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5:00 PM
you're welcome!
 
@BalarkaSen OK how about this question
I have a geometric line segment in $\Bbb R^3$, and I'm embedding a disc so that the line segment is a part of the boundary
How bad can that embedding be near the segment?
 
Extremely wild I think
Imagine a Cantor set many Alexander horns popping out as you approach the boundary
 
I guess I wanna $\epsilon$-approximate it with something nice
Those Alexander horns gotta shrink
 
My worry is that after approximating you only get some immersed smooth S^1 x I
 
I can triangular domainland so that the output of each section has tiny diameter or something
Oh yeah I'm assuming the boundaries of S^2 x I are already smooth so this won't change them too much. Otherwise the ribboned F-A arc minus a disc fucks us
 
5:09 PM
Yeah got your assumptions
 
Howdy, DogAteMy and a @Balarka.
 
So what's your worry
Heyo
Oh immersed
Sorry misread
Yeah 100%
 
hi @Ted
 
Hi @Lukas!
 
I got a job where I design exercise sheets for a masters algebraic number theory course, really fun stuff
 
5:13 PM
Can I always triangulate stuff so that no triangle is in any other triangle's convex hull or something @BalarkaSen
 
Writing (and grading) interesting homeworks seems to be something professors hate to do. I always enjoyed it, especially the former. Of course, if there's a text with excellent exercises, that makes it easy, but often there isn't.
When I taught out of Griffiths/Harris, I had to write lots of problems. Some were routine verifications, but some were interesting :)
 
Hm actually I don't think that's useful
 
@AkivaWeinberger Isnt it true that if you have an immersed disk bounding a knot in S^3 with an embedded collar then its actually an unknot.
a la Papakyriakapoulous
 
one of my least favorite jobs was grading homework for a prof who assigned small amounts of homework every monday, wednesday, and friday. he put no effort into it and it was obvious. students experienced whiplash. the scores provided no information.
 
Maybe something like this is true for an annulus as well
 
5:15 PM
@Ted I'm kind of sad that when we'll get to discrete valuations, I can't make exercises about how a vanishing order/pole order at a point on a Riemann surface gives you a discrete valuation
I find that really helpful to explain the terminology "ramification" as well
 
Agreed, that is the example that motivates everything.
You can give the algebraic set-up and put a remark about the complex curve case.
 
yeah maybe I'll do something like that
the thing is those students haven't necessarily taken complex analysis or algebraic geometry
 
Or maybe an optional problem ... "For those of you who've seen Riemann surfaces ..."
Yeah, but you don't need much. It's just a local holomorphic coordinate. You can just use that as an Ansatz.
You certainly don't need algebraic geometry for this. And everyone knows what a smooth manifold is since their first year in Germany ...
 
fin
hi teddd
 
Hi @fin
 
5:18 PM
4 mins ago, by Balarka Sen
@AkivaWeinberger Isnt it true that if you have an immersed disk bounding a knot in S^3 with an embedded collar then its actually an unknot.
Wait really?
Wow that sounds super useful
 
Sorry I couldn't do quantum mechanics out of context.
 
Yeah, that's the disk theorem [See en.wikipedia.org/wiki/Dehn%27s_lemma, although they state it for PL disks, it's true for smooth ones as well]
 
fin
@TedShifrin lmao dont worry
 
Yeah OK I'll look at a proof later but I kinda assume that sort of proof strategy is gonna work here
 
Yeah
 
5:30 PM
@LukasHeger sorry to ask you again but if $u_n$ converge weakly to u in W^{1,p}(\Omega)$ how to get that $\nabla u_n(x)\to \nabla u(x)$ a.e
 
Could anyone try to help me understand why in the proof of theorem 3.2 page 13 (here numdam.org/item/10.5802/aif.2098.pdf) is it so obvious that $\pi_{1}(C)\simeq H$? For what I have understood it should be $p*(\pi_{1}(C))\simeq H$
thank you in advance
Or if it is not that how can i obtain a surjective homomorphism from $H$ having one from $C$? For how it is left is should be trivial but I could not find any argument
 
if $u_n \to u \in L^p(\Omega)$ do we have that $\nabla u_n\to \nabla u$ dans L^p(\Omega) ?
 
ok im dumb, p* is always injective
 
0
Q: If an embedding of $S^2\times\{I\}$ has nice edges, are the edges ambiently isotopic?

Akiva WeinbergerLet $f_t(x)$, $~t\in I$, $~x\in S^1$ be an embedding of $f:S^2\times I\to\Bbb R^3$. (Thus for example the images of $f_{t_0}$ and $f_{t_1}$ are disjoint if $t_0\ne t_1$.) It's known that the images of $f_0$ and $f_1$ need not be ambiently isotopic. An easy construction of this is to "ribbon-ify" ...

@BalarkaSen Outsourcing to main
 
math.stackexchange.com/q/4288020/922120 Is there anyone can answer this question? I wrote the statement for general topological space but it may not true in general. As @love_sodam advised, I first tried to prove the statement of simplicial complex. But even though I restrict the space, I can't prove. And idea for this problem?
 
5:43 PM
Upvoted
 
Hm should I change $x$ to $\theta$
nah it's fine
 
Is there any method to solve (that is, finding nth term) a recursively defined sequence with variable coefficients?
Given $a_1$ and $a_2$, I want to solve $a_{n+1}=(1-\frac 1n)a_n+\frac 1n a_{n-1}$
 
6:26 PM
seems like diagonalization should work
 
what is the difference between "convex set" and "uniformaly convex" ?
 
6:49 PM
what do you mean by the latter?
 
@Koro usually, the moment you drop the assumption of constant coefficients you can't get an exact solution
there's some special cases of course, but there's no one-fits-all solution
that being said, let's see what mathematica has to say...
 
I guessed that there must be some series solution just as we have in case of an ODE with variable coefficients.
 
@AkivaWeinberger Yo
 
the analogue of 'series solution' for recursive sequences is just the sequence itself AFAIK
what you do still have is linearity, though
 
6:53 PM
@BalarkaSen Yo?
 
I'm still trying to solve the recursion though. Progress so far: I managed to show that $(n^2-1)a_{n+2}+a_{n-2}=n^2a_n$
 
Suppose there's an annulus $A$ bounding the trefoil $K$ and an unknot $T$ in $S^3$
 
hmm. mathematica does come up with an exact solution, but
 
Consider the disk given by $A \cup C(T) \subset C(S^3) = D^4$
 
Hm wait is the complement of the annulus homeomorphic to the complement of either edge
What's C
 
6:55 PM
@Semi: In the last expression obtained: indices differ by 2
 
Cone
 
$a_n=A+B \Gamma(n,-1)/\Gamma(n)$ where $\Gamma(n)$ is the gamma function and $\Gamma(n,a)$ is an incomplete gamma function
so....yeah
 
in the original one, indices differed by 1 and 2
 
Oh is that a disc you can do Dehn to?
 
6:55 PM
@Semiclassical no way
 
ya
like, what
 
@Akiva If this was a locally flat disk bounding $K$ in $D^4$ we'd be done; the trefoil knot is not slice.
 
that said
 
I don't think gamma will be involved
I'm sure the exercise doesn't want that
 
the appearance of the constant $a_n=A$ solution is sound
 
6:56 PM
But it is locally flat, because we went one dimension up. Any embedding stabilized one dimension up is locally flat
 
What does locally flat mean?
 
so i think we can assume $a_0=0,a_1=1$ to get the second solution
 
See, eg, this
 
I think this is trying to say something in a language that I don't understand
1 min ago, by Koro
in the original one, indices differed by 1 and 2
 
You can find the definition of local flatness in the question
 
6:57 PM
@Semiclassical It is given that $a_1=a$ and $a_2=b$
 
in that case Mathematica gives $a_n=\Gamma(n,-1)/(e\Gamma(n))$ which is...interesting but wacky
let me put in Wolfram Alpha so we can make sure i'm solving the right thing
 
Ah OK
Cool
 
@AkivaWeinberger Alexander horned sphere complement is certainly not homeomorphic to sphere complement. Should be false for FA wild arc as well.
 
I'll take a look at the solution in the book
:'(
 
But this is specific to trefoil @Akiva
 
Ah fair right
@BalarkaSen Yeah
 
I was thinking there would be some way to solve this recursion.
 
subfactorial(n-1)/n!, like wtf
 
Shall I write a comment?
 
what's a subfactorial??
 
7:00 PM
good question. let's find out
 
Sure, go on. But I didn't specify trefoil in the post
 
I never heard of that :(
 
Yeah that's why comment
 
i've heard of it, somethin somethin derangmenets
 
Balarka, your knowledge is immense :)
 
7:01 PM
oh, and i said something wrong
should've said a_n=subfactorial(n-1)/(n-1)!
or, using the traditional notation: a_n = !(n-1) / (n-1)!
huzzah notation
 
@semi: I observed that while writing $a_n$ in terms of a_1 and a_2, sum of coefficients of a and b = denominator. That's the conjecture which seems true if we write first few terms
 
Oh nevermind it's unclear that it's locally flat after stabilizing one dim up
 
quick sanity check: if a0=0 and a1=1, then a2 = (1-1/1)*a1+1/1*a0 = 0 and a3 = (1-1/2)a2+1/2*a1 = 1/2
 
Only after 2 dim up I can say it's locally flat
 
which does match what i'm seeing in the solution
 
7:04 PM
I take it back
 
the weird thing is that i know i've seen !n/n! before
OH
Wikipedia reminds me of the following identity: $\displaystyle \frac{!n}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}$
I bet the latter is what they want as a 'solution'
so $\displaystyle a_n = \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}=\sum_{k=1}^n \frac{(-1)^{k-1}}{(k+1)!}$
and i bet i know a good way to get to this representation too. one moment of algebra
@koro got it. you can rewrite your original equation as $a_{n+1}-a_n=-\frac{1}{n}(a_n-a_{n-1})$
 
yeah, just took a look at the solution in book.
@Semiclassical that's exactly how it was done. We can find nth term now.
 
right
that gives a closed-form in the sense of a finite summation
 
hmm
 
and that closed form happens to be equivalent to the subfactorial/factorial ratio
(you could even take it to define the subfactorial)
this does give one really cute implication
what happens to $a_n=\sum_{k=0}^{n-1} \frac{(-1)^k}{k!}$ as $n\to\infty$?
 
7:15 PM
1/e
 
yep
so if you start with $a_0=0,a_1=1$, your sequence converges to $1/e$
whereas $a_0=1,a_1=1$ gives $a_n=1$ for all $n$
from that you can get the limiting behavior for all $a_0,a_1$
 
so $\lim a_n=a+(b-a)(\frac 1e-1)$
in general.
 
e^-1, not e-1
don't think you need the last -1, but maybe i misremembered your initial conditions
 
no, I may have made some mistake
i think you're right
 
note how the problem only worked b/c we had that cute way to rearrange it
if you made small readjustments, like, change a coefficient from 1 to 2, then that wouldn't happen anymore
hence what i said earlier about the lack of a general method: usually, you need to spot something clever
 
7:22 PM
I drew this in class this morning:
 
Does that look phallic to anyone else?
 
@Semiclassical here is the final answer: $\lim a_n=a+(b-a)(1-\frac 1e)$
 
7:24 PM
oh, i get it. they told you $a_1=a$ and $a_2=b$
i was working with $a_0,a_1$
so naturally mine looks a little different
 
I stated above that $a_1=a$ and $a_2=b$
 
yeah
i missed that
 
yours' probably looks like this $b+(b-a)e^{-1}$
 
with a_0, a_1
 
7:26 PM
$a_0+(a_1-a_0)/e$
easy to check in mathematica, though
$a_n\to a_1+(a_2 - a_1) (1-1/e)=a_0+(a_1-a_0)/e$
@Koro out of curiousity, did they give any context for said recursion? you can intepret $a_n$ as a probability if you restrict to $a_0=0,a_1=1$
 
@semi: no, it's an exercise from sequences chapter. Next exercise is solving: $a_{n+1}=n(a_n+a_{n-1})$ :)
 
suppose you have $n$ guests each wearing a different hat. they put their hats in a box and select them anew at random. then $a_{n+1}$ is the probability no one gets their own hat
 
An extremely simple ODE of the form $y' = f(x)$ which only requires integration to solve is called a Pure-Time ODE. Is there a name for a comparable PDE, of the form $u_x = f(x,\ y)$ or $u_y = f(x,\ y)$?
 
@Koro gotcha
 
@Semiclassical I have read this, I think -derangements? But I don't remember them much currently :(. This will make more sense when I revise those concepts.
 
7:34 PM
yeah, derangement
!n counts them
 
something like bar and star method to find number of non zero integer solutions to x+y+z=6 etc.
or may be not as i don't remember currently. :'(
 
well, i certainly don't :P
i think i can see how to get through your next problem (with some assistance from Wolfram) but i probably shouldn't spoil it for you :P
 
$a_n=n!$
 
yep, that's one solution
 
with given $a_1=1$ and $a_2=2$
in this $a_1, a_2$ were given :)
 
7:48 PM
ah, cute
not too hard to spot it in that case
is that the only case they ask for?
 
yes
 
mmkay. then i feel alright pointing out how to do the general solution
namely, make the substitution $a_n=n! \,b_n$
in that case the recursion becomes $(n+1)!\, b_{n+1}=n(n!)\,b_n+n! b_{n-1}$, which upon dividng out $n!$ becomes $(n+1)b_{n+1} = n b_n+b_{n-1}$
 
nice, that removed factorials
 
at which point we only need to notice that this means $(n+1)(b_{n+1}-b_n)=-(b_{n}-b_{n-1})$
so we're back to a scenario like in the previous problem, where we can compute $b_n$ as a finite sum
 
yeah :) Thank you so much.
 
7:52 PM
np
 
But I am surprised that there is not a general method to solve recursions with such variable coefficients
 
i think the 'general' method may amount to converting such recursions into an ODE
which may or may not be any easier than the problem you started with
 
In mathematics, a recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms of the same function are given; each further term of the sequence or array is defined as a function of the preceding terms of the same function. The term difference equation sometimes (and for the purposes of this article) refers to a specific type of recurrence relation. However, "difference equation" is frequently used to refer to any recurrence relation. == Definition == A recurrence relation is an equation that expresses each element of...
 
as in, let $A(x)=\sum_{k=0}^\infty a_k x^k$
 
For constant coefficients we have a method as linked above.
 
7:56 PM
then the recurrence relation $a_{k+1}=f(k)a_k+g(k)a_{k-1}$ should be equivalent to some second-order differential equation on $A(x)$
 
@Semiclassical yeah, something inspired from this series solution
 
what i dont' remember is what said ODE will look like
and it may depend on which generating function you choose to use. it's not uncommon to write $A(x)=\sum_{k=0}^\infty a_k x^k/k!$ isntead
which, given how much simpler $a_k/k!$ was in the second example, would probably work out better. hmmmmmm
i have been successfully nerd-sniped
on the note of actual math i should be looking at, a problem I'm looking at today
and a ping to @anakhro b/c it seems in the vein of symplectic eigenvalues which came up a few months ago
 
@Semiclassical :)
So you'll find the series solutions to recursions with variable coefficients now?
 
lol, not right now :P
suppose i start with a 2-by-2 real symmetric but non-diagonal matrix $M$. what I want to do is find some transformation $M'=S M S^T$ for real matrix $S$ such that $M'$ is diagonal. but i don't want $SS^T=I$ like one usually does; rather, i want $S\Omega S^T = \Omega$ where $\Omega=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
hence I want to diagonalize $M$ using a so-called symplectic transformation
turns out this is basically something i already knew about in physics called a Bogoliubov transformation, i just didn't realize it
and the energy levels you get from that, i.e., the eigenvalues of $M'$, are thus the symplectic eigenvalues of the original matrix $M$
 
So your M is not similar to M'
 
8:05 PM
nope
they're congruent but not similar
and in such a way that you preserve $\Omega$ rather than the identity matrix
more generally you do this for $2n$-by-$2n$ matrices and replace the $1$'s in $\Omega$ with $n$-by-$n$ identity matrices
the demonstration in the book starts with $M=\begin{pmatrix} a + b & b/2 \\ b/2 & 0\end{pmatrix}$
 
How do I show that $\partial_{\mu}V^{\nu}$ is not the component of a $(1,1)$ tensor?
here a $(1,1)$ tensor is just a bilinear map $T:T_p^{*}M\times T_pM\rightarrow \mathbb{R}$
@Semiclassical
 
8:29 PM
Hello. Sorry for disturbing, but I'm learning math and I'm stuck at something I have never seen and I don't find any resource online. It is about: 0 ≤ x ≤ 2. Does this mean that 0 must always be less than or equal to x and that x must always be less than and equal to 2?
 
yes @BahSoh
 
Thanks
 
9:15 PM
wow, xander really drew that in front of people.
 
@leslietownes I mean, it didn't look problematic until I got to the end....
 
when you drew the vein, yes
 
Yeah, it is kinda veiny, in't?
 
they should get that dom(f) cap dom(g) thing checked out
shouldn't be that color
 
9:20 PM
xander do you know where the name 'elko' comes from? (yes, i read a blog post)
i ask because i used to live across from elko st in long beach, and while living there i came to learn it was one of maybe five things in the US called elko
and nobody knew where the name came from
my theory is it's just one guy who had a being or entity named elko visit him in a dream once, and he just traveled west, periodically stopping to name things elko
 
The apocryphal story is that the town is named "Elk" (for the animal) with an "o" added to the end.
But I honestly have no idea.
 
to add to the mystery, google maps calls the street E Elko St although it's Elko on the sign, and there's no W Elko St (or natural candidate for what it would have been) and never has been
anyway, if anyone figures that out, let me know
 
@leslietownes I currently live on the intersection of Greer Dr and W Greer Ave.
There is no E Greer Ave (so far as I know).
 
maybe google owns a portal from one to the other
 
And USPS seems to think that the drive should be W Greer Dr.
 
9:25 PM
i think that's a western thing, dating from when there were fewer colonializers than streets to name
 
Though there is no E Greer Dr.
 
maybe there's a guy named William Greer (or Elko)
even where i grew up in the bay area there are a few streets with similar names, and people with those names still live there
i'd find it embarrassing
 
Heh.
I work in Snowflake.
The town is not named for a weather phenomenon.
 
but the rich variety of cocaine cultivated there
 
Rather, it was settled by two Mormon families: the Snows, and the Flakes.
I have lots of Flakes in my classes here.
And a few Snows.
 
9:29 PM
part of my elko theory was that elko was a mormon. they did so much traveling at roughly the right time
 
Elko is kind of at the border of the Zion Curtain, so it is possible, I suppose.
Though I think that Elko was a railroad town (built on the transcontinental), rather than a Mormon settlement.
 
that's the competing theory. there are several railroad elkos.
so maybe someone worked for the railroad had elkos on the brain
 
According to Great Basin College (gbcnv.edu/howh/elkohist.html): "The town was probably named by Charles Crocker, superintendent of the CPRR. He was fond of animal names and simply added an "o" to Elk. "
 
might even explain elko st in long beach. it almost abuts what was a passenger rail line until about 1950
there's an elko SC and elko GA on railroad tracks
sorry to turn this into the elko conspiracy theory chat, i just can't get over it
the subdivision i grew up in had a whole set of streets named after ivy league colleges, which only gets funnier as the cheap tract housing ages and looks more and more decrepit
come see me on princeton lane, between the house with chevy trucks on blocks and the house with ford trucks on blocks. teatime is at 4pm sharp
 
Heh.
I've seen places like that.
 
9:36 PM
it's also common for spooky apartment buildings in LA county. you'll see something and it will be called something like The Buckingham Palace Arms
i believe they've got arms
the academia stack exchange is something else. what's the mirror image of aspergers, where you are so hyper-aware of potential social cues that it shuts you down
and so many posts like "my advisor kicks me in the head every time i meet him. every time i meet him. he says he's going to kill me. should i switch advisors?"
grad students should get hazard pay
 
@leslietownes Seems healthy.
 
it seems like there are two options for a phd advisor, completely awesome person and serial killer
 
Yeah, that sounds about right.
 
9:54 PM
and two kinds of phd students, the ones who will help their advisor move, and the ones who will help their advisor move a body
 
Though that body may belong to the advisor themselves.
 
there was a funny moment once in grad school where my soon-to-be advisor told the class he was going to have someone fill-in on the next lecture because he had to be in court, and some student said "jury duty?" and my advisor said "no."
we never got an explanation
 
when i was teaching, one time a student asked to take the midterm at an alternate because she had a court date. i said, not really caring one way or the other, jury duty? and she said "no, i witnessed a killing." and i googled it and it was true.
the lesson is, never ask anyone if it's jury duty
 
10:09 PM
it's too hot today.
 
We agree again.
 
It is a balmy 65°F here. I'm quite happy.
 
it appears to be low 70s where i used to live in oakland. that's my comparison to everything. near perfect weather, never hot, never too sunny. never too cold either.
 
It’s pretty dreary here
 
it appears to be 90F outside. i'm surprised our AC hasn't kicked on.
i lived quite happily in an apartment in oakland without heating or air conditioning for almost 10 years. there was some wall-mounted thing to do heating but i never needed it.
 
10:14 PM
46F , overcast and drizzly here
 
my comfort zone is about 30-70. below or above that range you will hear me complaining.
 
its warm outside 24c (~75f). i am suffering.
just resolved a customer issue that has been going on for days. turned out to be a goof on their part. sometimes time billing can be a blessing.
just settling into a nice psq and it got deleted.
 
ashes to ashes
 
the great think about sunday mass (i am reminded by your ashes comment) in the desperate environment of all boys schools was that girls turn up in their sunday best.
 
10:30 PM
my daughter saw a dead bird a few weeks ago and said 'dead birds are paper' and we didn't understand that until later she said that they're like paper in that when you're done with them you throw them out. it put me in the mood of contemplating the whole cycle.
 
wow. very pragmatic of her. i suspect we were entertained with the notion of bird heaven at that stage.
 
nothing but ashes and dust await us in this house.
 
our house is ashes & dust, with a few termites in between. the cost of fixing it up reminds me that i should not have taken time off to be with the kids.
its a house or life insurance catch 22.
 
i hear that. we have a termite abatement thing we are negotiating with the neighbor and maybe the previous owner of the house. i expect this dispute to go on forever, like in bleak house.
 
@leslietownes if I have a good coat and gloves I’m ok into the teens I guess
 
10:36 PM
@leslietownes sorry to hear it.
 
Zero is my “oh hell no” temperature
 
into the teens sounds like a recipe for arrest.
centigrade is so much better from that perspective.
 
It isn't really cold until the vodka freezes.
 
i used to shovel my driveway in iowa in a t-shirt. even at 0F. but with gloves. i'll be honest, it was a toxic masculinity thing.
i'd also usually retreat to a hot bath for several hours after.
 
@leslietownes I've worked up enough of a sweat while shoveling the sidewalks to remove my coat in ≈20°F weather.
 
10:37 PM
i can find it cold in albany in 50fs but have been in -25c in ireland in shorts and was fine. some mental thing
 
But that was after a significant amount of time spent shoveling.
 
yeah, i found that the low temperature caused me to do the work faster.
 
thats what i hate about snow camping, you shovel, get dripping wet, then freeze in your wet clothes.
ideally, snow camping would have a warm shower nearby.
 
my daughter's never going to know what any of this is. thanks, climate change.
 
i am sure your wife is able to freeze with a single glance?
i find i trigger that response in women frequently
 
10:41 PM
yes, that's usually how it works
 
my son has done some snow camping in his early teens thankfully.
 
he can tell future generations what ice looked like
 
my daughter has been laid up with a flu/cough thing for 2wks now, i suspect related to rowing in the frosty morning air.
no global warming evident in that part of the globe at present.
 
oh, she rows. my wife does too, but wouldn't go out in the cold. (it's never cold)
 
she used to be temperature sensitive and my son not, that seems to have switched.
 
10:44 PM
we live near the rowing area constructed for the 1932 olympics. i don't think it gets below 50 near the water.
 
nice
 
11:10 PM
Can someone native in English help me? In a discrete math video, I didn't understand what was said and the subtitle doesn't help.
if 4, then...?
 
if phi, then psi. phonetic names of greek letters.
sometimes pronounced 'fee' and 'see,' because nobody except people who went to really good high schools knows how the ancient greeks would have said those letters.
in chatjax, if $\phi$ then $\psi$
he seems to be writing something closer to $\varphi$ but the same letter
 
Wow. Thank you
 
happy to help.
 
11:30 PM
Hi Ted :)
 
Hi dc3.
 
@leslietownes the ancient greek pronunciations are pee and psee
 
Just a casual hello. Nothing to ask today....it's a stats day. Life good as it could be during these times?
 
"fee" and "see" are anglicizataions of the modern Greek pronunciations
 
I always said "fie" and "sai" especially "sai" because it reminded me of Raphael.
 
11:33 PM
yes, I say /fai/ and /sai/ too
 
The Ninja Turtle......not the scullptor
 
I like to say "puh-sai".
 
I say psee.
 
Lol.....is that geared to certain students when they tick you off?
Xander?
I guess both of you the question applies too
 
Because it annoys both the pretentious folk who are inclined to begin a sentence with "Well, actually..." then reference the fact that they took ancient (or MODERN!?) Greek in high school, and it annoys everyone else, too.
 
11:35 PM
there's actually people that take Greek?.....outside of somewhere that has Hellenic roots>?
 
@dc3rd One of the guys in my graduate cohort had a BS in ancient Greek.
(yes, a BS, not a BA)
 
there's a certain gif that shoots a look of inquisitiveness that would be perfect for this....it's with scowled eyes
 
i sometimes phoneticize french words to bother people. not enough people know greek to make it worthwhile.
 
@leslietownes Oh, I do that all the time!
And then I end whatever phrase I am horribly mangling with "...as the Norwegians say".
It has gotten to the point where whenever I say "Sest la vai", my daughter finishes my sentence with "...as the Norwegians say."
 
that's great. my daughter tends to repeat my uses of profanity. i have to watch myself very closely.
 
11:45 PM
Heh.
My reaction to Katja's use of profanity is typically to laugh, and then lecture her on the grammatically correct use of whatever word she is attempting to use.
 
Two troublemakers.
 
a lot of people randomly cross the road outside of crosswalks in my neighborhood, which is fine but sometimes i grumble about it. i'm not generally aware of what i say when i do it, but one time i was in the car with wife driving, and a guy jaywalked in front of us, and my daughter said "THIS f---in' guy."
if it's contextually appropriate and grammatical it's really hard to scold.
 
Hey, he's walkin' he-ah!
 
@TedShifrin are нра́виться and нра́вится pronounced the same?
@XanderHenderson then they should certainly know better that "CoRrEcT PrOuNuNcIaTiOn" isn't a thing because there are so many dialects and periods of greek
 
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