Mathematics - 2020-04-15 (page 1 of 2)

CaptainAmerica16

00:04

If $x \in f(E \cup F)$, then $x \in E \cup F$, right?

Thorgott

unless $f,E,F$ are supposed to be very specific things, no

CaptainAmerica16

Could you explain a little bit?

Ted Shifrin

@CaptainAmerica16: How can $x$ be in the domain and in the codomain at the same time?

CaptainAmerica16

Oooh

Thorgott

you can look at a function with disjoint domain and codomain

Ted Shifrin

00:10

Here's a hint for you, @Captain. I always tried to train my students to use $x$ in the domain and $y$ in the range. Or if $f\colon Y \to Z$, then use $y$ in the domain and $z$ in the range, etc.

CaptainAmerica16

Yeah, that definitely makes things clearer.

Ted Shifrin

It makes it easier to follow, too.

CaptainAmerica16

I'm trying to figure this out. I'm overthinking it somehow.

Thorgott

Try plugging in the definitions one by one

CaptainAmerica16

hm...

Thorgott

00:19

Start with $y\in f(E\cup F)$. What is this statement equivalent to by definition?

Ted Shifrin

You need to know, as a rule of thumb, that you prove two sets are equal by showing inclusion both ways. Or you write a chain of $\iff$'s.

CaptainAmerica16

I need to think about this for a minute.

Ted, you mean show that if x is in f(E U F) then x is in f(E) U f(F). And then show that if x is in f(E) U f(F), then x is in f(E U F). That'll show that they're equal.

Ted Shifrin

Use the right letters, please.

Yes, that's what I mean, though.

CaptainAmerica16

lol, sorry

Ted Shifrin

You might find that your arguments here are $\iff$ ... but not always.

CaptainAmerica16

00:54

@Thorgott Well, we know that $f(E \cup F)$ is defined as ${f(x) : x \in E \cup F}$. So I guess we could say that if $y \in f(E \cup F)$, then there exists an $x$ that $y = f(x)$. I don't know if that's what you were looking for.

Thorgott

Yes, but you don't just want any $x$. The $x$ must satisfy a certain condition.

CaptainAmerica16

The only thing I can think of is that x must be in $E \cup F$

Thorgott

right

now what does that mean per definition of union

CaptainAmerica16

then $x \in E$ or $x \in F$

Thorgott

so what do you get in the overall equivalence?

CaptainAmerica16

01:04

I think this is where I can start to use f(x), right?

$f(x) \in f(E)$ or $f(x) \in f(F) \rightarrow f(x) \in f(E) \cup f(F)$

I feel like that's jumping the gun, though

Thorgott

That's true, but I don't see how that continues from what we have so far.

CaptainAmerica16

exactly

give me a sec

maybe $f(x) \in f(E \cup F) \rightarrow x \in E \cup F$

Thorgott

that's sadly not always true

CaptainAmerica16

0-0

Thorgott

@CaptainAmerica16 you have an equivalence in this message. now try rewriting this using the definition of union

CaptainAmerica16

01:18

the equivalence is $y=f(x)$ so $f(x) \in f(E)$ or $f(x) \in f(F)$

Thorgott

no, that's what you want to prove

CaptainAmerica16

Good grief.

@Thorgott ok, I'm trying to find an equivalence the $y \in f(E \cup F)$, right?

using the fact that $x \in E$ or $x \in F$?

Thorgott

yes

but keep in mind that when you want an equivalence to $y\in f(E\cup F)$ you always assert the existence of a certain $x$

CaptainAmerica16

If $y \in f(E \cup F)$, then there exists an $x \in E \cup F$ such that $y = f(x)$.

This is the assertion, yeah?

Thorgott

not only "if then", it's an "if and only if"

CaptainAmerica16

01:33

oh wow

Thorgott

because that's what $f(E\cup F)$ is defined to mean

CaptainAmerica16

Yeah, that makes sense.

I don't know why I can't pinpoint this equivalence thing. I feel like I have it, but I'm just not stating it correctly.

CaptainAmerica16

01:54

@Thorgott I'm gonna take a mini-break. I can usually figure it out if I redirect my attention for a bit. Can I ping you if you're not here later on?

Thorgott

I'm about to go to sleep, so just ping me (though I assume someone else can take over)

CaptainAmerica16

Ok, np. I really appreciate the help. You cleared up some things I obviously misunderstood :)

Adam L

02:14

the things that lead you to select the user name captain America?

CaptainAmerica16

@AdamL I just really like Marvel.

Adam L

That's not the same thing as selecting a user name of their worst hero ever

CaptainAmerica16

@AdamL I beg to differ.

Adam L

well you don't have to beg fella everyone thinks that is tops just wear a helmet everywhere you go ok

CaptainAmerica16

;-;

I don't think that. Cap is #1.

Adam L

02:18

emojis don't parse here champ

CaptainAmerica16

I could ask you why your profile pic is a donkey named Adam.

Adam L

Captain Actual America Overweight, Hopelessly In Debt

►

gl

CaptainAmerica16

@AdamL Wow XD

Bob

03:11

Hi

I posted a question on the electrical engineering stack exchange

it consisted of a problem from a text book and my solution to it

The question got voted down because somebody said I made no attempt to solve the problem.

Any thoughts?

Knight

04:06

@Bob They want you to show some attempts in order to guide you well enough. It could be that your “my solution” a copy-paste thing from textbook solution. You should show your own solution, what YOU tried and then you will be helped.

Ted Shifrin

05:06

To clarify, did you make it very clear that it was your solution and that you sought feedback?

Omnomnomnom

@CaptainAmerica16 Actual the pic is a llama named Carl

CaptainAmerica16

05:25

@Omnomnomnom I've never seen that llama before

@TedShifrin I just realized what you meant by a chain of $\xleftrightarrow$'s.

Um, I don't know how to make that arrow with mathjax...

Leaky Nun

06:18

$\iff$ \iff?

skullpatrol

CaptainAmerica16

@LeakyNun thanks

skullpatrol

06:39

Out of curiosity @CaptainAmerica16 what was the first "if and only if" formal proof that you were shown in school?

CaptainAmerica16

@skullpatrol None, self studying

I'm still in HS

skullpatrol

You should have been shown one in algebra1, namely the zero-product property. @CaptainAmerica16

CaptainAmerica16

We weren't taught proofs in school. What do you mean?

Leaky Nun

that's the $xy=0 \iff x=0 \lor y=0$ thing?

I've never heard of the name "zero-product property"

@CaptainAmerica16 fun fact: when you solve an equation, you're really finding simpler equivalent conditions

skullpatrol

Yes, if a and b are real numbers: ab=0 iff a=0 or b=0.

Leaky Nun

06:47

so each step is actually an $\iff$

extraneous solutions show up iff one your steps is just an $\implies$

CaptainAmerica16

huh, never thought of it that way

Leaky Nun

e.g. let's solve $\sqrt x = 1$

square both sides is an $\iff$

because A=B implies A^2=B^2 but not the other way

I guess this example doesn't work

I can't be bothered to come up with one that works

skullpatrol

lol

CaptainAmerica16

lol

Leaky did you see the problem I was working on earlier?

Leaky Nun

7 hours ago, by CaptainAmerica16

If $x \in f(E \cup F)$, then $x \in E \cup F$, right?

this?

CaptainAmerica16

06:49

I don't have to do a series of equivalences, do I? Can I just go through inclusions?

Yeah, that

Leaky Nun

7 hours ago, by CaptainAmerica16

or maybe this one

CaptainAmerica16

Wait, not not that

yeah, that lol

Leaky Nun

sure, use inclusions

and draw a picture

CaptainAmerica16

That way is a lot more intuitive for me

Ok, it's almost 3 am. I'm going to settle this once and for all in the morning.

skullpatrol

cya, pal

CaptainAmerica16

06:53

cya, thanks for the help guys.

Anjani Gupta

If f is analytic on and inside some simple closed contour, there are only a finite number of zeroes. What if there is a pole of f interior to the contour as well?

Leaky Nun

@AnjaniGupta if the pole is $p$ then $(z-p)^n f$ is analytic

Anjani Gupta

07:14

Ok this argument will work if I can ensure that that this new analytic function is non zero on the contour..

Leaky Nun

Hong Kong: the lack of exponential growth

Anjani Gupta

Great, HK had controlled it !

*has

Balarka Sen

07:36

they have dealt with SARS before obviously they'd be the ones most equipped and experienced to handle this kind of thing

im sure the fact you told me about everyone wearing masks when they go out helped immensely, @LeakyNun

like how thats built into habit now

Leaky Nun

yeah

skullpatrol

Meanwhile, in the states they are bearing arms against mask wearing Asians.

Balarka Sen

Idiots

skullpatrol

While trump wants to lift the lockdown in two weeks.

Balarka Sen

There are some people who are rightly criticizing the lockdown. It's impossible to do a complete lockdown without making the economy crash down, and a semi lockdown is only as good. Testing and segregation + hygiene actually has nearly the same effect as a total lockdown and is less damaging to the country

Of course, lockdown, especially after the disease has spread immensely, is also a convenient way to cover up the government's incompetence, so naturally more popular

Everyone's handling it terribly and the effects will go a long way. We'll see.

Alessandro Codenotti

07:52

@Balarka Say I have an embedding $S^2\to\Bbb R^3$ such that the exterior has nontrivial $\pi_1$, can it be finitely generated? Or somewhat nice?

How many nonisomorphic groups can be realized this way?

Balarka Sen

That's a good question. The exterior has to be a homology ball, for sure. So $\pi_1$ is a perfect group.

Let's see what else I can deduce off the top of my head

@AlessandroCodenotti Let me one-point compactify to an embedding of $S^2 \to S^3$ and look at the exterior instead. The fundamental group doesn't change in this operation because I'm adding a 3-ball. In any case, the exterior in $S^3$ definitely has zero $\pi_2$, because take any element of $\pi_2$ and take a smooth representative so that it bounds a ball in both components, therefore null in the exterior.

Alessandro Codenotti

Makes sense

Balarka Sen

Let's fix notation; $M$ is the exterior. We deduced $\pi_1 M$ is perfect, $\pi_2 M = 0$. Consider $\widetilde{M}$, the universal cover, which is a noncompact 3-manifold, so by Hurewicz $\pi_3 M = H_3 M = 0$, and all subsequent higher homotopy groups are zero by Hurewicz.

Then $\widetilde{M}$ is contractible, so $M$ is a $K(\pi_1 M, 1)$ space. That implies $\pi_1 M$ is torsion-free, because otherwise it wouldn't have finite homological dimension

It's a torsion-free perfect group. Hrm.

Alessandro Codenotti

You lost me in the argument, but the final result is nice

Balarka Sen

Yeah I don't want to elaborate just yet, not sure if this helps.

OK, if $\pi_1 M$ was finitely generated then $M = K(\pi_1 M, 1)$ would have a CW structure consisting of a finite $2$-skeleton.

That should lead us somewhere.

Wild sphere exteriors in $S^3$ should not have CW structures which are anywhere close to finite

Bruh

What are examples of finitely generated groups $G$ with $H_k(G) = 0$ for all $k$?

I don't know any example lol

Astyx

08:21

$\{0\}$ ? :)

Balarka Sen

Haha

Nontrivial

Astyx

Are you sure there are any ?

Balarka Sen

Nope. But I also don't see why not

Astyx

Is $H_k$ the same homology as in topology ?

Balarka Sen

Oh no group homology

Tor_ZG(Z; Z)

Higman group

In mathematics, the Higman group, introduced by Graham Higman (1951), was the first example of an infinite finitely presented group with no non-trivial finite quotients. The quotient by the maximal proper normal subgroup is a finitely generated infinite simple group. Higman (1974) later found some finitely presented infinite groups Gn,r that are simple if n is even and have a simple subgroup of index 2 if n is odd, one of which is one of the Thompson groups. Higman's group is generated by 4 elements a, b, c, d with the relations a −...

This is the one

Astyx

08:26

Oh cool

Kevin. S

09:08

If I have a path-connected topological space $X$ and $A\subset X$, a retraction of $X$, We can get an isomorphism between the fundamental group of $A$ and $X$ and also $\pi_1(A)$ is a subgroup of $\pi_1(X)$ because $A$ is a subspace of $X$, Now, the question is can we rewrite $\pi_1(X)$ as $\pi_1(X)=\pi_1(A)\times \pi_1(X)/(\pi_1(A))=\pi_1(X)\times \pi_1(X)/(\pi_1(X))=\pi_1(X)\times \{1\}=\pi_1(X)$? Seems like this is correct, but I'd like to verify it. Thank you very much.

I still think there is something wrong in my statement, because of this quotient: $\pi_1(X)/\pi(X)$, is it really isomorphic to the trivial group, or do I need extra condition to make it true...?

Balarka Sen

(1) Just because $A$ is a subspace of $X$ doesn't mean $\pi_1(A)$ is a subgroup of $\pi_1(X)$. Take $X = \Bbb R^2$ and $A = S^1$, the unit circle on the plane. (2) If $A$ is a retract of $X$ that does not mean $\pi_1(A) \cong \pi_1(X)$. In fact, being a retract is what tells you $\pi_1(A)$ is a subgroup of $\pi_1(X)$. (3) If $A$ is a deformation retract of $X$, then $\pi_1(A) \cong \pi_1(X)$.

Kevin. S

@BalarkaSen OK I see, thank you, and what about my weird equation to rewrite $\pi_1(X)$? I'm afraid it isn't correct...

Balarka Sen

Well none of what you wrote makes sense unless you mean $A$ is a deformation retract of $X$

Did you mean that, first of all?

Kevin. S

@BalarkaSen I said in the first sentence that A is a retraction of X..

Balarka Sen

Retraction is not the same as a deformation retraction.

See point (2) and (3).

courge9

09:17

Can the product of an unbounded and a bounded operator be bounded?

Leaky Nun

@courge9 0

courge9

I interpret that as a "no"?

Leaky Nun

it means take 0 to be the bounded operator

courge9

Oh, I'm an idiot

Alessandro Codenotti

It can happen for nonzero operators as well

Wait what do you mean with product

courge9

09:19

usual product of operators, i.e. concatenation of maps

with obvious choice of domain

Alessandro Codenotti

Look at the map $T:\ell^2\to\ell^2$ given by $(x_n)_{n\in\Bbb N}\mapsto(\frac1n x_n)_{n\in\Bbb N}$ and its "inverse"

Leaky Nun

I guess you can expand a lot and then contract by the same amount

yeah exactly

courge9

Ah, ok. So if we know that the product AB of two operators A,B is bounded and one of them is bounded, this doesn't have to imply the boundedness of the other

I thought so, but for some reason I couldn't come up with an example

thanks!

Kevin. S

@BalarkaSen Oh!! I get that!

Mad Spaces

11:48

Can some assist me solving with induction this inequality $(\frac{n+1}{2})^n$$>n!$

For $n$ equals 2 it works. ) the inequality works for n bigger 1 of the naturals.
Consider now $(\frac{n+1+1}{2})^{n+1}$$>n+1!$ = $(\frac{n+2}{2})^n *(\frac{n+2}{2})$$>n!*(n+1)$

How do i proceed to show now that it works since i $(\frac{n+2}{2}) < (n+1)$ i dont know

Well i proved it alright, but it aint no induction: can you check if this argumentation is true?

we know for atleast $n$ = 2 this works. suppose there is an $n_0$ such that this inequality doesnt work. Since the Rationals are odered ( thus should apply that for $(n_0)!$ its bigger than $(\frac{n_0+1}{2})^{n_0}$ right? well then you can say that $(n_0)!$=$(n_0-1)! * n_0$ right? we can disolve the left side using the generalization of the binome formela or whatever you call it. Then devide both sides with $n_0$ and you get on the left side the binome formel for $(\frac{n_0+1}{2})^{n_0-1}$

But our assumption was that $n_0$ is the smallest number for which this doesnt work and now we have the same equation for $n_0-1$ and we can continue on doing such and making the same assumption until we reach 2. We know for 2 it works, thus there exists no such element $n_0$

VIVID

12:20

@MadSpaces Assume that it's true for $n$.

Mad Spaces

12:37

i assume the following , there exists the smallest number $n_0$ such that from this number on the ineq doesnt work. Why should i change the assumption to $n$? Could you please elaborate.

Balarka Sen

12:48

System Update With Glenn Greenwald - Edward Snowden, Andray Domise and Cassie King

►

Very crucial conversation

Kevin. S

Another Question: We can understand the quotient of group as identifying the elements of the equivalence class, right? Then, If a homomorphism $f:G\to H$ maps a group G to its normal subgroup H surjectively, then the group $G/(G/ker(f))$ can be understood as: First, identify all the kernel elements so there remains an identity element and the rest of the elements of $G$. Then, we identify the elements of $G/ker(f)$ in $G$ so that only kernel elements remains in the final result?

Balarka Sen

? G/ker(f) is not even a subgroup of G

feynhat

I think he is identifying G/(ker f) with f(G), which is in H.

Alessandro Codenotti

And that is not a subgroup of G, as Balarka was saying

Balarka Sen

But he is writing G/(G/ker(f))

feynhat

12:51

@AlessandroCodenotti maps a group G to its normal subgroup H surjectively.

Balarka Sen

Oh fine.

feynhat

@Kevin.S Since it is surjective you might as well write G/(ker f) as H.

Alessandro Codenotti

@feynhat Oh ok but then that's a weird way to write G/H

feynhat

@Kevin.S ^

Kevin. S

Ok, fogive me, I'm just having fun with groups..

Thanks to all of you.

Balarka Sen

12:55

Small note: If a group $G$ surjects to a normal subgroup $H$ of $G$ "naturally" (i.e., the surjection is identity on $H$, so you can think of this as a retraction - in the sense of topology - of $G$ to $H$) then the short exact sequence $1 \to \ker f \to G \to H \to 1$ splits, the section given by inclusion of $H$ in $G$, and $G$ becomes a semidirect product of $N = \ker f$ and $H$.

Kevin. S

@BalarkaSen I'll try to understand that, but now it sounds really abstract to me.

Balarka Sen

This is one of the few equivalent ways of understanding a semidirect product.

Alessandro Codenotti

I think it is actually an honest retraction on the level of Cayley graphs

Leaky Nun

"semidirect product is just a section"

@AlessandroCodenotti nobody thinks in Cayley graphs

maybe even deformation retract

Alessandro Codenotti

May I introduce you to geometric group theory

Balarka Sen

12:59

Garbage, @Leaky

Leaky Nun

which one is garbage?

Balarka Sen

The deformation retract comment

Alessandro Codenotti

It's not a deformation retraction, not even for direct products

Leaky Nun

I guess I'm thinking of one generator for the other subgroup

Mike Miller

@BalarkaSen Agree, Cayley graphs are garbage

Leaky Nun

13:00

I mean all I have is the picture of the Cayley graph of $D_{n,2n}$ in my head

Balarka Sen

Garbage, @MikeMiller

Mike Miller

@LeakyNun Garbage

Balarka Sen

^

Alessandro Codenotti

That's a lot of garbage

Leaky Nun

Balarka Sen

13:03

Oh. Kevin was surjecting to a normal subgroup of $G$, right? So if that's done naturally it's all very sad, because $G$ then becomes $\ker(f) \times H$

You shouldn't do that.

@Alessandro There should be a way to understand $\Gamma(N \rtimes H)$ as a twisted graph product of $\Gamma(N)$ with $\Gamma(H)$, and the projection to $\Gamma(H)$ should indeed be a retract.

Alessandro Codenotti

I think we talked about that once already

Balarka Sen

Yeah

Kevin. S

Just enjoy

Alessandro Codenotti

here chat.stackexchange.com/transcript/message/53440825#53440825

Leaky Nun

@AlessandroCodenotti garbage

Balarka Sen

13:08

How much hyperbolicity is lost when you take product of hyperbolic groups?

$F_2 \times F_2$ is of course not hyperbolic; you have embedded $\Bbb Z^2$

Mike Miller

I guess you need to start studying the metric geometry of $\Bbb H^n \times \Bbb H^m$

Balarka Sen

Maybe it's relatively hyperbolic, relative to those torii

@MikeMiller Yeah sounds cool I don't know

I am betting serious money $F_2 \times F_2$ is hyperbolic relative to $\Bbb Z^2$

@Alessandro prove it

Alessandro Codenotti

Sorry, I'm a PDEs guy until Friday

Mike Miller

PDEs on cayley graphs

Alessandro Codenotti

I know a guy who studies differential equations on graphs

But he's a computational math person

Balarka Sen

13:14

LOL

You know there's a thing called discrete Laplacian yeah

That stuff is pretty dope

So it seems hyperbolic surfaces are essentially like fishnets

the angle $\theta$ between the two strands change cell to cell, and the curvature of each cell is $-\theta_{uv}/\sin(\theta)$

Well, the $\theta$ satisfies $\theta_{uv} = \sin(\theta)$, rather, since curvature is $-1$

The sine-Gordon equations

Mike Miller

13:30

This makes me uncomfortable

Balarka Sen

Lol

user131753

13:47

0

Q: Do we, the community, need a HNQ block list?

Short Version The purpose of this post is to have a vote count of the community support/opposition of the recent HNQ block list proposal. If you don't want to read the long version of the post then please go directly to Aim of this post and vote accordingly. Long Version In the last couple of ...

user131753

Please do consider casting your vote on exactly one answer.

MathStudent

13:57

Hi @MikeMiller remember proving to me the thing about positive definiteness? And how @TobiasKildetoft said that if X is not invertible that Q is positive semidefinite at most?

Mike Miller

Yup

MathStudent

I´m pretty sure that @TobiasKildetoft was wrong because my X doesn´t necessarily need to be quadratic. In fact, I´m quite sure it´s not quadratic but a n times p matrix with p < n. And what I actually need is for X to have full rank p. Your argument can of course be done in the same way with X having full rank instead of being invertible (which would be the same for a quadratic matrix)

Mike Miller

You need to tell me your setup again for this to make much sense to me.

And I guess when you say a quadratic matrix, you mean n x n (what I would call square)?

MathStudent

14:14

@MikeMiller $Q = X^T D X$ with $D$ being a diagonal matrix with only positive diagonal entries

@MikeMiller And yes. Sorry, English isn´t my native language

Mike Miller

Oh. I agree with your conclusion.

No need to be sorry, I just wanted to be sure I understood you.

MathStudent

Awesome, thanks!

Now I just need to know whether I can assume that the design matrix $X$ has full rank in a logistic regression model but I´m guessing that you probably can´t answer that

Mike Miller

Not at all!

MathStudent

I think the answer is yes in my case though. Because in particular I am interested in the situation of $p = 1200$ and $n = 4000$ and $X$ is a random matrix composed of i.i.d. standard normal entries.

Also, the paper I'm looking at is looking at the MLE and that wouldn't exist without $X$ having full rank I think

Maybe someone else in here has a clue about that

Mike Miller

You are free to ask, but few people here know about statistics

MathStudent

14:20

Good to know, thanks

Tobias Kildetoft

14:59

@MathStudent Is $X$ the covariance matrix?

MathStudent

@TobiasKildetoft Did you mean covariate matrix? If so, yes, otherwise no

Tobias Kildetoft

@MathStudent Right, that was the word I was looking for

In that case yes, you can assume it has full rank unless you explicitly try to do something degenerate (like fitting more parameters than you have data points)

Knight

If triangle $BOC$ makes an angle of $\theta$ with the triangle $ABC$ (in 3D space) then my book simply writes $$area (BOC) = area(ABC) \cos \theta$$ And I want to know how?

Well if a triangle makes an angle of $\theta$ with some other triangle then that means their normals make an angle of $\theta$ with each other but I don’t think we can conclude something about areas just from this

MathStudent

@TobiasKildetoft thanks! could you tell me why this is? cause it´s unlikely that it doesn´t have full rank? still technically possible though, right?

Tobias Kildetoft

15:17

@MathStudent Right. I am not sure I ever saw a precise statement about this, since my knowledge is from ML courses, which tend to be less rigorous in that department

MathStudent

@TobiasKildetoft Okay, thanks!

nbro

Anyone interested in stochastic approximation theory?

1

Q: Is RL just a less rigorous version of stochastic approximation theory?

After reading some literature on reinforcement learning (RL), it seems that stochastic approximation theory underlies all of it. There's a lot of substantial and difficult theory in this area requiring measure theory leading to martingales and stochastic approximations. The standard RL texts at...

reinforcement-learning comparison

Tobias Kildetoft

16:35

The comment thread on Not Even Wrong on the latest ABC post is now over 100 comments. And probably more than 50% of them have actual mathematical content from people who have studied the IUTT papers deeply. I find that rather amazing.

infinity

Hi ! can someone help me with this question?

Tobias Kildetoft

@infinity isn't $\sigma_A$ continuous?

infinity

it is the set $\{ \lambda : x-\lambda $ is not invertible $\}$. what do you mean by continuous? @TobiasKildetoft

i know this set is compact and not empty

Tobias Kildetoft

16:51

I mean the map $\sigma_A: A\to \mathbb{C}$.

infinity

i didn't know that it is continuous. does it follow from $x\to x^{-1}$ is continuous?

Tobias Kildetoft

I don't even know how the map is defined

infinity

it also not to $\Bbb C$ but to the set of subsets of $\Bbb C$

Tobias Kildetoft

Ahh, woops, I read the definition wrong. Sorry

infinity

no problem :)

geocalc33

16:59

3

Q: Maximal extension of domain of $f(x)=\sum_{n=1}^\infty n^{\frac{1}{\log_n(x)}}$ using analytic continuation

Given $$ f(x)=\sum_{n=1}^\infty n^{\frac{1}{\log_n(x)}}=\sum_{n=1}^\infty e^{\frac{\ln^2(n)}{\ln(x)}} .$$ By inspection this is a sum of nonlinear hyperbolas, over $n$. It's because $\ln(x)\ln(y)=\ln^2(n)$ is a hyperbola after a coordinate transformation $u=\ln(x)$ and $v=\ln(y).$ I think this n...

sequences-and-series logarithms analytic-continuation split-complex-numbers

This is a question but I welcome feedback because I don't know what the answer is

ted said that you have to use a riemann surface to do it

"More specifically, we associate the split-complex numbers and the split-quaternions as the semi-Riemmanian manifolds $R^{1,1}$ and $R^{2,2},$ respectively, in a the same manner that we associate $\Bbb C$ with $R^2.$"

I could use some help understanding that quote..

Ted Shifrin

17:20

This is saying what I said yesterday. It's the Lorentz plane, $\Bbb R^2$ with the indefinite metric $ds^2 = dx^2 - dt^2$.

CaptainAmerica16

@TedShifrin I figured out that thing from yesterday. I prefer the "inclusion both ways" approach.

geocalc33

@TedShifrin I'm still confused I suppose

Ted Shifrin

@CaptainAmerica: In the case of that particular exercise, you're saying definitions that give you equality. Did you try doing the same exercise for intersection instead of union?

CaptainAmerica16

@TedShifrin No, I haven't tried that yet.

Ted Shifrin

17:35

Also, learn to draw pictures and to think about examples.

Anonymous

Is there any easy way to see that $A_5=A_4S_5$ where $A_n$ is the alternating group of degree $n$ and $S_5 \in \mathrm{Syl}_5(A_5)$? Basically, I guess I'm trying to understand why any even permutation on $5$ letters can be expressed as a product of an even permutation on $4$ letters with a 5-cycle.

CaptainAmerica16

Lol, Leaky mentioned that. For some reason, I only remember to do pictures after banging my head against the wall for an hour.

Ted Shifrin

It's useful to draw pictures for mappings. You'll see that I do that constantly in those YouTube lectures.

But you should also be thinking about basic examples — either with precalculus (like $f(x)=x^2$) or with discrete sets.

@SanchayanDutta: Please do not write $S_5$ for a Sylow subgroup. It already stands for the symmetric group, and since you're doing $A_5$ this is beyond confusing.

Leaky Nun

@CaptainAmerica16 don't bang your head against the wall

Ted Shifrin

Leaky, good advice. He'll break the wall and it will be expensive.

Anonymous

17:40

@TedShifrin Uh, sorry! Could you suggest a different notation?

Ted Shifrin

Pick any letter you use for subgroups.

Edward Evans

Evening

Ted Shifrin

hi @Edward, @MikeM

Edward Evans

Heya @TedShifrin

just spent the day beside the lake doing infinite Galois theory! Super productive today lol

Anonymous

@TedShifrin Well, I guess $H_5$ then. $H_5 \in \mathrm{Syl}_5(A_5)$ and $A_5=A_4H_5$. Do you have any suggestions on how to attempt the problem, btw?

CaptainAmerica16

17:43

@LeakyNun XD

It was a figure of speech.

Ted Shifrin

It's OK (to me) just to say that you fix a subgroup $H$ of order $5$.

Why not use the standard fact that $3$-cycles generate the alternating group?

Anonymous

@TedShifrin Okay, sure. I'll go with that

Anonymous

@TedShifrin Oh, and then?

Anonymous

It's not immediately obvious to me how to apply that fact here

Ted Shifrin

Can you show how to get an arbitrary $3$-cycle as such a product?

Anonymous

17:50

@TedShifrin Oh, good idea! I'm thinking

Anonymous

@TedShifrin Okay, so any 3-cycle is an even permutation on 3 letters. And $A_4$ is the even permutations on 4-letters. So any 3-cycle should be an element of $A_4$ (?) I don't see any need for $H_5$ here. Maybe I'm missing something

Ted Shifrin

Yes, you're missing something huge.

CaptainAmerica16

@TedShifrin Tbh, the proof for intersection is much more immediately obvious. But maybe that's because I did the other one first.

Ted Shifrin

Well, let's be careful. You're saying $f(E\cap F) = f(E)\cap f(F)$?

CaptainAmerica16

um...

i just did the one in the book. it was that, but with a subset

proper subset

lol, wow. let me do this one too

Anonymous

17:56

@TedShifrin I guess $H_5$ being a 5-cycle helps us to make a choice of which 4 letters out of 5 to permute (i.e., the letters on which $A_4$ can act)

Mike Miller

Lay every definition out carefully

What does it mean for $y$ to be in $f(E \cap F)$

Ted Shifrin

@SanchayanDutta: So assume $A_4$ acts on $1,2,3,4$. Don't be difficult.

Stan Shunpike

@TedShifrin hey Ted!

Ted Shifrin

@MikeM: Please use the letter $y$. I'm trying to train @CaptainAmerica.

Hi @Stan. I'm fine with that.

Stan Shunpike

ok cool

Ted Shifrin

17:58

LOL, thanks, @Mike :P

Stan Shunpike

grad school requires a lot of self-study i'm finding :') which is good. i'm a lot better a math then when i started the year

Mike Miller

most study is self-study!

CaptainAmerica16

@MikeMiller Well, from what I sorted out yesterday. The first thing I think of is that y=f(x) for some $x \in E \cap F$.

Leaky Nun

διάγραμμα @CaptainAmerica16

Transcript for

$Mathematics$ Mathematics