00:00 - 23:0023:00 - 00:00

12:05 AM
It's odd when an OP says "I didn't realize that the upper semicircle would indeed produce a greater surface area than the lower one until I read your comment" but no voting ensues.

What an odd sentence, regardless.
Context lacking.

Yeah, I was testy with an OP who complained that I had answered the question he asked, but not the one he meant to ask.

@TedShifrin surfaces of rotation.
@TedShifrin weren't you reading their mind?

So I surmised, but not the $x$-axis, clearly.
I said I might answer the new one if he posted it. But this OP sorta sends danger signals.

12:09 AM
use a 10 foot pole

If I answer, I’ll use Maurer-Cartan forms, which of course will be beyond OP’s scope.

the 10 foot pole also keeps you out of their grasp

As opposed to 10 meter stick

10 ft poles have no residues, oddly.

12:19 AM
@copper @robjohn Leslie responded. He's got a bad cold and is swamped with work, so is conserving time/energy. He doesn't miss us.

@TedShifrin Thanks for the update, glad it is routine.

@copper.hat don't they bump their heads on the backboards?

:-) my son is the tallest in both sides of our family, his only issues so far have been finding a comfortable fit when driving

The continuing weather changes are giving me a sinus headache.
@copper.hat And probably will gruesome more.

@copper.hat I'm not tall (only 5'10") but my leg-torso ratio is such that when I sit I am as tall as my dad who was 6'1". That does interfere with getting into cars sometimes.

12:23 AM
for me, i get congested a few days after rain.

My wife and I had terrible headaches over the weekend from the pressure changes and wind

my dad was the opposite, he was tall (by then standards) but when he sat down he was short

Oh yeah, pressure changes with Santa Annas … maybe that too.

12:24 AM

@mohan10216 yes, we've seen that and it has been reported.

Oh ok.

this is not the forum for reporting it

then where?

For that, you'll have to call our customer service hotline

12:24 AM
@mohan10216 meta.SE

That works too

Robjohn, did you see Pedro’s query to you/Davide?

Gonna take Galois theory and functional analysis this semester

@TedShifrin on your post? No, but I will look

Yeah, he didn’t ping anyone.

12:26 AM
The textbook for Galois theory (Galois Theory by Ian Stewart) is a book I have owned since middle school

@TedShifrin I have responded.
@TedShifrin I saw it when I was on my phone this morning and so it was not highlighted when I got to my computer. Thanks for reminding me.

Tanx.

Gotta take an anxious pup to the park.
BBL

@TedShifrin Well, at the time…
In any case, not super worried about the course
Teacher seems super nice as well

12:32 AM
I meant for the course, not younger you!

Functional analysis will be more new, but I'm excited

Where’s geometry/topology?

Also taking Japanese level 4 (continuing that) and a course titled Music In Japan
and possibly auditing Intro to Sociolinguistics

Nice.

@TedShifrin Unfortunately, Intro to Lie Groups conflicts with Music in Japan

12:33 AM
Eastern music very different from western.

1 hour later…
2:02 AM
Bro, my mind had a literal eruption just now. Clicked on this random bprp video in my recs and said, "area under the tangent line", and it clicked in my mind. We can use $\frac{d^2}{d^2 x} f(x)$ to compute the integral of $f(x)$.
*to compute $f(x)$, my bad.
Much easier to work with. We just need to know how the area changes over time and then the conjunction of the area under two tangents over time, and then we can take the limit as those tangents approach another tangent for every point on $f'(x)$ to then get an integral, $f(x)$.
It sort of just falls into place, doesn't it?
Stupidly simple, too.

2:29 AM
First, your second derivative notation is wrong. Second, this has nothing to do with the area of that precise triangle.
So, no, not to me.

Yeah, I realized that after the fact. Nevertheless, yes, I'm aware any given $f(x)$ has nothing to do with that particular triangle. I'm attempting to generalize to any triangle defined by $f'(x)$ and $f''(x)$.
Where the three vertices are the origin, and the two intersection points of the tangent line.
Besides, I got a better idea anyways
We create a recursive formula that takes the area of the previous triangle, then computes the area of the current triangle, then subtracts their conjunction. If the formula has a known explicit formula, then we have something that at least resembles $f(x)$.
Ideally we have for the recursive formula $a(n)$ that $a(0)$ is the actual area under the curve of $f'(x)$ between two values of $x$ if not an approximation so that we can take a limit and have the true $f(x)$.
So, e.g. we have already computed in closed form the integral $a(0) = \int_0^{\epsilon} f'(t) dt$ or an approximation. We then construct the triangle as detailed above for some $x > \epsilon$ to get an area; compute the conjunction of $a(0)$ and this newly constructed triangle and subtract it from the sum of the two areas to get an approximation of $\int_0^{x} f'(t) dt$. That's what I meant to say.

I won’t engage, but start by specifying the hypotheses you need for “the triangle” to make sense.

Good idea.
Unlike the paper I'm trying to write, I'm having a hard time finding the words to describe it succinctly without being verbose.

2:46 AM
Hey

o/

Gonna try to finish this math assignment by midnight.

What, it's due tomorrow?

Ye, @3:05 pm

breh

2:49 AM
It's not as bad as it sounds. I have a rough outline of all of the problems, but I need to refine them because I'm shaky on one of the definitions.
Unless I don't finish by midnight, then it will be as bad as it sounds.

Is the grading system as cringe as the grading system I dealt with in HS?

wdym?

Some weird, arbitrary ratio of test to homework scores.
Such that not doing homework brings your grade down substantially even if you ace all the tests
Because our school system totally isn't rigged against those who want to learn...

I have to check the weighting, but this particular class leans heavily towards homework grades.

Ah, so about the same cringe system.

2:53 AM
ye, I guess so

Not sure what part of practice belongs in one's skill evaluation...

I guess it's more for the students who aren't likely to practice. That way they'll still end up learning from the accountability.

Having any sort of weighting difference where practice mutually affects overall grade whatsoever undermines the purpose of tests and examinations in the first place which exist... to test and examine a person's competence in the subject according to some ideally objective and well-designed metric.
At that point, you may as well just equate the two and say that tests and exams are just homework that weighs more.

Yeah, I don't know what the ideal situation would be.

Tests and exams, one metric; practice, another metric. Simple.

3:00 AM
Hm, simple.

My grades should reflect my competence, not the amount of work I've accomplished.

That much I agree with.
My stats class has weekly hw due on Friday, and I feel it's frustrating to have to rush through it without a chance to go through the textbook chapter and do the problems there first.
I try to do it simultaneously, but there just isn't enough time. I end up having to finish the chapter over the weekend.

I see... so at least over here, elementary, middle, and high school spoonfeed us information, and then university slaps you with the responsibility of learning to find information by throwing you a textbook heavy enough to kill a man.

That's basically how it is where I am as well.

I think that's enough of me ranting about American public "school" for now.

3:07 AM
ok

I don't want to get too cynical. Pleasure chatting. I need to get ready for bed anyways.

Alright, goodnight. Maybe we'll talk tomorrow.

Good night!

3:57 AM
Hey, @Ted! How is your evening?

4:20 AM
@TedShifrin A lot of the post-WWII stuff sounds to my ear very similar to Israeli music of the same era
Globalization and all that

4:51 AM
@robjohn alive and kicking … you?

2 hours later…
6:36 AM
@TedShifrin pretty much the same, just slow in responding ;-)
who cleaned out the room? Five is the lowest population I've seen in a while.
3

6:48 AM
There is a well-known inequality relating the arithmetic and geometric mean. It follows from this that if the arithmetic mean is less than a certain number, then the geometric mean is also less than this number.
My question is: is it also possible, with some sum, to estimate a lower bound for the geometric mean?
I mean $(abc)^{\frac{1}{n}}≥?$. And this question should also be the sum of the elements.
I determined that this could be the following: $(abc)^{\frac{1}{n}}≥\frac{a+b+c}{n+1}$ but this is only on condition that $a>>1,b>>1,c>>1$

2 hours later…
8:47 AM
@dtn this cannot scale properly. increase $a,b,c$ to $\lambda a,\lambda b,\lambda c$ then $(abc)^{1/n}$ increases to $\lambda^{3/n}(abc)^{1/n}$ while $\frac{a+b+c}{n+1}$ increases to $\lambda\frac{a+b+c}{n+1}$
The geometric mean can be bounded below by the harmonic mean

9:03 AM
Hello, if I have two real power series $\sum_{n=0}^\infty a_n x^n$ and $\sum_{n=0}^\infty b_n x^n$ that both converge and agree in the open interval $(1,2)$, which does not contain their center $0$, does it follow that $a_n=b_n$ for all $n$?

@abenthy yes

But the identity theorem requires that I have a sequence $(x_n)_n$ with $x_n \neq x_0$ and $x_n \to x_0$ where both series agree, where $x_0$ is the center, doesn't it?

there is a theorem which says you can center a series at any point of convergence and it will converge to the nearest end of the interval of convergence of the original series

@robjohn Ah i see, thank you.

@abenthy see this answer, or if you are comfortable with complex analysis, this answer

9:17 AM
Great, makes sense to me now

Anyone uses greasemonkey or chatjax++ here?
in MathJax, 19 hours ago, by Wolgwang
in The h Bar, 10 mins ago, by john
i've installed greasemonkey and chatjax++ but still see the latex code, even after refresh

I just use chatjax (of course ;-)

3 hours later…
12:05 PM
@TedShifrin It's true if we assume the condition to hold for all $y\in \mathbb{R}$, you have to use compactness of $[0, 1]$.

I received a notification that "you haven't voted questions in a while, questions need votes too!". What's up with that?
3

12:23 PM
How do I find total number of anti-symmetric relations on a finite set A containing exactly n elements?
I think this should be $3^{\sum_{j=1}^n (j-1)}$

anti-symmetric like $<$ or like $\leq$?

Suppose that $A=\{a_1,a_2,\cdots, a_n\}$. Then by definition of anti-symmetric relation(let's call it R), if $a_i R a_j$ and $a_j R a_i$ then $a_i R a_i$.

so weakly anti-symmetric, yes?

not sure if I have heard that term.
$\le$ yes.

$aRb$ and $bRa$ then $a = b$

12:32 PM
yes.

if we draw a matrix of this relation, then for every point on the diagonal we can have either $1$ or $0$ so that's already $2^n$ choices we are making

Given any $(a_i,a_j)\in A\times A$ and if R is anti-symmetric relation, we have three cases:
1. $a_i R a_j$ and $a_j Ra_i$ so that $a_i Ra_i$
2. $a_i R a_j$ and $(a_j, a_i)\notin R$.
3. $a_j Ra_i$ and $(a_i, a_j)\notin R$
4. $(a_i,a_j)\notin R$ and $(a_j,a_i)\notin R$

isn't it $2^n$ for diagonal and $3$ cases for every two element set $\{i, j\}$, $i\neq j$

So we choose any $(a_i,a_j)$ in $^{n^2} C_2$ ways and then 4 possibilities. So why is $^{n^2} C_2 \times 4$ not correct?

seems like $2^n\cdot 3^{\binom{n}{2}}$ for me

12:37 PM
you are right, of course.
But I don't understand why the wrong answer I state is not correct.

the implication is stronger actually
we don't have just $a_i Ra_i$, we have $a_i = a_j$
maybe that's why

You are right, I messed up $(1)$.
@Jakobian: I got it, thank you so much :). I think I was getting confused because I misunderstood the definition of anti-symmetric relation.

Can anyone recommend me a book for basic topology

general topology?

Like, one that has simple definitions and introduces definitions without assuming familiarity.
is general topology the most basic one?

12:49 PM
wdym simple definitions

Or should I get "Real Analysis a long-form mathematics textbook" by Jay cummings since it has a little bit of topology as well
I mean explanations.

if you're looking for something really dumbed down, then I don't think you'll get that from a decent book
Munkres is usually what people recommend and it seems to be fine
I used Dugundji myself, it's ok if you look past the second part of the book about algebraic topology
There's also book by Engelking and Sieklucki, but don't confuse it with Engelking's General Topology which is more of a reference book
a lot of books have literally the same things written in them just in different words
I don't think you'll ever escape the "unintuitivness" of certain topics unless you dive into it

When you read these books how do you break down the notation so you can understand it says?
Is that just intuition and experience

idk, is there that much notation in topology?
for me a lot of it is just passed down from set theory

Oh so I should look at set theory first then, huh.
Man i'm learning things so out of order i'm starting to look like a fool...

12:56 PM
well, yes, topology draws a lot from set theory
it's not strictly necessary to learn, but things up to ordinals and cardinals helps sometimes, especially with examples of some topological spaces
and in more advanced level you often see very advanced set theoretical things such as forcing applied

so what I should really be asking for is...Please recommend me a basic but lengthy(covers a lot) book in set theory.
So something like Stewarts calculus

doesn't have to cover a lot, but it helps to be on a decent level
isn't that a calculus book?

Yeah, I only know a little because of stat and prob.
It is, but I mean the coverage of content.

that's bad place to learn set theory

I want something that starts like it(basic) and gradually builds up to really complex stuff.

12:59 PM
I mean, not bad, probably sufficient for analysis and stuff

well, not SO complex

why do you want to learn topology anyway

...I really don't know actually. I want to learn it mostly for the satisfaction and new knowledge i'll gain.

alright, then learn set theory first
then general topology or metric topology
second one might be more intuitive for a beginner
you don't have to learn all the tools of general topology from the start, a lot of universities make you learn about metric spaces first, so you get more intuition for how things work, at least in theory

yeah. So what I want is a book(on set theory) like Stewarts in terms of content and understandability.
I have no clue what metric spaces are =)

1:03 PM
if you'll want to learn about metric spaces, then probably even less set theory is needed
metric spaces are a more analysis-part of topology, kind of

man you sure know a lot. Are you a grad student?
PHD?
Maybe a thesis on the topic of linear algebra...

why do you need to know

You don't have to say ya know
i'm just curious
you probably shouldn't feed my curiosity by answeing the question

I like topology, that's all

we'll(I will) only end up diverging from the topic of books
Ok then
So do you know any good book for set theory?

1:08 PM
not for beginners, no

But can you recommend me a book to read after reading some first year course notes on set theory? Any at that level?
Like if I read some notes...

from set theory?
I have a graduate level book from set theory in mind, so not really

But I will be in like 2 years...
But thanks for your time and help tho =)
Gotta go study for my IB exams now...
guess this stuff can wait since it's not tested
Bye!

bye

2:03 PM
Let the function $K:\mathbb{R}^d \to \mathbb{R}$. Let $\rho$ be a probability density on $\mathbb{R}^d$ with finite second moment. What conditions can I possibly put on $K$ to get the following to hold ?
For a fixed constant $C$ independent of $\rho$...
I can only think of bounded $K$...

you're integrating wrt $\rho$?

yes $\rho$ is a probability measure
(sorry I wrote density before)

what about, say, setting $\rho$ to be invariant with respect to translations
I know there's no probability measure like that
Maybe you could try $K(x) \leq C||x||^a$ and chose $a$ later

I cant see how to get a $C$ independent of $\rho$ unless $K$ is uniformly bounded. Here $\rho$ is arbitrary.

If it's supposed to hold for all $\rho$, then we can try setting some $\rho = \delta_y$ for example, this gets you $K(x-y)\leq C(1+||x||)$
and this means $K$ needs to be bounded
so unless you're working with some subset of probability measures, $K$ needs to be bounded, or just $C$ needs to depend on $\rho$

2:18 PM
mmm
guess I cant do better than bounded then

2:51 PM
Are there other important functional limit theorems besides the sum,product,reciprocal and sandwich?

3:26 PM
@Prithubiswas Probably... but that depends on what you mean by "important", "functional", and "limit theorems".
And I suspect that categorizing these results is maybe missing the point a little.

3:42 PM
Hi all, I want to discuss a question with a user in a room, as I don't want to go into the extended comments. Can someone please create a room for me?
With the title, "Triangular Billiards Problem"
Oh, I figured it out!
Nevermind, everyone.

3:59 PM
@XanderHenderson Oh ok. Thanks for the help =)

4:52 PM
@robjohn The MathJax issue has been fixed in most places. It's still present in a few lesser important places, like when you view the list of all your actions.

@RandomVariable let me take a look...
@RandomVariable the main page has the proper CSS now. :-) where is a page that it doesn't wrap correctly?
My "all actions" page looks okay
and the CSS looks okay there.
perhaps it's been fixed there since you looked. If you have that page open, you might need to refresh.

5:09 PM
Suppose that I have been given a polyhedron, how would I know how many vertices it has in a base?
For example: I have been given dodecahedron, then how do I know the base polygon?
For example: in a tetrahedron, the base polygon is a triangle; in a cube, a square etc.

@robjohn Look at the third page of your "All actions" page.

I know the formula: V-E+F=2, where V= no. of vertices, E= no. of edge, F= no. of faces but for a dodecahedron, I have only that $F=12$.

@RandomVariable I see the problem. Let me check the CSS for that page.
@RandomVariable I have passed that on to Davide Cervone.

5:34 PM
I think there's no way to find the shape of base of a polyhedron.
except guessing and drawing a lot of pictures.
For example: for a dodecahedron, one will start with a triangle and try make 12 faces and see if they get polyhedron with that or not. If not, then start will be with square and then from a pentagon and so on until that polyhedron is found.

@robjohn It appears to be the "All actions" page, the "Responses" page, and the "Votes" page.

Hi all, which is the correct definition for a complex number
This i^2 = -1 Or i = sqrt(-1) ?

@CroCo Yes.

Yes what?

Yes, it is correct.

5:49 PM
you mean both correct?

Sure. or either. or neither. And there are other correct definitions, as well. Which definition is most "correct" will depend on what you want to do with those definitions. Without further context, it is impossible know which definition is appropriate.

@RandomVariable I don't think anyone else can view a "Votes" page, but I do see the issue there. I just can't point to it and expect anyone to see it.

@XanderHenderson this is a political answer. If it is a precise definition, why we need a context.
The reason I'm asking because I've heard i = sqrt(-1) is bad or not precisely correct.

@CroCo It may also be worth pointing out that a definition cannot be "correct" or "incorrect". A definition just tells us how we are going to use a word or notation. Definitions can be more or less useful, but we can define words to mean whatever we want them to mean.

@robjohn I don't think anyone else can view a "Responses" page either.

5:55 PM
@CroCo The definition is "bad or not precisely correct" in what context?
What about that definition causes problems?
For whom?

runs

Unfortunately I don't remember the justification why it is bad. That's why I'm asking.

@Koro Well, there are only 5 regular polyhedra. That's pretty easy to show. Of course, there are lots of semi-regular ones, eg en.wikipedia.org/wiki/Archimedean_solid

@CroCo Every definition has context.

@RandomVariable ah, let me check

5:58 PM
@XanderHenderson but it seems to me the second definition may break the radical concept but the first has no problem because this is the way we define it. I may be wrong.

@CroCo I don't understand.

@RandomVariable It may be because I am a mod, but I can see your responses page. Maybe I can check on another site.

@PM2Ring Thanks. :) I didn't know it was so complicated.

Also, defining $i$ by the property that $i^2 = -1$ could cause problems of its own, as that equation has two solutions. One then has to make a choice about which of those solutions is actually $i$ (though the choice is irrelevant... something something automorphisms something something field extension something something).

@robjohn I can't see your "Responses" page.

6:00 PM
So, again, the context in which the definition is meant to be used matters.
In any event, I need to go teach a class now.

The question was to find order of group of rigid motions of dodecahedron. @PM2Ring

@Koro Every edge is shared by 2 faces. And in a regular polyhedron, k faces meet at a vertex for some k>=3. And the angle sum at a vertex must be <360°, so there aren't a lot of options. 3, 4, 5 triangles, 3 squares, 3 pentagons.

@CroCo you might be interested in math.stackexchange.com/questions/144364/…

@RandomVariable Yeah, there isn't even a tab for it when I go to a site where I am not a mod.

@CroCo sqrt is a multi-valued function so how are you defining it, I mean what branch are you taking?
Unless that is clear, I don't think the second definition you said would make sense.

6:03 PM

@PM2Ring so given any "___" hedron, I can find the base shape?

@Koro I come across the second definition in some books but not sure if they are sloppy definition to infer the first definition I've posted.

There is some continuity issues with the second definition.
The link I posted clarifies that.

@PM2Ring Given the base shape, I could find the order but I think Dummit &Foote expected reader to know this while writing the exercise.

6:11 PM
@Koro Sort of. You can expand that equation V-E+F=2. Let n be the number of edges on a face, and k as above. Then E = Fn/2 and V=Fn/k. You can simplify that to give an equation of n in terms of F and k, and then try various k until you get an integer for n.

Thanks. I see some light now. I was thinking that we'll have to draw everytime.
So suppose that I want to try that for an octahedron. An octahedron has 8 faces.

@Koro Ok. If they're asking about symmetry groups of the 5 regular polyhedra (aka Platonic solids), they expect you to know their basic structures.

$8n/k-8n/2=-6\implies n(-1/k+1/2)=3/4$
$k=4, n=3$ works.
but that looks wrong.

Here's a semiregular solid called the small stellated dodecahedron. You can think of it as a dodecahedron with a pentagonal pyramid stuck to each face. Or as 12 pentagonal stars stuck together.

In the calculation above, n is correct but k is wrong @PM2Ring. Can you please help me understand why so?

6:23 PM
@Koro k is correct. In an octahedron, 4 triangles meet at each vertex.

@PM2Ring Oh yes. I misread 'vertices' as 'edges'. Thanks a lot :).

:)

I think I should study what semi-regular/regular mean.

In 4 dimensions, there are 6 regular hypersolids. In all higher dimensions, there are only 3.

For now I'm concerned only with R^3. :)

6:26 PM
Good idea.
In a perfectly regular solid , all faces are congruent regular polygons. And each vertex looks like any other vertex. In contrast, the solid in my avatar image, the rhombic dodecahedron, the faces are rhombuses, and all the vertices aren't the same. Some have 3 edges, some have 4.

6:59 PM
@PM2Ring simplex, hypercube, are you counting a hypersphere?
or is there one I am missing?
dual to the cube? does that exist in higher dimensions?

There's an entire section in my algebra book on symmetry groups of the regular polyhedra. (And in another section the classification based on group actions.)
@Croco $\sqrt{-1}$ is not well-defined.
So you have to say $\pm i = \sqrt{-1}$, and then we don't know which is which.
Who knows which is which, anyhow.

@Odestheory12 Exactly. And, quoting Asaf's very find answer: "The question whether something has a root or not must include a setting. Definitions do not appear magically, they require some preexisting framework."
CONTEXT MATTERS! :D

7:17 PM
People in algebra use the notation $\mathbb{Q}[\sqrt{d}]$ where $d$ can be negative
just notation like any other

What is with people these days? I get it that once the OP corrected his post, it looked like I was insisting that $\setminus$ be $-$. I don't give a flip. Is it true that only those of us over 60 write $A-B$ instead of $A\setminus B$?
I don't object to the notation, @Jakobian. I'm just pointing out that $\sqrt$ need not be a function, and is not working over any field other than $\Bbb R$. The field extension doesn't care which root you adjoin, of course.

I'd say everyone writes $A\setminus B$ but I'm French

Yes, you write $]a,b[$, too. Do you correct me if I do not?

@TedShifrin Yikes! So many flags on the comments there!

LOL, really? I'm flagged for rudeness?

7:30 PM
I wouldn't correct you, I'd ask for clarification if the notation is ambiguous for me

Lads, I'm trying to prove the continuous mapping theorem - $X_n \rightarrow{\text{p}} X \implies h(X_n) \rightarrow{\text{p}} h(X)$ for a continuous function $h$. My idea was to just use the fact that $h$ is continuous to bound $|X_n - X| < \delta$ and then use $|h(X_n) - h(X)| < \epsilon$ to complete the proof and show convergence in probability. Is this right?

I wonder if $\setminus$ really is an ageist thing. In my mathematical upbringing, I believe I only saw $A\setminus B$ in the context of quotient objects although $B/A$ is more common for that; it depends on who's acting on which side.

@TedShifrin No, the comments have been flagged as "no longer necessary". It looks like your correspondent deleted their comments, then flagged your responses as no longer necessary, as the inciting comments are gone.

I dont know how to do the rightarrow wiht a p on top of it.

Ah, I would have removed my comment immediately 2 days ago if the OP had let me know he'd responded.

7:32 PM
But that's what I mean

I shall remove mine.

For what it is worth, in the area I work in $A - B$ is the Minkowski difference, i.e. $$A-B = \{ a-b : a\in A, b\in B\}.$$ Hence I use $A \setminus B$ for the relative complement.
But I have zero problem with $A - B$ as long as there is no chance of ambiguity (and there is none in the question being asked).

@Xander: Yes, I agree completely. And I would not use $A-B$ if it were ambiguous.

And I have seen $A\setminus B$ and $A/B$ to denote left- and right-cosets (or something like that). Context matters. :/
People who are overly prescriptive about notation piss me off.

Munkres's Topology book still uses $-$, not $\setminus$.
Yes, I have used left and right quotients throughout my career. Because of Chern, my principal bundles seem to have the group action on the opposite side from almost everyone else. :)
So I do have to write $H\backslash G$ instead of $G/H$. Sigh.
But the spacing is all wrong with setminus.
So I write \backslash.
Anyhow, I agree that that algebraist is pissing me off and calls me rude for saying he was "ridiculous."
@Govind, ah, you mean $\overset{p}{\to}$.
I might do $\overset{p}{\longrightarrow}$.

7:38 PM
@TedShifrin Ah I see, cheers

The second looks a little better.

@TedShifrin Right. So it makes perfect sense that you would prefer $A-B$ for the relative complement, as that avoids ambiguity in your chosen field.

Honestly, @Xander, it comes from having been taught set difference that way, and no book I taught out of made me switch. None.
I always adopt the text's notation to minimize anxiety for the students. (We've had this discussion zillions of times before.)
@Govind: So what does convergence in probability actually mean? What does $|X_n-X|<\delta$ mean?

Convergence in probability means $\forall \epsilon \lim_{n \rightarrow \infty} Pr(|X_n - X| > \epsilon) = 0$

Now keep going.

7:44 PM
this is called convergence in measure in measure theoretic context iirc

Right, @Jakobian.

Okay, since $h$ is continuous, for any $\epsilon >0$ I can find $\delta > 0$ s.t. $|X_n-X| < \delta \implies |h(X_n) - h(X)| < \epsilon$

Wrong.

Is it uniformly continuous?

no

7:47 PM
So $h$ is a function mapping what to what?

The question doesn't explicitly state a domain and image
it just says $h$ is continuous

What does $|X_n-X|<\delta$ even mean?
The domain seems to be a certain space of random variables.

The difference between the images of the random variables is always less than $\delta$?

So $X_n$ and $X$ are functions on our probability space, mapping to $\Bbb R$? And $h$ is a function on the space of random variables. Surely your course/textbook has made these things precise.

Oh I guess $h$ is a function from $\mathbb{R}^n \rightarrow \mathbb{R}^m$ where $X$ and $X_n$ are in $\mathbb{R}^n$

7:54 PM
Random variables are usually functions.

I'm just speccing the lecture notes and it says $X$ and $X_n$ are $\mathbb{R}^k$ - valued random variables

I'd split $\{|h(X_n)-h(X)|\geq \varepsilon\}$ into a set where $h$ is uniformly continuous, and a set of small measure
@Govind75 and we can use this where it's uniformly continuous

I'm gna do some more reading on the topic before I go further with this question I reckon, cheers for the help though lads

8:39 PM
Ah, almost done writing the rough draft for the paper.
Just have to finish writing about how to compute.

9:16 PM
Why is it that if I take an integral like $$\int_{0}^l sin(\frac{n\pi x}{l})sin(\frac{m\pi x}{l})\mathrm dx$$ and I calculate the integral, I get a formula which is undefined at $n=m$ for $n,m\in \Bbb N$, but if I set $n=m$ to be equal integers in the integral itself I get a finite value?
I feel like in this case, since the integral is equal to whatever it evaluates when you work it out, you end up with two equivalent $f(n,m)$, one of which is undefined at $n=m$ and another which is not
I feel like this relates to the Dirac delta function

10:04 PM
@Charlie at $m=n$ the integral is $l/2$. the integral is $0$ for $m\ne n$, assuming $m,n\in\mathbb{Z}$

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