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00:00 - 19:0019:00 - 00:00

12:00 AM
I will email you better
I like the geogebra background
hi @BalarkaSen
how his face gently blends in
There are $4^2$ people in this chat. This is an even number. This implies that an even number of people are in this chat
hi @Adeek
12:01 AM
Who is that guy?
My proof has a major flaw
12:15 AM
Hey user do you know how to define repulsion mathematically?
@Daminark Thank you so much
@Ultradark Me?
12:32 AM
Can someone explain why $f(x)=\sin{1/x}$ is not uniformly continuous on (0,1]? I'm thinking intuitevly, since sine never gets bigger than 1, there must be a delta for which all epsilon are satisfied for, no?
12:50 AM
@user76284 yes
1:14 AM
@Daminark I sent you one final thing
I want to know your final opinion
i will probably use this final
1 hour later…
2:28 AM
@Ultradark Did you have a particular kind or example of repulsion in mind?
Repulsion between electric charges? Or in general relativity?
2 hours later…
4:24 AM
13 hours ago, by Silent
Does there exist a function $f:(0,1)\to\Bbb R$ that is twice differentiable such that $f$ bounded, $f'$ bounded but $f''$ unbounded?
@Silent I think $\begin{cases}x^2\sin(1/x),&x\ne0\\0,&x=0\end{cases}$?
10 hours ago, by Rithik Kapoor
is $\lim_{n \to 0} \frac{1}{n}\sin \left( \frac{x}{n} \right)$ a function?
10 hours ago, by Rithik Kapoor
also is it a space filling curve?
@RithikKapoor The limit does not exist for any number other than $0$
so no, it is not a function
and therefore it is not a space-filling curve
As an example, take $x=\pi/2$. Consider what happens when $n$ goes on the sequence $1,\frac12,\frac13,\frac14,\dots$
You get the sequence $\sin(\frac\pi2),2\sin(\pi),3\sin(\frac{3\pi}2),4\sin(2\pi),\dots$
which is $1,0,-3,0,\dots$
which has no limit
4:47 AM
Is the transitive closure of set membership describable in ZF? That is, does there exist a formula $\phi$ in the language of ZF such that $\phi(x,y) \leftrightarrow x \in^* y$?
5:06 AM
A: Can post-selecting on the screen in the Delayed Choice Quantum Eraser experiment be used to predict the quantum-eraser measurement results?

Emilio PisantyYou're over-interpreting these sketches - they are only sketches, and their specific details can't really be used to make any real predictions. Here is a more accurate version of those sketches, with a proper underpinning on a solid model of the experiment's behaviour: Mathematica source via ...

What infinite sets tend to look like to me: $R_{01}, R_{02}$
They tend to give me that impression of having a brush like structure
6:02 AM
@user76284 that's the same as $x \in \displaystyle \bigcup_{n=0}^\infty \cup^n y$
i.e. whether $x$ is in the transitive closure of $y$
6:24 AM
@LeakyNun Is that definable within the language of ZF itself?
I think so
Of course it's possible up to any finite $n$, but I'm not sure about whether there is a single formula that captures all $n$.
Like how it's impossible to say there's an infinite number of objects in first-order logic.
7:14 AM
1 hour later…
8:25 AM
$0 = \{\}$
$f(\omega) = \{\omega(0),\omega(1),...\}$
2 hours later…
10:51 AM
Given a profinite group $G$ I can define an inverse system on the sets of $p$-Sylow subgroups in each finite quotient of $G$ by just restricting the system on $G$ to those sets... does that make sense? In that case the inverse limit of such a system is a pro-$p$-Sylow subgroup of $G$, in the sense that its image in each finite quotient of $G$ has index prime to $p$ (since the image in each case is a $p$-Sylow subgroup of the relevant factor)
(And by restrict I mean that the morphisms in this system are just given by the restrictions of those for the factors of $G$)
11:10 AM
I guess that makes sense, and it means that the conjugacy of p-Sylow subgroups extends to the profinite case as well, by the connecting maps
If $F$ is a field, and $E$ a subfield of $F(X)$ containing a rational polynomial, must $E$ contain an honest-to-god polynomial?
My suspicion is that $E$ must contain a polynomial, but I'm having trouble proving the claim.
@user193319 I wonder if $E$ is somehow a localisation of $F[X]$ and whether or not this helps
Suppose $f(X),g(X) \neq 0$ and $f(X)/g(X) \in E$. If either $\deg f = 0$ or $\deg g = 0$ (but not both), then obviously $E$ contains a polynomial
Hmm...localization...possibly...I hadn't thought of that.
Possible way to construct counterexample: if I take a proper rational polynomial and look at the subfield generated by it, will it contain a polynomial?
E.g., the subfield generated by $f(X) = \frac{X}{X+1}$
11:50 AM
I can show that $F(X^2) \cap F(X^2-X)$ doesn't contain any polynomials, but does it contain any rational polynomials?
How about this: suppose $f(X) \in F\left( \frac{X}{X+1} \right) \cap F[X]$. Then $$f(X) = \sum_{n=0}^k a_n \left(\frac{X}{X+1}\right)^n$$ for some $a_n \in F$. We can turn this into a relation in $F[X]$ by noting that $f(X)(X+1)^k = \sum_{n=0}^k a_n(X+1)^{k-n}X^n$. Let $\deg f = m$. Then $\deg f(X)(X+1)^k = m + k$ while $\deg \sum_{n = 0}^k a_n(X+1)^{k-n}X^n \leq k$, so $m+k \leq k$ ?
Wonder if that works..
12:11 PM
I don't understand. $1 - X/(X+1) = 1/(X+1)$, and inverse of that is $X + 1$. That's a pure polynomial contained in $F(X/(X+1))$.
@BalarkaSen Ah, very good. What about $F(X^2) \cap F(X^2-X)$?
In fact, by Luroth's theorem, any proper extension squished between $F(X)$ and $F$ is $F(g(X))$ for some rational function $g$. You should be able to prove using that that $E$ always contains a pure polynomial.
@BalarkaSen the dream
Take the field E fixed by the F-automorphism F(X) -> F(X) given by mapping X -> X/X-1.

Pretty sure one can show that no polynomials are fixed
12:18 PM
That's the same thing as X -> 1/X isn't it, if you do a linear change of coordinates, X = Y + 1.
Let's say F = C. This automorphism, X -> 1/X, corresponds to the biholomorphic automorphism of P^1 switching the two poles. Any function which has a pole at 0 and \infty are fixed, eg X + 1/X. That's what E entails
No polynomial is there certainly; they don't have a pole at 0
E = F(X + 1/X) I am sure, because X + 1/X is "degree 1" and Luroth
@user193319 So I suppose that's your exampl
@BalarkaSen Oh yeah - I just happened to know this example specifically from a pset in a class that I was TAing lol
Ah, so $E = F(X+\frac{1}{X})$ doesn't contain polynomials. Thanks!
@BalarkaSen When you speak of "poles", are you taking $F = \Bbb{C}$?
Ya but it doesn't matter, I just like to be overly geometric
No, geometry is good. I need to build up my geometric intuition. Thanks for the counterexample!
Basically the problem becomes immediate if you think of $F(X)$ as the field of functions on $\Bbb P^1_F$, you're looking for a subfield containing no function which just has a pole at infinity (these are precisely the polynomials).
12:29 PM
What is $\Bbb P^1_F$?
projective line on $F$, the analogue of Riemann sphere for arbitrary fields
Oh, I see. Thanks.
1 hour later…
1:43 PM
@user76284 If you go through the proof that $\mathrm{trcl}(x)$ exists in $\mathsf{ZF}$ (show that $\bigcup^nx$ exists via the recursion theorem, use replacement on $\omega$ to show that $\{\bigcup^n x\mid n\in\omega\}$ exists and then take the union) it looks like this should be unfoldable to some $\in$-formula, even though it's going to be kinda painful
Way easier than that $\phi(x,y)=\forall z(z \text{ is transitive})\land x\subseteq z\rightarrow y\in z$ should hold iff $y\in\mathrm{trcl}(x)$
Which, to be 100% pedantic, can be expanded to the $\in$-formula $\phi(x,y)=\forall z\forall a\forall b(a\in b\land b\in z\rightarrow a\in z)\land\forall w(w\in x\rightarrow w\in z)\rightarrow y\in z$
2:14 PM
Here's a cool one: Let $k$ be a perfect field and $p$ a prime. Show that there exists an algebraic extension $k^{(p)}/k$ such that each finite subextension has degree prime to $p$ and such that $k^{(p)}/k$ has no non-trivial finite extensions of degree prime to $p$.
2:30 PM
In mathematics, what precisely is the meaning of

Or you mean greater than or equal to?
Not greater than or equal to.
You mean this: ⇒
That's implication
2:32 PM
Does it have any other meaning?
A ⇒ B == A implies B == if A then B
It has whatever meaning the person who uses it intends (which is typically what I write above)
Okay, thanks a lot :)
@SebastianAlexanderBNielsen Your description has the shrugging person with their right arm missin
You might want to fix that
I fixed it ... people won't get confused by a confused text emoji missing his left arm anymore.
Stage left arm
2:43 PM
@SebastianAlexanderBNielsen Thanks, that was important (obviously)
3:05 PM
@AlessandroCodenotti Is there a typo in your formula? $z$ is free in the second conjunct. And is the outer $\rightarrow$ inside or outside this conjunct?
Did you mean $\forall z ((z \text{ is transitive} \land x \subseteq z) \rightarrow y \in z)$
^Check out this link
Two essays on how quantum computing works
Highly recommended
You can search an unsorted database of size N in $O(\sqrt N)$ time
The no cloning theorem has a great easy proof.
That is, if you have a non-zero Hilbert space and a state $|\psi\rangle\in H$, you cannot find a linear $T\colon H\to H\otimes H$ such that $$T|\psi\rangle = |\psi\rangle\otimes|\psi\rangle,\quad\forall |\psi\rangle\in H$$
For such a commonly known theorem, it's quite easy.
3:25 PM
Hmmm... I'm having a lot more trouble than I care to admit trying to prove (p ∨ q) ∨ r ↔ p ∨ (q ∨ r). In a discrete math course I would likely just write out the truth table, but I'm trying to write a program that illustrates the proof. :/ Something is not clicking.
@Dair are you wanting something akin to a deduction proof, or like Coq?
@anakhro Lean specifically, but yeah more in the vein of Coq.
Unfortunately I don't know how Coq and Lean work all too well.
Same. I'm trying to get better at understanding it.
But if it is anything like a deductive argument, then you could try showing two things:
$$(p\vee q)\rightarrow (p\vee(q\vee r))\tag{1}$$
$$r\rightarrow (p\vee(q\vee r))\tag{2}$$
And it follows from that.
3:32 PM
There is something like that, but the tutorial I have uses or.intro_{right|left} and or.elim... The or.intro* stuff seems to make sense, I'm having some trouble internalizing what or.elim does exactly.
(2) will just be two introductions of $\vee$
What an elimination of $\vee$ does should be the following:
$$\frac{p\rightarrow r,\quad q\rightarrow r,\quad p\vee q}{r}$$
Which will be handy for (1).
thanks, i'm probably going to look at this for a while haha... I think I might need to draw it out before I code it out. :P
Yeah, try making a deductive proof first, then translate it to make sure it works.
4:19 PM
ok thanks @anakhro I got it figured out lmao.
The proof I coded is stupidly long lol, almost certainly can be improved haha.
Shimura passed away.
A week ago
Yeah, pretty crappy, but he lived a long time.
4:51 PM
all these old mathematicians are dropping like flies lately
@Eric: I'll try to drop like a cockroach for you :P
@TedShifrin So in case of a nuclear holocaust you'll still be alive?
I doubt it.
but cockroaches can survive a nuclear fallout.
4:56 PM
this bit is dead end it
back to logic proofs. cya.
Hey @Ted
RIP Shimura
5:15 PM
24 mins ago, by Érico Melo Silva
all these old mathematicians are dropping like flies lately
The Prophecy is realising...
Only 2 days left
preparing the hyperbolic manifold made of Dedekind finite sets interfaced with a projection of the subspace of the amorphous manifold in 40 hours...
I would assume the death rate of mathematicians will only increase, assuming the birth rate of mathematicians is increasing.
just no more another age of boomers please, they tend to screw everything
"metrically amorphous" manifolds??
hmm... I wonder, how can one define an amorphous metric...
probably some map $d$ such that it satisfy the axioms of a metric, but having some extra structure that scrambles the details of it just enough that $d$ is not any arbitrary metric, but otherwise cannot be pinned down exactly
Will explore this later...
@TedShifrin hi Ted
can you please help me with the question on the final I didn't know the answer to?
one possible way is to consider an equivalence class of metrics and the topologies they indluces
Theorem. Let $R$ be a ring such that $R$ is a subring of a division ring $D$. If for all $d(\ne 0)\in D$ either $d\in R$ or $d^{-1}\in R$ then $R$ is a local ring.
5:28 PM
My proof. It suffices to prove that $R\setminus U(R)$ is an ideal of $R$ where $U(R)$ denotes the set of all units of $R$. For this we first show that if $x\in R\setminus U(R)$ then $Rx\subseteq R\setminus U(R)$. So let $x\in R\setminus U(R)$. Clearly then $x\ne 0$. Suppose that there exists $r\in R$ such that $rx\in U(R)$. Consequently, since $D$ is a division ring, we have $x\in U(R)$, a contradiction. So $Rx\subseteq R\setminus U(R)$ for all $x\in R\setminus U(R)$.
Let $x,y\in R\setminus U(R)$. Let $x^{-1},y^{-1}$ denote the inverses of $x$ and $y$ respectively in $D$ (clearly both $x$ and $y$ are non-zero so that they are invertible in $D$). Then note that by hypothesis $xy^{-1}\in R$ or $yx^{-1}\in R$. Since $1\in R$ it follows that either $1+xy^{-1}\in R$ or $1+yx^{-1}\in R$. In the former case, $$(1+xy^{-1})y\in R\setminus U(R)\Rightarrow y+x\in R\setminus U(R)$$ and in the later
$$(1+yx^{-1})x\in R\setminus U(R)\Rightarrow x+y\in R\setminus U(R)$$
So we are done.
Is this proof ok?
hi @ÍgjøgnumMeg, @GFauxPas. What's the question?
5:34 PM
let $f$ be a positive smooth real valued function on $[a..b]$;
let $E$ be the surface generated by rotating $f([a..b])$ around the $x$-axis in $\mathbb R^3$
find the normal vector field to $E$
You're in $\Bbb R^3$?
The normal vector field? What is "the"?
well find one of the two unit normal vector fields
So can you give me a cartesian equation for the surface? Or a parametrization?
BTW, why are you using Maple notation for an interval $[a,b]$?
just a habit of mine from editing a wiki that used that notation
It's horrible notation unless you're working in Maple.
5:37 PM
Can someone give an intuition when a point moves across a circle with a unit speed why does acceleration always points towards the center of a circle no matter how itpoint travels over it, clock-wise or counter-clockwise
I'm talking about f(t) = (cos(t),sin(t)) and f(t) = (-cos(t),-sin(t))
@famesyasd: Draw the velocity vector field around the circle and see how it changes in a little time $\Delta t$.
I think the parameterization is $x = f(t) \cos \theta, y = f(t) \sin \theta$, or maybe I have $\sin$ and $\cos$ switched
By the way, your second parametrization is wrong. It still goes counterclockwise.
You're rotating around the $x$-axis, @GFauxPas, so the circles are parallel to the $yz$-plane.
let me draw a picture for myself
gonna assume $a > 0$ to make it easier to draw
5:39 PM
Notice that the velocity vector is the position vector rotated $90^circ$, @famesyasd. So that tells you that its derivative is ...
@TedShifrin yeah, but I hoped to find some other reason to see why it should be this way
I got Tu's differential geometry book from the Yellow Sale but I have to do other stuff before I read it. :(
okay so if I drew it correctly I'm thinking it's
$z = f(t) \cos \theta; y = f(t) \sin \theta$?
Well, @famesyasd, I told you to draw pictures. That's intuitive. But look at it this way: If the acceleration were NOT pointing inwards, the particle would fly off the circle.
I would switch $y$ and $z$, @GFauxPas, but that's fine.
okay, I don't know what $x$ is
5:42 PM
Duh. $x=t$.
this would also work differently if the motion weren't uniform, e.g. $f(t)=(\cos \omega t,\sin \omega t)$ with $\omega$ a function of time @famesyasd
Because you're looking at the graph of a function of $x$. So $(x,f(x))$ becomes $(t,f(t))$ if you're putting in $t$ in your formulas.
ah, gotcha
that's not a bad exercise, insofar as it shows the importance of $\omega=1$
5:44 PM
okay so I've parameterized $E$
If you have a parametrization, how do you find the normal vector?
solve $\langle n(v), T_vE \rangle = 0$?
How do you find a vector in $\Bbb R^3$ orthogonal to two vectors?
oh, cross product
Easier yet, can you give me a cartesian equation that says you have a circle of radius $f(x)$ parallel to the $yz$-plane at $x$?
5:50 PM
gonna be something squared + something squared $=f(x)^2$, thinking about what the somethings are
Hi, DogAteMy.
What do you call the surface $z=ax^2-ay^2$?
Specifically as opposed to the case $z=ax^2-by^2$, $~a\ne b$.
A hyperbolic paraboloid, I think. I call it a saddle.
$y^2(t,\theta) + z^2(t,\theta) = f^2(t)$?
I'd call it a "square" hyperbolic paraboloid
5:51 PM
Oh, there's no particular name that I know, DogAteMy.
and the other one a "rectangular" hyperbolic paraboloid
@GFauxPas: We want just an $(x,y,z)$ equation. No parameters.
but I dunno if that's standard
It's not.
5:52 PM
(Or maybe "right" because if you intersect it with its tangent plane at the origin, you get two lines intersecting at right angles?)
I'm not particularly interested in names for things, DogAteMy.
so just $y^2 + z^2 = x^2$?
Anyone? any feedback?
I'm not sure
5:59 PM
@GFauxPas: You yourself told me the right-hand side should be $f(x)^2$.
LOL, so?
hi demonic @Alessandro
so $y^2 + z^2 = f(x)^2$?
@user76284 yes
6:01 PM
Right, @GFauxPas. Now how do you get a normal vector?
oh it's a circle at each slice
so it's just gonna be pointing at the circle
didnt think it could be that simple
Don't you know how to find the normal to a level set $F(x,y,z)=0$?
its the gradient of $F$
6:03 PM
Now understand why your formula is incorrect.
so $2\left[ {-f'(x), y,z }\right]^T$?
uh oh, Ted is growling at me :(
A smack is next.
And, even when you fix this, of course you don't have a unit vector (but I don't really care about that).
wait I meant
$-ff'$ for the first coordinate?
6:07 PM
$(\text{normalizing constant}) [-f(x)f'(x), y, z ]^T$?
With $x$'s in $f$ and $f'$, yeah.
Now you can check that you get the same answer working with the parametrization.
okay so if I did it that way I need 2 vectors at each point
with cross product
And how do you get a basis for the tangent plane if you have a parametrization?
the kernel of the derivative matrix
6:11 PM
people with the blue patterns for pictures remind me of Mike. :(
the image of the derivative matrix
the columns
The kernel of the derivative is the implicit case where you took the gradient as the normal, @GFauxPas. Yes, the image.
lesse then
posted a long ass probability answer lol
my first answer in the probability tag im sure
6:16 PM
Mazltov, a @Balarka :)
waiti no
ignorte that line
well golly jee
turns out out the theorems are consistent
Yes, I can prove that :P
thanks for your help Ted
You're welcome.
6:28 PM
@TedShifrin I have a question for you. Say $X \subset M$ is a Whitney stratified subset. Denote $TX \subset TM$ to be the union of the tangent bundles to each strata in $X$.
how do I know whether the normal points out or in without drawing a picture in the first method?
I know cross product gives you the right hand rule pointing out
But be careful. It depends on the order of the basis vectors.
Whitney condition (a) basically says something like $TX$ is "topologically stratified", i.e., if $S < L$ are a pair of strata in $X$, $TL$ contains $TS$ in the frontier in $TX \subset M$.
What does condition (b) mean in terms of $TX$?
In the implicit case, you have to think about in what direction $F$ increases ...
I don't remember the condition, @Balarka. It's been a decade or more.
6:30 PM
which is towards the area of steepest ascent, right?
the gradient
But you have to look and see if your $F$ (or its negative) increases as you move "out."
Obviously, you could take any nonzero scalar multiple of $F$ and the answer changes.
@TedShifrin I can tell you if you want, but if you're busy/not interested, that's completely fine.
thanks for your help Tecd
I used to know this stuff, a @Balarka, but now I've forgotten everything I once knew (meme, anyone?).
Hahah. I know, and I still think you might have valuable intuition about these.
6:34 PM
hi chat
Hi @SemiC
I need to either sleep more or drink more coffee
Cigarettes. Middle ground.
(and/or exercise more)
@BalarkaSen riiiiight
6:35 PM
NOOOOO cigarettes.
hi @Semiclassic
i've managed to entirely avoid the appeal of cigs
Lung cancer is so attractive.
oh, truly
did u quit or did u never use to smoke
6:36 PM
worse than lung cancer is copd because in lung cancer you at least die at the end
Smoking is for chumps.
and emphysema is really how i want to spend the end of my life /s
@TedShifrin I know you hate smoking
I am a hateful person.
it was a question to semi
@anakhro come at me bro
6:37 PM
OK. :)
just so long as you hate the right things
@BalarkaSen never started
@BalarkaSen I don't believe that you smoke.
@anakhro OK :)
But I do
by contrast, I was willing to drink alcohol after i graduated college but wasn't during it
6:38 PM
@BalarkaSen no way jose
You are way smarter than that.
@Semiclassical I have probably drunk less alcohol after finishing university than I did during :)
(in total I mean)
@TobiasKildetoft on my part it probably helps that i didn't go to parties etc when i was in high school, so drinking didn't get normalized for me until college
and at that point i still stuck to "leave it be" until i graduated
i mean, it depends on how much you do, but cigarettes seem like a far worse lifestyle choice than alcohol
Depends on how much you drink.
<--- still drinks
6:40 PM
I don't want to try drinking tbh
at least alcohol is only addictive to some percentage of people
I think you're more of an addictive personality, a @Balarka. :)
As long as you don't over-drink, yes, alcohol is less damaging. However, people who over-drink are not only damaging to themselves but often to others as well.
Cigarettes is just a dumb idea in general because it's just self-damaging.
6:42 PM
yeah, that's true. "some drinking" is better than "some smoking" but "a lot of drinking" is probably worse than "a lot of smoking"
Well, second-hand smoke can kill, too.
yeah, they're not great habits either way
Passive smoking is also not quite good. Far less damaging than active smoking but still not good on a regular basis
That's what second-hand smoke is? Or do you mean something else?
@anakhro I'm an accelerationist. If I have to die I'll die faster.
@TedShifirin I thought second hand was sharing
Passive is inhaling smoke
6:44 PM
that's what second hand is too
Sounds like something a person would say when their actual reason for smoking is much less inspired, @BalarkaSen
sharing is just smoking except you didn't pay for it :)
On the bright side I'm not doing crack cocaine
Consider that
just because there are worse things you could be doing, doesn't make what you're doing better :P
6:45 PM
Thanks, a @Balarka. No meth, either.
Yes, comparing bad things to horrible things makes people ignore the fact you are doing bad things.
"tbh, Judge, consider that I could have killed him instead of just badly injured him!"
To which the judge responds, "Oh you're right! Good point! No jail for you!", right???
Did this used to be a math chatroom?
(That's the craziest syntactical structure I know in English. WTF.)
Problematic grammar there @TedShifrin
6:48 PM
Mathematics? I thought it said methamphetamines!
Wait, that's actually grammatically correct?
I think that's how it goes, though, a @Balarka. I just commented.
I am pretty sure it is grammatically correct.
Well, I will sit corrected if someone has a source. :)
6:49 PM
Though it's probably debatable whether it is chatroom or chat room.
I would have thought "did this use to be" was the correct term. But then again, "used to be" can't really be put into any other tense.
There are some locutions in English that are just crazee.
This summer I hope to get around to reading those characteristic classes notes you gave me, Ted. I had written a super condensed set of notes last summer as a speedy introduction to forms and integration (assuming only the definition of manifolds) for some friends, and I think I will do something similar for characteristic classes and other stuff in that vein this summer.
nothing too strange about this, but I heard someone using the word "licit" sensibly during a talk yesterday
6:56 PM
That's a lot more ambitious, @anakhro.
I remember going on a hike during a college retreat in Vermont discussing words in English that have only negated form. We thought of a number.
Flammable and inflammable don't count.
licit is an example where the negated form is far more accessible in my brain than the positive form, despite both being sensible
I'm imagine there's cases where the original form is now archaic whereas the negated form has stuck around
I'll try to remember a few more.
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