Mathematics - 2024-08-23

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Lucas Henrique

01:55

hey chat. any geometers around?

i'm working with an moduli space (in algebraic geometry) that is naturally in bijection with the following space: for a specific open $U \subsetneq \mathbb{G}(k,n)$, it is in bijection with the set $(B, V)$ of ordered bases $B$ of $V\in U$. looking up on the internet, I found out about the tautological bundle of Grassmannians and the concept of frame bundles. the frame bundle of the tautological bundle seems to be exactly what I'm looking for. does someone know the sheaf version of this?

that is: since the topology of the Grassmannian is the (Zariski) projective one in A.G., what is the most reasonable way to construct the frame bundle in the case of algebraic varieties/schemes?

4 hours later…

Soham Saha

06:00

user image

user image

Has anyone seen this fractal before?

Please note the differences in the scale

nickbros123

06:58

@BinkyMcSquigglebottom Antony??

1 hour later…

Soumik Mukherjee

08:04

@SohamSaha that doesn't seem to be a fractal

Soham Saha

@SoumikMukherjee Please see the two images. One is on [0,1), while the other is on [0,.28571), and they are graphs of the same function. The self-similarity is apparent.

Soumik Mukherjee

08:31

No, I am saying that you can't call those things fractals

Soumik Mukherjee

08:44

@XanderHenderson is that a fractal?

Soham Saha

09:23

@SoumikMukherjee I don’t think there’s a well accepted definition for fractals, but still, let’s wait for Xander Henderson’s opinion…

Soumik Mukherjee

09:53

@SohamSaha What is the function btw?

1 hour later…

Pizza

11:04

Hi 👋

Soham Saha

@SoumikMukherjee Sorry, can’t share the function right now. I was just curious whether structures similar to these one have already been studied previously, or if it could be something new…

Pizza

11:18

$$\int^{2\pi}_0 \int^1_{1/2} r^5 \sqrt{1+\frac{1}{r^4}} dr d\theta$$

Is there any way to simplify the integral?

$\sqrt{1+ \frac{1}{r^4}} \approx 1 + \frac{1}{2r^4}$

could I do it?

Actually, I just multiply, I don't need to do that.

Soham Saha

11:53

@Pizza Try substituting $u=r^4$ in the inner integral.

Pizza

@SohamSaha I did it like this: I multiplied and I found $r^3 \cdot \sqrt{r^4+1}$, then I substituted $u = r^4+1$ and found that $dr = \frac{1}{4r^3}du$

So $\frac{1}{4} \int \sqrt{u} du$

That should be correct, right?

Soham Saha

12:08

@Pizza Yes

Pizza

@SohamSaha thanks !

Soham Saha

No problem :)

rohit1729

12:39

Let $f(x, y, c_1) = 0$ and $f(x, y, c_2) = 0$ define two integral curves of a homogeneous first order differential equation. If $P_1$ and $P_2$ are respectively the points of intersection of these curves with an arbitrary line, $y = mx$, then prove that the slopes of these two curves at $P_1$ and $P_2$ are equal.

Mats Granvik

Do you know how to compute the Determinant in a 3 dimensional matrix or even higher dimensional matrices?

This is the matrix generating function:
$$\frac{\zeta (\text{s1}) \zeta (\text{s2}) \zeta (\text{s3})}{\zeta (\text{s1}+\text{s2}+\text{s3}-1)}$$ for which I know the GCD matrix for. I want to compute its eigenvalues, therefore I would like to know the determinant in dimension 3.

Joe

You can compute the determinant of an $n\times n$ matrix in terms of the determinant of an $(n-1)\times(n-1)$ matrix (the "Laplace Expansion"), or you can use the Leibniz formula for the determinant. I'm sure there are many explanations of this online.

user123234

can someone help me here:

1

Q: How can I prove that a complex Brownian motion is not pointwise recurrent?

Let $B=B^{(1)}+iB^{(2)}$ be a complex Brownian motion starting from $y\in \mathbb{C}$, i.e. we work under $\mathbb{P}_y$. I want to show that $B$ is not pointwise recurrent which means that it does not returns to a single point $z\in \mathbb{C}$ with probability one. I want to use the conformal ...

complex-analysis probability-theory stochastic-processes stochastic-calculus brownian-motion

Joe

See here for instance.

Mats Granvik

@Joe Yes for the n by n matrix. I am looking for n by n by n matrices.

Joe

12:49

Oh, I see.

Mats Granvik

The motivation is this n by n by n matrix/sum in Mathematica 14:
In[512]:= Clear[a, n, k, m]
nn = 40;
g[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]]
s1 = 12;
s2 = 12;
s3 = 12;
Sum[Sum[Sum[g[GCD[a, b, c]]/(N[a, 20]^s1*b^s2*c^s3), {a, 1, nn}], {b,
1, nn}], {c, 1, nn}]
N[Zeta[s1]*Zeta[s2]*Zeta[s3]/Zeta[s1 + s2 + s3 - 1], 20]

Out[518]= 1.000738441321476615

Out[519]= 1.0007384413214766152

Jakobian

@SohamSaha there isn't

And Xander would tell you the same thing

rohit1729

I took $x^2 + c$ as the family of curves. This is clearly not true for this family of curves. led me to think if it is really true? Can someone check if I'm interpreting the question wrong or the question is wrong?

Soham Saha

13:16

@Jakobian Have you ever seen such a fractal before?

Joe

I've never seen a triangle before.

psie

13:48

I have some fundamental doubts. Consider the following definitions in my book:

A family $(U_{\alpha})_{\alpha\in A}$ of sets is said to cover a set $S$ if $S$ is contained in the union of the $U_\alpha$'s. An open cover of a metric space $X$ is a family of open subsets of $X$ that covers $X$. $X$ is compact if every open cover has a finite subcover.

So is it true that the finite subcover, or the open cover for that matter, will always equal $X$? By definition of cover, $X$ is contained in the cover, and by definition of open cover, the sets are subsets of $X$, so their union is contained in $X$.

Feels odd somehow. I thought a finite subcover/open cover could not be contained in $X$.

Xander Henderson

@SohamSaha You have provided two pictures of functions. Without knowing how those pictures were created, who knows?

But, also, you have asked two kind of useless questions: first off, suppose that someone says "Yes! I've seen those before!"---then what? What does it matter if someone has seen it before?

Second, who cares if it is a "fractal"? A generic set is fractal. Being fractal is not special. Being not-fractal is special. And that assumes that we have an agreed upon definition of what "fractal" even means, which we don't.

So, like, what do you actually want to know?

Joe

@psie: There are two slightly different definitions of "open cover". Are you familiar with what it means for a subset $E$ of a topological space $X$ to carry the induced topology?

psie

@Joe kind of. The induced topology are the sets of the form $E\cap U$, where $U$ belongs to the topology of $X$.

Joe

Yes, that's right.

Xander Henderson

14:05

@psie To sets $A$ and $B$ are equal when $A\subseteq B$ and $B\subseteq A$. So it is entirely reasonable to have a collection of subsets of $A$ whose union is all of $A$. Or maybe I'm misunderstanding your question?

Joe

Okay, so the definition of "open cover" that you have for metric spaces $X$ is as follows (although it works just as well in the case that $X$ is a topological space): if $S\subseteq X$, then an open cover of $S$ is a family $(U_\alpha)_{\alpha\in A}$ of open sets in $X$ such that $S\subseteq\bigcup_\alpha E_{\alpha}$. If $S=X$, then as you have noted, it will be the case that $S=\bigcup_\alpha E_{\alpha}$.

So yes, in the case that $S=X$, it will be the case that $S$ is both a subset of $\bigcup E_{\alpha}$, and a superset.

Does that make sense? The only caveat here is that some authors define "open cover" slightly differently.

psie

@XanderHenderson Indeed, it is reasonable.

@Joe Makes sense. Thanks.

Joe

(Some authors define open covers as follows: if $Y$ is a topological space, then an open cover of $Y$ is a family of open sets of $Y$ whose union is equal to $Y$. Then, they would say that if $Y$ is a subset of a topological space $X$, then an open cover of $Y$ is just an open cover of $Y$ with the induced topology. Thus, for them, the union of an open cover of some set must always equal that set.)

But maybe it's not so useful to pay attention to this now. It's a very small difference, and the definitions are essentially equivalent.

psie

Ok, subtle differences that seem just...annoying :)

Joe

Yeah, in any case, the point is that if $(U_\alpha)_{\alpha\in A}$ is a collection of open sets of $X$ which cover $Y$, in the first sense, then $(U_{\alpha}\cap Y)_{\alpha\in A}$ covers $Y$, in the second sense.

psie

14:20

ok 👍

Soham Saha

14:33

@XanderHenderson Firstly: I have been working on this function for some time now. I tried to find references online for fractals which might appear to be similar _in appearance_ to the graph of my function. So, I am not looking for any proof of it being a fractal, just asking whether **anyone has seen anything that looks like my image**.
And why is it required to know "how those pictures were created" if I am just asking if anyone had seen anything similar before?

Secondly: If you will refer to my comment below the image, I _never asked_ if it is a fractal. I was just asking if anyone had …

Xander Henderson

Okay, so what if someone says "Yes, I've seen things like that before?" Then what?

Soham Saha

I will ask for a reference to where they've seen it previously, and study the resource if available.

Xander Henderson

Like, I have no idea what those pictures are supposed to represent. I may have seen something superficially similar which is completely unrelated and useless to you, but it is also possible that whatever thing you did is related to something which looks completely different.

So, without knowing what those are pictures of, the answer is "No, I've never seen anything like that."

Which is likely what everyone here is going to say.

You've given no context.

Soham Saha

Why is it not possible that nobody has ever seen anything visually similar to my image?

Xander Henderson

I didn't say it was impossible.

Soumik Mukherjee

14:39

I have seen the graph of y=x, which is similar to your image

Soham Saha

@XanderHenderson I think this was pretty definitive

Xander Henderson

What I said was that, lacking context, having seen something visually similar is unlikely to be useful.

@SoumikMukherjee Case in point.

@SohamSaha The word "likely" does have meaning in that sentence.

In any event, the answer I gave is "Lacking context, I have no idea what that is." I'm not going to engage anymore, since you seem unwilling to provide any context. I can't help you.

Soham Saha

Actually, I am not "unwilling", but I believe that this might be something new. So, I would wait till I have published something on it before providing details about the function.

This being a public chatroom, I am afraid of my idea being stolen.

Xander Henderson

That seems like a silly thing to be afraid of, but if you are so paranoid that you aren't going to tell us anything about that image, then there is likely nothing that we can do to help.

I wish you luck.

Soham Saha

Is that silly? I don't know. I am just starting to study math seriously, and am just an amateur wrt most people here. Isn't it possible for ideas to get stolen? I have no idea about the value of my work, but still...

@XanderHenderson Thanks.

Is it considered safe in general to discuss ideas that you may have discovered on a public place?

Xander Henderson

14:47

@SohamSaha I mean, almost no one cares about whatever idea you have. The idea that someone is going to steal your niche idea in some tiny, specific micro-specialization in some distant branch of mathematics is kind of laughable.

Does it happen? Sure. But very rarely. Most people would rather focus on their own work, rather than learn whatever they need to understand what someone else is doing and then scoop their work.

Moreover, if there were actually a priority dispute, one could always point back to the chat and say "I had the idea first, here are the chat logs." But, like, no one cares.

Soham Saha

Ok. Then here goes a description of my function.

$f_{a,b}(x)$ is defined on $\mathbb{R}^{+}$ as "take in x, express it in base $a$, and read the represented value in base $b$". To put it more formally: $$f_{a,b}(x)=\sum_{i=-\infty}^{\lfloor \log_ax \rfloor}b^i\lfloor a\cdot\text{fractionalPart}\left(\frac{x}{a^{i+1}}\right) \rfloor$$ where the $\text{fractionalPart}$ function calculates $x-\lfloor x\rfloor$

$f(x,2,-2)$ is similar in appearance to Cantor dust: desmos.com/calculator/ua48uatbpk

While, $f(x,2,3)$ is somewhat of an inverse to the Cantor function (devil's staircase): desmos.com/calculator/yhteyr69aq

Sorry, notation mistake. $f(x,a,b)=f_{a,b}(x)$

$b\in\mathbb{R}$ and $|b|>1$ to be sure about convergence

Not sure about what restrictions to place on $a$ though.

The graphs posted previously are of $f_{3.5,5.6}(x)$. Made in PARI/GP

Xander Henderson

15:09

Okay, well, sure. Any time you round down, you are going to see some kind of cyclic behaviour (whether truly so, or only probabilistically so). That's not super surprising.

Soham Saha

Not probabilistically. Try simplifying $f_{a,b}(ax)/b$

$f_{a,b}(a^kx)/b^k=f_{a,b}(x)$ for any integer $k$

Soham Saha

15:22

Does anyone think that this is interesting enough to write a paper about? Also, I am just a school student. Is it possible for me to publish a paper anywhere?

Xander Henderson

A series, absent context, is not really of interest to anyone.

What does this series represent? What is it for?

Why would anyone want to study it?

Soham Saha

@XanderHenderson Base conversion?

Relation to previously discovered fractals?

Xander Henderson

@SohamSaha "Fractal" is not a useful category.

Almost everything is "fractal". Labeling something "fractal" doesn't really given you an insight into its properties.

Cathartic Encephalopathy

does this proof look correct?

https://math.stackexchange.com/a/4962110/688539

Soham Saha

@XanderHenderson So you don't think that this is anything of interest?... Sigh

I was thinking that perhaps the different fractals that were produced were worthy of study by themselves.

Xander Henderson

15:31

@SohamSaha You have not convinced me that it is interesting. So, no, I do not think that what you are doing is interesting. But I am only one person. Maybe someone else will find it interesting. But it is up to you to sell it, and to convince others that there is something worth looking into.

Soham Saha

Right. Are you aware of any way how a school student can publish?

Xander Henderson

@SohamSaha Again, "fractal" is not a useful category. If you give me a completely arbitrary set in a metric space, it is almost completely certain that the set will be "fractal" for any reasonable notion of what "fractal" means. A "prevalent" set of subsets of $\mathbb{R}^n$ is "fractal".

@SohamSaha MAA has some journals which are aimed at high school and college students. But surely you have some kind of advisor who can help you out?

Soham Saha

Ummm. No. I am just a school student.

So no guidance about those matters.

Cathartic Encephalopathy

indian school system moment

Xander Henderson

I don't know what "school" means. But you should have teachers who know something, or can put you in contact with someone.

Cathartic Encephalopathy

15:34

I think maybe in US there is more options to get ahead if you want to get ahead

ap courses, local unis and all. Here it is like, there are some unis which are any good and those are not always in the state one lives in. Also scholarship and young scientist programs. Heck, admission to best unis depends on the stuff you do in your teens ( on an interest basis) while here it is entirely based on success in exams

Soham Saha

@XanderHenderson To put in Indian terms, I am a class 12th student (age~17). I have tried talking to teachers or other people, but not found any reference yet.

Xander Henderson

Ultimately, if you want to publish, you need to read. Find journals which publish papers in areas related to your own work. Once you have something finalized, submit your work to those journals. But read, read, read those journals to figure out (a) what kind of material their editorial staff find interesting and (b) to see how your own work fits into established knowledge.

Jakobian

Read, read, read until your eyes bleed

Soham Saha

Thanks a lot @XanderHenderson and @Jakobian for your advice.

Jakobian

I've only cracked a joke

Soham Saha

15:48

But it was a good quote. Worthy of pinning up in my room. ;)

Jakobian

No problem I guess. I'll take the whole credit

Soham Saha

Kinda poetic you know, "read"/"bleed"

Joe

16:10

Suppose $X$ is a compact Hausdorff space, and $C(X)$ is the ring of continuous maps $X\to\mathbb R$. For each $p\in X$, write $\mathfrak m_p$ for the maximal ideal of functions which vanish at $p$. Is the localisation of $C(X)$ at $\mathfrak m_p$ isomorphic to the ring of germs of functions at $p$?

I have read that "localisation" often corresponds to concentrating one's attention near a point, so I was just wondering if this was an example.

Thorgott

16:24

yes, this is correct and indeed a classical example

Jakobian

If $I_p$ is the ideal of functions equal to $0$ on some neighbourhood of $p$, then the ring of germs at $p$ is $C(X)/I_p$?

Thorgott

no, localization means inverting the complement

also, as for the hypotheses, normality of $X$ should suffice

Jakobian

@Thorgott yes, I was just asking if that's what the ring of germs at $p$ is

Thorgott

no, that quotient is $\mathbb{R}$

Jakobian

it's not

it will only be $\mathbb{R}$ if $\mathfrak{m}_p = I_p$

Thorgott

16:39

ah, misread, you're right

Jakobian

Now, why are the two isomorphic? Does it have to do with $I_p$ being the intersection of prime ideals contained in $\mathfrak{m}_p$

Thorgott

I have an explicit argument in mind. If $f\in C(X)$ and $g\in C(X)\setminus\mathfrak{m}_p$, then $g$ is invertible on a nbhd of $p$, so you can interpret the formal fraction $f/g$ also as a germ of a function at $p$. If, conversely, a germ is given by a function $f$ on a nbhd $U$ of $p$, use Urysohn to pick a function $\rho$ equal to $0$ outside $U$ and equal to $1$ on a nbhd $p\in V\subseteq U$ and interpret $\rho\cdot f/1$ as an element in the localization. These are inverse isomorphisms.

Joe

This was the argument I had in mind. Although, I think you need to check that if the formal fraction $f/g$ equals $f'/g'$, then $f(x)/g(x)=f'(x)/g'(x)$ for all $x$ on some open neighbourhood of $p$.

Thorgott

yeah, there's a number of detail checks I've left out for laziness' sake

Joe

No worries. I think this part is easy enough to check using the definition of localisation.

Thanks for the help.

@Thorgott: Do you know of any theorems like this which are actually useful in say, differential geometry? I was thinking along the lines of something which uses the theorems about localisation to prove things about the rings of germs of functions.

PM 2Ring

17:00

@Pizza That's an elliptical helix. It looks like this: sagecell.sagemath.org/…

Thorgott

@Joe I don't think so, the ring-theoretic properties of the ring of germs of functions are not particularly relevant in differential geometry. Instead, I think this observation is mostly useful as a perspective, specifically when you learn algebraic geometry and try to establish the analogies to differential geometry.

zeta space

17:27

user image

Soumik Mukherjee

@SohamSaha are you applying for isi/cmi?

PM 2Ring

@zetaspace Nice. What is it? How did you make it? It looks related to the Mexican Hat function, which is used as the logo on physics.stackexchange.com

sagecell.sagemath.org/…

Sahaj

Elon Musk renamed twitter to "X". One of his many children has the first name "X". He's fond of unknown variables in mathematics according to his former partner too. Weird obsession.

His AI company is called "X.AI"

Further, he announced the company on 12/07/23, since 12+7+23=42, the "answer of life"...

PM 2Ring

17:44

@zetaspace On second thoughts, that curve looks rather Gaussian.

Pizza

@SineoftheTime hi!

Sine of the Time

hi

Pizza

I didn't get an answer about yesterday's thing... anyway

48 mins ago, by PM 2Ring

@Pizza That's an elliptical helix. It looks like this: https://sagecell.sagemath.org/?z=eJwdjLEKAjEQBft8xSMWJrLiqSA2%2DQs7EQkxetHzNu6ton69nM0UwzCvKG6qU2%5FeCEg8OPXmg4DlYniIupXHDEPpR%5F1FgJoaJd6zSknH2rG6%5FZvwIXwPBKeE%2DXpWC2GkJwOMzbFy6XUIm6YhJO5Ygp1sz40lnMdZ2Mkz%5F2Pha05auA%2DWRVu%2DSKxtSZagbb7nYE9RbtabHy33N34=&lang=sage

Sine of the Time

yes, that's an elix, but it's not what is written in the problem

As I said yesterday, you don't have to focus on the single exercise

Pizza

oh yes indeed I let it go

copper.hat

17:50

people swoon over musk because he has a lot of money

Sine of the Time

are you studying a particular topic or are you doing exercises?

Xander Henderson

@copper.hat He has trouble finding pants.

Because, you know, his... wallet... is so thick.

Pizza

@SineoftheTime I'm doing a bit of all kinds of exercises

copper.hat

@XanderHenderson :-) used to be a running joke of my brother's, pointing to his wallet and saying "see that, haven't been able to bend that for weeks"

Xander Henderson

Heh.

Pizza

17:53

I had a question, but should the Hessian be calculated only in the case that we are in an open domain?

Soham Saha

@SoumikMukherjee Yes.

Soumik Mukherjee

Great

copper.hat

The Hessian can be defined on domains that are not open, but things become messy.

PM 2Ring

There was a question on the main site a couple of days ago about a helix that follows a parabola, but it got closed for lack of details. math.stackexchange.com/q/4961198/207316

Soham Saha

@SoumikMukherjee You are from where in WB?

copper.hat

17:55

the close brigade are out in force again.

Soumik Mukherjee

@SohamSaha Barasat, you?

Soham Saha

Burdwan

Sine of the Time

@Pizza do you mean, when $f$ is defined on a compact, you don't have to compute the hessian for the points on the "edge"?

copper.hat

it is time for proof of the single prime conjecture.

Pizza

@SineoftheTime yes

Soham Saha

17:56

@SoumikMukherjee Bye for now, way past bedtime. Good night (at least IST)

Soumik Mukherjee

Good night

Sine of the Time

you use the Hessian to determine the nature of a point, when you want to find the max/min of a function over a compact set you usually just evaluate the function in the points

you don't use the hessian

Pizza

Oh ok, thanks!

PM 2Ring

@copper.hat I understand the desire for clear questions, with context. I vote to close plenty of low quality questions. But it annoys me when a question does have enough clarity & context for those familiar with the field, and it gathers a few good answers, but it still gets closed anyway.

Binky McSquigglebottom

Hi

Sine of the Time

18:01

@BinkyMcSquigglebottom hi

Pizza

so like in this case I did it like this: Determine the absolute extrema of the following function:$$f(x, y) = e^y (x^2 - 4xy)$$in the square with vertices $(0,0), (0,1), (1,0) , (1,1)$

Binky McSquigglebottom

Fix latex!

👍🎉

Sine of the Time

$Q=[0,1]\times [0,1]$

copper.hat

@PM2Ring i get the need to filter lazy stuff out. but adding pointless steps to ward off the closers is silly.

Pizza

I calculated the partial derivatives and found the critical points, I evaluated the function at the vertices of the square and checked the function on the edges

then I compared the values and determined the absolute minimum and maximum

Sine of the Time

18:06

sounds good

Binky McSquigglebottom

but to solve differential equations of the type $x''+x'+x=f(t)$ , can we always use the Lagrange/variation of constants/Wronksian method?

And also similitary methods

I mean they are all the same method right?

PM 2Ring

Here's a helical catenary tube surface thingy. :)

sagecell.sagemath.org/…

Pizza

18:24

@BinkyMcSquigglebottom no, you have to find the particular solution with the similarity method, it is done in a different way

Binky McSquigglebottom

Yes

Thanks

Jakobian

19:24

@Thorgott If $X$ weren't normal, then I suppose we can't identify the ring of germs with $C(X)/I_p$, right?

Because by the ring of germs I think we mean here equivalence classes of functions defined on a neighbourhood of $p$

and they don't necessarily extend

Jakobian

20:00

I'm just not sure if the "ring of germs" here refers to those "global" based on functions $X\to Y$, or those "local" ones that are based on functions $U\to Y$ where $U$ is a neighbourhood of $p$

and I don't think the approaches are equivalent

at least when it comes to spaces with bad extension type properties

Joe

@Jakobian: Given two functions $f,g\in C(X)$, say that they are equivalent if there is an open neighbourhood of $p$ on which $f$ and $g$ coincide. The germ of $f$ at $p$ is the equivalence class of $f$.

Jakobian

@Joe I see, but the two definitions are equivalent in this case. I was more asking because I was curious which definition is more important

Joe

@Jakobian: Oh, I see. Well, I believe that in general, when authors define the equivalence relation on some space of continuous/differentiable/smooth functions, it is not assumed that they have the same domain: they both just need to be defined in an open neighbourhood of $p$. Not sure if that answers your question.

Jakobian

It doesn't

Joe

I don't really know what you are asking then.

Jakobian

20:15

I am asking which definition leads to a better theory

if we were to care for more general topological spaces

but let's just say we care for the "global" definition, then indeed this is the same as $C(X)/I_p$

can we use some general results about localizations to prove this? As to not require normality?

Joe

Hmmm that looks interesting. I'm not sure.

Jakobian

@Thorgott but perhaps we just need to modify this proof. If $X$ is Tychonoff, then I believe that taking a cozero set $p\in V\subseteq U$, say $V = \{g > 0\}$ then we can use a similar argument

So we probably don't need normality

say, $0\leq g\leq 1$ and $g(p) = 1$ and then take $2\max(g, 1/2)$ so that this is $1$ on a neighbourhood of $p$

1 hour later…

PM 2Ring

21:33

@SophieSwett This looks like a good equal-area projection: en.wikipedia.org/wiki/Lambert_azimuthal_equal-area_projection You probably need to centre the projection on the centre of your triangle.

Your 72° equilateral spherical triangle has sides of length ~63.4349488°. Centred at the north pole, its vertices have latitude ~52.6226319°

Thorgott

21:55

@Jakobian yeah, I suppose Tychonoff suffices

@Jakobian I don't have a counter-example locked and loaded, but yes

user 20458579510081670432

wb

Thorgott

germs are always equivalences classes of functions defined in a nbhd of $X$, this yields the general and good theory

psie

I am reading a proof of the fact that the metric space definition of continuity of a function is equivalent to the sequential definition. Consider the forward direction, which is proved by contraposition.

> Suppose $f:(X,d)\to(Y,\rho)$ is not sequentially continuous, i.e. there is a sequence $(x_n)\in X$ such that $x_n\to x\in X$ as $n\to\infty$, but $f(x_n)\not\to f(x)$ as $n\to\infty$. Passing to a subsequence, we can assume $\rho(f(x_n),f(x))\geq\epsilon$ for some $\epsilon$. ...

Maybe this is obvious, but I don't see it. Why do we need to pass to a subsequence?

Jakobian

@Thorgott I see, thanks

@psie That $f(x_n)$ doesn't converge to $f(x)$ only means that $\rho(f(x_n), f(x))$ doesn't converge to $0$

However, it can have $0$ as a limit point, so we need to take a subsequence

For example, the sequence that alternates between $1$ and $0$ has $0$ as a limit point

psie

@Jakobian so you're saying $\rho(f(x_n), f(x))$ could alternate between $0$ and $1$, depending on the behavior of $f(x_n)$ and ultimately $x_n$?

Jakobian

22:09

@psie that was just an example but it could

I'm watching a video about a guy trying to convince people that hairdryer marks are marks from aliens. The fact that people believe this kind of things fills me with dread

you make yourself a persona and people believe you just because of the persona

humanity is doomed

psie

You were speaking of earworms the other day. I currently suffer from one.

I guess there is nothing you can do about it.

Jakobian

I read that they pass if you are doing something which occupies you

mentally that is

psie

yeah, I see how that could work

Jakobian

how I read it is that they usually arise in neurotic people because they have this space for thought and that's why they occupy this space, often with negative thoughts

and for similar reasons earworms arise

psie

the thing is, I am partly not actively listening to the song. Others are, which I have a hard time doing anything about.

PM 2Ring

22:22

This guy, Milo Rossi, aka Miniminuteman, is pretty good at debunking conspiracy theories. m.youtube.com/@miniminuteman773/featured Check out some of his shorts.

Joe

@psie: We don't "need" to pass to a subsequence, in the sense that you could prove the result without doing this, but the point is that if $f(x_n)$ does not approach $f(x)$, then there is an $\varepsilon>0$ such that for all $N\in\mathbb N$, there is an $n\ge N$ such that $|f(x_n)-f(x)|>\varepsilon$. Thus, we may choose an $n_1$ such that $|f(x_{n_1})-f(x)|<\varepsilon$, an $n_2>n_1$ such that $|f(x_{n_2})-f(x)|<\varepsilon$, and so forth.

PM 2Ring

@psie Maybe try listening to a radically different genre of music to the earworm.

Jakobian

@PM2Ring I still like Trey the Explainer more

psie

@Joe ok, but if $|f(x_n)-f(x)|>\varepsilon$ for $n\geq N$, how come we can choose an $n_1$ such that $|f(x_{n_1})-f(x)|<\varepsilon$ and so forth, or did you mean to write $>$ instead of $<$?

Jakobian

No absolute values, this is not $\mathbb{R}$

psie

22:33

right, some metric then, but I realize I misread what Joe typed, I have to read it again

Jakobian

@Joe The first thing you said is false. The second thing you said is also false. What do you mean?

Oh no okay, you said for all $N$ there is $n\geq N$. My bad, the first thing is true

psie

@Jakobian I misread the same thing :)

Jakobian

Okay the second thing is that you are describing process in which we could take such subsequence

Joe

Sorry, I meant to write $>\varepsilon$ in my second sentence, and of course those should be of metric distances rather than absolute values. I'm a little tired...

Jakobian

If someone isn't aware that you can take such subsequence, then an explicit argument helps, sure

However if someone already seen such argument then it's better to blackbox it and just assume such sequence can be taken

user10478

22:38

Does anyone see why none of my four attempts at writing the Lagrangian form of this program yield the same answer as the non-Lagrangian form? i.imgur.com/8tYQrcJ.png

PM 2Ring

@Jakobian Fair enough. Milo is very snarky, which can get annoying after a while.

user 20458579510081670432

22:54

Yeah, "snark" isn't a very attractive feature for most people.

Especially in written form.

Jakobian

23:16

@PM2Ring It's also that critiques of someone often aren't very educational

There is more appearance of content than there is content in those type of videos

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