1:13 AM
@Jakobian Possibly as many as 144.
@XanderHenderson is there an inside joke here? please explain
1:32 AM
Btw, are you going to watch the fencing at the Olympics?
π€Ί
π₯π₯π₯
2:07 AM
@user85795 gross meaning g.co/kgs/weXZwmX
Look at the nouns.
@user85795 Likely not. It depends on how much NBC let's out of their sandbox.
2:54 AM
TIL, thnx.
> (n) "a dozen dozen," early 15c., from Old French grosse douzaine "large dozen;"
4 hours ago, by Thorgott
very gross, disgusting even
could be 12(gross) = 1728
+1 ;-)

5 hours later…
7:50 AM
If $f_n:X\to Y$ is a sequence of homeomorphism between two metric spaces $X$ and $Y$ which converges locally uniformly to $f:X\to Y$, then $f$ is also homeomorphism?
8:21 AM
one: X = Y = real numbers, f_n(x) = x/n?
@XanderHenderson shouldn't my dominant hand be better at gross motor tasks as well?
soumik: i have no idea, but part of it might be - you feel slightly more in control with your dominant hand free. your non-dominant hand might as well be steering the bike because if you needed to quickly react to something with one hand, you'd probably want it to be your dominant one.
9:15 AM
@Thorgott I would change $\Sigma_2\times S^1$ to a mapping torus $M$ of $\Sigma_2$ and further require $\Sigma$ is transverse to the suspension flow of $M$.
@leslietownes ooh maybe that is the reason, thanks
Otherwise, $\Sigma_2\times\{1/2\}\cup\Sigma_2\times\{-1/2\}/\sim$ by connected sum along a round disk on $\Sigma_2$ whose boundary is nulhomotopic on $\Sigma_2$ is transverse to $S$ but embedded in $\Sigma_2\times (-1,1)$.
Requiring transversality with the suspension flow abandons this.
Also found a fact that could be helpful: $M$ has a natural taut foliation $\mathcal{F}$ with leaves $\Sigma_2$. In this case, if $\Sigma$ is incompressible, meaning $\pi_1(\Sigma)$ injects to $\pi_1(M)$, then $\Sigma$ can either be homotoped to a leaf or can be homotoped to intersect $\mathcal{F}$ in only saddle tangencies.
10:03 AM
@onepotatotwopotato Limits of homeomorphisms are hard to be made into homeomorphisms, you can have piece-wise linear mappings for $X = Y = [0, 1]$ and the uniform limit not a homeomorphism
maybe better to describe this pictorially, basically you have $f_n$ and $f$ piece-wise linear on $[0, 1/3], [1/3, 2/3]$ and $[2/3, 1]$
$f(0) = f_n(0)= 0$ and $f(1) = f_n(1) = 1$
$f(1/3) = 1/2$ and say $f_n(1/3) = (2-1/n)/6$ and $f(2/3) = 1/2$ and say $f_n(1/3) = (2+1/n)/6$
then $f_n$ converges uniformly to $f$, each $f_n$ is a homeomorphism but $f$ fails to be injective, namely on $[1/3, 2/3]$ where its constant
sometimes you can assure its a homeomorphism
the weird $G_{1/i}$ are maps which are surjective but not $1/i$-maps (i.e. fibers have diameters $\leq 1/i$

1 hour later…
11:31 AM
Do you agree with the examples of the indeterminate forms $0^0$ here? I don't know, but I'm not fully convinced. For example, $\lim_{x\to 0^+} x^0$, we know $x^0$ is already $1$, before taking the limit
11:41 AM
@SineoftheTime Yes? That's the point being made by the Wikipedia article. They aren't claiming that $\lim x^0$ is an indeterminate form---they are using that limit to help explain why $\lim f(x)^{g(x)}$ is indeterminate if both functions go to zero.

2 hours later…
1:53 PM
$0^0$ is in fact undefined. In many situations , it is considered to be $1$ due to usefulness.
@XanderHenderson thanks. I misinterpreted
2:10 PM
$0^0$ is undefined due to calculus students
Everyone else usualy defines it as $1$. Unless you have some odd exception like the one Xander likes to talk about. But whatever
2:23 PM
@Jakobian Most people don't use $0^0$ at all, and leave it undefined, because there is no need to define it one way or the other.
When someone needs to have a definition for that notation, they'll define it in whatever way makes the most sense in the context in which they are working.
@XanderHenderson most people do use it, in analysis, e.g. $x^0$ in Taylor expansion
Most people aren't analysts.
Its something almost everyone goes through.
You don't need to be an algebraist to use algebra
Sure, but most people are not working in situations in which they need to care about the definition of $0^0$. Most people leave it undefined, because it doesn't matter to them.
I also think that it is unwise to make strong statements about the definition of $0^0$ in a setting where the primary audience is students who are just learning the ropes. The people who care about having a definition for $0^0$ are typically mathematically mature enough to know how to define it in the context they need. People here, who are asking questions about that definition, are likely to be confused by declarations like "most people define it to be $1$".

2 hours later…
4:27 PM
People who are learning the ropes need to be aware of the different conventions about value of $0^0$ and why most people set it to $1$
And saying everyone here among the people still learning isn't mature enough to discuss definition of $0^0$ is belittling. I have faith that most people here are mature enough
4:42 PM
@Jakobian That isn't what I said.
What I said is that the people who actually need to have a definition for $0^0$ are going to be able to define it themselves, and people who are already confused by it (i.e. just learning about it) don't need to be confused by having people tell them "Well, most people define it to be $1$."
It is extra information which isn't useful to someone who is just learning the idea.
And, again, the most common convention is to simply leave it undefined. Because for most working mathematicians, the value of $0^0$ is simply irrelevant.
I mean, students already show a great deal of confusion about the entire concept of "indeterminate forms". That, already, is a hurdle that students have difficulty going over. It is a pedagogical work to explain to students that $0/0$ is not the value of the limit, but that you can have limits "of the form" $0/0$, which require more careful analysis.
I really think that it would benefit you to actually teach students. I suspect that these are conceptual difficulties which you have never had in your life, hence you have lack that first person experience of being confused. It might help your outlook to see the confusions which are actually present in a lot of classrooms.
And, by the way, if you think that it is belittling to acknowledge that students have difficulties, I find it equally belittling to state "$0^0$ is undefined because of calculus students". It is like you intellectually understand that there is real confusion here, but you dismiss it because, well, they're just students---who cares?
(This is probably no what you meant, but it certainly comes across that way.)
@XanderHenderson this is as much of a hearsay claim as saying most people define it to be $1$
@Jakobian So what you are saying is that I shouldn't assert that $0^0$ is generally left undefined?
4:57 PM
Well yes. And I shouldn't say people define it as $1$
Okay. I am entirely willing to accept that compromise. I assume that you will also cease emphatically proclaiming that "most mathematicians" define it to be $1$?
Okay. We agree.
Excellent.

1 hour later…
I am trying to show $\overline{\varphi_X(t)}=\varphi_X(-t)$, where $\varphi$ is the charactersitic function, i.e. that it is Hermitian. Recall $\varphi_X(t)=E(\cos tX+i\sin tX)$, and then, out of the blue...cold turkey, the author simply writes $$\overline{\varphi_X(t)}=E(\cos tX-i\sin tX),$$i.e. the author did $\overline{\int f\, d\mu}=\int \overline{f}\, d\mu$. Is that allowed?
@psie recall the definitions
$\int f := \int \text{Re}(f) + i \int \text{Im}(f)$
yeah, okay, thanks, I see it now
@XanderHenderson I was not expecting that headline...
@MikeEarnest Indeed.
6:24 PM
Also, I see a long argument here about zero to the zeroth power. I don't have anything to add, except I would suggest anyone who is interested watch this excellent video by Field's medalist Richard Borchards about the topic. youtube.com/watch?v=r52bmLFIhpg I found it to be a very refreshing explanation of the situation which shows the benefits of both perspectives.
6:41 PM
Does anyone know how to derive the following estimate: $$\left|e^{ix}-1\right|\le \left|x\right|?$$
Context; I'm reading a proof of the fact that the characteristic function is (uniformly) continuous.
@psie $|e^{ix}-1|\leq |x|\cdot |e^{ix_0}| = |x|$ for some $x_0$ between $0$ and $x$
from mean value theorem for vector-valued functions
ah ok, need to look that up, thanks π
not sure if there isn't some more elementary approach
7:28 PM
@XanderHenderson now that heβs dropped out, would there be primaries again to select the candidate? Or would it just be Kamala Harris?

1 hour later…
8:42 PM
@Sahaj In the US, primaries are a party affair. Technically, your are not voting for a candidate, but rather for a representative to the party convention. If the candidate drops out (or if no candidate wins on the first round of voting at the election), those people are free to vote however they like.
Biden has endorsed Harris, so one presumes that she will get those votes at the convention (frankly, that is probably what is best for appearances of party unity), but it could be REAL wild.
But, ultimately, if it's up to the Democratic party.
9:05 PM
I'm reading through a proof that the Gaussian $N(m,\sigma^2)$ has characteristic function $$\varphi_X(t)=\exp\left(imt-\frac{\sigma^2t^2}{2}\right), \quad t\in\mathbb R.$$ I quote the proof from the texbook:
> We may assume that $\sigma>0$ and replacing $X$ by $X-m$ that $m=0$. We have $$\varphi_X(t)=\int_\mathbb{R}\frac1{\sigma\sqrt{2\pi}}e^{-x^2/(2\sigma^2)}e^{itx}\, dx.$$ Clearly it is enough to consider the case $\sigma=1$.
1. Why can we assume $\sigma>0$?
2. Why is it clearly enough to consider the case $\sigma=1$?
1. Isn't it standard?
In probability theory and statistics, a normal distribution or Gaussian distribution is a type of continuous probability distribution for a real-valued random variable. The general form of its probability density function is f ( x ) = 1 2 Ο Ο 2 e β...
@Jakobian you mean standard deviation? but the standard deviation can be negative too.
no
no to both
In statistics, the standard deviation is a measure of the amount of variation of a random variable expected about its mean. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range. The standard deviation is commonly used in the determination of what constitutes an outlier and what does not. Standard deviation may be abbreviated SD, and is most commonly represented in mathematical texts and equations by the lowercase Greek letter Ο (sigma...
ok, yeah, I was trippin, had to look into my statistics book too :) the variance is nonnegative, and the sd is simply the square root of that
so why is the author even writing that "We may assume that $\sigma>0$" if this is a positive parameter?
one reasons that comes to mind is sometimes constant r.v. are considered normal
but I don't know
9:25 PM
@Jakobian ok, so a constant r.v. has $0$ variance?
@psie yes, $0$ variance iff constant a.e.
@Jakobian and what's the pdf/law of such a random variable? trying to figure out what $$\varphi_X(t)=\int_{\mathbb R}e^{itx}\, P_X(dx),$$ would be in that case...
@psie no pdf
law is $\delta_t$ for some $t$
its a discrete random variable
$\varphi_X(t) = e^{itc}$ if $X = c$ a.e.
its an integral of a constant
yeah, the dirac/delta measure picks out the value $c$
ok, so to summarize: if $\sigma=0$, and $m=c$ say, we just get a constant a.e. random variable equal to $c$, and computing the characteristic function of this random variable is kind of trivial, but it does give us the formula of the characteristic function of a normal distributed random variable, so we could say the constant a.e. r.v. is a normal one.
that's the next theorem though...that the characteristic function determines the law of a random variable...so saying the constant a.e. r.v. is normal requires a proof
9:48 PM
@psie i.e. the formula is the same, yes.
Now that doesn't necessarily means its okay to call constant r.v. to be normal, but...
@psie this theorem has nothing to do with what you said
@Jakobian if two random variables have the same characteristic function, they have the same distribution...so if the cf of $X=c$ agrees with the cf of $Y\in N(c,0)$, that theorem applies, right?
what?
If we were trying to look from the perspective of "let me see what r.v. I obtain by plugging those values into my characteristic function" then you would be correct, but we are not doing that
10:03 PM
@Jakobian hmm, that's exactly what the theorem says, isn't it? If we have the cf of $X=c$ given by $e^{itc}$ and the cf of $Y\in N(m,\sigma^2)$ given by $e^{imt-(\sigma^2 t^2)/2}$. Clearly, they agree when $\sigma=0$ and $m=c$, and if they equal, then $X=Y$, where equality means they have the same distribution.
anyway, maybe time to call it a day, getting late
@psie how do you know the cf is given by that when $\sigma = 0$
aren't you tryin to define what $Y\in N(m, 0)$ would mean based on the cf?
no one is doing that here except for you
ah, yeah, good point, my argument is flawed
I am saying that constant r.v. is normal by definition. Nothing to prove
it just so happens that the cf is "correct" then and that your theorem tells us there could be no other correct choice
still. Nothing to prove
there are other ways for why someone would like constant r.v. to be normal, not just the cf
for example, $Y\in N(m, \sigma^2)$ is of the form $Y = m+\sigma X$ where $X$ is a standard normal distribution i.e. $X\in N(0, 1)$
isn't it natural to demand that this holds when $\sigma = 0$ as well?
another reason: wouldn't it be nice for linear combinations of jointly independent random variables to be normal?
yeah, it simplifies things
i.e. there are other, algebraic reasons for why would someone like to define $Y\in N(m, 0)$ to be constant
10:20 PM
yeah, thanks for the conversation Jakobian, I am leaving for today
you're welcome