Mathematics - 2023-09-23

Jakobian

12:13 AM

@AlessandroCodenotti Let $G$ be a topological group. Does the property, all homomorphisms $G\to \{0, 1\}$ are constant, where $\{0, 1\}$ is a discrete two-point topological group, have a name?

this is an analogue of connectedness for topological groups

hmm... I suppose $\{0, 1\}$ is not very relevant as a group?

perhaps change it to homomorphisms $G\to\mathbb{Z}$

even if I only demanded that for algebraic homomorphisms, what would that condition even signify

X4J

12:36 AM

@TedShifrin Oh. I don't think this applies, anyways, I could've proved that $\lim_{\k \to \infty} ka_k = 0$ which also makes sense I guess

sunny

@sunny mustn't a modulus of continuity be finite-valued? how come the answer defines it to be infinity for $s>\delta_1$?

Ted Shifrin

1:11 AM

Not according to the definition in that post.

sunny

strange

Ted Shifrin

The concern is with values near $0$, not large values, anyhow.

sunny

So how does one resolve this?

Ted Shifrin

Huh?

sunny

I thought the modulus of continuity is defined in $[0,\infty)$, not $[0,\infty]$.

Ted Shifrin

1:20 AM

Not according to the post. We're talking about values here, not domain.

sunny

ok

Ted Shifrin

You can read the wikipedia post. It explains why in some settings $\infty$ needs to be a value.

sunny

Ok, I'll take a look

@TedShifrin could we define the modulus to take another value than infinity when $s>\delta_1$?

It seems like infinity is a convenient choice, but I just wonder if it's possible.

Ted Shifrin

Did you read the wikipedia post? In some situations, it's necessary. Of course, not when you have a compact metric space or a uniformly continuous function.

sunny

Ok, I found the remark, thanks.

X4J

1:34 AM

Does anyone know where I can find real analysis 2 theoretical exercises for infinite series/improper integrals? The university I study at doesn't provide much

EM4

2:19 AM

I believe Rudin does, but I am not that sure.

X4J

2:30 AM

Yeah it does, just checked. Thanks

EM4

no problem :).

Alessandro Codenotti

4:33 AM

@Jakobian I'm not sure what this condition gives. Surely it is satisfied by some big classes of groups (torsion, divisible) but apart from that I have no idea

Koro

4:44 AM

why? how?

Sine of the Time

5:13 AM

if we consider the Cauchy problem $ y'=f(t,y), y(t_0)=y_0 $, there are theorems about existence and uniqueness of the solution. Is there a theorem regarding the uniqueness if $f$ satisfies the Holder condition?

Ted Shifrin

What do the theorems say and what condition did you have in mind?

Sine of the Time

Cauchy theorem: $f$ is Lipschitz the solution exists and it's unique. Peano's theorem: if $f$ is continuous, then the solution exists

Ted Shifrin

5:31 AM

Lipschitz in the $y$ variable. So what Hölder condition?

Sine of the Time

yes. I was just wondering if there's a theorem where $f$ satisfies the Holder condition

Ted Shifrin

What condition? Be explicit. I’ve asked twice.

Sine of the Time

something like this en.wikipedia.org/wiki/H%C3%B6lder_condition

for some $\alpha$, for example 1/2

Ted Shifrin

Don’t you know a counterexample? What’s the non-Lipschitz famous example?

Sine of the Time

$f=\sqrt{|y|}$

Ted Shifrin

5:42 AM

Yeah. So isn’t that Hölder as you suggested?

Sine of the Time

ok, thank you. I was trying to prove Cauchy theorem with a function satisfying holder condition, but the iterative method did not work

probably is something related to the convergence of picard's iteration

Ted Shifrin

Well, since you have an easy counterexample, it can’t work.

Koro

6:14 AM

p iff q. I always forget which is "only if" and which one is "if".

q implies p is "if"

p implies q is "only if"

leslie townes

i understand what you're saying, but it doesn't exactly make it less confusing when you're swapping the order of p and q there and then not writing what you mean inside the quotes

haha

despite this confusing stuff about mapping between language and symbols, a lot of people (maybe most people?) love symbologizing because it makes them more comfortable somehow

PM 2Ring

6:38 AM

It will be the equinox shortly, ~2023-Sep-23 06:50 UTC. ssd.jpl.nasa.gov/api/…

Prithu Biswas

6:54 AM

Can I ask a question about linear algebra from axler?

Thomas Finley

8:14 AM

$\frac{x}{x+1}\lt \log (1+x)\lt x$

Is there a way to solve this using Mean-Value Theorems?

leslie townes

by "solve" it, do you mean prove it? it doesn't hold for x = 0. if x isn't 0 it might be helpful to note that there is c between 0 and x with (log(1+x)-log(1+0))/(x-0) = 1/(1+c)

Thomas Finley

8:41 AM

@leslietownes yes, by solving , I meant proving. So, this inequality is valid only if $x\gt 0$ ?

X4J

9:00 AM

This is a question is real analysis 2, infinite series:

Let $\{\alpha_k\}_{k=1}^{\infty}$ be a sequence of positive reals.<br>
Assume that for each $k$ : $\alpha_{k}\cdot \alpha_{k+1} $ < $\alpha_{k+1}^{2}$ and $\alpha_1 < \alpha_2$. <Br>Prove that the sum $\sum_{k=1}^{\infty}\alpha_k$ is finite.

Does anyone intuitively see why this works?

I struggle to even intuitively understand why the assumption that $\alpha_1 < \alpha_2$ is necessary (although I'm aware the result cant be deduced without it)

Soumik Mukherjee

9:42 AM

@PrithuBiswas just ask

one potato two potato

11:41 AM

do probabilists care about a situation where R.Vs are not independent?

Prithu Biswas

11:58 AM

@SoumikMukherjee Is there a way to prove that "Every subspace of a finite dimensional vector space is finite dimensional" using induction?

There is an answer that seems to use quotient spaces, but I am not familiar with them.

Thomas Finley

2:14 PM

0

Q: Prove that, $\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a)$ such that $f'(x)$ is continuous at $a.$

Prove that, $\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a)$ such that $f'(x)$ is continuous at $a.$ My solution is as follows: We have, $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{f(a-h)-f(a)}{-h}=\lim_{h\to 0}\frac{f(a)-f(a-h)}{h}.$$ We note th...

Need some help with this...

Jakobian

@onepotatotwopotato of course

Just the case when things are independent is usually the easiest to handle, that's why you see theorems stated for independent R.V.'s

Thomas Finley

@Jakobian Do you remember we were discussing a limit problem? I got it. We can solve simply by using a ratio test for sequences. @ParamanandSingh gave me that brilliant idea :D

Hi @SouravGhosh Haven't seen you in a while? How are you doin?

Jakobian

@PrithuBiswas Let $V$ have dimension $n$ and suppose that every vector space of dimension $< n$ have finite dimensional subspaces. The $V_0\subseteq V$ be a subspace of $V$. If $V_0 = V$ then $V_0$ is finite dimensional, if not then take some $x\in V\setminus V_0$. Extend $x$ to a basis $e_1 = x, ..., e_n$ of $V$. Then $V_0\subseteq \text{span}\{e_2, ..., e_n\}$ which is a vector space of dimension $n-1$, hence $V_0$ is finite dimensional by induction hypothesis

something like that?

I do use the property of basis extension though

well I suppose the proof I tried sketching is wrong

you need to actually choose $x$ so that its independent of $V_0$ somehow

冥王 Hades

2:36 PM

So, a friend of mine is starting linear algebra and asked me if I knew any good books for him

should I recommend Strang’s Linear Algerba?

user 726941

Yes.

Jakobian

This looks like a good question for someone experienced like @TedShifrin

Thomas Finley

@TedShifrin Can you please have a look at my post?

Ted Shifrin

@Hades First, there are different Strang books. Second, what are your friend’s interests and proof interests?

Not now, Thomas.

user 726941

For those starting linear algebra your friend won't find much better.

冥王 Hades

2:42 PM

@TedShifrin To be honest he is just trying to get through it, he isn’t particularly interested or great with math. It’s for his civil engineering major

Sha Vuklia

Jakobian

I feel like books take time, and someone "not particularly interested" and "civil engineer" should rather watch a series of lectures about linear algebra, maybe use a book as a reference.

user 726941

@ShaVuklia is that a question

Sha Vuklia

is there a clever way to row reduce this matrix to echelon form?

actually, all I need is that this matrix is not invertible (and it has to be done by hand, and no use of determinants is allowed), so I'm hoping to spot a linear combination

the context is: someone's first introduction to linear algebra (like 2nd week)

Ted Shifrin

Yeah, baby Strang is fine. Not adult Strang.

@ShaVuklia Yes. First get a leading $1$ by combining some pair of rows.

Thomas Finley

2:48 PM

@TedShifrin No problem! Thanks for responding!

Sha Vuklia

@TedShifrin I've done that, I'm just finishing my tedious computation now, in the hope that it's not that bad (+that I'm not making mistakes)

Ted Shifrin

In my books I went to great pains to make sure the computations were not horrendous. Some authors don’t do that.

one potato two potato

@Jakobian because it seems like many important named theorems assume independence condition.

Just a soft question. I don't actually care if they care or not.

Jakobian

@onepotatotwopotato you're probably talking about something like, Borel-Cantelli or law of large numbers?

There are versions of those which weaken the assumptions, its not that probabilists don't care about when we have lack of independence, its like in any other subfield of math, they care about results and if they can apply them. Usually the easiest ones to state and prove are when assuming independence

Sha Vuklia

if I could use transposes, then the computation is alright

one potato two potato

2:54 PM

Yeah but I once took a seminar in probability theory and the speaker always assumed i.i.d condition.

Sha Vuklia

since I'm not interested in the solution, just in the rank

one potato two potato

and emphasized independence condition is important and once we don't assume independence, things get ugly.

Sha Vuklia

I have to see if my student will have the theory to understand that (they just treat a bit of linalg)

Jakobian

well, independence always makes things simpler

but if you look at real life, things are often not independent

one potato two potato

stay independent.

Jakobian

2:56 PM

probability is a science that is close to applications, so they do have to care about what happens even if random variables aren't independent

@TedShifrin If we say that a manifold $X$ is connected whenever every smooth map $f:X\to \{0, 1\}$ is constant, what does this concept lead to? Is it just connectedness again?

I'm wondering about connectedness in the category of smooth manifolds

Thorgott

yes, that is connectedness

Jakobian

ah, cool

@Thorgott what about analytic manifolds then? Since they're more rigid

Ted Shifrin

The smoothness there is redundant. True iff continuous.

Jakobian

I'm talking about this because I discovered this thing called uniform or Cantor connectedness

and I wanted to test how this work in other categories

Ted Shifrin

The underlying topological space has to be connected. Smoothness gets you nothing.

Manifolds are always locally path connected, so you can play topology games. But the smoothness doesn’t matter.

Jakobian

3:14 PM

Yeah, I suppose thats true, smooth maps are in abundance so thats why. But what about analytic manifolds and analytics maps between them then?

Cantor-connectedness, for metric spaces, has interesting characterization of not being able to write $X = A\cup B$ for non-empty $A, B$ with $d(A, B) > 0$ for instance

so I'm thinking if there would exist something like this, that would resemble connected spaces from topology

Ted Shifrin

The identity principle holds, just as in complex analysis. An analytic function that is $0$ on a subset with a limit point still is constant on the connected component.

Alessandro Codenotti

@Jakobian yesterday's asker came back with a very nice question

Thorgott

5:05 PM

it's the same for analytic manifolds or also for any other category of topological spaces where the maps can be locally tested for

Jakobian

5:49 PM

@Thorgott hmm... but its not true for uniform spaces

aren't uniformly continuous functions local?

for example $\mathbb{Q}$ is Cantor-connected

Koro

So X_i's are given to be independent and identically distributed RVs and have mean 0, and T= X_1+...+X_n.

Then $E(T^3) E(X_1^2)= E(T^2)E(X_1^3)$.

Thorgott

you're right, I was only thinking manifolds

$\mathbb{Q}$ doesn't work cause it's not locally connected

Monty

given $f,g\mathbb{R}^n\to \mathbb{R}^n$ Is there a nice formla for $\nabla (\nabla f g )$

something like $\nabla ( \nabla f g )= \nabla f \nabla g + ...?$

Koro

I noted that $E(T^3)= E(X_1^3)+...+E(X_n^3)$. Then, I said since X_i's are identically distributed, $E(X_1^3)= E(X_i^3)$ for all $i$.

similarly, for E(T^2).

Jakobian

correct

Koro

5:58 PM

Is that correct?

Jakobian

@koro

Koro

ok, thanks.

My exam was better than I expected.

EM4

6:14 PM

your exam was on what subject?

Koro

it was on probability theory

EM4

oh nice .

Ted Shifrin

6:37 PM

@Monty What do you mean by this?

Koro

where can I learn about vector bundles?

i know fibre bundles

Ted Shifrin

What do you want to learn?

Koro

6:58 PM

I want to do these

Monty

@TedShifrin like $\nabla f$ is a matrix, so will $\nabla (\nabla f g)$ since $\nabla f g$ is a vector. But cant figure out how chain rule will apply to $\nabla (\nabla f g)$

Ted Shifrin

@Koro That’s all fiber bundles. All you need about vector bundles is the definition (local trivializations are vector space isomorphisms on each fiber).

Koro

why is a fiber a vector space?

Ted Shifrin

@Monty I still don’t understand. You’re doing derivative of the product $fg$? $\nabla$ either means gradient or covariant derivative.

That’s the definition, Koro.

Koro

and what is 'vector bundle of dim n with a metric'.

Ted Shifrin

7:06 PM

You have a continuously/smoothly varying inner product on each fiber. Your exercise told you that.

Koro

first para in the image makes no sense to me.

'the map form U to...' in the third line, what is that?

Ted Shifrin

Existence is immediate from partitions of unity, since you have obvious metrics on each local trivialization.

Koro

(they made a typo: the map from U to..

Ted Shifrin

He told you in the parentheses.

Cotton Headed Ninnymuggins

The probability of getting strictly more heads than tails in 9 coin flips is just 0.5, right?

Ted Shifrin

7:08 PM

Inner product on $\Bbb R^n$ is given by a positive definite symmetric matrix.

Assuming fair coin, sounds right, @Cotton.

Cotton Headed Ninnymuggins

Yes, fair coin

Ted Shifrin

Anyhow, you can look at volume 1 of Spivak’s 5 volumes, Koro.

Koro

how did they go to R^n?

sure, I'll take a look.

Ted Shifrin

He does tangent bundles mostly.

Monty

@TedShifrin Is it not common to use the notation $\nabla f$ to mean the matrix of first derivatives of the vector $f$?

thats how i was using it

Ted Shifrin

7:12 PM

Column vector? That’s the gradient. Row vector? That’s the derivative, which I write $D$.

Are you wanting the matrix of second derivatives?

Monty

ok then $D(Df g)$

$f,g$ vector valued functions, so $Df g$ is a vector, is there a neat way to write out $D(Df g)$

Ted Shifrin

I still don”t understand. Your notation sucks. $D(fg)$? Or $gDf$?

Monty

$gDf$

but i think of them as column vectors so its $Df g$

if I apply a derivative to a product i always put brackets.

Ted Shifrin

Well, what you’re writing no one would know how to parse.

$f,g$ vector valued. Be specific.

Mapping what to what?

Monty

i did say $f,g\mathbb{R}^n\to \mathbb{R}^n$

$f,g:\mathbb{R}^n\to \mathbb{R}^n$ *

Ted Shifrin

7:17 PM

So $Df$ is $n\times n$ and $g$ is $n\times 1$. OK, so $(Df)g$ makes sense.

You’re asking for the derivative of that.

Product rule. $(D^2f)g + [Df)(Dg)$, but the first term is tricky.

$D^2f$ is a tensor of rank 3.

Monty

yes exactly

it was that first term i had no idea how to handle

Jakobian

@Thorgott I guess you're right

It's interesting to me that what lead to formalization of topology is the concept of connected spaces

Koro

7:33 PM

not the Ko(two dots on o)nisberg bridge?

Seven Bridges of Königsberg

The Seven Bridges of Königsberg is a historically notable problem in mathematics. Its negative resolution by Leonhard Euler in 1736 laid the foundations of graph theory and prefigured the idea of topology.The city of Königsberg in Prussia (now Kaliningrad, Russia) was set on both sides of the Pregel River, and included two large islands—Kneiphof and Lomse—which were connected to each other, and to the two mainland portions of the city, by seven bridges. The problem was to devise a walk through the city that would cross each of those bridges once and only once. By way of specifying the logical task...

Jakobian

topology didn't really exist when this problem was stated

so I'm not sure what you're asking

Koro

but it is said to have started the topology.

solving the bridge problem required the ideas of 'connectedness' which were made precise after this.

you may see this:

1. History of Algebraic Topology; Homotopy Equivalence - Pierre Albin

►

Jakobian

You're missing the point

I said formalization of topology

Koro

and Marco Manetti's topology book

Jakobian

I never said some topology ideas didn't exist before that

Topology is considered to be formalized only in 1914 by Hausdorff

Koro

7:39 PM

(going deeper into the wiki page, it seems that only 3 of the bridges remain.)

Jakobian

@Koro either way, I see this more as a graph theory problem

Koro

I probably misunderstood your comment in response to which I'd commented.

After learning some graph theory, I know that's one of the graph theory problems.

related to Eulerian graphs.

Jakobian

The reason for why topology got formalized is partially because they wanted to formalize the concept of connected subspace

It's Hausdorff that's considered to be the father of topology

Monty

anyway thanks @TedShifrin i knew there was a bit of a nasty term there

Sha Vuklia

Let's say $\alpha=0$. For $\overline\lambda=-i\beta$, we would get the solution $w(t)=\tilde x(t)-i \tilde y(t)$, where $\tilde x(t)=\cos(\beta t)u+\sin(\beta t)w$ and $\tilde y(t)=-\cos(\beta t)w-\sin(\beta t)u$. Since the solution space has dimension 2, and $x(t)$ and $y(t)$ are linearly independent, we should we able to find $\mu,\lambda\in\mathbb R$ such that $\tilde x(t)=\lambda x(t)+\mu y(t)$ (for all $t$).

If I take $t=0$, this yields $u=\lambda u+\mu w$. By linear independence of $u$ and $w$, we must have $\lambda=1,\mu 0$. However, for $t=\pi/(2\beta)$, we would have $\tilde x(\pi/(2\beta))=w=-\lambda w+\mu u$, which yields $\lambda=-1,\mu=0$. This is a contradiction. Where do I go wrong?

I see my mistake

EE18

7:56 PM

I am trying to prove "Show that a nonempty set X is countable if and only if there is a surjection from $\mathbb{N}$ to X."

Sha Vuklia

it should be: $\tilde x(t)=\cos(\beta)u-\sin(\beta)$

then we have $\tilde x(t)=-x(t)$ and $\tilde y(t)=-y(t)$

so no contradiction

leslie townes

i was going to say, i thought i spotted a sign error but couldn't be sure

EE18

(Only the "if" is hard for me). This is my work so far, and I have shown that $X$ injects into $\mathbb{N}$. I am trying to show that this contradicts that $X$ is uncountable, but am struggling to see how to do so.

Sha Vuklia

I honestly already checked it several times, so I started to get convinced I got the functions right

leslie townes

this is extra fun when you're speaking and (1) a "sign" error sounds a lot like a "sine" error, and (2) the sign error might well be a sine error

EE18

7:58 PM

(Also, $X$ is infinite)

Sha Vuklia

looool

a sine sign error

leslie townes

ee18: without digging into the proof (so much text!) if there is a surjection from N to X, then any left inverse to that surjection [which exists at least with choice] will be a bijection from X to N (demonstrating that X is countable). this is maybe a cleaner approach, rather than deducing things from uncountability.

Jakobian

@leslietownes I don't think there is a cleaner approach because I think this requires axiom of countable choice

leslie townes

sometimes a proof by contradiction only introduces length to a proof. particularly if a notion is defined 'piecewise' (here, countability maybe being given in the alternative as 'finite or [something specific to countably infinite sets]') where the negation of the piecewise thing gives you a bunch of extra conditions that might not help.

jakob: yeah, that feels right.

Jakobian

Oh. You wrote "This is maybe a cleaner approach" and not "There might be a cleaner approach". I understood it as the latter

leslie townes

8:05 PM

oh, haha.

Ted Shifrin

@Monty I suggest writing it out in coordinates explicitly. We’re differentiating $\sum \frac{\partial f_i}{\partial x_j}g_j$

Contradicting uncountability sure feels like a double negative.

EE18

Gotcha, thanks v ery much leslie!

your comment re proof by contradiction comports with this (math.stackexchange.com/questions/192496/…) link which im gonna follow to get that result

leslie townes

8:22 PM

ted: honk

Ted Shifrin

The geese have invaded?

sunny

Let $x=(x_1,x_2,x_3)$ and consider the vector-valued function $f(x)=(x_2,x_3,1-x_1^2)$. I'm looking for a domain for which $f$ satisfies a Lipschitz condition. In other words, I'm looking to bound $$\lVert f(x)-f(y)\rVert.$$ As far as Wikipedia is concerned, there is no mean value theorem for vector valued functions depending on a vector, right? So how can I bound the above?

$x\in\mathbb R^3$, of course.

Ted Shifrin

Of course there’s a mean value theorem.

sunny

Ok, then Wikipedia has disappointed indeed :)

Ted Shifrin

It’s an inequality, which is really what the single-variable one is.

I bet you didn’t read carefully. Besides, this function is so simple, you can use what you know.

sunny

8:40 PM

hmm, ok, for the mean value "inequality" then, do I need to compute $\lVert f'(x)\rVert$?

Jakobian

I'm not sure what you mean by "domain for which $f$ satisfies Lipschitz condition"

I'm pretty sure this is going to be locally Lipschitz, but not globally

Ted Shifrin

Yes, sunny, but I still suggest you stop and use your brain instead.

sunny

@Jakobian I mean a subset of $\mathbb R^3$ where $f$ satisfies $\lVert f(x)-f(y)\rVert\leq M\lVert x-y\rVert$

Ted Shifrin

@Jakobian But there’s a simple description of the optimal subsets .

冥王 Hades

11:45 PM

Mathjax bugging out for the millionth time

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