12:26 AM
@robjohn Yeah, I mean exactly one of each.
@Thorgott Nope, I think a typical egg has two local maxima of curvature (one of which is the global minimum) and two local minima of curvature.
The global maximum is the "little end" and the other local maximum is the "big end". The local minima are near the "little end".
Actually, the "big end" has approximately constant curvature over much of its length, so it's not clear what the local extrema there are.

@TerranSwett: Look for the four vertex theorem. It's proved in my differential geometry text (free pdf in my profile).
A closed convex curve must have at least four vertices.
A vertex is a critical point of curvature.
(We're talking here obviously only about plane curves.)

12:45 AM
@TedShifrin Yep, that's what I was looking for. Thank you!
I had an idea for a proof by contradiction, but I'm not sure it would actually work.
Suppose you do have an oval with only one minimum and only one maximum. Then, by using the intermediate value theorem, we can find that there are two points on the oval whose tangent lines are parallel and which have the same curvature. Consider the two portions of the curve delimited by those two points.
In one portion, the curvature is everywhere greater than the curvature at those points; in the other portion, it's everywhere less than the curvature at those points.
Yet it seems like the portion with greater curvature must have endpoints which are closer together than the portion with less curvature, which is a contradiction.
I'm not sure if that's actually true, though.

@TerranSwett the best looking egg I've seen has been made of four circular arcs, two with greater, unequal, curvatures and two with lesser, but equal curvature on the sides.

I wonder if you can use the vertices of an arbitrary symmetrical trapezoid as the endpoints of those arcs.
I'm thinking... yes, you can, in infinitely many ways for each trapezoid.

1:05 AM
@Huy I don't think that is known, but alternative expressions using elliptic integrals are known.
@robjohn Recently found out about Chandrupatla's algorithm for root-finding, which may interest you. Essentially it chooses root-finding methods based on how close the previous midpoint was to the previous secant line between endpoints.
x.x I feel like I was born a few decades late! Really nice simple 1D root-finding methods are often amazingly recent discoveries...

2:04 AM
Let $G \leqslant K^{\times}$ be a multiplicative subgroup of the group of units of a field $K$. Then $\sum_{g \in G} g = 0$ necessarily.

**Proof.** If $\sum_{g \in G} g = u \neq 0$ then $u \in K^{\times}$ so that $0 = u^{-1}\cdot 0 = u^{-1} \sum_{g \in G} g = \sum_{g \in G} u^{-1}g = 1$.

3:02 AM
@robjohn i'm talking about $C^2$ curves.

sniff sniff

3:29 AM
@robjohn Sir, should I spend some more time in learning manipulative techniques ? My current level of manipulation is of Higher Algebra (Hall and Knight) level (I.e. I don’t get much problem in solving). Should I need to be more prepared?
(please don’t say “I don’t know that depends on you” like few other MSE users, you know what I’m asking and you know that your judgment matters)

3:43 AM
@TerranSwett Here is the egg shape.

2 hours later…
5:43 AM
Stephen Hawking married to Carl Fedrich Gauss

1 hour later…
6:55 AM
@TedShifrin with very minor modifications near four points, this could be made $C^\infty$.
but it does not have monotonic curvature. That is impossible.
I was just mentioning the nicest egg shape I had seen.

7:57 AM
@AbstractAlgebraLearner where do you have this problem from

1 hour later…
9:25 AM
Let's have a look at a *manipulation*:

*To find the minimum value of $\frac{(a+x)(b+x)}{(c+x)}$ we do something like this*
$$\text{Let} (c+x) = y \\ a+x \rightarrow a-c+y ~~~~~~~b+x \rightarrow b-c+y \\ \frac{(a+c)(b+c)}{(c+x)} = \frac{(a-c+y)(b-c+y)}{y} \\ \frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c)}{y} +(b-c)+( a-c) + y$$
Now:
$$\frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c) }{y} + (b-c)+(a-c) +y \color{blue}{+2\sqrt{(a-c)(b-c)} - 2\sqrt{(a-c)(b-c)}} \\ \frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c) }{y} + y \color{blue}{-2\sqrt{(a-c)(b-c)}} +(b-c)+(a-c) \color{blue}{+2\sqrt{(a-c)(b-c)}} \\ \frac{(a+c)(b+c)}{(c+x)}= \left( \frac{ \sqrt{(a-c)(b-c) } } {\sqrt{y} } - \sqrt{y} \right)^2 + (b-c)+(a-c)+ \color{blue}{+2\sqrt{(a-c)(b-c)}}$$
Hence, the minimum value of $\frac{(a+c)(b+c)}{(c+x)}$ occurs when $\left( \frac{\sqrt{ (a-c)(b-c) }}{\sqrt{y}} - \sqrt{y} \right)^2$ is least and therefore when it is zero.
Hence, the minimum value of $\frac{(a+c)(b+c)}{(c+x)}$ is
$$\color{red}{ (a-c) +(b-c) + 2\sqrt{(a-c)(b-c)} }$$
${\large \text{By mistake, at every LHS I have written c instead of x in numerators} }$

2 hours later…
11:46 AM
@AbstractAlgebraLearner I don't folow, why is $u^{-1}0=u^{-1}\sum_{g\in G}g$?
you probably want your subgroup to be finite, otherwise the sum doesn't make sense
and you want it to be non-trivial, otherwise the claim is simply wrong
the claim follows from the fact that $G$ is necessarily cyclic, not sure if there's an easier approach

12:04 PM
I'd really like someone to answer this question: math.stackexchange.com/questions/3788377/…

@Manan y?

interesting question

@SpecterProphet Although I haven't studied analysis yet, this question has dumbfounded me. Just like the asker, I'm of the view that such a function does not exist, but all sorts of weird functions exist in higher math so I'm curious to know what the answer actually is.

12:26 PM
I'd be very surprised if such a function doesn't exist

I'd be very surprised either way

enumerate the rationals as $r_1, r_2, \cdots$, look at $\sum_{k = 1}^n (x - r_1) \cdots (x - r_n)/100^{100^n}$. this should converge as $n \to \infty$ to some function like that
i guess the choice of the denominator really depends on your enumeration
u want to divide by something which grows much faster than $r_1 \cdots r_n$

12:41 PM
the derivative of that is gonna map rationals to rationals too or am I missing something

oh i misread the q
yeah thats not gonna work

I don't know whether something analytic will work

what's analytic, not rational and fixes $\mathbb Q$?

Balarka's example does just that if you do it properly

my guy works right

12:45 PM
ok yup

20

Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function?

an answer was posted

The question has been answered

1 vote in less than 10 seconds is a bit much lol

yeah its some fiddly cubic splines thing
i would be wary
GEdgar is usually solid though

12:50 PM
I agree and it checks out

@Thorgott Ah nice Pietro Majer upgrades this idea to construct an analytic diffeomorphism of $\Bbb R$ which takes any countable dense set to any other
That is extremely neat, I'll have to store that argument away in my brain

yeah, this is a problem that was posed by Erdös, I think

ah didnt know that

nice

1:17 PM
@MikeMiller I am getting terribly confused about something

Oof

Take a genus $2$ surface $S$, pick a meridian $\alpha$ around one of the holes. You can pick a hyperbolic metric on $S$ such that $\alpha$ is as long as you want, yeah? (In fact a lot more, this $\alpha$ is part of Fenchel-Nielsen coordinates for the Teichmuller space of $S$)

Are FN coordinates surjective onto $(0,\infty)^{3g-3}$?

Yeah I thought that was just a homeo

I wasn't sure if it was homeo to the whole space or had convex image or omething. Anyway

1:22 PM
So fix a basepoint for $\alpha$, travel sometime for more than half of the circle so that you have an arc of $\alpha$ which is a local geodesic but not a minimal geodesic.

Sure

Like you do to get non minimal local geodesics given any closed geodesic really
Ok so call this $\gamma$, and it's locality constant can be as large as I want it to be since I have control on the metric of $S$. It's like half of the length of $\alpha$ or something

I don't know what a locality constant is, does that matter?

Nah not really, basically just for how much your local geodesic is an actual minimal geodesic
Forget it
Here's my point
$\gamma$ is very far from the minimal geodesic joining it's endpoints, which is the other arc of $\alpha$, right?
Very far as in their images are very far away in Hausdorff metric

Yeah

1:31 PM
so why is this theorem (1.13.(1)) consistent with this example
seems to say in gromov hyperbolic spaces local geodesics with sufficient locality constant is very close to the minimal geod joining their endpoints
i must be insane or something

Is a hyperbolic surface (-1)-hyperbolic?

ah no its some badass constant, log(2)-hyperbolic

1-hyperbolic maybe
Ok but it doesn't depend on the lengths or anything that's good

yeah, its just the constant of hyperbolicity of H^2 the universal cover
so thats why im confused
@MikeMiller also true, log 2 is just the optimal guy. anyway nobody cares

I thought it would maybe be because when you jiggle the metric to get your example to have appropriate length for $\alpha$, it would jiggle the constants in this theorem so as that your "$2\delta$-neighborhood " is huge or something
But $k>8\delta$ so that seems false

1:35 PM
yeah
it must be something to do with the constants tho
you must be right

Nah let's think of it in a silly case
Take $p$ and $q$ to be antipodes on your geodesic
Then both sides are geodesic segments
And this is saying they're both in a neighborhood of the other

Seems nonsense to me.

Lmao yes
I dunno whats happening

Is the problem that when you make $\alpha$ abnormally long the geodesic from $c(a)$ to $c(b)$ is actually totally distinct?
Winds around the surface instead
Since maybe that's a shorter path

1:37 PM
Um um
That would be very surprising
But that can happen actually cant it

Yeah I don't see any reason why not. I'm thinking on the universal cover; $\pi_1(S)$ is a discrete set hardly contained on a geodesic
You're looking at a line in $\Bbb H$ along which $\Bbb Z$-many elements of $\pi_1(S)$ appear, and seeing that one of those is closer than another
But maybe $\pi_1(S)$ is way closer in a different direction

ur saying you cant make this tubish part in the hyperbolic surface around the meridian extremely fat while keeping the median a geodesic?
@MikeMiller yeah ok im starting to buy it

I know very little hyperbolic geometry, I'm just saying guesses about what's possible

do you think i know hyperbolic geometry fam

at least fuzzy hyperbolic geometry
So what this would mean is that your "actual" geodesic segment loops a lot and crosses over $\alpha$ many times
Enough that a $2\delta$ neighborhood must contain all of $\alpha$ I think.

1:41 PM
Yeah

I wish I had a picture lol
I wonder what Fenchel-Nielsen looks like in $\text{Hom}(\pi_1 S, PSL_2 \Bbb R)$

No I see what you are trying to draw. It's just bizarre to me that the metric has to deform to that
to make it fatter along $\alpha$
@MikeMiller Farb-Margalit would have it maybe, we can check

Now that I think of it it's going to be the same
Just some function on $PSL_2 \Bbb R$ that you evaluate on certain loops I'm sure
Stupid question, is a representation $\pi_1 S \to PSL_2 \Bbb R$ determined by its image in $\Bbb H$?

As in compose with the projection PSL2R -> H?
hm
But I mean, any translations along the same geodesic have the same projection under PSL2R -> H (which is the unit tangent bundle projection), yeah?

1:57 PM
How do we take a hyperbolic structure $\pi_1 S \to PSL_2 \Bbb R$ and translate it along a geodesic?

ok my last comment is irrelevant plus i understand now what you want to see
let me think
i have never thought about pi1S -> PSL2R -> H
ok so given a faithful rep pi1S -> PSL2R i get a hyp structure on S by quotienting H^2 by the image of this rep

I guess that's not quite right, since it doesn't make sense to quotient by the image of this rep
That's probably my issue
We only know the action on a single point (or the orbit through a single point), but that probably doesn't tell us the orbit through other points

why not? if the rep is faithful, the image is an embedded pi1S in PSL2R i.e. pi1S acting by isometries on H^2
trying to understand

You can't lift that back up canonically to the actual subgroup of $PSL_2 \Bbb R$; every element can lie at a different point in the circle-fibers (so long as they multiply correctly)
So you don't know just from the image what the action is
Say you're acting on the point $p$ with stabilizer $\Gamma_p$. If you have two representations $\pi_1 S \to PSL_2 \Bbb R$ which differ by a rep to $\Gamma_p$ then they should have the same image
But $\Gamma_p$ is not the stabilizer of every point, just that point. If you act somewhere else then $\Gamma_p$ will actually contribute to the action

OK I see

2:13 PM
Basically your $\Gamma_p$ homomorphism rotates (in some sense) the orbit elsewhere
A twisted rotation I guess depending on what point in $\pi_1 S$ you're looking at the image of

@MikeMiller Ok yes I understand what you're saying now

Twisted rotation maybe not my best choice of word
I mean twisted by the representation to $\Gamma_p$ lol instead of uniformly rotating the whole orbit

2:29 PM
If I have a ring endomorphism $f : R \to R$, what is meant by an $f$-linear endomorphism of an $R$-module? o.O

@Mike Do you happen to be a coalgebras expert (my question is actually very basic) or are you the sane kind of topologist?

He's a coalgebra expert

I can't even tell whether you're ironic

I know what a coalgebra is

err
I'll screenshot it
but idk if the context will help

2:38 PM
I have a coalgebra $C$ (with comultiplication $\Delta$ and counit $\varepsilon$) and a vector subspace $V$. I define $C(V)$ as $\{x\in C\mid ((\Delta\otimes I)\circ\Delta)(x)\in C\otimes V\otimes C\}$, and I want to show that $C(V)$ is the biggest subcoalgebra of $C$ contained in $V$

It's just going to be a blank page that only says "$\varphi$ is an $f$-linear endomorphism of the $R$-module $M$" and nothing else

loool
The bottom: a $\varphi_L$-linear endomorphism $\varphi_M : M \to M$ which commutes with the $\Gamma_L$-action

I think it should boil down to the fact that if $x\in C\setminus V$, then $\Delta(x)\in (C\setminus V)\otimes (C\setminus V)$, does that seem sensible?

God this is too many symbols

(I'm trying to show that $C(V)\subseteq V$)
@MikeMiller Agreed

2:41 PM
What does this dualize to
Ah ok

Uhm I'm not sure

Dually this would be the set of $xvy$ so that $x \in R, v \in V, y \in R$
The usual way you set up a two-sided ideal
(Sums of those, of course)

Wait what's $R$ now?

Some ring
I don't want to call an algebra/ring C
Anyway, I was just seeing if the dual picture helped me
@EdwardEvans Read the rest of the book to find out
I suspect it's not a module homomorphism
I suspect $f: M \to M$ is $\varphi: R \to R$-linear if $f(rm) = \varphi(r) f(m)$
"Twisted linear"

Look at this nice ring: $\mathcal{O}_F^{\sharp} := \widetilde{\Theta}_{\mathbb{C}_p}(W(\mathcal{O}_F)_L \otimes_{W(\mathcal{O}_{\hat{L}_\infty^\flat})_L} \mathcal{O_{\hat{L}_\infty^\flat}})$

2:49 PM
No

hahahaha

Do something better with your life

@Mike thanks, I also thought that might be the case but I wanted to be sure
slash
I'm still not sure, but I'll just keep reading

@AlessandroCodenotti Hm ok. Anyway the idea for this should be to use that $L\circ(\varepsilon\otimes I)\circ\Delta=I$, where $L:K\otimes C\to C$ sends $k\otimes c$ to $kc$

Yeah I tried to do this in sumless notation and it became too much a mess for me sorry

2:53 PM
No problem, thanks for looking into it

Yeah, I just can't do this abstract stuff, I need elements
And the sumless notation got very ugly
@AlessandroCodenotti No dude this is straightforward I think

Neither can I
@MikeMiller Unexpected plot twist
I don't think it's supposed to be hard, it's left as an exercise in some notes, I'm just not really seeing it

$$(\epsilon \otimes I \otimes \epsilon) \circ [(\Delta \otimes I) \circ \Delta] = (\epsilon \otimes I) \circ \Delta = I,$$ right?
So if $[(\Delta \otimes I) \circ \Delta](x) \in C \otimes V \otimes C$, applying the counit on the left and right you see that in fact $x \in V$
I might have some symbols wrong up there but the point is that $\epsilon$ cancels out the comultiplication on each side
Therefore $C(V) \subset V$
This is the dual of the argument that $I \subset RIR$ --- in that case you apply the unit on both sides to see that $1i1 = i \in RIR$

3:11 PM
I'm a bit confused now though because this seems to show that if $x\in V$,then $x\in C(V)$
Because $(\epsilon \otimes I \otimes \epsilon) \circ [(\Delta \otimes I) \circ \Delta](x)=x\in V$, but for $\epsilon\otimes I\otimes\epsilon$ of an element of $C\otimes C\otimes C$ to be in $V$ the central part should be in $V$?
(there are a bunch of multiplication maps $K\otimes C\to C$ and $C\otimes K\to C$ left implicit here and there)

@AlessandroCodenotti I don't think so
Just because $(\epsilon \otimes I \otimes \epsilon)(\sum x \otimes y \otimes z) \in V$ doesn't mean that $y \in V$
Maybe things cancel out after you apply the counit

Ah ok makes sense
I forgot about the sum

For instance, suppose $x$ has $\epsilon(x) = 0$. Choose $y \not\in V$ and $z$ so that $\epsilon(z) = 1$. Then $x \otimes y \otimes z$ is not in $C \otimes V \otimes C$, but $$(\epsilon \otimes I \otimes \epsilon)(x \otimes y \otimes z) = \epsilon(x) y \epsilon(z) = 0 \in V$$
No need for the sum to appear here. Just the fact that $\text{ker}(\epsilon)$ is nonzero, which it had better be if your coalgebra is interesting!

Fair
But so is it actually true that if $x\in C\setminus V$, then $\Delta(x)\in(C\setminus V)\otimes(C\setminus V)$?

3:54 PM
I don't know I didn't want to think about that. I would bet no
Actually what does that even mean
I know how to take tensor products of subspaces, not subsets
If you mean "sums of tensors..." then you inevitably have just rewritten $C \otimes C$

Hi! I'm currently a computer/electrical engineer, highest degree conferred MS. I'm considering a change to mathematics, with a goal of teaching at the college/community college level. I'm 40 years old, is this just a pipe dream at my age?

4:09 PM
Is there a Latex template for a Cayley Table?

Hm good point. I'm not sure what I wanted to say either now lol

Ahh, finally found it

4:59 PM
I’m looking for a classic introductory Real Analysis book. Something of early 20th century or late 19th century.
Just like for Algebra the classic one in Hall and Knight, for Geometry we have SL Loney, for Calculus we have Joseph Edwards’. So, in a similar fashion what would you suggest for Real Analysis?

why would you deliberately go out of your way to read dated texts

@AlessandroCodenotti You were probably imagining everything was simple tensors and that you could express the result as a simple tensor between things not in V
I think maybe try doing this with a subspace of a group algebra to see what happens in practice
Something in k[S_3] maybe

I’m old and I know none of these books.

5:21 PM
Hiya @TedShifrin , I asked a question earlier about some terminology. How would you interpret an "$f$-linear endomorphism $\varphi : M \to M$" where $f : R \to R$ is a ring hom and $M$ an $R$-module?
Just $f$-semilinear?

I’ve never heard of this term. But it seems that scalars pull out, using $f$ of the scalar in the image. I have no
idea what you mean by semiliear.

f-semilinear means $\varphi(rm) = f(r)\varphi(m)$
so that's what you're saying I guess
I just encountered it in one of the first definitions in a book I'm starting on and it threw me straight away hahaha

Yup, that’s what I’m saying.

okay thanks :)

Hmm, I was not aware that I had coauthored a paper titled "On good Formula Not Shown-filtrations for rational G-modules". But if it is in Google Scholar, it must be the correct title :)

5:36 PM
lool

Scholar also seems fairly confused with a couple of my papers, having several of them split up as if they were not the same paper
Really not sure what happened. The correct title is further up on the list, with an equivalent reference, so Scholar ought to be able to tell that they are the same paper, except $(p,r)$ has been replaced by Formula Not Shown.

Oy.

1 hour later…
6:54 PM
Take any field $K$
take any subgroup of $K^{\times}$
$G$
then is $\sum_{g \in G} g = 0$ always?

You asked this yesterday and I saw a proof, didn't I?

if the subgroup is infinite, this doesn't make sense
if it's finite, then yes, except for when it's the trivial subgroup

@AbstractAlgebraLearner why aren't you listening to the people on the question you asked on main? lol

7:10 PM
@Thorgott what about when $K = \Bbb{C}$
Then you have convergence etc
I think if $G \leqslant \Bbb{C}^{\times}$ and $-1 \in G$ then since $-a \neq a$ unless $a = 0$ then you can somehow form a set and its negative?

Can someone decode the question:
There is 1-to-1 correspondence between maximal ideals in $R$ and homomorphisms into $\Bbb Z_2$ which are not identically zero. Where R is an ideal in Boolean ring $(R,\cdot, +)$.
Is it asking for a bijective function on $R$ to homomorphism $\varphi : X \to \Bbb Z_2$ ?

you obviously won't have convergence in the vast majority of cases and even if you do, it won't necessarily converge to $0$

Thanks @Thorgott

this has specifically something to do with the behavior of finite subgroups of the multiplicative group of a field, so you should understand that

@AbstractAlgebraLearner Take $K = \Bbb R$. Each pair of real numbers $x$ and $1 - x$ cancels to $1$, so your sum won't converge. Or maybe actually take the pair $x$ and $2 - x$
or
$\pi - x$ or smth idk

7:17 PM
Why would you pair that way? with $1 - x$?

Why not?

@flowian it is asking for a bijective function between the set of maximal ideals in $R$ and the set of non-zero homomorphisms from $R$ into $\mathbb{Z}_2$

You got all the numbers you want

I see that now

@Thorgott cheers Thorgott

7:22 PM
@Thorgott pairing $A = \{ x : x \in \Bbb{R}\}$ and $B = \{1 - x : x \in \Bbb{R}\}$ would yield $\sum A + \sum B = 2 \sum_{r \in \Bbb{R}} r$ since $A$ covers $\Bbb{R}$ exactly once but so does $B$, and so by symmetry we have $2\cdot 0 = 0 = \sum_{r \in \Bbb{R}} r$
So as you can see your counter argument has failed

lol

but have you considered that $\frac{A+B}{0}=1$?

clearly it's $-\frac{1}{12}$

I don't get why I would have to do that

@Edward oh, I see what you're saying, so if $G$ is an infinite multiplicative subgroup of $K^{\times}$, then $\int^{p\text{ prime}}g=-\frac{1}{12}$

7:29 PM
bruuutal
hahaha

Haha, keep laughing :|
^_^

You have various (probably fairly advanced) mathematicians telling you your question doesn't make sense for $G$ infinite
have some humility
(not claiming to be such a mathematician myself btw)

7:43 PM
If: By removing 1 point from the set of points that make up a sphere does not change the cardinality
Then: Why is the punctured sphere homeomorphic to the plane, but not also the sphere? Can it be bijective at least?
I.e. is it the 'continuous' restriction that falls apart? Or the bijection altogether?

as your argument shows, they are all equicardinal
being homeomorphic is a lot more restrictive than having the same cardinality
and of course depends on more than just the sets themselves, namely the topology on them

The summation over the points of a sphere also equals zero :D

@Threnody $[0,1]$ is in bijection to $\Bbb R$, as well as $\Bbb R^n$ for any $n$, or even if you like to the space of sequences $(a_n)$ with $\sum |a_n| < \infty$, etc ...
It's not very hard for things to be in bijection
Convince yourself that most functions that exist are not continuous

@MikeMiller I can see that, yes
Hmm, I see... the 'continuity' is what seems to make the distinction
Can we say there are more functions than continuous ones?

Way more

7:57 PM
Sorry... non-continuous, I guess

if you formalize it correctly, yes

Interesting :)

8:17 PM
Way, way, way, way ... more.
Most functions are in fact nowhere continuous!

On the other hand, there is the following curious fact: If $f\colon\mathbb{R}\rightarrow\mathbb{R}$ is any function, there exists a dense subset $D\subseteq\mathbb{R}$ such that $f\vert_D$ is continuous.

Which is subtly but importantly different from saying $f$ is continuous on $D$. :)

very much so

Huh... different?
$f\vert_D$ is $f$ restricted to $D$... saying it is continuous means it is equal to the limit at every point in $D$...
is 'continuous on D' defined differently?

8:36 PM
Yeah, $f\vert_D$ is a function with domain $D$, $f$ is a function with domain $\mathbb{R}$. To say that $f$ is continuous at a point $x\in D$ means that for all sequences $(x_n)_n$ of real numbers with $x_n\rightarrow x$, we have $f(x_n)\rightarrow f(x)$. To say that $f\vert_D$ is continuous at $x\in D$ means that for all sequences $(x_n)_n$ of numbers in $D$ with $x_n\rightarrow x$, we have $f(x_n)\rightarrow f(x)$.
So in the second case, we are considering possibly way less sequences.
An instructive example is the indicator function of the rational numbers. $\chi_{\mathbb{Q}}\colon\mathbb{R}\rightarrow\mathbb{R},\,x\mapsto\begin{cases}1,&x\text{ is rational},\\0,&x\text{ is irrational}\end{cases}$. This function is nowhere continuous. But $\chi_{\mathbb{Q}}\vert_{\mathbb{Q}}\colon\mathbb{Q}\rightarrow\mathbb{R}$ is a constant function (with constant value $1$), so is very much everywhere continuous.

Anyone know what a clustering algorithm is

@TobiAkinyemi Maybe an algorithm like KNN? that classifies data into the nearest clusters? that's all I know unfortunately

I'm trying to do a continous version of clustering
i.e. find the peaks of a 2d plane
however, its not actually continuous, I have a 3d plot (in a computer)
idk if this is a solved problem

@Thorgott I see.. so saying it is continuous on D also specifies from where we can pull our sequences?

To say it is continuous on $D$ just means that it is continuous at every point in $D$. The point is that the domain of the function determines which sequences we have to take into account.

8:43 PM
@TobiAkinyemi hmm, no idea, sorry, although i'm just an undergrad :P
@Thorgott aha... I see

8:58 PM
there's a model of the hyperbolic plane in the first quadrant? I feel like nobody uses this and just uses the half plane model or poincare model or Klein model

9:10 PM
what is "geodesic flow" in fairly simple terminology? Is it basically a flow along geodesics?

9:50 PM
0

Let $K$ be a field of characteristic $\gt 2$ that is not necessarily finite. Start to define the notion of arbitrary summation of certain subsets $X \subset K$ via: If $-X \subset X$, then $\sum\limits_{x \in X} x := 0$. Call any such set $X$ arbitrarily summable to zero. In a field of charact...

Application of previous discussion here
Do I win $1 million (?) If I were in charge, I'd give it to you Thanks man :D It's shocking as to why this approach hasn't been fleshed out more by previous mathematicians It's the next to obvious approach you certainly outdid yourself with that one @Thorgott Ikr My greatest hope for mathematics is that all complicated proofs eventually become only 20 pages or less :D "Proofs as programs" (type theory) leaves a lot of room for optimization @Jack 10:20 PM can't wait to classify the finite simple groups in less than 20 pages math.stackexchange.com/questions/1221256/… how do we know in this question that$1 - xy \in m\$ is not a unit in the last sentence in the answer?

proper ideals can't contain units

oh its because it might contain 1 = uu^{-1}?

which implies?

the ideal (1 - rx) = R, so not proper

10:25 PM
yup

since if it contains a unit u, it might absorb u^{-1}I

10:45 PM
@geocalc33 No. it's a flow up on the unit tangent bundle.