00:00 - 22:0022:00 - 00:00

12:04 AM
I cannot remember what is true about actual Hilbert spaces versus what we sometimes call Hilbert spaces but actually aren't. Do $U$ and $U^\dagger$ always have the same domain in a Hilbert space?

12:14 AM
Hello!!
I have posted a question about expected value. I have the calculations but I miss something since y result is negative, which cannot be. Could you take a look at the edit part of my question?

Your payoffs are incorrect. If you get 0 matches, you should lose 1 euro, not have a payoff of 0. The way you have structured the game mathematically, the player wins either 0 euros, 1 euro, 2 euros, or c euros. So if c is also positive, then the player can never lose; at worst they 'push.' That's why you're getting that 'c' has to be negative to have 0 expected value.

12:34 AM
So do we have that if none of the dice shows the number that we have chosen then the profit is Payout−stake=0−1=−1 euro ?
Then at the expected value is the term $\binom{3}{-1}$ defined? Is it defined with negative numbers?

So is it correct that I get a negative number for $c$ ? What do you mean by "push" ? @KevinDriscoll

Yes if none of the dice match your number, the payoff is -1.

I do not think you understand where these binomial coefficients are coming from / what they represent. The small x represents the number of dice that match your chosen number, not the payoff. So we don't need to worry about what happens if we try to take 3 choose -1.

"Push" means that you get your stake back. Neither win nor lose money.
Is the correct answer for part c of your question a negative number? No, I don't think so.
Given the way that you set up the equations, did you have to get a negative value for c? Yes.

kevin: if you're a physicist, not necessarily by definition, although, a lot of the time for operators of interest.

Kevin used to be a regular in the anon/Pedro days, before leslie and copper showed up.

the good old days, people call them.

So at (b) my equation is correct, isn't it? At (c) we have to replace only the coefficient by "c" not all the "3" by "c", right? That is my mistake, right? It should be $$E[X]= 0 \Rightarrow 0\cdot p_X(0)+1\cdot p_X(1)+2\cdot p_X(2)+c\cdot p_X(3)=0$$ right? @KevinDriscoll

12:50 AM
Indeed, I have been sometimes, back when I was but a lowly grad student. I expect to be treated much more harshly now that I'm just some rando with a PhD.
Nope, I think the payoffs are still incorrect.

Oh yes at (b) we have $$E[X]=(-1)\cdot p_X(0)+1\cdot p_X(1)+2\cdot p_X(2)+3\cdot p_X(3)$$ and at (c) we have $$E[X]= 0 \Rightarrow (-1)\cdot p_X(0)+1\cdot p_X(1)+2\cdot p_X(2)+c\cdot p_X(3)=0$$ right? @KevinDriscoll

@leslietownes For practical purposes here, I am neither a physicist nor a mathematician. Closer to the awful hybrid mathematical physicist. And the cases where the domains fail to coincide actually matters to me sometimes. For example, my entire thesis is on some systems where the Hamiltonian fails to be essentially self-adjoint.
Yeah, I think that's right.

well that's weird. i assumed physicists always wanted their hamiltonians to be self-adjoint. i think you're a mathematician.
unless we're in a multiverse where we're all evolving accoridng to different self-adjoint extensions of the same thing. which is actually my theory of the universe.

I mean if it can be self-adjoint, then by all means let's go with that. Sadly when you take the limit that the "short-range" physics of atoms gets squished into a smaller and smaller region while holding the "long-range" physics fixed, it turns out that the result is a non-self-adjoint Hamiltonian

@KevinDriscoll I did the calculations for (b) again. It is correct that we get a negative number for the expected mean profit, or not?
I have $$E[X]=(-1)\cdot p_X(0)+1\cdot p_X(1)+2\cdot p_X(2)+3\cdot p_X(3)\\ =(-1)\cdot \binom{3}{0}\left (\frac{1}{6}\right )^0\left (\frac{5}{6}\right )^{3-0}+1\cdot \binom{3}{1}\left (\frac{1}{6}\right )^1\left (\frac{5}{6}\right )^{3-1}+2\cdot \binom{3}{2}\left (\frac{1}{6}\right )^2\left (\frac{5}{6}\right )^{3-2}+3\cdot \binom{3}{3}\left (\frac{1}{6}\right )^3\left (\frac{5}{6}\right )^{3-3}\\ \\ = -\frac{17}{216}$$ @KevinDriscoll
If that is correct, that would mean that we expect that we lose the game, right? @KevinDriscoll

1:04 AM
Which it should be, if you think that the short-range physics is not uniquely determined by the long-range physics. The fact that the resulting model is not essentially self-adjoint corresponds to the fact that there is some range of "short-range" physics that can correspond to the same long-range behavior. And the degrees of freedom that you have to choose are precisely the self-adjoint extensions.
I haven't done the algebra, so I'm not sure if the payoff is negative or not. But it wouldn't surprise me. Real life gambling games are always set up so that in the long run the player loses money to the "bank."

Ok! At (c) I get now c=20, positive!! So it must be correct now @KevinDriscoll

Probably so. Getting an integer for an answer when it could be any number is usually a good indicator that you've worked things out properly.

negative "a little bit" is also a very realistic expected return for a real-life gambling game or something that is approximating one.

Thank you for the clarifications and your help!! :-) @KevinDriscoll

No problem

1:08 AM
it's like playing with a cat. if you don't let it capture the toy often enough, they lose interest.

Yeah, a house edge of ~0.7% is pretty good. You'll usually only do better playing Blackjack or Video Poker optimally. If you get enough comps you might be able to turn a profit!

all the popcorn shrimp i can eat at the buffet?? count me in!
at least i didn't take my education and go into designing casino stuff.

2 hours later…
3:01 AM
I have been watching talks on symplectic and contact geometry recently trying to get a flavor of what's going on, might it be useful for me to study more deeply, etc. So far I have noticed that questions of which manifolds have symplectic/contact structure, if you have almost-symplectic/contract structure can you extend it to genuine symplectic/contact structure, and if two symplectic/contact structures agree on some nbhd or submanifold, say, do they agree everywhere get mentioned all the time.
Clearly they are foundational to the field in the same way that we spent a lot of time in differential topology asking when a manifold has a smooth structure and how many does it have up to diffeomorphism and so on.
From many of the talks I get the sense that people are trying hard to prove *positive* answers to these questions. In some sense, they want manifolds to have these structures, they want to be able to extend almost-structure to genuine one, and for agreement somewhere to imply agreement everywhere.

If such things were true, it seems like you could do a lot constructively by defining structures locally someplace and then guaranteeing that it extends to the whole space.
On the other hand, when I listen to Yakov Eliashberg it seems pretty clear he wants these things to be false. And I *think* I understand why he wants them to be false.

If you can't homotope almost-structures to genuine ones, presumably there is some characteristic obstruction that prevents it. We already know about some of them: these volume and no-squeezing constraints. And the point seems to be that if there are more, then they are novel in the sense that they lead to invariants that are different from the ones that we know about. That's good because presumably the invariants are topolog
Can anyone comment on if I am getting the right gist here? Positive results are nice because they allow you to convert local constructions into global ones. Negative results are nice because presumably they lead to invariants that you can use to classify things.

3:26 AM
beats me. :)

1 hour later…
4:28 AM
Hey, does anyone understand what this problem is asking?
It's just telling me to do two different integrals, right?
I just can't tell if the two integrals are meant to be related to each other.

yes, you'd set that up as a sum of two integrals.

Oh, the integral over C1 + the integral over C2
Alright, thanks

the paths they've chosen are somewhat arbitrary. if you vary them you get different results. they probably chose those C_1 and C_2 because the integrals don't get too ugly.

Ah, I see

it looks like maybe in (b) they're setting up an example that shows how the value of the result can depend on the particular way you travel from (0,0,0) to (1,1,1).
but the stuff in the screenshot applies to (a) only.

4:33 AM
Oh yeah, I just scrolled. The stuff in b is for problem 16, below what I have here.

it drives me crazy that the "(a)" and "(b)" and the figure annotations are in times new roman and yet the question is in a sans serif font (maybe a variant of arial).

Lol, yeah. The formatting for this textbook is nutty. I don't want to buy the physical copy though because even my prof said this isn't the best calc text out there.

electronic books in general are absolute garbage in terms of formatting. not just math books. even from otherwise reputable publishers.
it's like people decided, you know what, let's jump forward, but also go back decades in terms of typesetting quality.

4:49 AM
Hello
How are you leslie

i'm OK. the holiday was a nice break.
how are things with you?

Agreed, the inconsistent fonts are a nightmare. Very possible this happens because in some LaTeX package, the test body font is controlled in a different place to the enum fonts and someone didn't know or couldn't be bothered to change them all

The cat :/
I am fine......

the kid and cat are great. they spent a lot of time playing with each other this morning.

No movies watched :/

4:56 AM
the kid is in day care for most of the day during work weeks so it was fun to hang out with her while the sun was still out.
i was gonna watch a movie but then i couldn't decide what to watch, so i didn't.

:/
What movie you was going to wacth!

i never settled on one.

I love adventures movie, not fan of horror movies
action movies are fantasy :/
also like comedy ones

i was looking for a good horror movie, but couldn't find anything that appealed to me. a lot of the more recent stuff is too pointlessly violent, and i've seen all of the old stuff already.

:/
For horror movies, you should watch Japanese one
I found they are good with these genre

5:03 AM
yes, there is a lot of good stuff from japan.

The mirror movie forbade me from looking into mirrors for 4 months
OCULUS
leslie watch OCULUS also I don't recommend :/

if anybody else has anti-recommendations, i'm all ears. :)

6:03 AM
@leslietownes Not a movie, but a show -- Squid Game is pretty wild.
Violent and disturbing, but not needlessly so. The show makes its point clear, I think.

seen it!

I watched a show called See. It shows a world in which everyone has lost the power to see so when they are told that in past people could actually see, they would think of that to be a myth.
and one in horror genre is: Servant, though I don't know if anyone would like but I liked it.

hm, i think my wife subscribes to the thing i'd need to stream those.
we've unified many aspects of our finances, but not our streaming subscriptions.

6:26 AM
@leslietownes Ah, darn. lol

I agree with copper.

it peaked too soon for me. episode 2 was the most interesting. of course i did binge the whole thing.

I found the 6th episode to be the saddest.

6:56 AM
i watch all sorts of crap just to relax before going to sleep.

7:08 AM
and roam mse for convex psqs.

even two answers for once question, but the op likes their own answer which is a bit incomprehensible to me.
sometimes i just have no clue what the op is trying to do.

7:30 AM
perhaps, the op is trying to convey their frustration...

7:51 AM
this is quite possible. it is my main use of the internet

intension is everything

I'm doing a reading about Godel's incompleteness theorem for my math history course and I just got up to the part where it discusses statements that are "formally undecidable". I'm frustrated because I learned about one of these statements in a math class this semester and I can't remember the class or the statement -_-
It's at the tip of my brain, so I'm 99% sure this event actually happened lol
Maybe I'll remember what I'm thinking of once I get some sleep.

how many classes did you take?
:)

I'm taking 5 -- 3 of which are math courses.
Or at least math-related. We don't do too much actual math in my history course lol

8:02 AM
I guess I'll have to.

as long as you took only finitely many classes, the process of elimination will terminate
no undecidability there

Lol, that's a relief.

When checking if the expansion point x=x0 for ode, to solve using series method, is ordinary or not, we normally just check if B(x) or C(x) are analytic. Where the ode is y''+B(x) y' + C(x) y=0. But what if there was an f(x) on the RHS? do we need to also check if that is analytic or not at expansion point? Here is an example, using first order. Given y' + x*y = 1/x^4 and the expansion is around x=0. Clearly if we just look at y'+x y=0, then x=0 is ordinary point.
... Do I need to worry about 1/x^4 or not? Currently I only consider the homogeneous part of the ODE to determine if the point x0 is ordinary or not.

2 hours later…
10:38 AM
is this true without more assumptions on $f$? if $f$ is entire, and $f f' f''$ never vanishes, then $f = \exp(az+b)$ for some constants $a,b$. I know this is true if we also assume $f$ is of finite order (by hadamards factorization theorem), but I don't know if this is true in general
ideally (if it were true) one should be able to show $\frac{f'}{f}$ is forced to omit three points, it already omits $\infty$ and $0$
so one would hope there would be a way to say if $\frac{f'}{f} = C$ for some constant $C \neq 0, \infty$, then we contradict our hypothesis.. but i'm beginning to believe there may be a counterexample to this statement if $f$ does not have finite order

3 hours later…
1:29 PM
this chat room deserves emote functionality.
hey all!

1:46 PM
Hey.
Why is $vol(w_1 \times w_2) = vol(w_1) * vol(w_2)$ where as $w_1,w_2$ cubes? Seems simple but how to prove?
Okay nvm i can just understand as multiplication of the sides

1 hour later…
3:12 PM
Morning
How is the Riemann Zeta function usually evaluated?

2 hours later…
4:44 PM
@AMDG You can do it with the reflection formula to get something with real part greater than one then evaluating the infinite sum
but there's probably a more efficient way

Yeah, I mean I'm trying to implement Gamma, and using the reflection formula and Zeta seems like a better and better idea, but I'd prefer to implement Zeta using Gamma instead, and then I want to see how to implement the incomplete Gamma functions.

What does it mean to converge weakly in L^p ?
I have a sequence of locally integrable functions

@Derivative I've been trying to find a closed form of Gamma through the reflection formula that Euler found. I haven't made significant progress. It just seems like it would be possible to manipulate this into Gamma(x) and then generalize to the complexes. desmos.com/calculator/gm1ttzfqs5
However, I don't know enough about Gamma in principle to efficiently find such a closed form with what is given here. It would seem that we have all the pieces, and I just don't know how to fit them together.

hmm I'm not sure if that's possible

Yeah, but look onwards from formula (39) here: mathworld.wolfram.com/GammaFunction.html#eqn39
It's just teasing at a closed form
This is a really convenient form, too, since I can easily compute secant; the whole motivation here, of course, is uniform convergence across the entire domain of Gamma(z).
*cosecant

2 hours later…
7:02 PM
Been messing around some more. It looks like a plausible path forward is through derivatives here and manipulating digamma since the derivative of Gamma here is a sum of digamma functions multiplied by the original gamma product. That makes things a bit easier since the integral of these sums are sums involving log gamma.
Though it does look like digamma can be implemented as a piecewise function for $x<0$ and $x>0$ with the former being what appears to be a form of tangent.

jakobian: do you know anything about duality in L^p or L^p_loc spaces, or is this arising somewhere where maybe nobody bothers with that layer of abstraction

7:30 PM
can harmonic functions in two different coordinate systems coincide?

mm, different coordinate systems for the same space? yes. the laplacian is coordinate independent.

found another ripe convex psq

one way of seeing this for harmonic functions is noting that harmonic = continuous + mean value property on balls, with the right hand side not depending on what coordinates you choose
but that's probably not the easiest proof route

@leslietownes I wrote a differential equation to describe this let me dig it up

geocalc: the laplacian is also something like div grad, or -div grad, if you know coordinate independence of those things

7:39 PM
it means i need to pay more attention to the rest of my life.

Leslie, $\nabla^2_A f = \nabla^2_B f$ where $A$ and $B$ are different coordinate systems in $\Bbb R^2$

copper: don't we all

surely a gradient is independent of coordinate system?

for example one can fix $B$ as standard cartesian coordinates and fix $A$ as something else
I think Leslie and I are talking about different things

we're talking about the same thing. the laplacian is one thing. youcan express it differently in different coordinate systems, but it's not an artifact of any particular choice.
you originally asked about harmonic functions. those are characterizable by a mean value property which is obviously coordinate independent, although this doesn't address the more general issue of the laplacian, which i noted is \pm div grad, which is hopefully coordinate independent.
then you said it's coordinate independent using nablas.
same thing all the way through.
i forget why we didn't use nablas in my geometry class.

7:44 PM
okay

i guess people use nabla for a connection. maybe that was why.
i'm just talking about the nabla notation now, if that isn't clear.

here is a particular example
$u\frac{\partial}{\partial u}\left(u\frac{\partial f}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial f}{\partial v}\right)=\frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$

i'm stepping out, connections are outside my domain so to speak.

on each side is a laplacian

we're not in the world of connections. we're in R^2 with two different coordinate systems.

7:46 PM
i am getting tensor and tensor by the moment

i was just commenting on a vague memory of why one might not use the nabla symbol for "grad" if one is operating at a higher level of abstraction

I'm looking for all solutions to that PDE

geo: are u and v the same on both sides? asking only because this doesn't appear to involve two coordinate systems.

yeah u and v are the same on both sides so I am mistaken about something
okay so I derived the expressions in two different coordinate systems and then made the variables match

ok. 'making the variables match' is not likely to have intrinsic meaning. when one says the laplacian is independent of coordinates, this doesn't mean that the symbolic expression for it 'matches' after you symbolically swap in the old variables for the new ones. it's a fine equation to consider but maybe we are exiting the world of intrinsic stuff.

7:52 PM
oh okay
thanks for helping me understand what's going on better

e.g. x^2 - y^2 is a harmonic polynomial but if i pretend x is r and y is theta and use the expression for the laplacian in polar coordinates r, theta and 'relabel' i get something that isn't 0.
which in a way is good, because having fixed any one set of coordinates the laplacian ought to be only one thing in those coordinates and not two things.

yes

maybe a g--m-t-r would have smarter things to say about it but that's my own mental image.

that's actually a pretty cool way to construct PDE's

8:10 PM
moving from odes to pdes is like stepping from a umbrella drink beach with azure waters to the centre of an active volcano.

I'm going to plug this pde into an online pde solver

This math forum is a joke. One asks a legitimate question on how to solve an ode and show their attempts and gets downvoted for no reason.

the world is a joke by that measure

earth is a joke

even a joke is a joke
surely this can't be your first downvote?
on another note, as part of continued life procrastination, wonderful little article in the economist about ivermectin. the title is "Ivermectin may help covid-19 patients, but only those with worms". the punch line: “Ivermectin doesn’t treat covid,” Dr Bitterman wrote. “It treats parasites (shocker) that kill people when they get steroids that treat covid.”

8:37 PM
nasser, is there a particular post you have in mind? i agree that often it's annoying when people downvote without any comment.
there's also the phenomenon of the downvote pile-on, where one or two downvotes on a post seem to make more downvotes more likely. i haven't seen this much lately but i remember it from my earlier times of being active on mse and math overflow

i had a few serial downvoters over the years, but i think there is some sort of algo that detects & reverses in severe cases.

in the early days of math overflow, one or two downvotes on something would often lead to a pile-on effect even if the question was at least marginally appropriate. or so it seemed to me. people were really nervous, perhaps excessively nervous, about MO turning into a homework site. there may have been some overcorrection there.

yeah, i'm almost tempted to post there :-)

of course anybody 'we all knew' could post the vaguest question imaginable, or respond to a question with a non-answer, and get upvotes and responses.
MO seems to have settled down.
copper i think if someone goes around upvoting or downvoting the same user too rapidly over a given time period the algorithm automatically flushes the toilet.

i must admit the gamification got to me early on. i think i posted some utter rubbish.
one of my earlier posts was a batman post. another was a 'without a calculator' calculation.

8:54 PM
i do see math.stackexchange.com/questions/141639/… which is unforgivable.

Wait a minute. His hermitian inner product is conjugate-linear in the first variable, not the second? Ugh.

i hadn't even noticed that. i was blinded by the < > < > <

Oh, I was so upset I didn't even notice that.

now we're both twice as upset.

Time to eat Peking Duck.

9:00 PM
Awesome, will update, it offends my sensibilities too. Surely conjugate linear in the first variable is more consistent with the $\langle f, x \rangle$ notation for funtionals?
I am not particularly attached to either other than muscle memory.

My hermitian inner product has the bar on the second factor. Always. This is why we write $dz\wedge d\bar z$ in complex analysis.

but that will bump it to the front page. then you will be accused of attempting to attract internet clout to old posts.

ohhh, that is sooooo awesome

copper: do you mean the $\langle x, f \rangle$ notation?

sounds of liquid splattering on ground
very $\lambda$ of you.

9:02 PM
i wouldn't use langle and rangle for that pairing although a whole lot of people do. they're wrong and i'm right.
tune into my podcast, leslie speaks, for a 2 hour explanation of why

the hidden rules of mse

I don't use $\langle\rangle$ for duality pairings, ever.

ted is my guest on that episode of leslie speaks.

LOL
A Ted talk will cost Leslie money.

S1:E3 when angles are just wrong

9:06 PM
we also discuss which natural supplements sold by our advertising partners are most effective in fighting the coronavirus and undoing the effects of vaccines.

new on mathflix

And how many cases of omicron are already in the US?

ivermectin as an ed medicine. works in cows.
because of inappropriate use of function application, no doubt

Now leslie is making me curious to go see what my first answers were. I have zero recollection.

i wonder what latex transgressions Ted is guilty of?

9:09 PM
After typesetting four books, I would expect LaTeX transgressions are minimal.
It looks like this cherry was my first answer.
I missed the symmetry solution. Bad Ted.

Nice. Hardly counts as bad. Could be worse, you might have used < somewhere.

Wow ... so many answers that have never been accepted.
I think I posted things as answers that today I would just post as comments. Comes of having a lot of rep now.

ingrates. wait,... i see the convex question ray in the sky, gotta go
i'm expecting a major censure for my last answer.
echos of lou reed sounding in my mind...
i want to send a note to an old(er) friend who helped my mom a lot over the years. i am marveling that there are still places where a name, town & country are sufficient to get delivered.
no need for that modern eircode nonsense.

Wow, just a name and town? Must be a tiny village of eld.

i got a note from se about getting some swag (no idea why) on 10/13 and i responded. however, my mse emblazoned private jet has not shown up yet...

9:23 PM

@TedShifrin partly that and partly everyone knows everyone in that area.

I don't think at any point in my life I've lived in such a place.

that would be sooo cool. up there with the pic of my daughter in the back of a UCLA police car.

the USPS used to be pretty good with that. at least if someone had a distinctive name. my dad would sometimes get mail with just our hometown (pop. 60,000ish at the relevant time) on it, because there was no one else with his name in town.

Boy, your son will feel so left out.

9:24 PM
its odd, i had many gay friends growing up but none ever declared themselves. probably for above reasons.
small town in everyway.

How are you sure, then?

they told me

wasn't it criminalized until relatively recently? as in some parts of the US?

need to look that up...

"None ever declared themselves" <---> "they told me" ... Ted is confuzled.

9:26 PM
they didn't report to city hall and file the paperwork

i mean publically

or to customs on returning from abroad. i'm bringing back one homosexual

Ah, well, telling close friends is still a lot more than some people do, even now.

nobody was out when i was a kid, and this was like, i dunno, five or six decades after your youth, copper.

i can see why
i attended a few lectures by david norris who almost single handedly overturned the relevant laws.
i am sure there was a line of reasoning somewhere in my brain that brought about the above comments but any such line is lost to me now.

9:31 PM
I was wondering how we leapt from mail delivery to closeted gay friends.

ah i remember. a friend from a tiny town of 4 houses.

Ah.

there was no secret as such. just not said.

Define the function $v(x)=$\begin{cases} u(x) & x_n \geq 0 \\ -u(x',-x_n) & x_n\leq 0 \end{cases}$$

they somehow knew where to deliver the postcard to "closeted gay friend", town, ireland

9:33 PM
that could be ambiguous.

i need to prove is C^2 on the whole ball when u(x) is C^2 at the upper half of the ball

jam, some typesetting there and maybe other issues. the dependence on n is far from clear. the comma in the second part of the definition suggests u might be a function of more than one thing?
is x = (x_1, ..., x_n)? is the x', -x_n some notation for fiddling with one coordinate only of x?

$x_n$ is just the n variable

x' ?

the rest
$x_i$

9:35 PM
i worked with an economist who wrote "x/c_n" for "x, with the nth coordinate replaced with c_n". not the best notation, particularly if the thing you're putting in there isn't tagged with an n already.

also $u$ is zero at the $x_n=0$ so obvious my extension function is continuous by taking the limit

Looks remarkably like Schwarz reflection.

ok, so v(x_1, ..., x_n) is u(x_1, ..., x_{n-1}, x_n) on the set where x_n >= 0, and is -u(x_1, ..., x_{n-1}, -x_n) on the set where x_n < 0 ?

Ted thats exactly what im trying to do ( i think)

OK, and u is 0 where x_n is 0. that allowed you to use >= and <= in the definition.

9:36 PM
i can prove the continuity just the xn-->0
yes leslie

Do you know anything more about $u$? Like harmonic?

yeap u is C^2 and harmonic

2

I have a question about problem 9(b) in Chapter 2 of Evans' PDE book. It says if we have $u$ is harmonic in the open upper half ball $U^+$ and $u\in C^2(U^+)\cap C(\bar{U^+})$, $u=0$ for $x\in \bar{U^+}\cap\{x_n=0\}$ Then by the reflection principle we definev(x)=u(x)\phantom{fd} \text{ if }\ph...

You might want to give all information when you ask your question!

there wasn't any time for that.
life moves quickly, ted.

9:38 PM
To be more blunt: We will stop responding to your questions if you do not do so.

haha yes !!

we will deduct the fee from your account.

The question you posted is another question where $u$ is C at the boundary not C^2
im tryna prove something simpler given $u$ $C^2$ and harmonic at the upper closed half ball and zero at the base ($x_n$=0) prove the extension is also $C^2$ at the whole ball and harmonic
the question you posted lowers the smoothness of u at the shell of the ball
i think its just a multivariable exercise showing a function is differentiable 2 times at xn=0

the harmonic part is sort of crucial.

so the C^2 of u is not enough to show C^2 of v ?

9:44 PM
NO.

try with a single variable, $u(x) = x^2$.

you mean maybe reflect sqrt(x)
now that thing wont be differentiable at zero i guess

no, i mean the corresponding $v$ is $-x^2$ for $x<0$ and hence not twice differentiable at $x=0$.

ye exactly

Huh?

9:48 PM
same example. mine is not even once differenable

hahaha we are not making sense haha

maybe try constant functions.
ted, you have my permission to begin martinis early today

I might need to.

the surface of my tea is not harmonic

9:53 PM
copper do you have a tea to recommend? i am mostly drinking whole foods store label green up to about 3pm, then i switch to herbal 'tea' often just ginger.

what? isnt the surface of your tea just a plane?

i do not understand 'green tea'
@Jam i have a spoon in it.

oh no that ruins everything

my choices of tea would be barry's tea, lyon's tea, trader joes irish breakfast. cheap, plentiful and quick to make.
all my loose tea remains in cute little tins.

00:00 - 22:0022:00 - 00:00