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00:01
Does anyone want to buy my theorem for $9.99
you should raise your prices.
Here's the idea if you want to do it right algebraically. I told you this once before. If $(x,y)$ is in the $\delta$ ball, then $|x-a|,|y-b|<\delta$. So what's the smallest $y-x$ can be? $b-a-2\delta$? Now pick $\delta$ to guarantee that’s positive.
I was actually looking at the components, individually before I tried all the brouhahah, but didn't see how I could get $x, y$ together
The ball and the tilt make using the Pythagorean distance awkward.
Understand where I got that.
not immediately, but I'm looking at it
00:07
That's in line with triangle inequality skills you need for analysis.
Maybe @leslie will approve.
don't "look" at anything. that's what geometers want you to do.
joking aside, absolutely yes.
I removed geometry! Look at it!
Smallest $y-x$ is smallest $y$ minus largest $x$. @dc3rd must understand that!
I was going through the individual inequalities and then was going to combine them
OK
But understand my final statement, ultimately.
Will do....because as you might know from all your teaching experience. I never thought of it in terms of "smallest $y$ minus largest $x$ consciously. It usually would fall out a bunch of algebra, but being aware of it is more helpful.
00:18
That's why I enunciated it clearly for thee.
00:38
$y - x > b - \frac{\sqrt{2}b}{2} + \frac{\sqrt{2}a}{2} - (\frac{\sqrt{2}b}{2} - \frac{\sqrt{2}a}{2} + a) = b - \sqrt{2}b + \sqrt{2}a - a > a - \sqrt{2}a + \sqrt{2}a - a = 0 \Rightarrow y > x$

Smallest $y$ minus largest $x$........................ mathematics....bloody hell. 😓
@TedShifrin I cleaned it up a bit: not sure it helps much. cstheory.stackexchange.com/q/48736/61870
No worries if you don’t have time to have a look
This isn't right @dc3rd. You need a smaller $\delta$.
You made a serious mistake with inequalities.
Also, factoring is your friend.
Noooooo..... 😭.........I had chosen $\delta = \frac{\sqrt{2}}{2}(b-a)$, then expanded individually $|y-b|, |x-a| < \delta$ and looked at what the "largest" and "smallest" $x,y$ values I could use.
also with the idea that $b>a$ since the point is in my set
With that value, if you do arithmetic correctly, it can go negative.
My optimal choice for $\delta$ does not work without the geometry.
Find your algebra error. It's important.
ok. and I was going to go on to the next, but your stressing this point so I'll take it to heart.
a smaller delta you say.....hmmm.
00:54
Find the error in what you did. Important mistake.
well this distance of $||\mathbf{w}^{\perp}||$ is actually the distance to the $y = x$ line exactly, so if I choose something smaller from this I should be able to find something that works.
You certainly did not get my $b-a-2\delta$. For the fifth time, find your mistake.
I see what you mean about negative values could come about.
01:12
Simplify $\arctan(1/F_n)+\arctan(1/F_{n+1})$
@DogAteMy Make me.
$\arctan(1/8)=\arctan(1/13)+\arctan(1/21)$
Coincidence?
I don't know if this is what you meant @TedShifrin, but using the idea of $|y-b|, |x-a| < \delta$, after expanding and combining based on the idea of smallest $y$ and largest $x$, I arrive at $b - a - 2\delta < y-x$. As you hinted to. So I need this to be positive in order for everything to be satisfied. That means I need

$0 < b - a - 2\delta$. Rewriting I end up with $\delta < \frac{b-a}{2}$. So $\delta$ would have to be smaller than that to work
Curious now?
Yes, @dc3rd.
Go on, DogAteMy.
01:22
Go on what
So I'm going over all of the work I did......and I'm trying to attach the geometric idea I used with regards to $\mathbf{w}^{\perp}$. When I initially chose my delta I chose it based on the distance I got from $||\mathbf{w}^{\perp}|| = ||\frac{1}{2}(b-a,a-b)|| = \frac{\sqrt{2}}{2}(b-a)$ (after the algebra).....but this whole idea didn't show up in the thinking I just did......where's my disconnect?
The best $\delta$ requires geometric reasoning. I've said this all multiple times and I'm tired of typing on my iPad, so I'm done.
never look for the best delta unless someone is paying you to do it.
(It depends on the parity, turns out: $\arctan(1/F_{2n})=\arctan(1/F_{2n+1})+\arctan(1/F_{2n+2})$. Every other one works(
I see.....so in my math thinking I've always been under the impression that geometric reasoning and algebraic reasoning go hand in hand. So I should be able to get one right from the other without much issue. I guess my thinking is wrong on this @leslietownes ? I'll leave ted alone, I've grinded his gears enough for the moment :p
i should step back a bit. most of the stuff i say about algebra and geometry is, frankly speaking, shit.
there is a game in which most of what people want to do is a consequence of the triangle inequality. a lot of analysis is about that.
oh dear....I've broken the chat with my pedantry....... I know the triangle inequality game. I suppose I just have to be more at ease with the rules of engagement.
i should stop speaking as sarcastically as i do. there is a game. the game is getting the triangle inequality to do what you want it to do. it rarely has anything to do with the best delta and it is easier if you can just teach yourself to be dumb and find some delta that works for the inequalities.
you literally never want to optimize delta. ever, ever, ever. unless that's like your research specialty, like that guy korneichuk.
everyone else, you just forget about what's best and find something that works.
exact constants in anything are a whole research field. that is what you are doing when you try to find the best delta.
I didn't think it got that deep to find the best delta. I guess that is optimization theory?
there's verifying that things have the properties we want the things to have, and then finding the best delta for the given epsilon. which is definitely its own kettle of fish and sometimes called approximation theory.
01:50
i don't mean to discourage trying to find the best delta given an epsilon, sometimes that is the simplest path to finding a delta that will work. but please feel free to not look for the best one. you only need one!
@dc3rd if you pick $\delta$ as the mid point of $x,y$, then you can see that $(\infty,x+\delta)$ and $(y-\delta, \infty)$ do not overlap. So, if $(x',y') \in U=(-\infty,x+\delta) \times (y-\delta, \infty)$, then $x'<y'$. Since $U$ is open you are finished. If you really must, you can note that if the Euclidean distance between $(x,y)$ and $(x',y')$ is less than $\delta$ then $(x',y') \in U$ and so $B((x,y),\delta) $ lies in the set containing the points $x<y$.
This is the largest such $\delta$, not that it matters.
now if I could only show that $(1+\sqrt{2}i)^n$ is not real for $n >0$...
02:05
you took the 1-d counterparts and then built it up from there didn't [email protected] ?, Interesting. I do get a picture in my mind with it.
one way of looking at is is to project $(x,y)$ onto the $x=y$ line and this forms the corner of the box.
yea, that's what I did...I actually posted a pic of the idea a few days ago.
02:35
Hello!
I hope you are doing well, safe and healthy. I have quick question please about equivaraint location estimator $m(X_n)$ of a matrix $(X_n)$
Given I had, $B(X_n)\Sigma B(X_n)^-1 = I_p$, where $\Sigma$ is the covariance matrix and $B(X_n)$ is the correspondong eigenvectors for $X_n$
and,
$B(X_n)\Sigma_4 B(X_n)^-1 = D$, where $\Sigma_4$ is the 4th momentum covariance matrix.
Do you have what could be the equivariant location estimator in this setup? Any equivaraint location estimator $m(X_n)$ comes to your mind please?
Super quickie (just on my lunch break doing math)...does $i$ behave like an ordinary symbol for algebraic purposes? Specifically if I divide both sides of an eq by $i$ does it switch sign?
why would it switch sign???
it is just a number.
I thought so
Am i missing something or is there a typo here (this is supposed to be a sequence )
you're missing that $1/i=-i$
It's right. What is $1/i$?
02:48
Apart from i the rest of those are real numbers and psi is a function
Thor gave it away.
Ah thanks!
$\bar{i} = -i$.
:D
Oh i see why
0.5 - 1
Cheers
Still plenty of time for qm before lunch ends
03:04
I found out that affine equivariant locations is simply the mean E[y]
I feel sometimes mathematics just like to confuse readers
there is a lot of poorly written mathematics out there. many good authors tell you everything you need to know and nothing more. which doesn't make it easy to read.
2
@TedShifrin Any hints for the groups of order 8?
@leslietownes. Does they mean to leave you wondering or this is how they explain or they like you to go around crazy!
*Do (crazy)
i think it is mostly laziness and not thinking things through.
:/
03:14
p-groups, which include groups of order 8, are interesting. i would ask what the center is. i would try to think of the thing as a semidirect product.
good authors give you absolutely everything you need to know. but it's a skill. and for some people in math, not giving away everything is part of the 'fun' or how they think math should be done. which is one thing on math.SE where withholding information sometimes serves a purpose. it's different in a book.
or at least sometimes it is.
sometimes it's just, ha ha, i know something you don't know.
which for some may be an effective pedagogical strategy. it depends upon how you respond to taunting.
one time in my freshman college physics class we had a guest instructor from the former soviet union. you may know how the educational system used to work there. he posed what was obviously a trick question with a fairly obvious answer. it had to do with the velocity of fluid flow in the volga river. nobody raised their hand. he said "in [my country], several students raise their hand." he didn't understand that he was testing something about how students are socialized, not knowledge.
in late 20th century america, if someone came in trying to fool you, you don't outsmart them, you just let them wait for someone to put their hand up. psychology, not pedagogy.
03:34
@leslietownes. right!
it was different decades ago. if you didn't try to outsmart people in the USSR, maybe you'd spend your live shoveling mud or something. now it's at a point where a lot of people in the USA think, with no basis for the opinion, that they will just somehow win the game because they're american, without outsmarting anybody.
it's not a bad thing to be smart, that's all i'm saying. if you can manage it, do it.
@leslietownes. agree
Surely that is the last step?
just tidying up the mathjax copper
err, how does one relate x_{k,i} - a_i to epsilon at all? surely other choices are going on. it sounds like you make N of them and somehow win.
03:46
I'm trying to component wise convergence implies full convergence of the vector sequence. I'm at the last step. So I got:

$||\mathbf{x}_{k} - \mathbf{a}|| \leq \sum_{i}^{n}|x_{k,i} - a_{i}| < n \epsilon$ (with the triangle inequality used in the middle of all that. How do I get to my last step ?
well I was going to choose $K = max(K_{1}, \dots K_{n})$ for all of the individual sequences
note that n is not varying here. you are welcome to use epsilon/n throughout.
ah.....so in my statement of the proof I can say "let $\epsilon = \frac{\epsilon}{n}$?
oh, that pains me greatly. given epsilon, feel free to put epsilon/n into the various limit things. but do not write that equality. it makes angels cry.
which inequality are you talking about specifically?
You don't need to end up with $\epsilon$. You can end up with anything as long as it is clear that you can choose your starting quantity to make the ending quantity arbitrarily small.
03:51
this is also true. a lot of books make this huge thing about making everything less than epsilon, it's so not necessary.
but i've dealt with enough people teaching out of those books that i've learned epsilonology.
so what you're saying copper is since we can choose $\epsilon$ to be as small as we like this is sufficient because of all the other stuff I wrote prior.
copper.hat is giving you the grown-up, analyst version. it's fine to end up with 3 epsilon or n epsilon or 5M epsilon as long as the things that aren't epsilon are fixed and do not depend on epsilon.
well this is a shock to the system.....I've been indoctrinated with always less than $\epsilon$....
that's what the sheeple want you to believe.
the whole less than epsilon thing is partly a pedagogical device to keep people from losing track of what depends on what. it probably wouldn't do to wind up with "< n epsilon" if n was somehow chosen depending on epsilon. so they are trying to inoculate you against that.
admittedly a lot of the epsilon-delta stuff lends itself to confusion. even the people who came up with it in the first place sometimes confused uniform continuity with continuity.
which in this case $n$ is not dependent on $\epsilon$. Uniform continuity is seared in the head: delta depends on nothing.
04:01
for extra security, given epsilon, put epsilon/(300 thousand billion n) into every other limit that is assumed to exist.
belt and suspenders.
you take no risks....
If X is a compact Hausdorff space and A is a subalgebra of C(X,R) which does not separate points, what is the closure of A in C(X,R) then?
my guess would be it's just something that identifies points of whatever A does not disambiguate. i don't know if that's a quotient or subobject or whatever. subobject, maybe.
You know those functions that take in two real numbers and "zipper" together their decimal expansions, with appropriate choices on the expansions. Do those sorts of functions have a proper name?
@epsilon-emperor Why must there be a normal cyclic group of order $4$?
04:06
zipper functions seem as good a name as any to me. i don't know of a better one.
@dc3rd the point is that you want to show that something can be made arbitrailiy small.
I wonder if there would be a way to quantify how discontinuous a function is
If you can show that it is less than $52\epsilon+ 5 \epsilon^2$ then you are good.
Oscillation?
@Rithaniel for what purpose?
04:11
oscillation is the first thing that came to my mind.
it's maybe not the best choice of nomenclature. but it does quantify that.
04:35
@copper.hat For the purpose of perhaps generalizing some theorems that are true for continuous functions to discontinuous functions, probably
continuity tends to be a yes or no kind of thing. there are degrees of it within the realm of continuity. i don't know about degrees of discontinuity. which is not to say that theorems cannot be generalized. i am most familiar with non quantitative stufff.
e.g. 'this works for continuous functions and piecewise continuous functions' and you choose the pieces.
I mean, you could probably say "continuous except on a set of measure 0," as another form of semi-continuity, but I'm kind of envisioning some kind of quantity which could allow the space of all functions to be metrizable
i can't think of any result that is in any way "if this is x% continuous then we can do it" vs. not. the lebesgue spaces L^p have many discontinuous functions within them and great theorems about them, but none of the hypothesis hinge upon continuity.
It probably a pipe dream, though
in my own mind there are continuous functions, piecewise continuous functions, and then a kind of noise of other functions which live in integration spaces but may not be continuous anywhere.
and then the functions you just see in point set topology books.
those are the four kinds of functions.
04:51
Well, on $\Bbb R$, you can classify discontinuities — removable, jump, otherwise (I forget the "kinds"). Can you make meaningful statements of the sort you are contemplating if you allow some of those (removable being most tame)?
05:04
Not really. Trying to come up with a few examples, just now, and they're evading me
 
2 hours later…
07:06
If I am not brain dead then to know column vector b exist in span of c we start by doing span(c)=b right?
say c contains $c_1*v_1+c_2*v_2+c_3*v_3$ where $c_1,c_2,c_3\in\Bbb{R}$ and $v_1,v_2,v_3\in\Bbb{R}^4$ b is a vector $b\in\Bbb{R}^4$
It is 4×3 linear system so the solution will not be unique or will not exist
lol wrong
it's overdetermined 🤦‍♂️
07:22
@AndrewMicallef nope I dont see why
07:36
Oh i got it, nvm. Need to multiply by $i/i$
07:49
may I have a small hint for this question? imgur.com/a/APIt0gQ , so far I have gathered that since $\pi_1(X)$ is abelian, if $f$ is not null-homotopic, then $f_{\ast}(\pi_1(X,x_0))$ is a non-trivial abelian subgroup of $\pi_1(Y,y_0)$, and the latter is a free group, so this must be $\mathbb{Z}$
im not really sure how to then relate that immediately to $S^1$ :/
I feel like after collapsing a maximal tree in $Y$ we can reduce the problem to $Y$ is equal to a wedge of circles
08:27
another reduction I can think of is that since $X$ is compact and $Y$ is hausdorff, $f(Y)$ is contained in a finite subcomplex of $Y$, and so we can actually think of $f$ as a map from $X$ to a finite wedge sum of circles
09:09
In Lemma 1 here (mathonline.wikidot.com/…), equality holds iff f is continuous right? I'm not exactly sure about one of the two implications.
09:22
@epsilon-emperor if f is continuous then you are summing a bunch of zeros
Oh yes LOL
If f has jump discontinuities at exactly the points x_1,x_2, so on
Then equality holds
Is this a double implication though
i.e. are these the only functions which lead to equality
Would anyone here like to see this math.stackexchange.com/q/1770453/333392 In this question the matrix used for transforming Cartesian components to spherical components is not totally correct. There should be factors of $r$ and $rsin\theta$ in the components as well. On this modified matrix is the correct Jacobian and will give the correct transformation laws.
But the matrix used in the question looks like the matrix when the basis vectors have unit component. That will not give the correct tensor transformation although there is this business of taking unit basis vectors in vector calculus. Is that the issue here as well
Could any one confirm
09:45
@epsilon they don't have to be, for instance f can be constant for a while without a jump discontinuity, something like $f(x) = 0$ for $x \in [0,\frac{1}{2})$, then $f(x) = 1$ for $x \in [\frac{1}{2},1]$ and the partition $\{0,\frac{1}{2},1 \}$ will give equality , but $0$ and $1$ are not jump discontinuities of $f$
Got it, thanks
10:29
Can a continous function on a compact have infinitely many points of extrema?
CONSTANT function?
oops caps
yeah, but aside from constant functions?
yes
constant over a subset
What about sin and cos?
he said over compact
10:49
@famesyasd consider $x\sin(1/x)$ for $0\lt x\le1$ and $0$ for $x=0$
Oh...what does that mean?
that will oscillate quicker and quicker as $|x|\to 0$
No, not that, over compact?
Over $[0,1]$; that is compact.
So pver compact means within a finite domain?
10:59
In mathematics, more specifically in general topology, compactness is a property that generalizes the notion of a subset of Euclidean space being closed (i.e., containing all its limit points) and bounded (i.e., having all its points lie within some fixed distance of each other). Examples include a closed interval, a rectangle, or a finite set of points. This notion is defined for more general topological spaces than Euclidean space in various ways. One such generalization is that a topological space is sequentially compact if every infinite sequence of points sampled from the space has an infinite...
In $\mathbb{R}^n$, it means closed and bounded
In real analysis the Heine–Borel theorem, named after Eduard Heine and Émile Borel, states: For a subset S of Euclidean space Rn, the following two statements are equivalent: S is closed and bounded S is compact, that is, every open cover of S has a finite subcover. == History and motivation == The history of what today is called the Heine–Borel theorem starts in the 19th century, with the search for solid foundations of real analysis. Central to the theory was the concept of uniform continuity and the theorem stating that every continuous function on a closed interval is uniformly continuous...
Thanks, sometimes wikipedia is too much information to digest
$[0,1]$ is closed and bounded, therefore, compact.
Open and closed covers, that rings a bell. I feel like i was reading about that recently but failed to digest the importance / payload
11:21
If I had this function $f(x) = Ae^i$ and I took the complex conjugate of $f$ would that be $f^*(x) = A^*e^{-i}$ where $A$ is a constant complex number and $A^*$ its conjugate?
yes. That is a constant function?
I see no $x$ on the right side of the $=$
11:39
Oh, my typo, there isan supposed to be $ix$ in exponent
Sorry, im on my phone half the time and the chat window is pretty odd
I never see what i type till after it is posted
11:54
and $x$ is real?
12:49
hey chat, good morning
25
A: Why do we use the Euclidean metric on $\mathbb{R}^2$?

Alex BeckerThe Euclidean metric is special because it comes from what is called an inner product, and up to scaling it is the only metric that does so. This allows you to talk about angles between vectors in a sensible way, which you cannot do with other metrics. So really we don't choose to use the Euclid...

I know that inner products are good to talk about angles and distances. But is there a way to prove the converse assuming "minimal" hypotheses? If so, it's not hard to prove that the Euclidean distance is in fact $d(x,y)^2 = \sum (x_i - y_i)^2$ with a little linear algebra.
What I mean: if $d(x,y)$ is compatible in some sense with angles, then $d(x,y)$ comes from an inner product.
If a normed space has a norm satisfying the parallelogram law, then the norm comes from an inner product
if and only if
"if" being the much harder direction, of course
@LucasHenrique I see it as the most intuitive metric, a generalisation of Pythagoras theorem. However in reality it has no legitimacy as the "right choice", that will depend on the specifics whatever your doing.
13:27
@AlessandroCodenotti oh, yeah. that polarization identity
or something like that
@DanielAdams I mean, with some hypotheses, it "is" the right choice
for example, if you grant invariance under all orthogonal transformations, then by linear algebra the gram matrix of the inner product belong to the center of $\mathcal{M}_n(\mathbb{R})$
and thus $d(x,y) = t^2 (x-y)^T (x-y)$
by reality I guess im talking about 3-dim space $\mathbb{R}^3$ and time $\mathbb{R}_+$ which is curved and so doesnt take this distance.
oh, alright
13:51
@LucasHenrique it is translation and rotation invariant.
14:10
Could someone explain the first line of the proof? I understand everything else
Why is it enough to show that I_0 \cap (\bigcap_{n=1}^\infty G_n) \ne \emptyset?
I ask because: A set S is dense in M if and only if every non-empty open set intersects it
I_0 is a particular set, so not sure why the argument holds
A set is dense iff it meets every open set iff it meets every set in a basis, here they are using the basis of open intervals. Note that they are not doing this for a particular $x_0$ and $I_0$, it works for arbitrary $x_0$ and $I_0$ around it
it says I_0 is an open interval?
14:29
Ohh okay
Understand it now! x_o and I_o arbitrary means we have considered every open set
 
1 hour later…
15:31
there are a pair of mallard ducks in my front yard. i will keep the chat apprised of any changes to this situation.
In the UK the people would use "duck" as a greeting to one another
back in the day
e.g "how you doing duck"
@robjohn is that sufficient to prove that $d(x,y)^2 = t (x-y)^T (x-y)$?
15:46
the mallards are now in the neighbor's front yard. i can't tell if they're looking for food, or just futzing about
it's two male mallards, which is interesting. often you see them around females.
up to no good
i'm thinking of involving the long beach police department.
could be drones...
hi guys i know you've got a lot to do but there are these mallards, male mallards, kind of swanning around. on private property. they look like they might be acting with a purpose.
stand your ground
15:54
side note, the police department here is really weird. one time someone tried to break in while we were home. they only left after they saw that i was ready to stab them if they reached their arm inside the broken glass to unlatch a deadbolt. the police came an hour later. another time, a kid was drinking malt liquor on our porch for 20 minutes. we called the cops and like four SUVs showed up in 5 minutes. it was probably 5 hours of police time. they made him pour his malt liquor out.
there's no sense of proportion, is what i'm saying.
have had similar inconsistencies here in albany
i think it helps if you can connect your call to youths behaving irresponsibly.
i discovered that graffiti is not a cause for city concern unless it depicts genitalia in which case it will be addressed immediately.
there ought to be a separate, non-police number you can use to brush unwanted teenagers off of your porch. it's not a criminal thing, i just didn't want to get into it with someone and escalate a situation, which is probably what would have happened if i'd gone out there and not phoned somebody. i want somebody trained to de-escalate and shoo.
oh, that's quite funny
just buy a shoulder holster and casually stroll outside.
15:58
we used to live next to an alley and were always dealing with that. i should have invested in my own spray can. if it's not yet a problem i'll just supplement it with a depiction of testicles.
i am a fan of the broken glass theory of policing.
same with management in general.
our old house was something else. we always had weird people hanging out on our porch. it was close to the sidewalk, but very aggravating.
during the daytime?
yeah. it was not that great of a neighborhood. they were treating a bench on our porch like a bus bench.
sprinklers
one of those no soliciting signs with bullet holes in them.
spent cartridges
16:03
i did threaten somebody once. i told them to leave our porch, or at least stop smoking on it, and they said "call the police." i said "you're going to be the one calling the police if you don't get off of my porch." and got closer and conveyed an intention to throw them off of the porch. they left. i wouldn't have said it if they were bigger or visibly stronger than me, but they weren't. so hooray for me.
if someone did break into my house it's my wife they have to worry about. black belt in karate. some of it is goofy dancing and ritualistic but she broke somebody's nose once without even trying to.
it is aggravating to have to deal with such things.
the problem with such things is judging when to use excessive force.
i think overpolicing is sometimes definitely a problem. particularly in long beach, certain neighborhoods, the cops just punish anything. and our neighborhood it was more like camp do-whatcha-wanna. no rhyme or reason to it.
where we lived in ireland the nearest police (garda) were 30+ mins away at night, so anything that needed immediate attention was up to you.
but there were fewer layabouts around.
ha, i had a friend visit from germany. we were at my dad's house. there was a strange noise outside, and immediately several members of the family produced weapons. she had no idea what was happening. you don't need to be culturally one way or another to have weapons if you live that far from the police.
there were few guns in most homes in ireland.
16:11
you see this reflected in american law. a lot of law around creditors who have lent money that is secured by a property interest is very favorable to the creditor. "repo men" can mostly do whatever they want. in most states the test is whether what they do threatens a "breach of the peace."
even police in the usa do not seem to have appropriate situational training.
american law is admirably adaptive to circumstances. in a large city at night, walking into someone's driveway, no. in a rural area at night, walking into someone's driveway, absolutely yes.
yeah, that is annoying.
there's a dumb case where repo men repossessed a car that had a child in it, and the mother had no idea that they were repo men, she thought it was a kidnapping. the court held, no threatening of a breach of the peace. probably because the judge imagined it was the kind of neighborhood where people just do that.
different result in a rural area. perhaps it was race based.
i was at a shooting range a few years ago, and chatted with two groups, one were local police the other were active military. the active military were shooting as normal, the police were double tapping. i asked why, "to make sure we neutralise the threat'.
even a traffic stop is stressful here.
lights flashing, antagonistic enforcement,...
16:15
there's something toxic in the culture. and several members of my family are or were in police departments. there are good ones and bad ones, but the culture is, you don't tell on the bad ones.
i am not a defunder in case you are wondering, just a proponent of more community involved policing.
and this enormous fetishization of military stuff, military equipment, dumb expensive stuff that might have been needed for ISIS but probably not for main street.
don't get me started.
i mean really, double tap on the streets???
ok, off for my weekly 'run' in tilden.
i miss tilden. say hi to it for me.
:-) love the place. my mental health establishment
i love that puzzle.
This is actually the same as the Muddy Children puzzle
Best treated with Epistemic Logic
Check out "Reasoning about Knowledge" by Fagin,Moses,Vardi
They have a really cool discussion on this puzzle
Yeah nice puzzle
"It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics." i'd kind of like to see a solution to the puzzle that involves people doing genetics.
it has to be from first principles. first they identify the nucleic acid sequences that cause blue eyes, and then they use island technology to identify them.
the puzzle does exclude people with eyes of different colors. my wife has one blue eye and one hazel eye.
i don't know where she fits in
@leslietownes I think it is called heterochromia
16:26
another good puzzle is the one about prisoners and a lightbulb. most puzzles suck, but that one can stay.
words that sound the same but are spelled different are called homophobes.
it is called heterochromia, i just looked it up.
i'm thinking about an extended version of the puzzle where some islanders have heterochromia. could be a good generalization.
Um, homophones. Although people who hate homosexuals do sound the same.
They all sound like morons?
That's the best you might offer.
The best I can offer without getting banned
i can think of less delicate ways of describing them
16:38
Also hi @Ted @Leslie
lol
hey @Ted, are you familiar with Hopf surfaces?
good morning. the birds are chirping. the mallards have left the neighbor's yard. i do not know where they are now.
What's it like, 10am there?
Not intimately, @Thor. What is the question?
I was wondering how to compute their Dolbeault cohomology
16:44
Ah, to see that Hodge decomposition fails?
it doesn't fail as far as I've been told (though I've also been looking for an example of a complex manifold for which the Hodge decomposition fails separately)
but both vertical and horizontal symmetry of the Hodge diamond fail
We can look in $\Bbb C^2~0$ at forms invariant under $z\rightsquigarrow 2z$.
Ah, the obvious ones have poles.
I think this is somewhere in Kodaira's papers on classification of surfaces. Sadly, I no longer possess the complete works.
hmm, what's wrong with like $\frac{1}{z_1}dz_1$?
It has a polar divisor $z_1=0$.
That was the trap I fell into immediately.
But it's an invariant holomorphic form on $\mathbb{C}^2\setminus\{0\}$, no? $H^{1,0}$ should be $0$, so it has to descend to something exact on the Hopf surface, but I don't see where exactness comes from. Not sure how this relates to divisors.
16:56
For your info, though, here's an exercise I gave my complex geometry class 40+ years ago. I learned it from Kodaira. If $\phi_1,\dots,\phi_k$ are linearly independent holomorphic $1$-forms on a compact surface, then (a) they are $d$-closed; (b) $\{\phi_j,\bar\phi_j\}$ are linearly independent in $H^1(M,\Bbb C)$. Thus, $b_1\ge 2h^{1,0}$ and $h^{0,1}\ge \frac12 b_1$.
No, no, Thor, it's not holo on $\Bbb C-\{0\}$. When you pull out the origin, there's plenty of $z_1=0$ left.
Since $b_1=0$ for the Hopf surface, we know that $h^{1,0} = 0$ and $h^{0,1}\ge 1$.
@TedShifrin oh duh, I was thinking in C-0 and not C^2-0
Right, I know :)

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