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7:04 PM
@love_sodam huh? No!
Of course, your question might be ambiguous. Do you mean 0 everywhere?
Ola Ted, long time no chat
How's life with physics in the rear-view mirror?
Is that a question addressed at me? x'd
7:11 PM
If I interpret it correctly that you're asking me how life is after having abandoned physics, then it's better than it's ever been x'd
You interpreted correctly.
@ShaVuklia I have a string theorist friend. You know what he calls his cat?
Ah okay, ye it feels more like I've completely cut it out of my life, without ever looking back it at from a rear-view mirror x'd (Hence my confusion for a second)
Oh, wanna hear something that surprised me as a TA ;x
I actually like physics, but I only did 3 semesters at college.
7:18 PM
So I'm a TA for a first-year prob thy course, and ever since we've been dealing with (continuous) random variables, I have a weekly fear that my students will ask me how the things we're working with are properly, so I spend a lot of time refreshing my measure-thy skills, and while I'm enjoying that, I have yet to receive such a question (or remark) from a student x'D
Of course not.
@TedShifrin I would have like it if I was able to think as a physicist
None of my senior probability students 6 yrs ago asked anything advanced.
The regular stuff we did was hard enough for us all.
Well, it's not so much that I expect them to ask sth advanced (as in specific), but rather sth like "is this well-defined tho?"
They're beginners!
7:20 PM
Lol, ok x'd well, I'm ready if they ask, in any case x'D
It's good for you, regardless.
Ye, definitely x'D
7:46 PM
@feynhat what does he call his cat?
Edward Kitten.
(works better with physicists)
i never had to teach probability theory 'for real' although it was sometimes an adjunct to applied calculus courses. where the instinct was not to 'peek under the hood' but just grasp the language of it.
i should have guessed that.
that's funny except it gets less accurate as the cat grows older. maybe there's a string explanation for that.
i have no sense whatsoever of string theory. a friend of mine tried to explain it once, it didn't stick. he was patient and i think the explanation was probably very good. my brain just didn't light up in the right places.
2 hours later…
9:39 PM
@feynhat I was a math grad student at Princeton from 1981-1985 and Ed Witten was being talked about in the math department and I heard a lot about him from a physics grad student I knew. I don't know if he was there at the time, I think he went to Harvard after getting his PhD at Princeton.
. o O ( I talk to ghosts )
Howdy @robjohn. You started at Princeton as I started my 34-year stint at UGA. Would you believe I actually have never been back to visit Princeton since my college days? No one in the department would have seen fit to invite me (and quite appropriately so!).
@TedShifrin I have not been back since 1985, either. I didn't mind the snow in the winter, but I hated the mugginess in the summer.
I left as soon as I could when classes ended in the spring and waited until the last minute to go back in the fall.
there's actually a restraining order against ted. he can't go back. i have several of these too. whatever, i've been banned from better universities than berkeley.
If you thought Princeton was muggy, you should have tried 33+ summers in GA.
@TedShifrin There's a reason I've never been to Georgia.
9:46 PM
if you can eat meat the food in georgia is very good.
@Leslie Last time I tried to go into Evans when I was visiting Berkeley, it was a holiday and the building was locked. I sent a message on Facebook complaining to Ken Ribet (whom I've known for years).
Are you talking about Georgia, USA, or Georgia, Russia?
georgia, usa. i love southern cooking. my dad's mother was from mississippi, i have a weakness for it.
Never been to either, but I was referring to the USA
the summers are unpleasant. after iowa, michigan, and massachusetts i learned i'm just not supposed to leave california. no humidity please.
I was working some examples of interior points vs frontier points, before getting back to the question I was working on (which is driving me nuts), In particular I examined the set of integers $\mathbb{Z} \subset \mathbb{R}$. From applying the definitions, it seems like $int(\mathbb{Z}) = \emptyset$ and the frontier($\mathbb{Z}) = \emptyset$. Is this a correct assesment?
9:49 PM
my dad was supposed to go on some kind of friendship mission to georgia, the former soviet republic, and then like two weeks before the trip a coup broke out. that was the end of that.
dc3rd that seems right to me, bearing in mind that all of this is relative to a choice of topology.
no complicated topologies put on anything.
we called them boundaries. frontier is a gallic mistake that has been imposed upon the rest of the world via that imaginary guy bourbaki.
@dc3rd Isn't the frontier boundary $\mathbb{Z}$?
oh yeah, what am i thinking of. i'm thinking of where you can approach it by stuff that's not in it. frontier allows approaching by stuff in it.
i blame the french for all of this.
but for the frontier to be $\mathbb{Z}$ at least working with the definition and explanation @copper.hat had given me would mean that I can "approach" the frontier point from points in the set and from points outside the set
9:53 PM
my verbiage is a little muddled but i think robjohn is right. there's no need to use sequences of distinct points to get to the frontier.
@Leslie I am no fan of southern cooking, despite having lived there for 34 years. A few things I do like, but overcooked vegetables and fried everything — nope.
in this case the integers are always a distance of $1$ away, so I can't approach a frontier point from in the set
@dc3rd for nice sets, consider the frontier to be the boundary as you might imagine it the 'real world'. given that, what do you think the boundary of the integers would be?
not even the desserts? that's mostly where they shine. although i am not a huge fan of sweets either.
@dc3rd No, the integers are $0$ away from themselves
9:54 PM
just desserts?
when i make collards i do not over cook, and do not use bacon fat.
No, too cloyingly sweet, most of them, @Leslie.
ahhh, lard
i use a combination of olive oil, garlic, and lemon juice. it's not appropriate to any region it is just what works with me.
i love collard greens.
i do think southern cooking could do with using less pork/meat.
frontier, frontier, frontier. old habits...
10:00 PM
@copper.hat Well there is one of two ways I'm thinking about it, I'm thinking of the integers being a fixed distance apart, but any nbhd of an integer will also contain the reals.,.......so I'm going to say the frontier of the integers is the integers themself.
@dc3rd From Wikipedia: "The boundary of a set is empty if and only if the set is both closed and open", though no reference is given, I believe it to be true.
but what is this now about integers being $0$ away from themselves?
By "approach" I think they mean "in the closure"
take an open set around an integer. does it contain both an integer and a non integer?
@dc3rd The distance of any non-empty set from itself is $0$
10:02 PM
can you find an open set around a non integer that contains no integer?
guess then verify
@copper.hat yes, take half distances for instance
No one said a frontier point cannot be in the set. Indeed, this is an example of a set where every point is a frontier point and there are no others.
then that would mean the integers cannot be a frontier for itself
I'm going to set a bot on this room that changes every occurrence of frontier to boundary ;-p
10:04 PM
it means it CAN
i'll start using $\partial$ :-)
let me back up for a sec
@robjohn: I ordinarily say boundary, too, but I explained to copper that I specifically chose frontier here because I have found that students get super-confused when we get to manifolds with boundary. Typically, the whole manifold (and boundary) consists of boundary points as it sits in $\Bbb R^n$.
The joys of embedding.
just setting my boundaries...
the right thing to do would be never to get to manifolds with boundary. do we need them? i don't.
10:09 PM
@Ted if you ever come back to LA, we should go to Disneyland and visit Boundaryland.
okay so this is what is confusing me in this example: the definition of frontier point of a set $S$ is "every nbhd of $a$ contains both points in $S$ and points not in $S$. "

So in my mind I am picturing a number line, with the points as you would expect for the integers. I can take nbhds that don't contain any points of the integers except for the point $a$ itself. So are you saying the point $a$ itself is sufficient to satisfy the definition?
a neighborhood of 1 includes both 1 itself and stuff like 1 + e with e small that isn't an integer.
@dc3rd i don't understand what you are asking
this may have become somehow blended up with the definition of limit, where your 'neighborhoods' are punctured neighborhoods. that's not this.
there are only two classes of points to consider, an integer and a non integer.
pick one and decide if it is in the frontier or not.
10:11 PM
that's where my head was for a minute. if it wasn't confusing, then mathematicians couldn't earn the megabucks teaching this stuff. and oh, the royalties on the textbooks.
@leslietownes Sound of nail being hit on the head.
if it was like copyright the descendants of Fourier, etc would be making a fortune...
@robjohn Would you believe I've never been to Disneyland? I went one day to Epcot in Orlando, I think.
wait, my proof of Parseval is slightly different in Paragragh 4.11(ii)(a).
the US term is not quite that long but yes.
ted can pilot his solid gold yacht to long beach and we can all go to disneyland. i hear they have a version of "it's a small world" where it's just mathematicians from different countries and they teach math to you.
10:13 PM
I was in Epcot and had free tickets to Disneyworld & Disneyland (DAC) but gave them away.
before i had kids
i went to disneyworld once. i became violently ill almost immediately and spent the day in their infirmary. the cots were sub par. it was frankly a disappointment. and mickey didn't visit, or anything.
not a fan here.
You said that the integers are a frontier for itself. So let's take an integer, $a$ and draw a nhbd around it small enough so that the only integer in the nbhd is the centre of the nbhd. Does this then make it a frontier point?
how can you do such a thing? a neighbourhood is open.
that neighborhood will still contain both an integer (namely a) and nonintegers (namely stuff close to a). these are not punctured neighborhoods.
10:17 PM
Leslie is capturing the idea I'm trying to get across
any nbh of an integer must contain non integers. that is the point.
I agree, but I was not including the "centre" of the nbhd. I was thining of it in terms of the punctured disc as Leslie mentioned.
A frontier point ("boundary point") need not be a limit point, folks.
sry for the bad pun, it is in my nature
i get why this is potentially confusing because often in limits you don't consider the point at which you are taking the limit, at all, and you insist on only using other points. but in evaluating the boundary, that's not what you do.
10:19 PM
light humour is needed
Actually, I did not define limit points in my text. That's going too far into topology. Of course, I defined closedness in terms of limits of sequences, so the notion is there.
I was looking for limit point in the index too
Almost as if the definition of limit isn't quite right and you'd be better off considering the point as well
maybe it would help to think about how to think about a point of the closure of S could fail to be in the boundary of S.
but back to this...OK, this brings a bit of clarity to me. I was thinking about it wrong by not including the point itself
10:20 PM
This does make my point why Rudin's book is overkill; he does way more point set topology in chapter 2 than is needed for virtually 98% of the book.
sometimes less is more
from a pedagogogogogogical perspective
This is bad — if copper, leslie, and Ted are in agreement, the end of the world is nigh indeed.
you (ye in proper Irish English) might be spared, my rear tYre has a punctured neighboUrhood in the inner tube that needs to be completed.
10:22 PM
i'm going to get to work on some truly cranky opinions. this cannot be tolerated.
@TedShifrin Oh, one of the areas in Disneyland is Frontierland.
LOL, I wondered :P
wife to daughter: "do you want to go outside for a bit?" daughter to wife: "my mom said it's freezing out there." wife to daughter: "uh, i am your mom."
@dc3rd: Now that you're feeling better, did you understand why with $S=(a,b)$ the interior of $\bar S$ is just $S$? You kept claiming to prove otherwise. Do we have a proper example of an open set where a point of the interior of $\bar S$ is not in $S$?
@leslie Is your daughter already at the third-person stage for parents? How precocious.
@leslietownes maybe your daughter is practicing for depositions?
10:26 PM
No Ted, that's how I got to the example of integers. I'm trying to find some more elaborate examples. Can't lie it is messing with my head.
Wait 'til you get to the "Mommy says it's OK if it's OK with Daddy." "Daddy says it's OK if it's OK with Mommy" game ...
i'll wait for the day when my wife asks her something and she says "on advice from my dad, i will not answer that question."
I just made it one step easier. I said open set.
@dc3rd to be frank (which is odd, because i am joe) i think you need to focus on basic examples first?
@TedShifrin fixed point parenting.
@copper Just so long as you're not ted.
10:27 PM
oh but Ted I said $S = \overline{[a,b]}$ not just the open set $(a,b)$
@TedShifrin infinite recursion parenting
integers is not a basic example @copper.hat ?
There was a wonderful Foxtrot comic strip with that mommy daddy interchange. I put it as an exercise in the logic review of my algebra book.
@dc3rd are you comfortable with determining the frontier of the integers?
@dc3rd $[a,b]$ is already closed. What's the point?
10:29 PM
what is the frontier of the complement of the integers?
yes I am now. Taking into account that I should also consider the centre point and not look at it as a punctured disc.
If you have $S=\bar S$, then of course any interior point is a point of $S$.
i'm glad we are using the disc terminology.
i'm still struggling with tail fields.
the frontier of the complement to the integers is the integers
yes. good.
if you feel like stretching, show that the frontier of a set is the same as the frontier of the complement.
which is not too surprising if you think of the frontier as a 'fence' as i do :-).
10:35 PM
Am I just stupid, or are linear programming problems as hard as they seem to me?
hard to know really.
we never did them at school, and whenever I get into a situation where I need to maximize a linear form subject to linear constraints, I feel like it should be the easiest thing ever, but I end up filling a page and getting nowhere
there is a reason that they are usually solved on computers.
I taught linear programming at the end of a linear algebra course once. It was not hard. But the duality principle was very powerful.
for small ones there are a few tricks.
10:36 PM
so if I have say, two linear equality constraints and one inequality constraint, and want to maximize a linear function subject to that
in n (unknown) dimensions
should I expect that to be annoying to do in closed form?
I just read the second part of your post about the interior point of $\bar{S}$ not in $S$......I'm still thininking about it.
getting a closed form really depends on the specifics.
@dc3rd: I'll offer one hint. $\Bbb Z$ has a property that's quite different from the interval $(a,b)$. One is that it's "discrete" — each point has a neighborhood consisting just of itself. But what's another big difference?
maximising $x$ subject to $x \le 10$ clearly gives $10$, but it is easy to given examples without any obvious answer.
the problem is to maximize/minimize $\sum_i \lambda_i^{-1}\beta_i$
given $\sum \lambda_i \beta_i$, and $\sum \beta_i$
and the $\beta$-s are all positive
so there are some relationships between the coefficients
10:39 PM
and you are minisiming over the $\beta_i$s?
Funny you mention $\mathbb{Z}$ because it is from your original question where it came up for me to work with it specifically.
i mean in general there is no closed form solution.
How so? I was just piggybacking off your long discussion here.
Oh, 8b made you think of $\Bbb Z$?
in that particular example it may be easy to enumerate the extreme points and figure a solution.
10:41 PM
There are plenty of other examples for 8b.
Well when we talked yesterday, I went back and just tried to fiddle with some other sets. And then I said to myself "self" find the frontier and interior of $\mathbb{Z}$.
yes I did use integers for 8b
Ah, OK. It's perhaps the most obvious example for 8b, but what's another example? This is an important notion for later. Say in two dimensions. Not just a discrete set of points.
prove that the frontier cannot contain an open set.
that's probably a better way than what I'd been trying so far
was wondering if anyone can help me with a frobenius method problem
10:44 PM
I was wondering when I would have to go into higher dimensions.....not just a discrete set of points....well there goes my idea of discrete coordinates...
yeah :( wanted to make sure I got it right
The direction I was trying to lead you for 9a doesn't require any more dimensions. Working in $\Bbb R$ is fine, but for general purposes I wanted you to give me a (non-discrete) set in $\Bbb R^2$ every point of which is a frontier point.
diff eq is y''-2xy'+(mu)y=0 where mu is some fixed parameter >0. I got the recurrence relation a_(n+2)={ (a_0 (mu-2n)) / (n+1)(n+2) } and im not sure if its correct. I used the frobenius method
Damn, I haven't seen that stuff since 1974, when I TAed sophomore DE my first quarter in grad school.
10:48 PM
THis just came in my head, but if I restrict my set to being just the $x,y$ axes in $\mathbb{R}^{2}$ I could imagine that being a set where every point is a frontier point
There you go. Or just any subset of one axis.
sorry my bad I meant a_(n+2)={ (a_n (mu-2n)) / (n+1)(n+2) }
OK, good, @dc3rd. Back to 9a. Thinking of subsets of $\Bbb R$. And perhaps thinking about my query about how $\Bbb Z$ is different from an interval.
@corgilover123 i get $a_{n+2}= {2n-\mu \over (n+1)(n+2)} a_n$, but could easily have made a mistake.
I'm thinking about the 2nd difference of $\mathbb{Z}$ being different from $\mathbb{R}$
10:53 PM
but since the $\mu$ is on the same side as the $y''$ i suspect my sign is correct.
@copper.hat someone had also said its 2n-\mu but im not sure why mu is negative instead of the 2n
Can you decompose $\bigoplus_{x \in A^{z + 1}} \Bbb{Z} x$ inductively where $A$ is a finite set and $A^{z + 1} = A \times A \times \dots \times A$?
What is the letter $z$? Just an integer?
Yes an integer $\geq 0$
the balance equation is $(n+1)(n+2) a_{n+1} - 2 n a_n + \mu a_n = 0$ which follows directly from the ode.
10:54 PM
So what kind of decomposition are you talking about?
@corgilover123 wait, i think i made a big mistake
I get the relation a_(n+2) - 2*na_n + \muna_n = 0
I just want to write something like: $\bigoplus_{x \in A^{z + 1}} \Bbb{Z} x = (\bigoplus_{x \in A^z} \Bbb{Z} x) \oplus \text{(something)}$.
@corgilover123 no i didn;t
oh no I made the mistake
sry I made a silly mistake with the signs
10:56 PM
your original answer was correct except for the sign, i believe.
thank you very much @copper.hat
i am not taking responsibilty for homework answers :-)
So if $A$ is a single point, we just have $\Bbb Z^{n+1}$. (I refuse to use $z$ and $\Bbb Z$ at the same time here.)
if I were asked the find the radius of converge, would I just do the ratio test for each solution separately?
@copper.hat :)) youve helped thank u
i'm off for my mental health ride... needs to be really long...
10:57 PM
safe journey
Ride well, @copper. I was feeling very nostalgic for Berkeley today, as it was foggy as I went north to the Farmers Market. Very nostalgic.
Why this level of abstraction, @StudySmarterNotHarder? Why do we just not work with cardinality of $A$ and the cardinality of $A^n$ and say we have the appropriate $\Bbb Z^N$?
Why think of $\Bbb Zx$ with $x\in A^n$?
Maybe I'm missing some important idea.
I think you mean choose any set $B$ so that $|A^z \cup B| = |A^{z + 1}|$.?
I'm not coming up with anything special about the other property Ted, the only things that came to mind were each point is an equal distance from each other. The other thing was the integers are not dense, but we're not in that realm of thinking and it is not the clearest idea for me yet.
I'm just wondering why you care about the structure of $A^z$ rather than just keeping track of its cardinality.
Think about traveling in an interval or traveling in $\Bbb Z$, @dc3rd. What leaps to mind?
traveling in an interval is smooth, but traveling in $\mathbb{Z}$ would involve jumps.
11:02 PM
Hello everyone. Is it true that a function satisfying the hypothesis of the Schwartz reflection principle, has a unique analytic continuation?
@geocalc33 what's up?
@geocalc33 I got banned for two days, that's where I went all of a sudden
OK, @dc3rd ... leaps rather than bumps.
So does that suggest an idea?
@newUser Depends on the domain, doesn't it?
The extension from $D$ to $\bar D$ is unique, but can we analytically continue to all of $\Bbb C$?
@TedShifrin Thanks. Would the answer be affirmative if $D=\Bbb C^+$, where by $C^+$ I mean the upper-half complex plane ?
Well, in that case, when you do Schwarz reflection you have extended to all of $\Bbb C$.
Any two analytic functions with the same domain that agree on a set with a limit point must agree everywhere, right?
yes, right
11:08 PM
So, yes, if you can extend by Schwarz reflection, that extended function is the unique analytic continuation to $\Bbb C$.
then I don't understand the discussion of this paper arxiv.org/pdf/2011.11708.pdf in Appendix E, page 87-88. They have a function p(p0) = I \sqrt{(m-p0)(p0+m)} and they say that they can consider two different analytic continuations. here the complex variable is p0
I was thinking to apply the Schwartz reflection principle to the function $-I p(p0)$, since one of the hypothesis is that the function must be real on a subdomain of the real axis
There are two branches of the square root function here.
You have to make branch cuts.
what do you mean exactly?
the function $-I p(p0)$ is analytic in the whole upper-half complex plane, and it is real for $|p0|<m$. Therefore, I thought the analytic continuation is analytic in $\Bbb C$ minus the branch cuts
I'm not looking at the paper, and the notation is disgusting, so I won't. I'm saying that this is not about Schwarz reflection. It's about the point that you have make a branch cut in the plane to define $\sqrt z$ (e.g., you remove the negative real axis) and there are two different branches — i.e., two different holomorphic functions on the domain.
When you do something like $\sqrt{1-z^2}$ it's the same idea, but the branch cut either goes from $-1$ to $1$ or else has two rays from $\pm 1$ to $\infty$.
Regardless, there are two branches of the square root function.
That's what they're talking about with "two different analytic continuations."
what is the "leaps" idea supposed to be getting at?
11:15 PM
There's a gap in the set, isn't there? It's not a matter of a bumpy ride. It's a matter of having to leap to get from one point to the next. You have to leave the set.
@TedShifrin thank you. So, If I define $\sqrt{1-z^2}$ with the branch cut for $|z|>1$, is it true that there is only one analytic continuation?
There are still two holomorphic functions. You have to choose one of two values at one particular point, and then the rest of the values are determined. I don't consider this an analytic continuation issue. It's a branch issue.
@TedShifrin ok, thank you very much.
Good luck :)
I kinda think I see what you're hinting at, but let me attempt to unravel it more here. You had also mentioned that the nbhd's of the points of the integers are the points themselves. I also know that the frontier points of the integers are the integers themselves.
11:20 PM
That's not the property I'm interested in now.
I'm trying to get you to think about different sorts of sets ... not just intervals.
so are you saying that if we take the closure of $\mathbb{Z}$ the interior points of it can be the frontier points?...I may be thinking in circles..
Forget about $\Bbb Z$. Think about the need to leap.
as you said leaping makes me leave the set.
So give me other examples of sets where leaping is required.
well anything that is a subset of $\mathbb{R}$ would require leaping.
11:29 PM
does anyone have any experience with real analysis integration?
You told me that an interval did not.
sorry, I should be more clear..... I meant $\mathbb{Z}, \mathbb{Q}$ also require leaping.
Keep going.
the irrationals (does it have a notation?) $\mathbb{Q}^{c}$ perhaps.
the only set that doesn't require leaping would be $\mathbb{R}$ itself.
11:31 PM
Come on.
We just agreed an interval does not.
I could also mention $\mathbb{N}$
Starting in an interval, I can leap to something else. Damn, you refuse to open your mind. You keep being stuck in the same rut.
I'm being loose in when saying the "only set"
that is my bad and I should be more exact with my language
Forget integers, rationals, ... for now.
Focus on intervals (remember I said OPEN sets hours ago).
Hi. Anyone here?
Is anyone here?
11:42 PM
I suppose? Kinda-maybe here
So to bring everything in focus for me again. I am trying to obtain a counter example to the claim that all interior points of $\bar{S}$ are points of $S$....so using this suggestion of focusing on open intervals....
@Quin Ok
Keep reviewing. What else have we been talking about most recently?
interior points, frontier points, open/closed sets, convergence
11:49 PM
most recently we talked about the integers and its frontier. Also its differences from $\mathbb{R}$, which are its discrete and now also leaping in and out of the set, which make sit different from an interval
Leaping. Yes. Forget integers, etc.
Leaping + open sets
the thing that first comes to my mind with this is unions of open sets. since I am leaping from one open set to another
OK, give me a concrete example.
$(0,1) \cup (2,3)$
One leap will be enough.
Now test out our question.
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