Mathematics - 2023-08-09

Soumik Mukherjee

00:08

@Jakobian Oh, okay

@AkivaWeinberger I was thinking about the problem :P, but thanks for the hint

MathCrackExchange

02:56

@Shaun hey

:D

Damn might have to write an I2C driver at work. -_-

Shaun

05:02

@MathCrackExchange Hi :)

Dannyu NDos

Are homologies like $H_n(X; \mathbb{R})$ worth investigating when they're always free vector spaces?

leslie townes

05:39

maybe depends upon you mean by 'worth investigating.' e.g. if you only care about something that is guaranteed not to detected by those, maybe not.

but e.g. de rham cohomology is regarded as a pretty natural thing to look at, and is maybe even more 'concrete' in the ingredients of its definitions than singular homology is. and you can compute H_n(X,R) in terms of that at least if the space X is nice enough.

one potato two potato

What does it mean by free vector space? anyway, taking real coefficients ensures there're no torsion elements.

leslie townes

are they 'free' vector spaces, or only 'gratis' vector spaces? will i have to listen to a lecture from some open source guy as a condition of using them?

one potato two potato

06:03

The automatic group I said before, its origin is computer science. It's reasonable not to care but it can be used to study group theory in some sense

many reasonable groups are automatic groups so I guess it's one reason people study those

sunny

06:37

Consider $$f(x)=\sum\limits_{k=0}^\infty\frac{\arctan (kx)}{k^2}.$$ Apparently, the function is continuous on $\mathbb R$, however, I have a hard time showing continuity at $x=0$. For $x\neq 0$, we have $$\left|\frac{\arctan (kx)}{k^2}\right|\leq \frac{\pi/2}{k^2}.$$ This shows it's uniformly convergent on $x\neq0$ (and since the terms are continuous, the function is too according the uniform limit theorem), however, what about the point $x=0$?

robjohn

What is the problem at $x=0$? Why do you think it is not uniformly convergent there?

sunny

Actually, I have a minor typo, the summation should go from $k=1$ to infinity.

leslie townes

arctan is pretty chill at 0.

robjohn

real chill

leslie townes

i haven't seen more chill, actually.

sunny

06:47

maybe it's not differentiable at $x=0$?

but still, it could be continuous there

robjohn

it is differentiable there

leslie townes

well, the inequality you wrote above "for x neq 0" also holds at 0. so does that help whatever you are trying to do?

sunny

that helps :D

user 858770

professor Townes for the win

robjohn

Don't you have$$f'(x)=\sum_{k=1}^\infty\frac1{k(1+k^2x^2)}$$

sunny

06:50

not at $x=0$ :)

robjohn

That does have a problem at $0$

sunny

@robjohn how do you see this?

simply because it diverges at $x=0$?

user 858770

How do you pronounce "Townes" @leslietownes? Like Tow/ness or Towns

leslie townes

the latter.

user 858770

Thnx

sunny

07:30

According to this question, the inequality $$\arctan(x)>\frac{x}{1+x^2}$$ only holds for all positive $x$. However, does it also hold for all negative $x$?

leslie townes

no. lots of ways to see it. both functions are odd, so they swap places.

or, x/(1+x^2) eventually goes to 0 at either infinity, so it's gotta hug that axis in the way that strictly increasing arctan with its nonzero limits at infinity does not.

sunny

ok

Dannyu NDos

@onepotatotwopotato Sorry; all vector spaces are free.

sunny

I'm very confused about the last inequality in this answer, i.e. $$\sum_{k=1}^\infty \frac{1}{k}\frac{1}{1+k^2h^2}\ge\sum_{k\le \frac 1 h} \frac{1}{2k}.$$ Is this simply an index substitution? Which one?

If it is an index substitution, why is it an inequality and not an equals sign?

leslie townes

07:47

it's using (1) the fact that if k < 1/h then k^2 h^2 will be [positive and] less than 1, so that 1/(1 + k^2 h^2) will be greater than 1/(1 + 1) = 1/2, and (2) if you discard the terms for k >= 1/h the sum does not increase

definitely not just jiggling indices. it's looking at a particular partial sum of the series instead of the whole thing, and then using an estimate specific to that partial sum

sunny

@leslietownes got it, thank you!

Jakobian

@XanderHenderson about that question you mentioned me, what a way to start a sentence, eh?

rostader

08:05

Hi. I might have a trivial question but I cant figure out whether it's true or not.

Let X be a k \times d matrix (everything is real) with rank d and let W be a diagonal matrix with d non-zero positive elements and the remaining elements on the diagonal may be zero or positive (not certainly negative). Then wil (X^\top W X) be necessarily invertibile? Intuitive I think it should be

leslie townes

not in general. the product WX can have smaller rank than X e.g. if the nonzero diagonal entries of W don't 'match' where X puts stuff into the space that W acts on

e.g. X = a 2x2 identity matrix with a row of zeros under it, having rank 2, and W the 3x3 matrix with 0,1,1 down its diagonal, also having rank 2, but WX having rank 1 and X'WX having diagonal 0, 1 also having rank 1

Jakobian

@rostader it's true for X W X^T that rank(XWX^T) = rank(W)

rostader

08:22

@Ja

@Les

@leslietownes The counter example is indeed corrext. W needs to have non zero elements corresponding to the linearly inderpendent columns

PNDas

09:45

How to find $\frac{\partial }{\partial y}\int g(xy)\,d(xy)$?

PNDas

10:09

Ignore the above question. Thanks.

mick

11:01

i lost 10 points rep , but did not get a message or reason ? no deleted user , no deleted upvoted answer , no downvote ??

I did delete two wrong answers though

and posted the correct ones ...

no big deal just 10 points

just saying

maybe undone upvote is not marked ?

I GUESS IT IS THAT

i noticed my mistake here and corrected it :

0

A: $f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$ and $\lim_{n \to \infty} \frac{f(n)}{\pi(n)} = 1$?

Let $x$ be a positive integer. I considered thinking about estimating $$\pi(x) + \pi(x/2) + \pi(x/3) + ... $$ The idea is simple. we take primes $p_i < x$. and we take $2 p_i < x$ and in general $$n p_i < x $$ Then we naively expect all numbers between $1$ and $ x $ to be of these forms. Hence th...

0

Q: Asymptotics for $f(x)$ such that $f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$?

Consider $$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$$ $$f(n) < \pi(n+1)$$ Where $\pi$ is the prime counting function and that inequality follows for instance from these here $f(x) + f(x/2) + f(x/3) + ... = x$ and conjecture A : $\pi(x/a) > \frac{\pi(x)}{a}$ $f(x) + f(x/2) + f(x/3) + f(x/4) + ......

one potato two potato

11:41

Is a 2-dimensional orientable hyperbolic orbifold a Riemann surface?

Thomas Finley

This is imho a trivial problem and if I am not mistaken, then, |x-1|<9/25 is the required condition.

The book provides a simpler condition i e |x-1|<=1 as an answer. But is my condition correct?

I just bashed with some inequalities

I cross verified and it seemed correct

But I need what other think bout this. I feel weird somehow.

Jakobian

12:23

am I crazy or does this proof not prove the proposition at all

Xander Henderson

You're crazy.

Now... let me have a look at that proof...

user4539917

You're not crazy.

Now... I think you lost sight of the forest because of the trees...

Jakobian

how did a whole proof disappear

how did this happen, I'm in shock, did anyone check this book before publishing

it's a good book and the proposition is pretty obvious but still

Peter

12:45

Proof wanted : dead or alive :)

sunny

I want to show the following function series $$f(x)=\sum\limits_{k=0}^\infty\frac{x}{2+k^3x}.$$ is continuous on $[0,\infty)$. Here's my approach in bounding the terms $$\left|\frac{x}{2+k^3x}\right|\leq \frac{x}{k^3x}\leq\frac1{k^3}.$$ This looks nice, however, when $x=0$ we have division by zero in the middle expression. How can we avoid this?

The index $k$ should go from $1$ to infinity.

When $x=0$, the inequality doesn't hold for all $k$. Is this a problem?

Jakobian

@sunny no, it's not a problem since $f(x) = x/2 + \sum_{k=1}^\infty ...$

@sunny even if division by zero occurs in the argument, the inequality still holds for $x = 0$, so there's no problem

@sunny if $k$ is from $1$ to infinity then I don't understand why the inequality doesn't hold for all $k$, it clearly does

sunny

13:03

I think you are right, my bad. Thanks for checking.

geocalc33

13:15

I found an answer I think is AI generated

sunny

@Jakobian from uniform convergence on $[0,\infty)$ of the function series, we can only conclude that the function is right-continuous at $x=0$, right?

Jakobian

@sunny it's not defined to the left of $0$ so it wouldn't make sense to ask for continuity from the left

Unless we define it on $\mathbb{R}\setminus\{-2/n^3:n = 1, 2, ...\}$ I suppose

But yes, in this case continuity on $[0, \infty)$ is equivalent to both-sided continuity at every $x> 0$ and right-sided continuity at $x = 0$

sunny

13:32

ok

geocalc33

13:44

A warped product metric on $(M, g) \times (N, h)$ is defined as $g + \varphi^2(x) h$, where $\varphi$ is a positive smooth function on $M$.

why is $\varphi(x)$ squared as opposed to cubed?

geocalc33

13:56

Just for a simple example: $ds^2=dx^2+\varphi^2(x)dy^2$ is a warped metric

geocalc33

14:27

anyone know if there is a name for the function $\varphi$?

Xander Henderson

@geocalc33 Yeah, it's called Phil.

geocalc33

@XanderHenderson lol

Xander Henderson

Honestly, I've never seen the construction you mention. However, I don't see any reason why $\varphi$ should be cubed.

And there may be an advantage to writing the metric as $(\mathrm{d}x)^2 + (\varphi(x) \mathrm{d}y)^2$.

Again, I don't know what you are on about, and Google isn't helping me much, but, like, squaring the term makes more sense to me than cubing it (given what little you've shown me).

geocalc33

@XanderHenderson the $\varphi$ is called the warping function I just found out

I don't know why it's squared but it's probably for a good reason

Thorgott

@Jakobian ok, I was missing the strategy of your proof, but I get that part now. but how does $\Psi$ preserving order of the roots imply that it agrees with $\Phi$ on the relevant elements?

Jakobian

14:41

@Thorgott $\Psi$ maps roots of minimal polynomials in $R$ to the corresponding roots in $R'$, in an increasing order

So if $a_1 < ... < a_n$ are roots in $R$, then $\Psi(a_1) < ... < \Psi(a_n)$ are roots in $R'$, but they are all roots (again from Sturm's theorem)

so $\Psi(a) = \Psi(a_j) = b_j$, since $\Psi$ is a bijection on the roots

minimal polynomials over $F$, that is

one potato two potato

pretty sure because of the change of basis

Jakobian

and hopefully you see now that this precisely means that $\Phi(a_i) = \Psi(a_i)$ for all $i$

so it has a little stronger property that it maps all roots to their respective positions

Thorgott

but you're talking about multiple minimal polynomials (those of $a,b,a+b$ and $ab$), what guarantees that potential overlaps between their roots are the same in both fields? is that a generalization of Sturm's theorem? also, I still don't see how to conclude from that. $\Phi$ is defined in terms of the order of the roots of one polynomial, but who says that the relative positions of the roots of these minimal polynomials are the same in $R$ and $R'$?

Jakobian

14:56

But you're just looking at one minimal polynomial at a time

$\Phi(\alpha)$ depends on what the minimal polynomial of $\alpha$ is

and for each $\alpha\in R$ we take a different minimal polynomial corresponding to it

I'm not sure what you mean by "who says that the relative positions of the roots of these minimal polynomials are the same in $R$ and $R'$"

$\Phi$ maps roots of minimal polynomials in $R$ to roots in $R'$ from definition

and it does so preserving their order

So e.g. you take minimal polynomial $p(x)$ of $a\in R$ over $F$, you take its roots $a_1 < ... < a_n$, $a = a_j$, and roots $b_1 < ... < b_n$ in $R'$, from definition $\Phi(a) = b_j$ and since $\Psi(a_1) < ... < \Psi(a_n)$ we need to have $\Psi(a_i) = b_i$ too so that $\Phi(a) = \Psi(a)$

you do the same for $b, a+b, ab$

"what guarantees that potential overlaps between their roots are the same in both fields"
the answer is, nothing, and I don't see how it's a problem

"is that a generalization of Sturm's theorem?"
no, I think you misunderstood me somewhere

Thorgott

15:44

oh, I think I get it now

I misunderstood you to take the roots of the minimal polynomials of $a,b,a+b$ and $ab$ simultaneously

but you're looking at them one by one

ok, the argument looks good to me now!

Jakobian

@Thorgott wdym? Elements $c_1 < ... < c_m$ are all roots of those polynomials

I'm looking one by one when trying to apply $\Psi$ or $\Phi$, if that's what you meant

Thorgott

right, you take all of them when defining $\Psi$, but then you look at them one by one when comparing $\Psi$ and $\Phi$

@Jakobian in this step

the crux is that $\Psi$ preserves the roots of polynomials individually each, allowing this comparison

Jakobian

ah, yes

I'm actually going to go through the whole first chapter of Morandi because I hate to leave something unchecked, as I already said, and primitive element theorem is something I never encountered before

it's in section 5 of that chapter

originally I thought the first 3 sections would be good enough, but I end up requiring more and more theory

@Thorgott could you look in corollary 4.10 in Morandi? He assumes that the extension is finite for some reason, I don't understand why

the corollary says that separable extensions are precisely the ones contained in Galois extensions, and that extensions by separable elements is separable

using characterization of Galois extensions as the ones which are normal and separable

Ted Shifrin

16:05

@geocalc33 If you actually sat down and learned differential geometry, you'd know why.

It's basically for the reason Xander said. You want to (locally) write the metric as a sum of squares in order to do any sorts of computations.

shintuku

@geocalc33 personwhostartsadifferentialgeometryreadinggroupsayswhat

Xander Henderson

16:21

@shintuku Hrm?

:P

Thorgott

16:39

@Jakobian these are all true without a finiteness hypothesis, perhaps he hasn't defined the terms in the infinite case or not proven necessary prerequisites in the infinite case?

Jakobian

hmm... no, it looks like the assumption is just totally not needed

I was wondering if there's any subtlety I'm missing though

Thorgott

17:34

I'm not seeing any

Mad

17:48

hello, is this statement true?
if a matrix times itself is the identity then the eigenvalues must be one or minus one, i am deducing this from my quantum mechanics lecture where a compex matrice has this attribute where then to " proof" they use this on an eigenvector to get a^2 = 1 since A^2 = Id
however i dont understand why a cant be an exponential "exp(i phi)
since that fullfills it

i am deducing this is not a true statement

robjohn

18:06

@Mad $e^{i\pi}=-1$

Mad

i wrote phi and not pi

any angle phi satisfies the relationship

robjohn

no, it doesn't.

Mad

The absolute value does, or is the squared in this case no to be considered as absolute value

robjohn

unless you have $A\bar A=I$

Mad

No i have $A^2 = 1$

robjohn

18:08

Then only $+1$ and $-1$

Mad

usually notation makes confusion

Z^2 is sometimes refered to as the absolute value

robjohn

for real Z, but otherwise you need $Z\bar Z=|Z|^2$

Mad

yes i understand what you mean, in some notation the bars are left out

robjohn

Then that can be very confusing

Mad

But in this case, you are right, it actually refers to A^2

Thank you for clearing it up

sunny

21:23

I'm looking for a sequence $(x_n)_1^\infty$, tending to $0$ as $n\to\infty$, such that $$\sum_{k = 1}^{\infty}\frac{\arctan(kx_1)}{x_1k^2} > 1, \quad \sum_{k = 1}^{\infty}\frac{\arctan(kx_2)}{x_2k^2} > 2, \quad \ldots$$ In other words, I want to show $\lim_{x\to0}f(x)$ does not exist, where $f(x)=\sum_{k = 1}^{\infty}\frac{\arctan(kx)}{xk^2}$.

Jakobian

note that $\arctan(y)/y\to 1$ as $y\to 0$

also note that we can take $x_i$ to be positive so that terms of the series in definition of $f$ are all positive

so... we can take some huge $N$ first, and then consider $\sum_{k=1}^N \frac{\arctan(kx)}{xk}\cdot \frac{1}{k}$

now consider small enough $x$ as well, and use that the harmonic series diverges

@sunny

sunny

@Jakobian thanks for the pointers, what does $\sum_{k=1}^N \frac{\arctan(kx)}{xk}\cdot \frac{1}{k}$ evaluate to?

Jakobian

huh?

sunny

ok, never mind. What would be our sequence $(x_n)$?

Jakobian

21:38

nothing explicit

sunny

21:56

@Jakobian ok, for some context, I'm looking for a different "proof" than this answer. I think it's not clear enough. But maybe that's as clear as it gets...

Jakobian

@sunny the argument I'm trying to describe to you works. I'm just looking for you to fill in the details

this is the approach to education of the type "I'll give you some guidance, but do it yourself"

sunny

sure :)

Xander Henderson

@sunny In what way is that argument unclear?

It shows that the limit which defines the derivative diverges at zero, via a fairly explicit computation.

Jakobian

I can see how someone can get intimidated if they're not familiar with using integrals to get bounds, and some weird $H$ symbol

Xander Henderson

@Jakobian Sure, which is why I asked sunny which part of the argument is unclear.

sunny

22:06

@XanderHenderson ok, I admit, it's pretty clear

Xander Henderson

@sunny Don't let me intimidate you into agreeing with me. If, after reading through the argument a couple of times, and trying to fill in the gaps on your own with paper and pencil, you are still stuck, explain your confusion.

sunny

Sure :)

robjohn

22:34

@sunny If it is still unclear, there is a very similar argument that relies on the Mean Value Theorem that shows the limit does not exist.

sunny

@robjohn If it's not too much of a hassle, I'm interested in hearing more :) I'm familiar with the mean value theorem of course, but I think some inequalities in the provided answer are not so straightforward

robjohn

okay, let me write something up

sunny

cool :)

Ted Shifrin

@sunny Can you tell us the values of $N(k)$ so that $\sum_{j=1}^{N(k)}\frac1j \ge k$?

This is essentially what you’re insisting on.

Jakobian

23:25

I'm reading about purely inseparable extensions. I'll probably never work with them, but the topic is pretty interesting, and how separable and purely separable extensions work together

I'm also beginning to think that field theory is just about defining a lot of different types of field extensions

Thorgott

purely inseparable extensions are very important, but only really once you have to think about positive characteristic

Xander Henderson

23:47

I am tempted to create an account on Politics SE just so that I can upvote politics.stackexchange.com/a/80732 . And that would be entirely for the line "I think that whole region colludes with map makers to help them constantly sell updated maps."

geocalc33

It only takes 2 seconds to create an account. Join us!

I can't find that line

Jakobian

Seems so. But I think I'm not going to study fields of positive characteristic, so this is purely out of curiosity for what people do in prime characteristics

Ted Shifrin

You can learn about tame and wild ramifications.

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