Mathematics - 2024-01-06

Jakobian

00:00

Yes. But other one where he explicitly states "this is a free but not transitive action"

psie

I have a stupid question maybe. I see how a function $f(x,y)$ on the unit circle induces a $2\pi$ periodic function, namely by the parametrization $f(x,y)=f(\cos\theta,\sin\theta)=g(\theta)$, but how does a $2\pi$ periodic function induce a function defined on the unit circle? Is it just the argument backwards? i.e. if $g(\theta)$ is a $2\pi$ periodic function, then we recognize the argument $\theta$ as an angle and write $g(\theta)=f(\cos\theta,\sin\theta)=f(x,y)$.

I feel like the argument backwards does not really work...

given $g(\theta)$, can we write it as some function $f$ that depends on $\cos\theta$ and $\sin\theta$?

Jakobian

@psie $f(e^{it}) = g(t)$

This doesn't depend on $t$ as $g$ is $2\pi$ periodic

And as $t\mapsto e^{it}$ is surjective onto the circle, it defines a map $f$ on the circle

psie

00:23

hmm, could you explain what you mean by that it does not depend on $t$? $g$ is a function of $t$, right?

leslie townes

it doesn't depend upon how the point on the unit circle is represented as e^(it)

if e^(is) = e^(it) then s and t differ by an integer multiple of 2pi so that g(s) = g(t)

psie

I see, got it

leslie townes

just the usual crap about 'can you define a function on a set in terms of how something on that set is represented' and the usual crap about the slight ambiguity of 'the angle' associated with a point on the unit circle

i guess if you wanted to write it as a function of x and y you could do that too with some arctan stuff (the choice of angle wouldn't matter for purposes of plugging into g) and handling a few exceptional cases

but in the galaxy brain POV its kinda weird to think of a function that is a priori defined only on the unit circle as "f(x,y)"

psie

shame, I thought I could circumvent the galaxy brain POV :)

Ted Shifrin

Munchkin happily back torturing teachers?

leslie townes

00:34

yes. she hates subtraction.

Ted Shifrin

No trickle down for her?

Jakobian

01:03

@leslietownes why

Ted Shifrin

Too mind-bending.

Jakobian

I never thought about those things much while growing up, they didn't bother me

leslie townes

01:53

it is less familiar and takes longer for her to do.

i also heard that her instructor assigns problems that even she doesn't know how to do, and is on probation for giving too many Fs.

like ted

Ted Shifrin

02:14

Wow ... impressive to be like Ted at pre-school!! Or is it now kindergarten?

Mys_721tx

02:47

So I started working on Moschovakis's Notes on Set Theory this week. Question x1.2 wants me to prove De Morgan's laws. For $C \setminus (A \cap B) = (C \setminus A) \cup (C \setminus B)$, I followed chapter 1 to have the following $C \setminus (A \cup B) \iff \exists x \in C \land \neg (x \in A \lor x \in B) \iff \exists x \in C \land (\neg x \in A \land \neg x \in B)$

Would it be proper to say $\exists x \in C \land (\neg x \in A \land \neg x \in B) \iff (\exists x \in C \land \neg x \in A) \land (\exists x \in C \land \neg x \in B)$?

Mys_721tx

02:59

I understand that $\land$ and $\lor$ distributes over each other but that one feels a bit sketchy.

leslie townes

03:23

ted: it's now kindergarten :o

Ted Shifrin

@leslietownes You’re getting old!

leslie townes

no, and i proved that with my second child, who is a baby. that's how that works.

Ted Shifrin

03:54

Oh, so that’s what that means. So you end up with countably many children?

leslie townes

05:30

and immortality.

PM 2Ring

05:43

@leslietownes Belated congratulations on the birth of your new scion. Here's another avian themed tune from my favourite Polish Texan:

Sarah Jarosz - Pay It No Mind - Live Performance from World on the Screen

►

Wishing @TedS a speedy recovery from his eye op.

An improvised piano piece from blind Californian multi-instrumentalist Rachel Flowers:

Hathaway - Rachel Flowers (In the Moment)

►

CroCo

06:21

hi everyone, I need to show $(\mathbb{Z},\oplus)$ forms an Abelian group where the binary operation is defined as follows

\[
a \oplus b = a + b - 1, \quad a,b\in \mathbb{Z}
\]
1- Associativity:
$$
\begin{aligned}
a \oplus ( b \oplus c ) &= a + (b+c-1) - 1, \\
&= a + b + c -2. \\
(a \oplus b) \oplus c &= (a+b-1) + c - 1, \\
&= a + b + c -2.
\end{aligned}
$$
2- Neutral element $e$: I need to find $e\in \mathbb{Z}$ such that $a\oplus e = e \oplus a = a$. This is what I've done
$$
a\oplus e = a + e -1 = a \implies e = 1

My question is is it possible for the neutral element to equal the inverse? Or this is merely a coincidence ?

sorry I think I've made a mistake in finding the inverse. This is correct approach the way I see it

3- Inverse element $a^{-1}$: I need to find $a^{-1}\in \mathbb{Z}$ such that $a\oplus a^{-1} = a^{-1} \oplus a = e$. This is what I've done
$$
a\oplus a^{-1} = a + a^{-1} -1 = e \implies a^{-1} = 2+a
$$

Is this the correct way to show that $(\mathbb{Z},\oplus)$ forms a group? I will handle the commutativity later.

sorry I can't modify the post. $a^{-1} = 2 - a$.

leslie townes

in your identification of the neutral element, the \implies is an \iff, and you need that. you don't want to show "hey, assuming that there is a neutral element, among the properties it must have is that it's equal to 1." you want to show that 1 is a neutral element, but that's the reverse implication from the \implies. it doesn't change the "work" but does change how you present it.

same with the identification of a^{-1} for a given a.

other than that it looks fine.

i guess if you aren't previously showing that oplus is commutative, you might also want to show or at least expressly remark that 1 oplus a is a (your calculation, with the iff, shows that a oplus 1 is a)

CroCo

06:40

@leslietownes thank you for being informative and helpful. I still don't understand "in your identification of the neutral element, the \implies is an \iff, and you need that.". The issue I'm having I don't know the neutral element a priori.

leslie townes

same with the inversion calculation, state both sides of it. for expository cleanliness, it also helps not to write "a^{-1}" until you know it exists. so e.g. your calculation might be somewhat clearer if stated in a form that a is in Z and b is an integer, then a oplus b = 1 holds iff b = 2 + a. this happens to have significance for inversion in your group but that is a consequence of the property of b.

croco: well, you found it. in your calculations, you found that e = 1 is going to do the job. if you like you can skip presenting how you found it, and present it to the reader as a direct verification by calculation: "Let a be an integer. Then a oplus 1 is [compute using definition] and 1 oplus a is [compute using definition]. Since a was arbitrary this shows that 1 is an identity element for oplus."

you might want to think of it in terms of: there are calculations you do to identify what the neutral element is going to be, and then there is what you would offer as a proof that the thing you found is a neutral element. they happen to be almost exactly the same thing, the only difference being what you regard as your assumption and what you regard as your conclusion

the way that you found that "if there were to be a neutral element for oplus, it has to be 1" can, if read it backwards, be a proof that 1 is a neutral element.

most of an exercise like this is expository in nature, it is less about the calculation than how you introduce the calculation, and relate it directly to the thing that you are asked to do. sometimes i think the difficulty of an exercise like this actually goes down if the binary operation is more complicated (or there are no neutral elements or these laws do not hold) so there is "more to do."

CroCo

You're saying things in a way that I'm starting to understand.

I'm reading this book "A book of Abstract Algebra" by C. Pinter. Most of the exercises like this and indeed some binary operations are involved.

something like this $a * b = \sqrt{ |ab| }$, on the set $\mathbb{Q}$.

thank you so much for this invaluable answer and happy new year to you all.

Mad

10:52

@TedShifrin Sorry for that! i will resend it more clean)

nickbros123

11:51

"Given an experiment, let C denote the sample space of all possible outcomes. As
discussed in Section 1.1, we are interested in assigning probabilities to events, i.e.,
subsets of C. What should be our collection of events? If C is a finite set, then we
could take the set of all subsets as this collection. For infinite sample spaces, though,
with assignment of probabilities in mind, this poses mathematical technicalities that
are better left to a course in probability theory" why is this so?

the book is Hogg, introduction to mathematical statistics

the idea of a sigma field can be extended to countable infinities right?

Jakobian

12:21

@nickbros123 Its the same issue as with Lebesgue measure not being defined on all subsets of $\mathbb{R}^n$

Sigma fields can be defined on any subset, the point is pretty much that its not as simple as with the standard classical probabilities taught in high school

the author is stating that they won't be concerned about such technicalities because that's not the point of discussion here

Thomas Finley

13:32

Let $f:I\to\Bbb R$, where $I\subseteq \Bbb R$ and $a\in\Bbb R$. Can we say, $$f'(a)=\lim_{h\to 0}\frac{f(a)-f(a-h)}{h}$$ ?

Thorgott

what's the definition of $f^{\prime}(a)$

Jakobian

@ThomasFinley $a\in I$?

@Thorgott how do we go about the inductive step here?

The statement used here is that action of $SO(n)$ on $S^{n-1}$ is transitive for $n\geq 2$

But my issue is that $\text{span}\{v_j, ..., v_k\}$ and $\text{span}\{w_j, ..., w_k\}$ are different linear spaces

how do we handle this without explicitly restating the proof of transitivity of $SO(n)$ on $S^{n-1}$?

Thomas Finley

13:51

@Jakobian Ah, pardon me for the typo. Yes, $a\in I.$

Jakobian

substitute $t = -h$

Thomas Finley

@Jakobian yeah, that's how I came to the conclusion.

But I was validating whether what I did made any sense

Jakobian

yes

it does

Thomas Finley

Thanks, indeed $$f'(a)=\lim_{h\to 0}\frac{f(a)-f(a-h)}{h}$$ holds true then.

Also, I was convinced somehow intuitively that this holds as we know that $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

From here, the conclusion (if I am not mistaken ) must seem obvious enough.

Thorgott

@Jakobian This doesn't make sense to me

why would $w_j$ have to be in the orthogonal complement of $v_1,\dotsc,v_{j-1}$ at all?

the statement we're trying to show is that $SO(n)$ acts transitively on $k$-dimensional subspaces, yes?

Jakobian

14:08

yes

Thorgott

just take two $k$-dimensional subspaces $V,W$, pick orthonormal bases $v_1,\dotsc,v_k$ and $w_1,\dotsc,w_k$ respectively, complete them to orthonormal bases $v_1,\dotsc,v_n$ and $w_1,\dotsc,w_n$ of $K^n$ respectively and take the matrix representing the linear isomorphism $v_i\mapsto w_i$ for $i=1,\dotsc,n$

if the determinant is $-1$, take $v_n\mapsto-w_n$ instead

Jakobian

Hmm yes, so instead of going by induction we prove that the argument that worked for the sphere works for every pair of $k$-dimensional subspaces

Thorgott

yeah, what we're really saying is that $SO(n)$ acts transitively on the Stiefel manifold $V_k(K^n)$

(for $k<n$, for $k=n$ one needs the oriented Stiefel manifold actually)

$V_1(K^n)=S^{n-1}$

Jakobian

that makes sense
would you classify this as an error in the text in need of correction?

Thorgott

and the $k$-Grassmannian is $V_k(K^n)/GL(k)$

@Jakobian error is hard to say cause it's just unclear to me what the argument is supposed to be exactly, but I would say it needs clarification

Jakobian

14:27

alright, I'll classify this as a potential error then

thank you

Soumik Mukherjee

The answer is 4 right?

Jakobian

1, 2 are always false, 3 is sometimes false and 4 is always true

Thorgott

I disagree

they are quantifying over all $f$

Jakobian

oh

Then $Y = 2^X$

Sorry, somehow I understood it as $Y = f^{-1}(0)$ for some $f:X\to \mathbb{R}$

Soumik Mukherjee

@Jakobian That's what I assumed as well

Thorgott

14:37

yeah, same

I disagreed with you while still thinking that cause 4. would still be wrong

but then I realized we both had this misconception lol

Soumik Mukherjee

$g$ is periodic and continuous, hence UC right?

Jakobian

$g$ is not periodic as far as I can see

Soumik Mukherjee

I thought $g(x)=1-\vert 2x^2-1\vert$ for $x\in [0,1]$ and we copy paste this graph for each $[n,n+1]$, so $g(x)$ is also periodic

Jakobian

14:54

$f(x^2) = f(x^2-k) = 1-|2(x^2-k)-1|$ for $k = \lfloor x^2\rfloor$

So the periodicity is disturbed by the square

It will be $g(x) = 0$ for $x = \sqrt{k}$, $k$ an integer, $g(x) = 1$ for $x = \sqrt{k+1/2}$

on the intervals $[\sqrt{k}, \sqrt{k+1}]$ it will first increase and then decrease

Soumik Mukherjee

Oh okay

Thank you Jakobian and Thorgott

Thomas Finley

18:01

If $$e^{a\arcsin x}=a_0+a_1x+a_2x^2+...+a_nx^n+...$$ then how to show that, $(n+1)(n+2)a_{n+2}=(n^2+a^2)a_n$ ?

psie

Consider the Dirichlet kernel $$D_n(x)=\frac{\sin\left(\left(n+\frac12\right)x\right)}{\sin\left(\frac{x}2\right)}.$$ I'm trying to understand the plot on wiki. Why does the value at $x=0$ increase as $n\to\infty$? It looks to me as if we have an undefined function at $x=0$, but maybe they are graphing the continuous extension. In that case, what is the continuous extension?

Jakobian

18:14

@psie $\lim_{x\to 0} D_n(x) = ?$

psie

@Jakobian yeah, hmm, do you use the small angle approximation? In that case, I'd say $2n+1$, right?

Jakobian

Limits like this should be straightforward for you

yes, its $2n+1$

psie

ok 👍

Sahaj

18:49

@ThomasFinley, try to find some identity followed by $f(x), f'(x), f''(x)$ where $f(x)=e^{a\arcsin(x)}$ and try to use that to make a recurrence for coefficients in its power series

Ted Shifrin

19:35

@ThomasFinley Notational suggestion: Do not use $a$ and $a_j$ in the same problem. You are going to mess it up. To have the $a_j$ as functions of $a$ is highly confusing. I recommend using $c$ for the parameter, for example.

robjohn

21:12

@psie It tends to the Dirac Delta when tested against smooth functions. The Cesaro means of the Dirichlet kernels are the Fejer kernels. They tend to the Dirac Delta when tested against $L^1$ functions.

@psie $D_n(0)=2n+1$

Xander Henderson

21:56

I like to pronounce "Fejer" so that it rhymes with "V'Ger" (from my favorite Star Trek movie).

Xander Henderson

22:32

I'm a little hurt that my previous comment earned no response. :(

Jakobian

22:45

@XanderHenderson Its pronounced like "feyer"

psie

23:01

@robjohn ok, thanks

so $D_n(0)=2n+1$ would be the continuous extension, makes sense

Xander Henderson

@Jakobian Yes, I know.

psie

23:53

I've proved, that for a positive summability kernel $K_n$ on an interval $I=(-a,a), a>0$, if $f$ is integrable near $0$ and continuous at $0$, then $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0).$$

It is then claimed that one can easily extend the above result when $f$ is continuous at some other point, say $s_0\in I$, by making a change of variables. They claim $$\lim_{n\to\infty}\int_I K_n(s)f(s_0-s)ds=f(s_0).$$ But when I make the change of variables $s_0-s=u$, then I get a kernel centered not at $0$ but at $s_0$. Is that a correct observation? I'd like to somehow use the result I've already proved above...

In other words, I'd like to show $\lim_{n\to\infty}\int_I K_n(s)f(s_0-s)ds=f(s_0)$ using $\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0)$.

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