values. It's ok for 10,000 or so values, but it's not great. My Numpy version is much faster.

Fdst Zfsy

16:19

3

Q: Solve the differential equation $(2x^3y+4x^3-12xy^2+3y^2-xe^y+e^{2x})dy+(12x^2y+2xy^2+4x^3-4y^3+2ye^{2x}-e^y)dx=0.$

Solve the differential equation $(2x^3y+4x^3-12xy^2+3y^2-xe^y+e^{2x})dy+(12x^2y+2xy^2+4x^3-4y^3+2ye^{2x}-e^y)dx=0.$ I tried solving the problem like this: We assume $M=(12x^2y+2xy^2+4x^3-4y^3+2ye^{2x}-e^y)$ and $N=(2x^3y+4x^3-12xy^2+3y^2-xe^y+e^{2x}).$ Now, we observe, $\frac{\partial M}{\parti...

ordinary-differential-equations solution-verification alternative-proof

Can anyone please help me get over this dilemma?

@robjohn Sorry, for responding this late, but no I am not clarified yet.

First of all, I want to know, whether this method is valid or not?

Second of all, for my previous of my two posts, I am getting flooded with ambiguous answers.

Ted Shifrin

17:07

No, your method is certainly not valid. If you have $(P_1+P_2)dx+(Q_1+Q_2)dy=0$, there's no reason that this should imply $P_1\,dx+Q_1\,dy=0$ and $P_2\,dx+Q_2\,dy=0$. The converse is true, yes. Just try a simple case: Can you solve $dx+dy=0$ by setting $dx=0$ and $dy=0$? Of course not.

Ted Shifrin

17:28

@FdstZfsy See my response above. I forgot to ping you.

robjohn

I believe what Fdst Zfsy was asking me was about this question.

Ted Shifrin

I was responding to the one he just linked.

robjohn

@FdstZfsy The problem with solving $x\,\mathrm{d}y-y\,\mathrm{d}x=0$ is that it we get to $\log(y/x)=c$, as you said, which gives $y=e^cx$. As was mentioned in comments, this implies that the constant is positive (unless one expands the underlying space to $\mathbb{C}$). One way to get passed this is to use $\log(|x|)+C$ as the indefinite integral for $\frac1x$. Then we get $|y/x|=e^c$, which gives $y=\pm e^cx$.

The '$\pm$' can be rolled into the constant, letting $m=\pm e^c$ and get $y=mx$ for any real constant $m$.

@TedShifrin I know

Ted Shifrin

Yes, it's interesting to throw away the denominator of $d\theta$ and think the solutions are different :) Of course, the arctan is only partly correct.

robjohn

at least one can get any real value for $y/x$

oh, no, that's not right. You're correct

Ted Shifrin

17:34

Funny: We turn a closed $1$-form (locally exact differential) into a non-exact differential because it's easier to solve. :) I don't usually think this way.

robjohn

Oh, wait. I keep flipping which side the tan is on. You can get all real constants.

shintuku

in a metric space for any open cover there exists a radius r s.t. any r-ball is contained in at least one element of the cover

how can i see the truth of this, i can't manage to convince myself

not a proof

why can't we have an infinite open cover whose elements become progressively smaller

robjohn

That is not necessarily true... cover the reals $\ge1$ with balls centered at $\log(n)$ with radius $\frac1n$.

shintuku

oh no, i forgot the extremely nontrivial and defining element of this problem

the space is compact

robjohn

Ah, then any open cover has a finite subcover

Then you can do it

shintuku

17:45

sorry again, i meant sequential compactness

the proof to go from sequential compactness to compactness uses the above mentioned lemma (to do a proof without contradiction)

it seems more intuitive to me that, given a sequentially compact space, represented by a square in R^2, there's a way to cover it sort of like a geometric series on a square

then the covering subsets get infinitely smaller, so there's no place we can find a minimal subset size

Sgg8

Hi. Can anyone help me with this question about Boolean function minimization I posted a while ago? math.stackexchange.com/questions/4484690/…

shintuku

i'm going to go search for some spaces that don't have equivalence between sequential compactness and usual compactness to see if that makes this issue clearer

Ted Shifrin

@shintuku You can’t. You need a metric space for the statement to make sense.

shintuku

18:01

hm, the notion of an r-ball

Ted Shifrin

Define ball without a metric?

shintuku

right

Ted Shifrin

Anyhow, you can easily do a proof by contradiction using sequential compactness. I assigned that every year in my multivariable math class.

shintuku

at the bare least, the above mentioned method will fail to contain the sort of limit to these covering attempts, so that specific example fails

Ted Shifrin

Huh?

shintuku

18:09

this sort of attempt to cover the square won't contain the 'limit', so it's not a cover

i'll see how i can use sequential compactness

thanks for the suggestion

shintuku

18:50

yep that's the proof, any attempt to cover a square this way leaves out the limit

PlaceReporter99

19:02

Time for a new field of calculus:
$\frac{d^{f(x)}}{d^{f(x)}x} f(x) = ?$

I call this “variable-squared calculus”

Alessandro Codenotti

19:31

@shintuku what you're talking about is called the Lebesgue number lemma

The proof if easy once you know the correct trick, you can read it from wikipedia

Ted Shifrin

@Alessandro I purposely didn't give the name because I wanted to let shin figure it out himself :)

Alessandro Codenotti

Woops

To be honest it's not the kind of statement I expect people to figure out on their own

I don't think I would have been able to figure out the proof myself when I first encountered this lemma (iirc this was in algebraic topology, I think it's used when proving excision for singular homology at some point? After doing some barycentric subdivisions maybe)

D.C. the III

@AlessandroCodenotti Ted made me suffer for a few days trying to do the proof when it was asked in his book.....😭😭

Alessandro Codenotti

With some hints it's doable

Also Ted is a far better teacher than I am, so I'd trust his opinion on this one

D.C. the III

Don't give him the satisfaction of exerting his brand pain. 😂

shintuku

19:43

i did the proof

before alessandro's tag sunglasses

you construct a sequence left out of a finite number of cover elements, then get their point of convergence with sequential compactness

and that point is outside any infinite covering attempt

thanks for the tag to the proper theorem tho

but uh i've been reading similar proofs i wasn't able to understand so it wasn't really my idea

Ted Shifrin

@D.C.theIII That might have been the finite subcover exercise, as I recall.

D.C. the III

I'll tell you now since I have the text open. I also have a question about how to assign variables.

Yes it was #13 - Sec 5.1

Ted Shifrin

20:01

I assigned both, but the particular struggle I recall with you ….

D.C. the III

Lol...think it had to do with the logic in deducing things

Ted Shifrin

Logic? What’s dat?

shintuku

it's a type of irish jig some lawyers do

it used to be spelled law-jig in middle english

it was popularized by the irish catolic church and their readings of ancient greek dancers like aristotle

D.C. the III

At the end of the last video yo set up the problem of finding the integral over the region between $y = 2x^2, y = x, xy = 1, xy = 3$. You did the variable assignment of $u = xy$ and $v = y/x$. I could see why you would choose $u = xy$, because it should give us simple constants when we look at things in terms of $u$, but how did you choose $v = y/x$?

if you don't want to remember all that @TedShifrin it was at this point: youtu.be/Rup0XQ8pzgM?t=2828

Ted Shifrin

To simplify $y=x$.

Sometimes the function one’s integrating is a hint, too.

D.C. the III

20:11

So what we want to do with our choices is simplify each of the boundaries of the initial region.

So in the $v = y/x$ case we are turning $y$ into a scalar multiple of $x$, but we will be using that scalar (in this case $v$) to be one of the boundaries of the simplified region.

let me see how you treat things in the next video and I'll come back if I have questions about it

shintuku

don't forget to try law-jig first

dancing releases endorphins which improves cognition

D.C. the III

over rated process, goal is to be an automaton

logic requires too much expenditure of energy.

Koro

20:43

user1294729

Can someone help me here:

0

Q: How can I prove this characterisation of measures being absolutely continuous?

Let us consider $T:[0,1]\rightarrow [0,1]$ be a continuous map. Let $\mu, \nu$ be $T$ invariant probability measures on $\mathcal{B}([0,1])$. I want to show that if there exists $D>0$ s.t. $\int_{[0,1]} f~d\nu\leq C\int_{[0,1]} f~d\mu$ for all $f:[0,1]\rightarrow \Bbb{R_+}$ continuous then $\nu ...

functional-analysis analysis measure-theory borel-measures absolute-continuity

Koro

what is this symbol on left?

so one of the 'teachers' here is also 'teaching' introductory Linear algebra course to first year bachelors students. He's 'teaching' them modules, group theory etc. with little to emphasis on matrices or linear maps etc.

he's 'teaching' introductory functional analysis to us also. he didn't talk about Fredholm theory (available in for example, Kreyszig's) but 'taught' Peter Weyl theory and group representations, Haar measure etc.

Sine of the Time

@Koro $\otimes$ ?

Koro

yeah

Sine of the Time

isn't it tensor product?

Koro

20:49

I mean the meaning of it here. What is it doing?

Also, what is I here?

Sine of the Time

don't know, I've seen sometimes indicates tensor product but in this case I've no idea

Koro

there is 'product of measures' where this symbol is used. But I wonder what measure I is.

Sine of the Time

@Koro "let us intoduce few notation" but does not specify I :(

Koro

that's how his notes usually are- lacking head and tail.

Sine of the Time

that's sad

Koro

20:51

@SineoftheTime exactly.

This is related to Haar measure.

Ted Shifrin

Think of as defining two measures on $G\times G$ …

Koro

Showing existence of Haar measure on a compact topological group in an introductory functional analysis course.

Ted Shifrin

That’s not unreasonable.

If you can’t do serious math in grad courses, what the hell.

Koro

the course looks like algebra. Haar measure, fine. But group representations, Peter Weyl theory? Why?

I mean do everything in class but the exam should be set from the official syllabus of the course.

Putting questions from these topics (not in syllabus) in exam is nonsense.

user1294729

@TedShifrin haven't you wrote a comment to my question?

Koro

20:55

there are other subjects too and only 24 hrs in a day.

Ted Shifrin

@user1294729 I commented on a comment that was removed. He suggested looking at the characteristic function of, say, an interval. It’s not continuous, but it can be approximated by continuous functions. Try playing with that.

Koro

I hadn't studied functional analysis before. I'm a beginner. I expect studying compact operators, open mapping theorem, Hahn Banach theorem, Uniform boundedness etc. But group representations, Peter Weyl theory, Measure theory intrusion whenever he can. Why?

He is offering a course next sem to us. Guess what no one is taking that.

I wonder what advanced functional analysis course would be like.

If introductory fa has Peter Weyl theory but no Fredholm theory in it.

user1294729

@TedShifrin okey but is this true what I did or is it compleatly useless?

Ted Shifrin

Well, you’re the expert, as usual.

user1294729

Are you talking with me? Why I'm the expert?

Ted Shifrin

21:02

No, not you.

user1294729

ah okey

Koro

I'm no expert. I'm confused.

I have too many 'why's with no answer.

Ted Shifrin

I missed that you were already doing the char function idea. But that is not continuous. I would start with showing it works for intervals and then play set games. Why does it work for intervals?

user1294729

I'm a bit confused what I can now assume and what I need to show I mean if I take an intervall $[a,b]$ I need to take the characteristic function of the intrval, otherwise I cannot assume the assumption.

D.C. the III

22:00

chaos

Ted Shifrin

Everywhere

D.C. the III

22:18

Do you come back to covering the cubical norm in the video lectures?

Ted Shifrin

22:48

No. I spent only half a lecture talking about the proof, if I recall.

D.C. the III

Haven't watched that vid yet. That's the next one to watch

GUess I will encounter it eventually in my travails

shintuku

23:10

or your trabajos

or your muehen

is there such a thing as having an intuition about the behaviour of a 10-dimensional manifold

Silly Goose

23:32

In the proof showing (i iff ii), are we implicitly defining like a new category whose objects are sets (morphisms from $C$) and whose morphisms are all isomorphisms of $C$?

then we are showing that the isomorphisms of $C$ are still isomorphisms in our new category. in particular, isomorphsims between sets and so bijections? (it is assumed that the original category is locally small)

Thorgott

23:48

I'm not sure you're envisioning this quite correctly, but something similar is true. Look up the Yoneda embedding.

Transcript for

$Mathematics$ Mathematics