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00:00
I do most of my staples at Sprouts. I get dry cereal and a few other things at Ralphs, which I can walk to for light small purchases.
it would only be about 20 miles :D but there's just lots of busy streets around the 405 on and off ramps. it's psychologically intense.
Well, there are good reasons LA is my least favorite city to drive in. Worse than NY. Worse than Paris. I am told Rio is the worst, but I've not been there.
hm. i thought boston was mildly worse than LA. LA style chaos driving, plus many streets not being designed with cars in mind. which is not to say that LA is a paradise of car friendly design.
the idea of driving in NY terrifies me, and i don't plan on finding out what it's like
I have to leave now, but I haven't done Boston since 1981 or so, so who knows.
When I lived in SoCal, I did nearly all of my shopping at WinCo.
WinCo: The Happiest Place on Earth™
00:14
that would be more of a drive for me, although, it is the happiest place on earth.
According to the Google, there is a WinCo in Lakewood. If you are in LA proper, that would be a bit of a drive.
it's not too far, but i'd have to drive right by a ralphs and a whole foods and two targets with grocery sections before i got there, so maybe not today.
i'm happy enough.
Yeah, but WinCo is cheaper than all of those, and better than most (Whole Foods might be better quality).
01:03
our target has better fruit than whole foods, for some reason. i'll never understand it.
01:50
hello, is there name for C1, C2, etc from analysis? I'm in the class of n-th differentiable functions
hm, there's no standard name for 'n-times differentiable' that i know of. C^n for having n continuous derivatives is fairly standard, but not so standard that i'd use it without defining it.
02:19
anza is covered in mosquito fern again
$C^{n-{1 \over 2}}$.
02:32
haha
 
1 hour later…
03:57
@leslietownes Totally standard in geometric analysis.
hm. does it come with a standard norm?
04:18
Nope, not unless you’re on a compact manifold,
standard norm seems almost redundant
 
1 hour later…
05:47
clifford algebra is crazy. Who invented this nonsense?
a big red dog
 
2 hours later…
07:32
@copper.hat what about a non-standard norm?
Then again, I’m usually unusual.
 
1 hour later…
08:42
The head-band look is making a come back!?
👏👏👏
09:19
For Passover/Easter
I see what you did there.
I colored some eggs ;-)
and added an appropriate letter to Pascal
:D
 
2 hours later…
11:20
What do you all think of a feature request for "voice chatrooms" like they have on Discord?
 
2 hours later…
Jam
Jam
12:54
How can i draw γ(t)= ((1+2cost)cost,(1+2cost)sint)
Finding its cartesian coordinates after alot of calculations ys and xs not seperable and cant find an already known expression
Looks like an ellipse with no constant a and b
Is there a clever way instead of calculating for every t?
13:12
Hi guys. Can someone help me to unstuck me from this problem?

I want to compute ln 2 with a error bound of 10^-5. But when I use a Taylor theorem, I get stuck into a loop where to compute ln 2, I need to know this result.

Here is the problem:

$$ f(x) = ln (\frac{1+x}{1-x}) \\
x_0 = 0 \\
x = 1/3 \rightarrow f(1/3) = ln 2 \\ \\

M (majority of the function)

|R_{k,0}(1/3)| = $$
I'll continue here
@robjohn :-)
@copper.hat good morning
M (majority of the function in [x_0, x] interval) = ln 2

But I don't know ln 2, so I'll use ln 2 < ln e = 1

$$ |R_{k, 0}(1/3)| < \frac{1 * 2 * (1/3)^{2k+3}}{2k+3} < 1 / 10^5 $$
$$ 3^{2k+3} (2k+3) < 2 * 10^5 $$
$$ (2k+3) ln 3 + ln (2k+3) < 6 ln 2 + 5 ln 5 $$
At this point, I don't know what to do
because ln 2 is what I want to compute
For people in doubt about how to read $|R_{k,0}(1/3)|$: This is the module of the remainder of the Taylor polynomial built around x_0 of k order computed in x value.
Notation: $|R_{k, x_0} (x)|$
13:33
@robjohn Good morning! I'm enjoying the rain here
@MatheusSousa Do you know the power series for $\log(1+x)$
@copper.hat I wish it were raining here
@robjohn Yes, I'll try with this function
$\log(2)=2\left(\frac13+\frac13\left(\frac13\right)^3+\frac15\left(\frac13\right)^5+\dots\right)$
the error is less than $\frac98$ of the first omitted term
14:25
@robjohn seems to have stopped
That's too bad. Just about to take the dog for a walk in the park. BBL.
14:37
Hey folks! I am writing my thesis, and I'm looking for a (single) word saying that something is "not equal to 1". Something along the lines of "nonunity." Any ideas?
14:53
sorry man, but you are guiding me to a wrong path.

$ ln (1 + x) = \sum_{n = 0}^{\infty} \frac{x^{n+1}}{n+1}, where |x| < 1.$

If I want to get ln 2, I need to choose x = 1. But when x is equal to 1, this equality between function and series is not valid.
Its better to get a series for $ ln (\frac{1+x}{1-x}) $ because whatever argument is in logarithm, |x| will be less than 1 for whatever value I want to get.
0
Q: t operator on hilbert space restricted to orthogonal complement

MathematicallyInterestedLet $T$ be a compact self adjoint operator on a hilbert space $H$. If $(e_n)_n$ is an orthonormal system of eigenvectors then setting $H_0= \{ z : \langle z , e_n \rangle=0$ forall $n$ $\}$, we have $T(H_0)=0$. My attempt: If $H_0=0$ then clearly we are done. So assume otherwise. If $z\in H_0$, ...

 
2 hours later…
16:40
@Jam Think in polar coordinates!
@AlessandroCodenotti I've shown that the space of irreducible subcontinua of a continuum is a a $G_\delta$ set of all subcontinua
Now, how do I show it's dense if our space is taken to be $[0, 1]^n$ where $n\geq 2$ or $n = \infty$?
They say I should use the pseudo-arc, but I'm not sure how to formalize my thoughts here
16:56
I'm trying to understand what's going on in the answer to [this question](https://math.stackexchange.com/questions/4429077/what-does-an-exponent-of-a-differential-form-mean). Trying to do the computation myself, I get

\begin{align*}
\omega \land \omega &= (dx^1 \land dx^2 + dx^3 \land dx^4) \land (dx^1 \land dx^2 + dx^3 \land dx^4)\\
&= (dx^1 \land dx^2) \land (dx^3 \land dx^4) + (dx^1 \land dx^2) \land (dx^1 \land dx^2) + (dx^3 \land dx^4) \land (dx^1 \land dx^2) + (dx^3 \land dx^4) \land (dx^3 \land dx^4).
17:27
I meant indecomposable subcontinua
It looks like what I thought was pseudo-arc doesn't have to be homeomorphic to the pseudo-arc
@Novice Review my lecture on wedge product. It’s all the basic rules.
So $\omega$ being a symplectic form is totally irrelevant in that answer?
Or maybe not totally irrelevant, but irrelevant to the computation at least.
17:47
Correct.
Thanks.
The computation ultimately is needed to show it is symplectic.
17:59
Hello
What does hyperplane section section of a variety X means? Does it mean it is given by a single equation? If $H$ is hyperplane section in $X$, does this mean that $H = V(h) \cap X$?
I think I ought to go back to thinking about the regular exterior and differential forms for a while before thinking more about symplectic forms, because I'm struggling to place the symplectic forms relative to the exterior and differential forms.
@JGuy It means $X$ is sitting in some $\Bbb P^n$, and you literally intersect it with a (linear) hyperplane in $\Bbb P^n$. You restrict the homogeneous linear equation of the hyperplane to $X$ and that's the equation.
@TedShifrin Yeah, so we we get the above. Ty!
@TedShifrin Mornin'
Morning, @copper.
18:07
Nice day with sprinkles of rain here.
You're welcome, @JGuy
Nothing resembling that here, copper.
Just mid-60's.
If X is a variety and $D$ is a divisor then for $m$ big enough we have $mH - D$ is very amply, so there exists effective divisor $D^{\prime}$ such that $mH - D \sim D^{\prime}$?
Very ample is a much stronger condition than that. Seems like you need to go master the basic definitions.
What you wrote is just the statement that $|mH-D|$ is nonempty.
yes
Excuse me. I want to ask, so $H$ is given by $V(h) \cap X$. Look at the homogeneous polynomial h, if we pick an affine chart and set one of projective coordinate equal to 1. If we convert back to Projective coordinates we get rational function $s$ on $X$ such that $div(s)_X = H + \text{other things}$?
18:31
Does anybody have a good visualization for almost sure convergence of a sequence of random variables? I feel like I have a decent visual understanding of convergence in probability, but almost sure convergence seems trickier.
be right back I will be back later.
hmm, i had the opposite issue, understanding pointwise ae. seems straightforward, whereas convergence in probability was harder to imagine.
Here's my tentative informal explanation of convergence in probability.
First keep in mind pointwise convergence of functions. Just imagine a simple example in $\mathbb R$. Now we want a concept like this in measure theory, except in true measure theory tradition we want to be able to permit behavior on sets of measure zero that we don't care about. Imagine a sequence of functions, like maybe $f_n(x) = (1/n)x$, converging to the zero function, except maybe for all $n, f_n(2) = 5$. So you have pointwise convergence except at a point.

To accomplish our goal, we come up with a concept of convergence of a sequence of functions "in measure", which essentially says
If $(e_n)_n$ is an orthonormal family of eigenvectors in a hilbert space $H$, can we always extend it to an orthonormal basis of eigenvectors?
Basically, convergence in probability seems relatively simple because it translates easily to general measure theory. However, almost sure convergence is not an applicable concept in measure theory in general.
18:43
@monoidaltransform That’s false in finite dimensions.
Convergence almost surely says that $\text{Pr}\left( \lim_{n \to \infty} X_n = X \right) = 1$. Two notable things have happened here, compared to convergence in probability: we're relying on the fact that we're working in a finite measure space, and the limit moved inside the measure.
I was thinking of trying to visualize functions on the unit interval, maybe. But ideally I'll find a visual example that illustrates almost sure convergence, but does not converge in measure (i.e. in probability).
19:14
as. convergence is essentially pointwise convergence, i am not sure i see the visualisation convern?
Well, correct me if I'm wrong, but the explicit definition of pointwise convergence, $\text{Pr}\left( \{ \omega \in \Omega \colon \lim_{n \to \infty} X_n(\omega) = X(\omega) \} \right) = 1$, basically says the subset in the sample space where $\{ X_n \}$ converges pointwise to $X$ has full measure. So the complement will have zero measure. But that sounds essentially the same as my intuitive understanding of convergence in probability: the set where convergence doesn't occur has measure zero.
So clearly I'm not seeing the distinction, i.e. the extra strength of convergence almost surely.
...that first sentence should have said "explicit definition of convergence in probability".
19:33
convergence in probability is much weaker.
Wait, it should have said convergence almost surely, not convergence in probability. Sorry.
Isn’t conv in prob just conv in $L^1$?
@Jakobian I expect $G_\delta$ to be the hard part
I already have that one
For density you can mimic the pseudoarc construction (with nested chains) inside any open set
19:39
I haven't learned much about $L^p$ spaces yet. I wonder if that would help shed light on this. copper.hat, what is your intuitive understanding of the difference between convergence almost surely and in probability?
try something like f_n(x) = n if 0 <= x <= 1/n and f_n(x) = 0 otherwise. converges pointwise to 0 almost everywhere. does not converge in L^1.
or something similar to that. a book that discusses L^p spaces and the various kinds of convergence would have a host of examples.
i don't know the probability languagee.
For being $G_\delta$ I used $\mathbb{C}_n = \{X : X = A\cup B, A\not\subseteq N(B, 1/n), B\not\subseteq N(A, 1/n)$, then using compactness of $C([0, 1]^n)$ and some more calculations I can prove $\mathbb{C}_n$ is closed, and $\bigcap_{n=1}^\infty \mathbb{C}_n^c$ is the set of all indecomposable continua
A Vietoris open sets tells you "continua that meet $U_1,\ldots,U_n$ and are contained in $U=\bigcup U_i$, you can build a pseudoarc in $U$ from nested chains long enough to meet every $U_i$
I think something like that should work
@TedShifrin no, on $[0,2 \pi]$ the functions $f_n(x) = \sin (nx)$ converge to 0 in $L^1$ but do not converge in probability.
Ah, so what is the correct defn?
19:41
Btw since you don't need hereditarily indecomposable there is an easier nested chain construction in Nadler's Continuum theory that produces an indecomposable but not hereditarily so continuum
yeah, 1.10 right
there is an exercise later which is to prove the same thing but for hereditarily indecomposable continua though, that will be rough
Yes I expect the $G_\delta$ to be much harder there
@Jakobian ah makes sense
I was thinking hereditarily indecomposable for some reason and it seemed difficult to me
@TedShifrin Ignore my earlier remark,.
i wish one could delete one's outright lies
I can trash it for you. But the usual defn has the epsilon in it. Is it equivalent?
$X_n$ converges to $0$ in probability iff for all $\epsilon>0$, $\lim_n P[|X_n| > \epsilon] = 0$.
So, by Chebyetchevved, $L^1$ convergence is stronger.
I should pay for my impatience.
19:56
Chebyetchevved is the weirdest transliteration of this surname
I wanted to make sure I covered all the bases.
I had a quick look for the Cyrillic bur was not sufficiently quick.
i know it's just a transliteration thing but i like to imagine that the various spellings are due to the heirs fighting over rights to the name and separately licensing the name for promotional purposes.
like when you go to the state fair and see al jardine and the beach boys "experience"
Chevychase was one of them
Something in my brain causes the most critical letter to be changed or omitted
I do have to say that the Chebyshev inequality is simple to prove, but it takes me ages to stumble upon it when proving something.
Does paying for your impatience mean I should not trash your gross stupidity? 😻🤷‍♂️
Yes :-)
I prefer the label impatient though :-)
20:03
Of course you do!
The "close it now" crew seems to be slacking as of late.
activity just seems down, generally.
must be busy fighting other battles
20:20
Hey chat
I'm back once again!
I am not here.
Parse: False
I do have an arse though.
Lucas only pretends to be here.
I'm not sure whether I'd pass the Turing test.
20:23
I never answer questions when I am turing around.
I do not see, in principle, why a computer cannot replicate a human's behaviour.
I'd like some mathematical advice (in the career sense): every time - I mean it, every single time - I try to prove that irrational rotations are dense in the circle I have a bad time. Since my first year up to my last year, and we're using this fact quite usually in the course of Intro. to Dynamical System
Isn't it a pigeonhole principle thing?
Yeah, and I kinda know how the whole proof goes
There are numerous proofs! I was shown several I hadn’t known by the students in my graduate differential geometry class at MIT in 1979.
Are they very different from the pigeonhole approach?
20:26
You prove the rotation is injective and somehow prove that the points need to be arbitrarily near somehow and then translate them by the number of times you're rotating them
But I ALWAYS miss the technical details
Point set. Measure. Elementary number theory. And one I can’t remember.
Does that make me a bad mathematician or just not fit to work with analysis?
It means you are holding yourself to an impossible standard.
It’s not really analysis.
(You can somehow see that somehow I don't really know how to prove it by the number of times I said "somehow" earlier)
20:28
ted and i agree again
@TedShifrin uhm, what would you call it?
convergence
pigeonology
haha
sincerely I hate pretty much everything that Dirichlet proved
it's that part of real analysis/number theory where everything is so tricky and you consider just the right intervals
you should establish an anti-dirichlet field of mathematics. dedicated to tearing down his legacy.
@leslietownes that's my brand-new project. thanks :)
hahahahaha
20:32
Dirichlet should be a chewing gum brand name.
@copper.hat that does comfort me.
that's another thing you could do, begin pronouncing dirichlet so it rhymes with the chiclet part of chiclets.
are y'all teachers?
Mathematics is just a depressing subject. All you learn is how much there is that you do not know and how much one can forget.
17
What was I saying...
I am not a teacher. I wish I was then I would have a retirement income.
@leslietownes Always a conundrum. A French name for a German.
@LucasHenrique I’m a bum.
20:44
I always thought that a conundrum was a musical instrument
I just voted to close a question. I have become one of them.
Part of the grinch horde!
I’m proud of this answer.
one side effect of people being gone is a lot of questions that maybe otherwise wouldn't be on the main page hanging around longer than usual.
proof by picture
You mean by thinking?
I mean the geometric picture is an effective means of communicating the answer and a general approach.
I feel mean sometimes when dragging folks through a solution process.
Its a beautiful day here atm
I managed globle in 6 this morning. Very happy at that. Especially after my recent 22 disaster.
21:05
I’m not doing so well with Anti-Wordle.
that one is not so fun for me.
crap, 3 for anti-wordle
I got 5 today. My best is 10. I am not so good at that.
21:22
My son has got 2 in wordle a few times. Usually my daughter is the word person in the family.
wordle in 2 is such a rush. i've done it twice.
Your son is the great guesser/poker player.
my daughter can tell me my score in wordle after a completed game. she can also name the letters i point at. she's not at the level of reading words or understanding the game.
I have 2 7/78 times.
Perhaps my strategy of always starting with matin is not the best. Surely some morning it will win.
21:25
I start with 3 vowels …
we play a domino-like game whose dynamic rules - placing tiles you have where they fit according to domino-like rules and drawing a new tile if nothing fits - she understands very well. but she doesn't understand the more abstract rules.
reminds me of the time when i missed the bart train and my 3 yo daughter learned a new word which she repeated most of the way home when we did catch the train, much to the amusement of the other passengers.
e.g. if there are no tiles left to draw, and no legal moves, an abstract rule determines who 'wins'. she wants to keep playing, even if she won.
playing is probably a better focus than winning atm
wait, you're a lawyer...
well, yeah, but then she should calm down when i collect the tiles and we start a new game. she wants to keep playing with the tiles she has even though there's no way to do that.
it's a very interesting moment in her development. a lot of stuff is right around the corner.
21:29
i read a book yonks ago called the society of mind my marvin minsky, i found it helpful in terms of kid's development.
another one called mindstorms by seymore papert
@robjohn nice egg :-)
both of my kids got infinite value out of a little toy i got for them in the cancun walmart of all places. it had a collection of (plastic) pretend food, everything from vegetables to pretend muffins. same for a little pretend doctors kid. that was helpful when they actually did go to the doctors.
@Copper Both MIT AI folks.
@copper.hat Thanks. I've been accused of being an egghead by better people...
Better than copper? Is that even possible?
silver.hat, my arch nemesis
21:44
Ah, gold.hat must be just around the corner.
:-). gold is not quite as conductive, so it is easier for aliens to scan my little brain
@copper.hat I used to work with Oliver Steele who was Marvin Minsky's nephew, I believe. Once, he was in town in Cupertino and we had dinner with Marvin.
I just looked it up, Oliver was going with Margaret Minsky at the time. They are now married.
So Oliver is Marvin's son-in-law
we worked together on QuickDraw GX.
Going to get food. BBL
22:06
@robjohn very cool!
that was when jobs got lost in the nExt thing
22:48
I think I am gradually convincing myself that the set (class?) of $\mathscr E, \mathscr B (\bar{\mathbb R})$-measurable functions is closed under pointwise limits. That is, if $(f_n)$ is a sequence of $\mathscr E, \mathscr B (\bar{\mathbb R})$-measurable functions, and the pointwise limit $f$ of the sequence exists, then $f$ is an $\mathscr E, \mathscr B (\bar{\mathbb R})$-measurable function.
Question: why is this result important? I have a feeling it's related to the Lebesgue integral, but I could be wrong.
to be useful with limit operations ($\sup, \inf, \lim, \limsup, \liminf$, differentiation, etc.) you need the limiting function to be measurable. It is a natural extension if the requirement that if $A_n$ are measurable sets then $\cap_n A_n$ is measurable.
You mean applying limit operations to the limiting function? Or to the sequence of functions? Because the proof I've been reading shows that sup, inf, limsup, and liminf are themselves measurable (roughly speaking), from which it's deduced that the pointwise limit is itself measurable, when it exists.
I mean if you apply one of those operations to a sequence of measurable functions.
I would encourage you to think in terms of sets first, and then characteristic functions, then linear combinations thereof and finally limits of these. It is a natural progression that you will encounter many times in measure theory.
In particular, the measurable sets are the 'germs' of the general progression.
23:06
Oh, well that seems weird. Because the proof I've read defines functions $\inf f_n \colon E \to \bar{\mathbb R}$ by $x \mapsto \inf \{ f_n(x) \colon n \in \mathbb N \}$ (and similarly for the other functions), shows they're measurable, and uses this to conclude that the pointwise limit is measurable. So unless I've misunderstood, it seems backwards to say that you need the pointwise limit to be measurable in order to appy those operations to the sequence.
All sorts of things follow from the limit result, measurability of the set of points of continuity of a function, the measurability of the Banach Indicatrix, etc.
@copper.hat I have seen this. However, I'm also cognizant of a book by Lang I'm planning to read, where he allegedly does all of this in Banach spaces (i.e. without using any order properties), so I am curious how he does that.
You lost me
To some extent, the whole point of measure theory, at least from an integration perspective, is to handle limit operations.
This is why Lebesgue integration is so much easier to apply than Riemann integration.
@copper.hat I will post screenshots from my book to try to clarify what I'm saying.
$\inf f_n$ is a single function.
23:11
@copper.hat Yeah, in 1997, our whole department was let go.
@robjohn That's tough.
What I'm trying to say is that this proof defines functions $\inf f_n, \sup f_n, \liminf f_n$ and $\limsup f_n$, and shows that they're measurable, and from this concludes that the pointwise limit is measurable.
ok, i know this, you were asking 'why is this result important? ' and i was trying to address that
my analogy was with sets, you need $\cap_n A_n, \cup_n A_n$ etc to be measurable, the functions thing is an extension of this
if you want intuition, it is easier to look at sets and their indicator functions, imo
@copper.hat Indeed. From 1989 to 1997 Apple had paid for my flights to/from LA, my hotel, and rental car when I was there 1 week a month. After the big layoffs, they found out they needed someone to maintain the code we wrote that was going to be used in the next Mac OS, but they were not willing to pay the travel, room, and car. To pay for all that would more than make up for the better pay that Apple was offering than down here. Not to mention the disruption of going up there every week.
23:16
@copper.hat Yes, but unless I've misunderstood, your reason why this result is important seems backwards somehow.
(and presumably I have in fact misunderstood)
@robjohn wow.
@Novice to be broadly useful, you need the limit functions to be in the same generally useful class of, say, measurable functions. It is analogous to completeness in a metric space.
i want with reasonable functions to be able to do $\lim_n \int f_n = \int \lim_n f_n$ and similar. if $\lim_n f_n$ is not measurable then i cannot integrate it.
Right, so this is important for something like Fatou's lemma, and maybe other results too.
probably the single most useful result is the DCT
Yes, so it's related to things like moving limits in and out of integrals. I think I'm starting to get an idea now. Thanks.
again, i would think in terms of sets and measure first, integration is one step away from this.
23:27
Well, I think I got what you wrote earlier, about how since countable unions and intersections of measurable sets are measurable, it's natural to want the limit of a sequence of measurable functions to be measurable.
Thanks for your help.
robjohn, I got full marks on that integration problem we discussed. Thanks.

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