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

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

1:11 AM
Not according to the definition in that post.

strange

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

So how does one resolve this?

Huh?

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

1:20 AM
Not according to the post. We're talking about values here, not domain.

ok

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

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.

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.

Ok, I found the remark, thanks.

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

2:19 AM
I believe Rudin does, but I am not that sure.

2:30 AM
Yeah it does, just checked. Thanks

no problem :).

2 hours later…
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

4:44 AM
why? how?

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?

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

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

5:31 AM
Lipschitz in the $y$ variable. So what Hölder condition?

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

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

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

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

$f=\sqrt{|y|}$

5:42 AM
Yeah. So isn’t that Hölder as you suggested?

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

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

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"

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

6:38 AM
It will be the equinox shortly, ~2023-Sep-23 06:50 UTC. ssd.jpl.nasa.gov/api/…

6:54 AM

1 hour later…
8:14 AM
$\frac{x}{x+1}\lt \log (1+x)\lt x$
Is there a way to solve this using Mean-Value Theorems?

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)

8:41 AM
@leslietownes yes, by solving , I meant proving. So, this inequality is valid only if $x\gt 0$ ?

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)

9:42 AM

2 hours later…
11:41 AM
do probabilists care about a situation where R.Vs are not independent?

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.

2 hours later…
2:14 PM
0

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...

@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

@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?

@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

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?

Yes.

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

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

@Hades First, there are different Strang books. Second, what are your friend’s interests and proof interests?
Not now, Thomas.

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

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

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.

@ShaVuklia is that a question

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)

Yeah, baby Strang is fine. Not adult Strang.
@ShaVuklia Yes. First get a leading $1$ by combining some pair of rows.

2:48 PM
@TedShifrin No problem! Thanks for responding!

@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)

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

@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.

@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

if I could use transposes, then the computation is alright

2:54 PM
Yeah but I once took a seminar in probability theory and the speaker always assumed i.i.d condition.

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

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

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

well, independence always makes things simpler
but if you look at real life, things are often not independent

stay independent.

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

yes, that is connectedness

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

The smoothness there is redundant. True iff continuous.

and I wanted to test how this work in other categories

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.

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

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.

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

2 hours later…
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

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

So X_i's are given to be independent and identically distributed RVs and have mean 0, and T= X_1+...+X_n.
Then $E(T^3) E(X_1^2)= E(T^2)E(X_1^3)$.

you're right, I was only thinking manifolds
$\mathbb{Q}$ doesn't work cause it's not locally connected

given $f,g\mathbb{R}^n\to \mathbb{R}^n$ Is there a nice formla for $\nabla (\nabla f g )$
something like $\nabla ( \nabla f g )= \nabla f \nabla g + ...?$

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).

correct

5:58 PM
Is that correct?

@koro

ok, thanks.
My exam was better than I expected.

6:14 PM
your exam was on what subject?

it was on probability theory

oh nice .

6:37 PM
@Monty What do you mean by this?

where can I learn about vector bundles?
i know fibre bundles

What do you want to learn?

6:58 PM
I want to do these

@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 $\nabla (\nabla f g)$

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

why is a fiber a vector space?

@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.

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

7:06 PM
You have a continuously/smoothly varying inner product on each fiber. Your exercise told you that.

first para in the image makes no sense to me.
'the map form U to...' in the third line, what is that?

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

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

He told you in the parentheses.

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

7:08 PM
Inner product on $\Bbb R^n$ is given by a positive definite symmetric matrix.
Assuming fair coin, sounds right, @Cotton.

Yes, fair coin

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

how did they go to R^n?
sure, I'll take a look.

He does tangent bundles mostly.

@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

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?

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

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

$gDf$
but i think of them as column vectors so its $Df g$
if I apply a derivative to a product i always put brackets.

Well, what you’re writing no one would know how to parse.
$f,g$ vector valued. Be specific.
Mapping what to what?

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

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.

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

@Thorgott I guess you're right
It's interesting to me that what lead to formalization of topology is the concept of connected spaces

7:33 PM
not the Ko(two dots on o)nisberg bridge?
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...

topology didn't really exist when this problem was stated
so I'm not sure what you're asking

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:

You're missing the point
I said formalization of topology

and Marco Manetti's topology book

I never said some topology ideas didn't exist before that
Topology is considered to be formalized only in 1914 by Hausdorff

7:39 PM
(going deeper into the wiki page, it seems that only 3 of the bridges remain.)

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

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.

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

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

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

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."

it should be: $\tilde x(t)=\cos(\beta)u-\sin(\beta)$
then we have $\tilde x(t)=-x(t)$ and $\tilde y(t)=-y(t)$

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

(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.

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

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

7:58 PM
(Also, $X$ is infinite)

looool
a sine sign error

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.

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

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.

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

8:05 PM
oh, haha.

@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.

Gotcha, thanks v ery much leslie!

8:22 PM
ted: honk

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.

Of course there’s a mean value theorem.

Ok, then Wikipedia has disappointed indeed :)

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.

8:40 PM
hmm, ok, for the mean value "inequality" then, do I need to compute $\lVert f'(x)\rVert$?

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

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

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

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

3 hours later…
11:45 PM
Mathjax bugging out for the millionth time