Is this a hyperbolic metric or a metric of a different kind? $ ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ I noticed that if you make the change $x=y$ then it's the Poincaré metric

you can write it as $ds^2=d(\ln x)^2+d(\ln y)^2$, so it's basically what you get if you drew gridlines on a log-log plot and then mapped back to the original space

so it's not hyperbolic?

wait how can you write it like that?

$d\ln x= dx/x$

and no, i don't see anything terribly hyperbolic about it

can you allow $x,y<0$?

yes. but that metric only makes sense away from the x-,y-axes

so there's no meaning to the distance between (1,1) and (1,-1) for instance

okay so $x,y>0$ then makes sense. And I can probably try to find the isometries of the space with this metric.

to get a sense of why you want to avoid the axes: the distance from (1,1) to (1/e,1) along the line y=1 is ln(e)=1. so is the distance from (1/e,1) to (1/e^2,1)

so the distance from (1,1) to (0,1) is infinite

oh yeah good point. So the familiar "translation" in $\mathbf R^2$ would look a little bit different here

I think

familiar translations are $(x,y)\mapsto (x+a,y+a)$ I think

yeah

for some constant $a$

the equivalent would be $(x,y)\to (ax,by)$ here

that looks like a dilation

Any metric $f(x) dx^2 + g(y) dy^2$ ($f,g>0$) is flat.

Time to learn basic undergrad diff geo.

that's what I'm doing

@TedShifrin the connection itself vanishes, right?

Yes, @Thor. Obvious change of coords makes it Euclidean.

Of course, that would be true for any flat metric, but here no trickery is needed.

@robjohn ok right!

@Rover Does that mean you understand, or just partly understand?

@BalarkaSen if you want to think of symplectic camel as some classical uncertainty (I don't know what you precisely mean by this) but one place to look for will be the corresponding Poisson manifold on the algebra of smooth functions. That's one classical way to rephrase uncertainty.

But then you would have to know some way to compare symplectomorphisms with Poisson morphisms. Every Poisson manifold is foliated into symplectic leaves so maybe some patching up of symplectomorphisms can be done I guess.

@SayanChattopadhyay I am certain that there are words in that statement.

@SayanChattopadhyay more words. I am sure of that.

is there a quick way I can show that a square matrix over a field cannot be similar to two different matrices in rational canonical form?

i think computationally its going to come down to extracting the invariant factors from a matrix in rational canonical form

actually, its pretty clear, becase if two matrices in rational canonical form are similar, then the matrices that detemine their characteristic polys are similar, and so each block in the matrices that determine their characteristic polys are similar, but then you can derive what a companion matrix is from its characteristic poly, so we ar edone

actually know, im assuming block diagonal matrices over a field can only be similar if they have the same number of blocks of each dimension, and that the blocks themselves are similar..

why is f(x,y,z) = z^(x+y) defined at z=0 but the equivalent f(x,y,z) = exp((x+y)ln(z)) is not defined there?

assuming x+y > 0

nvm, sorry!

$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\underbrace{\frac1{\frac{\mathrm{dx}}{\mathrm{d}y}}\frac{\mathrm{d}}{\mathrm{d}y}}_{\frac{\mathrm{d}}{\mathrm{d}x}}\underbrace{\left(\frac1{\frac{\mathrm{dx}}{\mathrm{d}y}}\right)}_{\frac{\mathrm{d}y}{\mathrm{d}x}}=-\frac{\frac{\mathrm{d}^2x}{\mathrm{d}y^2}}{\left(\frac{\mathrm{dx}}{\mathrm{d}y}\right)^3}$$

@robjohn in this how can we take dx/dy out of derivative?

I mean $ (\frac{d}{dy} \frac{1}{dx/dy})$

=$\frac{1}{(dx/dy)^2}$ $\frac{d}{dy} \frac{dx}{dy}$

I.e why $\frac{d}{dy} \frac{dx/dy}{(dx/dy^2)} =\frac{1}{(dx/dy)^2}$ $(\frac{d}{dy} \frac{dx}{dy})$

@Rover $$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac1{\frac{\mathrm{d}x}{\mathrm{d}y}}\right)=-\frac1{\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\frac{\mathrm{d}^2x}{\mathrm{d}y^2}$$

chain rule

$$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac1u\right)=-\frac1{u^2}\frac{\mathrm{d}u}{\mathrm{d}y}$$

@robjohn oh ok understood 👍🏻 thanks

time travel is weird

i am born -> i create time machine -> go to past and kill grandfather (well i would not do that this is just an example) -> father not born -> me not born -> i don't create time machine -> grandfather doesn't die -> father doesn't die -> i am born -> ...

@Euler2 Consider this scenario: You go back in time, then you killed your grandfather, you stay there, you had a child and then, a grandchild. Then, your grandchild mysteriously went missing because he(assume that grandchild is male) traveled back in time to kill you. Who is the grandchild? Is it you or not?

sometimes i wonder if time travel is even possible

Entropy increases over time, right?

yeah

What if time travel will not decrease entropy, but increase?

we cannot break the second law of thermodynamics

i don't even know if humans will remember the future when they travel to past

What if going either way in time increases entropy

Should we stay frozen in time?

consider this: I break a vase so I go back in time so that I will not do that. But when I go back, it is highly probable that I have forgot I will break the vase in future (humans don't remember, or rather, know the future) so I break the vase again.

After that it is my choice if I go to the past again or not

@Euler2 Is there free will, then?

My favorite time travel movie is probably Primer

idk if I knew everything then I would have made a time machine

btw we all are doing time travel

at the rate of 1 second per second

I think there can be a sort of "read only" time travel where we can go back to past but we cannot change anything

i don't think time travelling is physically possible

any probability/combinatorics people know how to tackle math.stackexchange.com/questions/4106131/… ?

it hasn't got much love sadly even though I thought it might be a known problem

I've seen involutions show up from time to time. is there a good reason to look at these maps ?

If i add a the complex conjugate of a function to the original function, do they sum to 0?

Ie does $f + f^* = 0$ where $f^*$ is the conjugate of $f$

Or do i just get double the real part of $f$ as a result?

(Not sure what the notation for that would be)

I gurss if $f \to a+bi$, then $f + f^* \to (a+bi) + (a-bi) = 2a$ ?

@Rithaniel Best Movie: Primer, best book: Orthogonal Series (okay 3 books, but it is worth it for the payoff)

@AndrewMicallef no

@AndrewMicallef yes

If a group $G$ is quasi-isometric to $\mathbb{Z}$ and $g\in G$ has infinite order, then why is the subgroup generated by $g$ quasi-dense in $G$?

I can show this if there somehow exists a quasi-isometry $f$ for which $f(g^n)=f(g)^n$ but not sure if this is necessarily the case.

I am having problem with digesting implication in delta epsilon definition of limits

why $|x-a|<\delta\implies|f(x)-L|<\epsilon$ and not $|f(x)-L|<\epsilon\implies |x-a|<\delta$ in limit definition

ah I think it's because of truth table of implication

am I right?

I am okay with quantifier here

@JarneRenders I don't have details, but the point should be that the the positive and negative powers of g define two ends of G. if <g> wasn't dense, we could find a sequence of elements whose distance from <g> grows unbounded, which would define an end of G distinct from the two ends previously defined, contradicting that G only has two ends, being quasiisometric to Z.

@user863565 because we have a specific geometric idea of what a limit should be. the first version models that, the second does not.

@Thorgott That's also pretty much how you prove that a f.g. group has two ends iff it is virtually $\Bbb Z$

Find an element whose power let you figure out where the ends are which generates a finite index subgroup

yeah, I think that's basically the second half of a proof

no clue how to get 2 ends => there exists element of infinite order, though

I can tell you about that later

But I'm busy for a while

user, it's just the definition. if you think about constant functions, |f(x) - L| < epsilon is not guaranteed to tell you anything at all about |x - a|

Is the 'natural map' just the inclusion?

the section O(D(f)) is defined as the localisation of A at the set of all g such that D(f) ⊆ D(g), btw

to be clear, i've already proved this under the assumption that the natural map is the inclusion, but it seems too easy for him to label it as a 'great exercise'

it's the map that operates as identity on representatives

so the inclusion right? otherwise, i don't know what you mean precisely

you have a canonical projection A->O_{Spec A}(D(f)) and it's easy to check that this extends to a map A_f->O_{Spec A}(D(f)) by universal property of localization

conversely, I don't know what you mean by inclusion

$\frac{a}{f^n} \mapsto \frac{a}{f^n}$

I wouldn't call that an inclusion, but your mileage may vary

alright, thanks

why not

what's going on is that we have multiplicative subsets $S\subseteq T$ and obtain a natural map $S^{-1}A\rightarrow T^{-1}A$. these maps are in general not injective, so I can hardly call them inclusions. this particular one turns out to be an isomorphism, but I don't believe there's an a priori sense in which this map can be likened to an inclusion.

ah

right, thanks

The abelian group $\{e^{i \pi k/2^n}: k, n \geq 0\} \subseteq \mathbb{C}$ is isomorphic to all of its nontrivial quotients

known as the Prüfer-2 group

is that what it's called

if i'm not missing anything, there's nothing special about 2. you just need a prime

(for context, i just imported about 15 years worth of notes onto this computer and i'm having fun rediscovering stuff i wrote down)

yes

they're called Prüfer p-groups for prime p

they play an important role in the classification of infinite abelian groups or something

it's the p-primary component of Q/Z i think

i'm getting to that in my notes

hahaha

infinite abelian groups frighten me. there is a long list of things that frighten me, but they are high on the list.

yeah

hello i think i have a brainfart: If looking at perrons formula the right hand side (with the integral) is differentiable (with respect to x) whereas the left hand side isn't even continious

they're a fun example of non-cyclic groups in which every subgroup is characteristic

it isn't clear to me that the right hand side is differentiable. yes you could push d/dx in there and write a symbol, but there's a lot going on. e.g. it's a PV integral, not an actual integral. so maybe there's no problem.

I don't see why the RHS there would be differentiable

thorgott and i seem to be on the same page.

i'll take that as a sign that i'm not delusional. about this.

when i say "maybe there's no problem" i mean with the equality, as opposed for example to the differentiation operation that you propose.

yet neither of us is an analytic number theorist

we can all agree it's a funny formula.

i have henry helson's book on dirichlet series around here somewhere. and notes on a talk of a generalization of something in there that he never published.

he was really into dirichlet series in his final years. for some reason

Still want to heard about two ended groups? @Thorgott

sure

So the idea is that $G$ acts on the set of ends and the kernel of the action has index at most two (we are assuming $G$ has two ends and we want to get that $G$ is virtually $\Bbb Z$), so assume that the action is trivial

yeah

Now we can find some finite connected $C$ in the Cayley graph whose complement has two connected components, say $W_+$ and $W_-$

And we can find a $g\in G$ with $gC\cap C=\varnothing$ since $C$ is finite

Now $gC$ is contained in one of those components, say $gC\subseteq W_+$ and $g^{-1}C\subseteq W_-$. But since $G$ acts trivially on the ends we also have $gW_+\subseteq W_+$

and in fact we get an infinite descending sequence $gW_+\supseteq g^2W_+\supseteq g^3W_+\supset\ldots$

which implies that $g$ has infinite order

ah ok, if g moves a separating ball off itself, it keeps shifting off the end it lands in towards infinity, hence has infinite order

makes sense

Write odd prime $p=a^2+b^2$ for $p=4k+1$, then must one of $a^2-b^2+2$ and $a^2-b^2-2$ a prime number?

thanks

@Thorgott yeah exactly

And then you take that ball and you show that $\bigcup_{z\in\Bbb Z}g^zC$ is the whole thing to show finite index

nice

In $\mathbb{R}^2$, does "reflection about the origin" mean the same thing as "180 degree rotation"?

good question. in R^2, i think so, yes, both are the map (x,y) -> (-x, -y).

in general, no.

the vibe of rotation is a different vibe. for example in R^3 rotations have fixed points and "reflection about the origin" (i.e. v -> -v) does not.

there's also stuff about, god help us, orientation.

Is there anyone who wishes to help me slightly with economics..

I need some help..I will post my queries elsewhere...

Not in this group..

Okie, yeah they would be played as two different animations in a Youtube video, hence a different vibe, but I was thinking my confusion over whether an odd function was comprised of a rotation or a reflection of its right half might not be based on an actual distinction.

Does it matter in terms of generalizing evenness and oddness into higher dimensions whether one thinks of the left half as a rotation of the right half VS a reflection of the right half around the origin?

'point reflections' are generally very different beasts from rotations. they tend to change orientation, which rotations as a rule do not do. a geometer could explain this more simply than i could.

okay, thanks for the help

Can someone recommend me a modern analysis textbook with clean notation and proofs that includes standard theorems such as the Leibnitz integral formula and integration by parts?

Could you please help me find the actual statement of the payley wiener theorem for the mellin transform in here. Can't seem to find it. It should be in section 4.

Quote: "Now the desired Paley–Wiener theorem for the Mellin transform can be stated as follows: Theorem 3 Paley–Wiener." And then they continue to prove that the Bernstein Space = Mellin-Bernstein space. Is my understanding correct: If f is in the Bernstein space (and maybe further not so important conditions), f is a mellin-transform.

@Thorgott Thanks for the hint about ends, hadn't heard about these yet, but I get the point. I have now constructed three sequences in $\mathbb{Z}$, where the distance of each of them with a common point grows unboundedly and the distance between each other grows unboundedly. I think it is obvious that this is a contradiction (but haven't written it down yet).

yeah, because you can transfer the situation to Z via the quasiisometry and in Z it's pretty clear that can't happen

hi every one

I need help for this sequence of function

if it converges in there, say to L(x), then L has to have some value at 0. i'd start there.

it looks like a perfectly fine convergent sequence in L^1 so you really need to work with the continuity assumption

so, no one knows what conditions f must satisfy?

is an involution the same as a convolution?

no

completely different things

someone writes $b * f $ and calls it a involution

hmmm

either there is some weird context in which this is taking place or they are simply misspeaking

I cannot stop laughing: hal.archives-ouvertes.fr/hal-01667383/document

ok, the pictures at the end got me

lmfao

@Thorgott Yes that's literally him (the author) in the last picture :'D ...

@leslietownes how to do with$\lim_{n\to\infty}\int_0^{\frac{1}{n^2}}|n-f(x)| dx$

what I f hear ?

You have to figure out what $f$ is first.

Or why it cannot exist.

@PolineSandra here is an indirect way: the sequence $f_n$ is Cauchy with $d$. suppose $f_n$ does converge to some $f\in C[0,1]$, then we will have $\int f_n \to \int f$ and we can compute that this value must be $2$. however, since $f$ is continuous on $[0,1]$ it is bounded. use this fact to show that $\int f < 2$ which is a contradiction.

I see. With $L^1$ convergence, I a priori do not know pointwise convergence a.e.

this sequence is monotonic, so monotonic + L^1 convergence should yield pointwise a.e.

@copper.hat I don't understand this: use this fact to show that $\int f < 2$ which is a contradiction.

$\int_0^{\frac{1}{n^2}} |f_n(x)-f(x)| dx+\int_{\frac{1}{n^2}}^1 |f_n(x)-f(x)|dx\to 0$ how you get what you say ?

assume $f_n \to f$ with $d$. show $\int f_n \to 2$. since $f$ is assumed to be in $C[0,1]$ it is bounded, that is, $f(x) \le B$ for some $B$. Now compute $\int f$ and show that it must satisfy $\int f < 2$.

@copper.hat yes I calculate $\int f_n(x) dx=2-\frac1n\to 2$

and $d(f_n,f)\to 0$ what I found by this

that is the straightforward part. now show with these assumptions that you cannot have $\int f = 2$.

choose $n \ge B$ then $f(x) = f_n(x)$ for $x > {1 \over B^2}$ and $f(x) \le B$ for $x \le {1 \over B^2}$.

but from where we get this idea to choose $n\geq B$?

trial and error. i want to show $\int f <2$.

maybe drawing a little diagram might help?

note, it is not true that $\int f <2$, which is why this leads to a contradiction.

but there is no way that $f$ can be continuous at $x=0$ and match $f_n$ for large $n$ and have $\int f = 2$.

This comes back to my point. You need to argue pointwise convergence away from $0$?

@TedShifrin you can do that. the above sidesteps the issue.

Why does $f=f_n$ as you said?

For $x>1/n^2$?

i am supposing that $f \le B$ everywhere and i choose $n \ge B$ then for $x \ge {1 \over B^2}$ i have $f(x) = f_n(x)$.

^^@robjohn, Can you insert an orange circle with you distinctive face in the image above, or something not gif? Always good to pay our respects to our Moms!

where is it mother's day today?

Planet x

Copper, I'm asking how you can assert that about $f$.

@copper.hat Preparatory suggestion for robjohn, to design his own version to "wear" around Mother's Day

I don't understand how to do

let me re group. i need some calories first.

ok I found a proof in French perhaps one can explain me it

I don't understand the idea of let $t\in ]0,1]$

@TedShifrin can you help me ?

$t$ is a device used to show that the hypothetical continuous $f$ would have to be $f(x) = 1/\sqrt{x}$ over $(0,1]$, which would make it unbounded and would contradict that it's continuous over $[0,1]$.

(if I understand correctly)

if there was any doubt about it, $]0,1]$ is the French way to say $(0,1]$.

@PolineSandra My earlier suggestion implicitly assumes the following, so it was not the short cut that I had implied (thanks to @TedShifrin for persisting). In general, if $f_n \to f$ in $C[a,b]$ with the $d$ metric then the limit is unique. In particular, if $f_n(x) \to g(x)$ pointwise, $g \in C[a,b]$ and $d(f_n,g) \to 0$ then we must have $f(x)=g(x)$.

Returning to our problem, suppose $f_n \to f$ in $C[0,1]$ with the $d$ metric. In particular, for any $a \in (0,1)$ we see that $f_n \to f$ in $C[a,1]$ with the $d$ metric. Let $g(x) = {1 \over \sqrt{x}}$, note that $g \in C[a,1]$. Since $f_n(x) \to g(x)$, and $d(f_n,g) \to 0$, we see that we must have $f(x) = g(x)$ for $x \in [a,1]$. Since $a$ was arbitrary, we must have $f(x) = g(x)$ for $x > 0$. Since $g$ is unbounded we cannot have $f \in C[0,1]$ which is a contradiction.

fwiw, this is more or less what the screenshot in French says

Makes sense. I was trying to avoid the mess.

Anyone here use an apple device as their main computer?

god no

Yeah me neither, one of my profs has said he's using "Notability" to produce all the videos for the course this semester, so the videos aren't available to non-apple users

Idk if there's some way to convert this so I thought I'd ask y'all nerds

Alas, probably not

@EdwardEvans Yes, since 1988.

No idea what Notability be.

Bleh

Prof's done a great job of making the course accessible

You could be sight-impaired.

@PolineSandra Round 2: Suppose $f_n \to f$, then we must have $\int f = 2$. Since $f$ is presumed continuous there is some $N$ such that $f(x) \le N$ for all $x$. Choose $n\ge N$ then $\int |f_n-f| \ge \int_{1 \over n^2}^{1 \over N^2} |f_n-f| \ge \int_{1 \over n^2}^{1 \over N^2} (f_n-N) \ge {1 \over N}+{1 \over n^2}(N-2)$ which contradicts $f_n \to f$.

@TedShifrin as in, I could have failed to read something?

I am curious why clicking on Poline's profile brings one to a page that says "Page not found"

@EdwardEvans No, even less accessibility!

Lol well I'm not, but I may as well be

i understand the need to make $$$ but i hate these private formats.

It is apple tbf, it's to be expected

Still, not sure why a prof would make his course unavailable to all but apple users

did you ask him/her if he could save it in an accessible format?

I use notability from time to time and am not aware of a screen recording mechanism that is built in (could be wrong) so i would guess (s)he is recording the screen. If that is the case, I would think it possible to convert the video to a format your computer can play

Yeah the answer is "We're producing the course with Notability this semester. We're looking for a way to make this accessible to Windows and Linux users." lol

wow

@Quin Hmm, the format of the files that have been uploaded are .note files, these aren't videos then?

ahhh yes those are not videos but notes. in that case, that is pretty poor form. There is a very easy solution in that Notability allows for PDF uploads (either instead of .note or in addition)

ergh what a poorly organised course this has been so far rofl

Thank you

yeah ask for them to upload PDFs. This is def possible as I've been able to do it for classes Ive taught. anyway, i gtg give a talk now hah. wish me luck!

Good luck! :P

@Thorgott (Replying to yesterday's limit definition) So how does the second one doesn't model the limit definition ?

I'm convinced $y = x^2$ has the same geometrical shape as $z = x + x^2i$. Why would someone not believe me? (looking to do a proof for fun)

@shintuku Nothing going on here. $z=x+iy$.

doesn't it designate a subset of $\mathbb{C}$?

Yes, but you're identifying the ordered pair $(x,y)$ with the complex number $x+yi$. Nothing else.

can anyone explain me if I reverse the implication in delta epsilon definition of limit definition then why will it not match geometric viewpoint ?

@TedShifrin oh, I get it

thanks!

What does that mean, @user863565. You mean why it's wrong for a limit?

@TedShifrin I am curious if our limit definition was modeled as $|f(x)-L|<\epsilon\implies 0< |x-c|<\delta$ in this Wikipedia definition

en.m.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit

what would happen

@user863565 a limit can always be defined as modeled by the epsilon-delta definition

chose any $\epsilon$ s.t. $|f(x) - L| < \epsilon$, and you can find $0 < |x - c| < \delta$, that's what it means for the limit to exist

I know that but I am asking if I reverse implication because I don't see anything wrong with it

Lots is wrong.

What happens to a constant function?

well, give a function and a limit that works one way, and try it! @user863565

What happens to $f(x)=x^2$ And $c=1$?

The question for you once you answer those ... is .... For what functions does your “definition” work?

Oddly I can find the limit 😅 with reverse implication

No, not with correct logic you can't.

Constant function has no limit at any point in the domain.

You better be super critical of everything you think and say.

ok I might need to think bit longer.

I was using truth tables of implication to think.

nods Cool.

ted's reading my mind again

i'm obsessed with constant functions and diagonal matrices. even better, a diagonal matrix whose entries are constant functions.