12:29 AM
@ManolisLyviakis It most certainly is NOT. It oscillates between bigger and bigger positive and negative values!
@TanMath No, a set with an operation with properties. The symmetries of sets are examples of groups.

@TedShifrin :P

Hi @Leaky

1:00 AM
I'm new in metrizability, are all order topologies metrizable?

27

From my perspective as an analyst, non-metrizable spaces usually arise for one of the following reasons: Separation axiom failure: the space is not, e.g., normal. This mostly happens when the space is not even Hausdorff (spaces that are Hausdorff but not normal are usually too exotic to arise ...

1:16 AM
thanks!

hi all

2:01 AM
@TanMath it's not "useful". It's just one way to visualize them geometrically. It doesn't give much insight.

2:55 AM
@TanMath everything is a set plus a whole bunch of other stuff that give that set it's context

@TedShifrin ah okay the symmetrics lead to specialized types of groups? But in general a group can be any sort of defined set with a defined set of operations?

And on that note if you start seeing all the people that talk to you and around you as sock puppets, you haven't taken your prozac

@TanMath groups have a specific type of operation. It is associative, has an identity element, and every element in the set has an inverse with respect to the operation.

Is Penrose diagrammatic notation something that's actually useful in math and physics?

Define "useful".

3:07 AM
Sorry I am asking such random questions, just trying to go through this book I am reading...

Can it be used to do things or make things clearer? Sure.
Is it necessary to the subjects? No.

Like it helps people obtain new results in math and physics
So it does make things clearer? Because looking at it now, it seems pretty confusing

Well it stands for something which can be even more confusing.

Also, it doesn't feel very useful (at least right now) like bra-ket notafion, which is immediately very helpful for understanding linear algebra in the context of QM
Do you know of any good tutorials on this notation? because Penrose just quickly skimmed over it at the end of a chapter in the book

I don't think you will be able to employ it meaningfully without some further learning of manifolds.
Otherwise it would just be notation for linear algebra stuff.

3:21 AM
Well just an intro to the notation would be nice, because it seems that Penrose will demonstrate its use throughout the book
If you know of any, otherwise it's fine if you don't...

3:43 AM
@Ted my $f_n(z)$ are $\sum_{k=1}^n e^{\pi i k^2 z}$ so the Weierstraß M-Test would show that $\sum_{n=1}^\infty \sum_{k=1}^n e^{\pi i k^2 z}$ converge absolutely uniformly on (compact) subsets of $\Bbb H$, which isn't what I want

4:17 AM
In my real analysis class, a metric space $(M,d)$ is said to be compact if $(M,d)$ is both complete and totally bounded. For completeness, we need Cauchy sequence converges, for boundedness, we need sequences in $M$ has a Cauchy subsequence.
Why do we require Cauchy sequence for these properties

@ÍgjøgnumMeg No, $f_n(z) = e^{\pi i n^2 z}$.
@Simple: Because you want a sequence that has a chance of being convergent if conditions hold. If you don't say Cauchy, you can have badly non-convergent sequences even if the space is good.

5:18 AM
@Simple simply because you wouldn't expect the sequence 0,1,0,1,0,1,0,1,0,1,... to converge

5:48 AM
I see, many thanks

2 hours later…
7:58 AM
in CRUDE, 8 mins ago, by skullpetrol
reheating old ideas may lead to an unexpected discovery

8:10 AM
@skillpatrol nice!

:-)

2 hours later…
10:12 AM
@BalarkaSen you'd like that

Very surprising identity

yet, so obviously simple once found...
"hidden in plain sight"

1 hour later…
11:38 AM
@Akiva related to some things you were talking about a while ago, I discovered that there is a notion of pseudofinite groups and pseudofinite fields

@skillpatrol if an online science magazine ive never heard of says its legit then that's all the evidence one needs really
remember when RH was proven
remember remember that skull ok im done
wait plastic can be converted into petrol ok now im finished

1 hour later…
12:50 PM
@Ted hmm well it looks like I did something different to what you were intending then hahaha

3 hours later…
4:04 PM
Hello. can anyone help me with the pde u_x+u_y=u^2 with initial condition u(x,0)=h(x). I solved it using characteristic method and got u(x,y)=h(x)/(1-yh(x)). but when I try to differentiate u and check that the result is really u^2, it doesn't work.

@BalarkaSen what is a batchmate?

Classmate.
Sometimes merely someone in the same year as you.
But usually synonymous to classmate in my experience.

5:09 PM
Hello!!

We have that the vectrs $\vec{v},\vec{w}, \vec{u}$ are linearly independent.

I want to check if the pair $\vec{v}, \vec{v}+\vec{w}$ is linearly indeendent or not.

I have done the following:

Let $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$.

Since $\vec{v}$, $\vec{w}$ and $\vec{u}$ are linearly independent, then $\vec{v}$ and $\vec{w}$ are also linearly independent and this means that $\alpha_1+\alpha_2=\alpha_2=0 \Rightarrow \alpha_1=\alpha_2=0$.

(I ask this way because if I tell you if it is right/wrong you never learn to be able to check your own work).

If Fermat were alive today and a user posting his conjectures on here as questions, how often do you think he would receive immediate comments like "please show you have at least made some attempt at proof"

Often. The point of this website is to help people, not give them answers.
Also, why would we make exceptions for an obviously proficient mathematician?

Yeah I know I was just making the point because it only just occurred to me
I mean 9 out of 10 times he didn't provide proof I can see why its annoying now at least

Since the three vectors are linearly independent, then if we consider just the two of them these are again linearly independent, right?
To me, it seems correct. But I want to have an confirmation.

5:19 PM
Yes that's true. Is there any particular part of your proof that you are skeptical about?

@anakhro Not really a particular part, I just wasn't sure if I can use that, I mean "$\vec{v}$, $\vec{w}$ and $\vec{u}$ are linearly independent, then $\vec{v}$ and $\vec{w}$ are also linearly independent"

@MaryStar try proving it, it's not that hard.

How to find the principal part of laurent series around the isolated singular point of $\frac{z}{e^z - 1}$ ?
for z = 0, there won't be any principal part, I can't find a way to find around $z = 2 n \pi$. One obvious way that we were taught is to use cauchy integral formula, but that seems circular to me.
Even if I go straight to find the Laurent series expansion and for negative power terms, I have to use $c_{-n} = \frac{1}{2 \pi i} \oint (z - z_0)^{n-1} f(r) dr$, for $n = 1$, the problem reduces to find the residue, which I have learnt to find via Laurent series expansion, but that's what is the original problem.
$r = \text{dummy complex number}$

5:38 PM
@GaloisintheField I don't see why American politics needs to be injected into direct discussion here, I'm sure many can agree that's a sock puppet performance that got old decades ago

@AjayMishra Let $w = z-2n\pi$ so the function becomes $\frac{w+2n\pi}{e^w-1}$ and $w=0$
now use the expansion of $e^w-1$ at $w=0$

shouldn't there be an $i$ in there somewhere

aha yes it should be $z = 2i \pi n$ instead
@ÍgjøgnumMeg how is modular forms

@Leaky it's alright, we introduced the Petersson inner product on thursday
which was interesting

what is it?

5:44 PM
take $f \in M_k$ and $g \in S_k$ (i.e. entire modular forms and cusp forms resp.). Then

$$\langle f, g\rangle = \int_{\mathcal{F}}f(z)\overline{g(z)}\Im(z)^k d\omega(z)$$

where $\mathcal{F}$ is a fundamental domain for the action of the modular group on the upper half plane and $d\omega(z)$ is
some
hyperbolic measure thingy
$\frac{dxdy}{y^2}$
which happens to be invariant under the action of the modular group as well
I think

who is $\mathfrak J(z)$?

just imaginary part of $z$

oh
would it be easier to state on the $q$-domain?
or is that only for $SL_2(\Bbb Z)$?

this one is only for SL_2(Z), I think you need to be more careful in congruence subgroups because the fundamental domain changes and you get more orbits under the action etc.
(I believe)
as soon as you state things for congruence subgroups instead of the full modular group the proofs get nasty
I mean we wanted to show that for $f$ a weight $k$ modular form wrt. a congruence subgroup $\Gamma$ that $g(z) = y^{k/2}\lvert f(z) \rvert$ is $\Gamma$-invariant and that if $f$ is actually cusp form then $g$ is bounded on the upper half plane
nope
not even gonna bother

@anakhro Ok! The same, with the similar proof, holds also for the pairs $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$ and $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$, right?

5:56 PM
All that does is either rename the variables in your proof or make the scalars slightly different.

@Leaky $$g(z) = y^{k/2}\lvert f(z) \rvert \leq \frac{\Im(A_\mu^{-1}\langle z \rangle)^{k/2}}{e^{2\pi \tilde{h_\Gamma}(A_\mu\langle \infty\rangle)\Im(A_\mu^{-1}\langle z\rangle)}}\left(e^{2\pi \tilde{h_\Gamma}(A_\mu\langle \infty\rangle) C}\sum_{n=1}^\infty\lvert a_n(A_\mu)\rvert e^{-2\pi \tilde{h_\Gamma}(A_\mu\langle \infty\rangle)nC}\right)$$
That's better

heh

hahaha
the lecturer was just like "and obv the terms in the brackets on the rhs are constant"
which is fair, there's no z

@ÍgjøgnumMeg How many days did it take to type that

@Alessandro it's been 86 years
@Gallifreyan tyty

6:05 PM
I know how to calculate the integral...But can't understand how to use the fact "Normal away from the z axis"

@loong What's your profile image? I recently saw some behind the scenes photos from a movie where the actors where holding a thing with the same pattern printed on it
Wait did he actually enter the room is my laptop drunk again?

@AlessandroCodenotti x-rite colorchecker

Just googled it, makes sense, thanks

basically how do we decide if normal is $r_r \times r_\theta$ or $r_\theta \times r_r$?

@Leaky ah, actually we did define the product for arbitrary congruence subgroups

$$\langle f, g\rangle_\Gamma := \frac{1}{[\operatorname{SL}_2(\Bbb Z):\Gamma]}\int_{\mathcal{F}_\Gamma}f(z)\overline{g(z)}\Im(z)^k d\omega(z)$$
where here $\mathcal{F}_\Gamma$ is a fundamental domain for the action of $\Gamma \subset \operatorname{SL}_2(\Bbb Z)$ on the upper half plane
and I think one needs to halve the index if the negative identity matrix is in $\Gamma$

6:15 PM
nice

6:28 PM
@AlessandroCodenotti you mean to tell me that if I show you a grid of the same dimensions, of equal uniqueness in terms of the colour of squares, then I show you another sometime later, you will be able to definitively say what if any alterations I have made between the two? lets test this
I have never met someone with photographic memory before this would be really be an amazing experience if you would agree

6:43 PM
@BalarkaSen hey man! how are you?

If $\Gamma$ is a discrete group, what is $\ell^{\infty} \Gamma$? Is it the set $\{f : \Gamma \to \Bbb{C} \mid f \text{ essentially bounded }\}$, where essential boundedness is taken with respect to the counting measure?
I know that the set of characters on $\Gamma$ forms a weak*-closed convex subset of $\ell^\infty \Gamma$. Characters are functions $\tau : \Gamma \to \Bbb{C}$ such that $\tau (1_{\Gamma}) = 1$, $(\tau (\gamma_j^{-1} \gamma_i))_{i,j}$ is a non-negative definite matrix for each $\gamma_1,...,\gamma_n \in \Gamma$, and $\tau(\gamma_1 \gamma_2) = \tau(\gamma_2 \gamma_1)$ for all $\gamma_1, \gamma_2 \in \Gamma$.

@Archer I take that to mean that, when considering the cone, you can take the local surface normal to point up and in (towards the z-axis) or down and out. That will change whether the flux is taken to be positive or negative

6:59 PM
@user193319 Essentially bounded and bounded are the same thing wrt the counting measure

Oh yeah, because a set of measure $0$ is the empty set.

7:15 PM
Let $X$ be a normed linear space, $Y$ a closed subspace, and let $Q : X \to X/Y$ be the canonical projection map. My book states the image of the unit ball in $X$ under $Q$ is mapped onto the open unit ball of $X/Y$. Does this mean that the image equals the whole unit ball in $X/Y$? I can see that the image is contained in the unit ball of $X/Y$; but I don't see the other inclusion.
That is, if $||Q \xi|| < 1$, why is $||\xi || < 1$, where $||Q \xi || = \inf_{y \in Y} ||\xi - y||$?

7:32 PM
well I guess there isn't any point posting questions on SE anymore if I know they wont get answered or attempted I guess that was a dick move to begin with

is that a swear word according to European children's books? It's confusing with English especially when you consider the name in the context of the geometry of the projectile used in the game of badminton for example

like it is a synonym for rocket penis how am I supposed to not laugh at that
shuttle cock*

My question is what this has to do with math

7:45 PM
does this look like a series for any known function? 1- 1/2*x + 1/3*x^3 - 1/4*x^4+....? If there a site that lists such things to search?

well at least im not being banned for 90 days and no I cant post my math because its disgusting

@Nasser that's log(x+1)
Just the power series.

like one of those conjectures you know is true but need another 5 years of reading abstract algebra for it to be not a conjecture

@Adam can you give an example?

@anakhro thanks. Yes, now I see it.

7:49 PM
well ive already formally stated the conjecture so yes I can give like 10,000 examples that's not the point

I mean, what is this conjecture, @Adam

and that will just get awkward silence im not a look how easy stuff is neutrinos just demonstrated it for us type of guy

well I was demanded relevance to mathematics so im explaining why I am not, it's pointless at this point in time
its nothing important

Can I know it? I want to try my hand at it

7:53 PM
@Nasser easiest way to confirm that it's log(1+x) is to differentiate it once

can u try both hands at the ones ive already posted?

where are they

all of them? well I mean that will take some time but yes give a counter example for that one and ill declare tonight a good game of badminton and award you the golden shuttle-cock
actually I think sleep will come if the laugh from that thought keeps this up

It looks a little too crazy for me sorry
Mainly I have no idea what these $\omega_{n,m,k}$'s and $v_{n,m,k,j}$'s are

1 hour later…
9:11 PM
@TedShifrin I'm back for more recommendations, where can I study series in normed vector spaces?

9:28 PM
Are you looking for more elementary stuff, or like functional analysis stuff?

9:55 PM

what?

TOO DARK

What's too dark

disappears into the darkness.

$P(2^{2^{2^s}},2^{2^{2^{-s}}})$ what function traces out these points

1 hour later…
11:18 PM
How can i show that the series $kq^k$ converges to zero for every $0 \leq q < 1$ for $k \rightarrow \infty$ ?

@T_01 what series tests have you tried?

well I'm trying with the standard $\epsilon$-definition. I am not allowed to use the logarithm or anything with continuos functions (including l' hospital or anything)
but this seems hard since I do not know how to pick $\epsilon$ according to the knowledge of $q^k$ converging to zero, since the factor $k$ changes, too?
or am i missing something obvious

Oh so you don't know the root test or anything like that?

Well, I know but we do not know XD
I'm doing a basic math course and there we did not introduce it yet

Well let's see. You know that x<1 for sure, right off the bat.

11:30 PM
what is x?

Sorry, q.

yes

Do you mean sequence and not series?

oohh
yes
i mean sequence, sorry. my english is bad

Much easier.

11:31 PM
can you give me a hint? :D

I thought you meant the sum $\sum_{k\geqq 0}kq^k$ :P
Which does converge for any $|q|<1$, but not to zero!

thought so

So what have you tried?

@anakhro i could show the convergence of this with the root test but i would have no idea how to get the limit value :D how does one do this?
@anakhro well, as i said, applying the epsilon definition

No, you cannot use the root test

11:33 PM
@anakhro ok. i did not try

So how far did you get with the epsilon definition.

nowhere at all

Okay, so what is the definition?

well
we want $kq^k < \epsilon$ for all $k \geq N$ for some $N$ dependent of epsilon
i mean we have $kq^k < \epsilon$ iff $\log(kq^k) < \log(\epsilon)$ iff $log(k) + log(q) < \log(\epsilon)$ ?... but i cannot use logarithm

Just checking: the only thing you have that you can use is the epsilon-N definition?

11:37 PM
i guess so? we also introduced the limit theorems like sum or product and stuff but $k$ on it's own is divergent so..
of course we also have the standard theorems like if we have a product of a zero-sequence and a bounded series this converges to zero
but $k$ is divergent so this does not help i guess
wait i did not try to show that it is monotone and bounded
okay it is not monotone obviously

It's not?

no it has a really strange behaviour, for q pretty close to 1 it first goes up very quickly and then falls down from a larger index (if my calculator is not lying to me)

Yeah, exactly, it goes monotone eventually.

hmm

Can you use monotone convergence?

11:46 PM
monotone + bounded => convergent, yes
or what do you mean?

Yes, that.
So if you prove it eventually is monotone, do you think you can use that?

so how do i find that point where it starts to be monotone "falling"
i think yes

You don't need to find that point exactly.

yeah, right
but something beyond that :D

Indeed.
I'd probably try that route.

11:49 PM
do you have a hint?

Well I think it is worth trying before getting any more help.
Since now you have a better idea.

@anakhro a smooth transition between a short elementary intro and anything that comes after would be nice

So something beyond the 9th chapter of Lang's undergrad analysis book?

Uhh I'm not sure what is that, just give me the elementary stuff

ok so i want to show that from some $k$ the following holds:
\begin{align} &(k+1)q^{k+1} \leq kq^k \\ \Leftrightarrow & q + \frac{1}{k}q \leq 1 \\ \Leftrightarrow & (1+ \frac{1}{k})q \leq 1 \\ \Leftrightarrow & 1 + \frac{1}{k} \leq \frac{1}{q} \end{align}

11:56 PM
@FuzzyPixelz that is the elementary stuff

this looks wrong
ah no it's not. And there is our dear archimedes
since 1/k gets arbitrary small, from some point this is true for every k, okay i do not know what i did on paper but this seems to be right