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11:00 PM
@Balarka graduated from the school of Tedsmacks, and now i'm the new student
 
@MikeMiller got any suggestions for classes to take in the math department if I want to continue optimization/ML?
 
I am not your guy for that question
 
Ah right - forgot. Abstract algebra
 
I don't even want to remind myself what a obnoxious and pompous ass I was a couple years ago. I deserved the smacks.
 
@Fargle what do u mean
 
11:02 PM
@BalarkaSen Ah yes, I have these same memories of puberty
 
@Balarka too bad im still going through that
anyways, time for some delicious italian meatballs!!!
 
@Zach You are nowhere close to my past self.
 
adios people
 
peace
 
11:03 PM
Hi.
 
ayy
 
can it be simplified this? $\dfrac{|x|c}{|x| + \lambda}$
I wonder if the result is $\mbox{sign}(c)(|c|-\lambda)^{+}$
 
@Krijn That maturity wasn't in any way triggered by biological reasons, certainly. [I like to give Tarkovsky credit for it :)]
 
No, you cannot eliminate the dependence on $x$. See, for instance, $c = 1, \lambda = 1$.
 
I mean I still see a lot of adult pompous asses around
 
11:05 PM
Ah, @MikeMiller I proved the equivalence between norms without using continuity.
 
I believe that you didn't use the word, but what I was trying to claim is that your proof would more or less be the same thing in different language.
 
Well, the problem I'm working on is, find the minimum of $h(x)=\frac{(x-c)^2}{2}+\lambda|x|$ where $\lambda > 0$, $c \in \mathbb{R}$
Yeah, that is hard to notice.
 
Anyway, I'm off to bed, good night!
 
Night
I should get some sleep too
 
I got $h'(x) = 0 $ if $x = \dfrac{|x|c}{|x| + \lambda}$ with $ x \neq 0$
But the answer of the problem it seems $x_m=\mbox{sign}(c)(|c|-\lambda)^{+}$
 
11:09 PM
They aren't equal
$\frac{1\cdot1}{1+1} \neq 1\cdot(1-1)$
I'm not sure what ${(\cdots)}^+$ means here
 
$(a)^+=\max(a,0)$
 
Ok then you can add the plus to above and it still holds
 
It seems they give a wrong solution.
Mhmm or not.
 
@BalarkaSen Is every compact totally disconnected perfect space the Cantor set? (Or a point?)
 
Well, if $x > 0$ then $ x = c - \lambda$, if $x < 0$ then $x = c + \lambda$
 
11:18 PM
whats the name given to the symbol like the = sign but is 3 lines instead of 2?
 
congruence?
 
/equiv
it means congruence or 'identically'
 
so its not some greek letter or anything?
 
no
do you mean $\Theta$?
or $\Xi$
 
yeah stuff like that
 
11:20 PM
\Xi is greek, \equiv is not
 
so you don't mean $\equiv$
 
google what capital Xi looks like if you don't have TeX installed
 
yeah i do mean that topo just wondered if it originated from some language
 
it's just the equals sign with another line to mean "a kind of equality but maybe not the one you're used to"
 
there we go
 
11:21 PM
@MikeMiller This is not valid since $x \neq 0$
 
it's called the "kindof equality, but not in the way you're used to" operator
 
But my fault, I didn't say it.
 
unless you meant $\Xi$... but come on that's not much like an equal sign really
though it does leave me wanting to say $\Xi \approx \equiv$
 
no it was to do with congruences that i saw the symbol
i believe Gauss used it originally
but don't think he gave it a simple name
 
fucking Gauss
always making life amazing but not simple
 
11:31 PM
@robjohn @DanielFischer @PedroTamaroff @mixedmath how much stuff these users have to still say before being banned for some years? @BalarkaSen @Krijn???
I've just received a message these one still say all kind of s*it on the main.
 
what are they saying
 
When will the ban come? Maybe a permanent one? Or I have to remove my my account completely?
Next time I removed my account.
 
where are the chat logs
 
what are you even talking about
and hey @Akiva
 
@Then What's going on?
 
11:34 PM
Hejhej
 
Stay civil, please.
 
don't think it happened in here pedro
 
@Akiva I have an easy problem that Ted just shared with me
 
interested noises
 
maybe you;d like to try and solve it as well
it's not that hard though, because even I could solve it
 
11:35 PM
Um
 
ok
find 2 mappings
from $\Bbb R$ to itself
such that
both are convex, simple curves
and they intersect infinitely many times
(can't be same curve)
 
The curves being the graphs of the functions?
 
yep
 
What does simple mean here
 
idk, just let them be infinitely differentiable
 
11:37 PM
OK
 
@Then Calm down.
 
I'm out now (hope for many other months)
Just don't distrub me. OK?
bye bye
 
I'll try not to distrub you
:)
 
bye!
 
11:39 PM
@ZachHauk $x^2-\cos x$ and $x^2$
 
yep, that was my solution
except with sin x
you did it much faster, though
it took me like 2 or 3 minutes
 
Graphing stuff on my phone helped
 
@Then come back we didn't throw a going away party yet
we even got pi
 
I don't know if I'm filled with rage or delight about this punny business
 
$\Huge\pi$
 
11:41 PM
can't it be both :D
 
$\tiny\pi$
 
Have a slice! $\Large\tau$
 
I can't eat it all
let me just take $\tau/4$
 
Wait, we've misspelled it
$\Large\pi e$
There we go
 
is that transcendental?
 
11:43 PM
About eight and a half
 
because the pie is so good it transcends this universe
 
@ZachHauk Wouldn't be surprised if that's an open question
In fact, I don't think we know whether or not it's even irrational.
 
wow, these people do math for a job and still suck at it /s
 
was going to order the monster group to play some music at the party too :(
 
doesn't that have an order of like 200,000?
Sorry dude... I think you should just invite the Klein Four Group
 
11:45 PM
Or, for the geography geeks, we could get St. Vincent and the Grenadines to play
@ZachHauk It's much larger, I think
 
oh that's what it was, if $e\pi$ is rational then $e+\pi$ can't be
 
808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000
 
ah yes,
 
It's the symmetry group of a 200,000-dimensional object, though, I think.
 
> 200,000-dimensional
pls no :'(
 
11:48 PM
Where spheres have negligible volume!
 
@Akiva can you give me another problem?
if you have any? :P
oh, the Crazy Ramanujan Kid just texted me
hey @heather, long time no see
 
@ZachHauk, hello, how are you?
did you make into the Math Camp?
 
parents won't let me go
 
oh, I'm sorry.
 
and, regarding how i'm doing, I'm just stressed... and a half.
how about you?
 
11:50 PM
pretty middling.
stressing too much about stuff i don't actually have to do
 
how is fizzics going?
 
rather well =)
 
whatcha studying?
 
quantum error correction
it's exciting stuff
 
sounds exciting
unless you make an error
 
11:51 PM
and i'm extra happy, because i wrote a practice paper (to get used to the style of writing a research paper) on it
 
then you have to use quantum error correction to quantumly correct your quantum error
 
@PedroTamaroff did I miss something?
 
hi @mixedmath another mod I've never seen
 
@ZachHauk lol
 
@ZachHauk hello
 
11:52 PM
do you guys just like lurk around here?
 
0
Q: Riemann zeta and quantum physics?

mickSometimes i read about connections between " advanced math " and quantum physics. But often I am skeptical. I can believe or understand the connections to calculus , vector calculus, differential equations , linear algebra. But when i read about connections with prime Numbers and the Riemann z...

 
@ZachHauk I infrequently write much in chat
At least, in this chat
 
-1
Q: Riemann zeta and quantum physics ??

mickSometimes i read about connections between " advanced math " and quantum physics. But often I am skeptical. I can believe or understand the connections to calculus , vector calculus, differential equations , linear algebra. But when i read about connections with prime Numbers and the Riemann z...

 
@mick please don't post twice
if someone knows the answer (@heather perhaps) they'll click it and answer it
 
I did not post twice !
I crossposted
To physics se
 
11:54 PM
uh... is that allowed?
 
(nope - and now ours is closed)
 
@ZachHauk Here's one that I think should be easy, but I'm not quite sure how to do it.

Suppose you have an infinite chessboard, with some squares painted blue, some squares painted orange, and some left unpainted. Suppose that no blue square is adjacent to an orange square (not even diagonally).

Also, suppose there are squares labeled X and Y. Suppose there's a path between X and Y that avoids blue squares (only goes through orange and unpainted squares), and a path between them that avoids orange squares. How can we prove that there's a path between them that avoids both colors (only goe
 
Well Maybe one of them Will be closed ... As long as i get the answer
Downvotes annoying me :(
 
man SE chat devs have too much free time... when you tagged mick it highlighted on my screen, does it work for you @Zach ?
 
@AkivaWeinberger well obviously both squares X and Y are unpainted
 
11:57 PM
I'm very impressed with @heather 's willingness to edit that question into a more appealing format
 
@ZachHauk Yes.
 
@Akiva maybe you could show that the complement of the blue and orange squares is totally connected
 

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