Mathematics

my best friend uses that all of the time. i don't know where she picked it up. we're in our 40s. i think it is the property of younger people.

i did not know the term 'kaomoji,' thank you. previously i just thought of it as 'one of those things that lauren does.'

HereToRelax

I think the facial features are brought to you from the japanese hiragana alphabet

i had a vague sense that that had something to do with it. i like how it aligns everything with the horizontal. so many "smileys" from the old BBS days, and now, require you to turn your head. :-)

i used to use (: because i'm left handed. it drove people crazy.

HereToRelax

12:26 AM

I think that's the russian way

That explains a lot about @Leslie.

i may have used it because it drove people crazy and me being left handed is just an excuse.

a little bit of me is the person who never sees a bull without wanting to wave something red in front of it. it's not out of malice, it's just an instinctual sense of humor thing.

i've read a number of profiles of comic actors who had this same pathology. they could never not be 'on' and it interfered with their ability to be real people. peter sellers and chevy chase are two examples.

i'm nowhere near as toxic as those people in other ways but i see the same thing in them that is in me.

if $f:\Bbb C\to\Bbb C$ is differentiable at $z_0\in\Bbb C$ then $\overline{f(\overline{z})}$ is also differentiable at $z_0$?

this is something i am guaranteed to mess up because of the number of order-2 things happening. the schwarz reflection principle comes to mind but this is not that. are you familiar with the cauchy riemann equations?

12:42 AM

I know C-R

the answer is no for very boring reasons

$f$ being differentiable at $z_0$ does not imply whatsoever that $f$ is differentiable at $\overline{z_0}$

yeah, wow, it may not even be defined there. this is why i was thinking schwarz reflection principle. if something is filling in potential gaps it would need to be known before the question can be answered.

Edward Evans

oh damn sry, was moving my laptop

@leslietownes I let the domain of $f$ be $\Bbb C$

@Thorgott ok but I want explicit example

At least $\overline{f(\overline{z})}$ satisfies C-R

@Thor How was your lecture?

12:50 AM

you can write one down yourself if you internalize what I said

If the domain is symmetric, the answer is yes.

@Ted you mean my talk? hasn't happened yet, it will be next wednesday

Oh, oops

but my function is $\overline{f(\overline{z)}$ not $f(\overline{z})$

Yes, it's holo if $f$ is and the domain is symmetric. I already said so.

Lucas Henrique

1:03 AM

hey Ted, lemme ping you again

@Ted, please take a look here

1:19 AM

@TedShifrin Is that due to schwartz reflection principle?

@TedShifrin you're thinking of something correct, but it isn't quite what sodam was asking. the answer to his question is negative.

He said domain $\Bbb C$.

for the sake of clarity:
False: if $f$ is differentiable at $z_0$, then $\overline{f(\overline{z})}$ is differentiable at $z_0$
Correct: if $f$ is differentiable at $z_0$, then $\overline{f(\overline{z})}$ is differentiable at $\overline{z_0}$

Oh, ok, Thor is right. I was making it holo on domain.

I should resign.

Schwarz reflection uses this.