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12:19 AM
@BalarkaSen So cockroaches are your first choice?
 
1:03 AM
I've recently learned about projective geometry. The question i have is regarding finding the equation of asymototes. The procedure I understand for fiding the equation of the asymptote of a curve $F(x,y) = 0$ when the point at infinity is real is as follows:
1. Homogenezie the equation $F(x, y) = 0$ by substiting $x = X/W$ and $y = Y/W$; simplify the expression
2. Find points at infinity by setting W = 0 and solving the equation for Y; let's say one of the points at infinity have the homogenous coordinates of [a, b, 0]
This procedure seems to still work when the points at infinity are complex - for instance, if i find the equations for the asymptotes of a circle $x^2+y^2=R^2$ using the procedure above i seem to get the correct equations of $x+iy=0$ and $x-iy=0$
But isnt the sitatuon when the points at infinity are complex entirely different as we have to treat the function as a complex function right? Is there an intuitive explanation why this procedure still works?
 
1:24 AM
David, if we’re talking about complex points at infinity, we’re now working in the complex projective plane.
 
Ooooh right right
I'm guessing the constructuon of that isn't as simple to explain as the real projective plane
 
Sure. It’s the same. Just remember that your real circle above becomes very non-compact working over $\Bbb C$, so asymptotes make sense.
 
From thinking about it myself, i couldn't really understand what a complex point at infinity really is as well; on the real projective plane, ppl talk about the points at infinity as representing the slope of the line
 
You can still think slope. It’s just a complex number (or infinity), hence a point in the complex projective line.
 
But what does slope mean when both x and y are complex? Am i thinking about it in a wrong way as in im trying to make it too geometrical?
 
1:29 AM
You can’t think of rise and run as real, of course.
This is a complex line, not a real one.
$w=cz$ with $c$ complex.
 
Considering i have no idea what non-compactness/compactness means, i'm just going to shelve this under "come back later" when i have learned more math
 
Non-compact means it goes off to infinity, in this case, unlike a real circle.
But projectively a circle, parabola, and hyperbola are all the same in the real case, anyhow.
 
@TedShifrin hmmm do you mean the circle goes off to on the complex projective plane? I understand the notion of a number going off into infinity - its value grows and grows - but what does it mean for a closed geometrical object like a circle to go off into infinity?
 
1:46 AM
It means “circle” is a misleading word. It becomes a hyperbola. Like $xy=1$, it becomes $(x+iy)(x-iy)=1$.
Think of this as analogous to the linear change of coordinates that turns $x^2-y^2=1$ into $xy=1$.
 
2:42 AM
Thanks a lot @BalarkaSen for the good wishes :-).
Given A,B subsets of R, sup(AUB)<= sup A+ sup B.
I think this is not obvious.
I want to prove that the sequence $(1+x^{2n})^{\frac 1{2n}}$ converges uniformly on R.
The sequence converges pointwise to f(x):= 1, |x|<=1 and |x| whenever |x|>1.
I am stuck at showing the convergence to be uniform.
 
@Koro bizarre. Why not max?
 
Fun fact, sup(A + B) = sup(A) + sup(B) is also true.
A + B = {a + b : a in A, b in B} being the Minkowski sum
I agree with Ted though that sup(A U B) is just bounded by (in fact equal to) max{sup(A), sup(B)}
 
Yes, that was what I would have typed at the computer.
Typing on phone/tablet sucks.
 
I agree
I don't like the mobile version of the chat
 
Half the time I can’t get the refer to or trash buttons.
And correcting mistypings is horrid.
 
2:57 AM
Word
 
@Koro. That inequality is garbage. Take $A$ positive and $B$ very negative.
 
3:10 AM
Heyo
 
Hi @AkivaWeinberger
 
Spent my plane ride reading about enumeration combinatorics
Look at this bonkers result
I guess I can copy/paste, actually
> A domino tower is a stack of horizontal 2 × 1 bricks in a brickwork pattern, so that no brick is directly above another brick, such that the bricks on the bottom level are contiguous, and every higher brick is (half) supported on at least one brick in the row below it. Let the size of a domino tower be the number of bricks. See Figure 1.7 for an illustration.
> [image]
> Remarkably, there are $3^{n-1}$ domino towers consisting of $n$ bricks. Equally remarkably, no simple bijection is known. The nicest argument known is as follows.
 
Why is it remarkable
 
Because it's such a clean result, and apparently there's no known clean bijective proof
It coincidentally comes out to $3^{n-1}$
 
Hm, OK
 
3:14 AM
(unless there's a simple proof everyone missed)
Incidentally
 
@TedShifrin I have been thinking about tangential Gauss maps recently.
 
Are you still thinking about the "sum of two arcs equals the sum of three arcs" puzzle
or can I spoil it
 
Yeah you can spoil it
I don't think I'll get it in the near future
 
So I have three different-ish answers but here's the simplest one to look at
 
@BalarkaSen what other Gauss maps are there, really?
 
3:16 AM
Line through the heart of a double spiral
 
@TedShifrin The normal Gauss map, I suppose.
 
Double spiral + line = red + blue + black
 
Oh cool
 
The fun thing is that the double spiral goes into the bad point and then goes back out again
 
It’s still the mapping to the Grassmannian of (tangent) hyperplanes.
 
3:17 AM
Near the bad point, one decomposition is two arcs that go in and out, and the other decomposition is two arcs that end at the bad point
 
@TedShifrin True, they carry the same information.
 
My thesis was all Gauss maps to incidence correspondences.
 
@BalarkaSen My first solution didn't have a spiral at all, the bad point was the rightmost point (so the lines didn't surround it), but in retrospect it's similar to this version sort-of "folded onto itself"
 
I like this solution, although I have not seen the earlier iterations
 
Here's iteration 3
So here's a double spiral
Here's another one
Together:
Let's do something weird and flip the colors in the top-right quadrant
Here's the black:
and the red.
Sorry for the image spam
 
3:23 AM
Neat pictures. I follow so far.
 
@BalarkaSen So the union of two double spirals equals the union of two wacky single spirals, which is pretty neat in of itself
To get an answer to the question, restrict to the portion of the curves that fit within the frame
Note that each double spiral is a single arc
The same is true of the red single spiral
 
@TedShifrin The inequality is not correct. But I think it is true if A, B have at least one non negative element.
 
but the black single spiral is cut in two pieces by the frame
 
@BalarkaSen yes.
 
So the second decomposition is actually three arcs
 
3:25 AM
Oh perfect
So the third arc is the wacky cuspidal curve on the top
Going from left edge to right edge of the frame
 
Yeah exactly
 
Cool
 
Also I don't know why the black single spiral and the red single spiral look so different to me
They're the same curve, just reflected on the diagonal, but my brain goes "wideboi, tallboi"
 
The uniform convergence for the question would follow from: $\sup_{x\in \mathbb R}|(1+x^{2n})^{\frac 1{2n}}-f(x)|\le \sup_{|x|\le 1} |(1+x^{2n})^{\frac 1{2n}}-1|+\sup_{|x|>1}|(1+x^{2n})^{\frac 1{2n}}-|x||$
 
@Koro Just go with the correct one that’s tight.
 
3:27 AM
@AkivaWeinberger Lol
So here is the situation for my Gauss map ponderings. Suppose $f : M \to N$ is a fixed smooth embedding of a closed manifold $M^m$ inside a closed manifold $N^n$, and let $G(df) : M \to \mathrm{Gr}_m(N)$ be the (tangential) Gauss map. Suppose $G_t : M \to \mathrm{Gr}_m(N)$ is a homotopy of the Gauss map starting at $G_0 = G(df)$, by rotating the planes of the Gauss map.
In general it's $G_t$ is not going to be the Gauss maps of anything. The planes have been disturbed too badly to integrate
 
@BalarkaSen It's also neat that it's not an accident of how I chose the frame / how I chose to crop it; any zoom will do basically the same thing because all that really matters is what happens near the center
 
@AkivaWeinberger Yeah it's about connectivity of the double spiral really
And that specific color exchange move somehow spits out a new component
I was confused for 2 seconds because it seemed ad hoc, and dependent on the frame
 
So somewhere around here I actually have an imgur album that explains my first answer
so I won't have to image spam
 
@BalarkaSen In fact, $G_t$ may not be even approximable by Gauss maps of any thing.
 
3:32 AM
However, if you allow mild singularities, it suddenly becomes true. There is a homotopy $f_t : M \to N$ of topological embeddings with so-called "wrinkle singularities" starting at $f_0 = f$ such that $G(df_t)$ is $C^0$-close to $G_t$ for all $t$.
Here is what wrinkle singularities look like:
The top row is a representation of the wrinkle as a movie of birth and death of a pair of semicubicle cusps
The caveat is that a wrinkle is only a neighborhood of the sphere (circle here, in dimension 2) traced out by the pair of cusp points.
This allows us to have wrinkles inside wrinkles:
something like this
 
@BalarkaSen I also have this solution for the same question but with "subsets homeomorphic to an open disk" instead of "arcs"
You'll notice that the solution from the Imgur album (and blackboard picture) are basically this but cut off on the left and "crushed" on the right
This is why I was surprised when you brought up that braid homology diagram thing @BalarkaSen
 
Oh I see
 
It was very close to my first solution
 
It's the simplest example of union of three connected sets which has a fourth one hidden I could think of
 
I have no,idea, Balarka, but it feels like you can integrate if you pass to appropriate blow-ups.
 
3:38 AM
That's the lemma really
 
And again, note that the decomposition into 2 arcs uses arcs that go into the bad point and back out again
 
I wonder how this relates to prolongation of the differential system …. If at all.
 
@TedShifrin You're right on the money. This is an example of an $h$-principle; there is a classical variant of this which is solved exactly by trying to approximate the formal 1-jet given by the disturbed plane distribution by holonomic 1-jets.
This one's slightly weird because the proof of this is some hands-on trickery. The usual machine doesn't apply
I want to understand this in a better framework
Oh, by the way, I forgot to mention, but note that the special thing about wrinkles is that -- just like cusps -- even though they aren't smooth the Gauss map is defined
Somehow the point is that a wrinkle allows for an abrupt change in the tangential Gauss map. You are in a totally flat, constant tangential gauss map, world, and suddenly you hit the wrinkle and do a little whoop with the tangent plane
And since you can wrinkle inside a wrinkle, by doing this many times, one should be able to approximate any plane field
But it's a fidgety idea and I'd like to write it down in a better way.
 
Sounds like you’re flirting with sub-Riemannian ….
 
It kinda does.
 
3:50 AM
@BalarkaSen I don't fully get what you're talking about but this reminds me of that one sphere eversion
This one
 
Yup.
 
(The wrinkles-within-wrinkles thing)
 
I think there is some serious analogy here with what I am reading, actually.
The loose charts I told you about allows you to have certain h-principles for Legendrian submanifolds. Namely, it gives you a purely algebraic topology criterion for when two Legendrians are isotopic through Legendrians if one of them has a loose chart
The main procedure feels like the one in this simple sphere eversion video. Take a smooth isotopy between the Legendrians (this has to exist first), then you take the {zig-zag} x I, helicopter it to somewhere whenever you encounter non-Legendrian-ness like the little loop of self-intersections in this video, then shrink it down so that it looks like a wrinkle
Use those wrinkles to avoid problems like tangencies
Then unwrinkle at the end
I'll draw some pictures one of these days
 
I believe it, in the sense that I don't think you'd lie to me
 
As long as I didn't lie to myself
By the way, a "Legendrian fish" contains a copy of the Legendrian zig-zag:
 
4:00 AM
Makes sense
Like an R1 Reidemeister move
 
R1 as in
Yeah exactly
 
Though what was the invariant you had? Crossings minus half cusps?
That doesn't seem to work here
 
Yup
Right, because you only contain the Legendrian zig-zag
As an arc
It's actually three zig-zags
 
Oh, right, the endpoints are different in the image
It's gotta return
 
Yup
Would you call it three zig-zags or two zig-zags or three cusps or zig-zeg-zag?
A Legendrian "M"
 
4:04 AM
A zigga-zigga-zah
 
Creative
 
I guess it's a three-long zig-zag
A zig-zag that's three cusps long
I suppose it's kinda a zig-zag-zig? 'Cause it goes back in the first direction
 
Aha
Yeah that's accurate
 
knowledge, wisdom, and understanding
 
Chochma, bina, daat
Do I pass the test?
Wait, that's wisdom understanding and knowledge
@leslietownes I assumed you were referencing Chabad because you seemed knowledgeable about Jewish things from that conversation we had a while ago but I realize there's no reason for that to be true
 
4:11 AM
i was most definitely referring to mysticism
 
that's as mystical as it gets
 
If you take a Legendrian devil and spin its front projection around its axis of symmetry you get a Legendrian 2-knot in $\Bbb R^5$
@AkivaWeinberger: This is a loose Legendrian, i.e., it contains a loose chart
 
Don't all the cusps need to be parallel?
 
Nah. Just not vertical
You can always arrange them to be parallel
 
4:16 AM
Ah OK
Sure
 
To see the loose chart, take the horns of the devil; after spinning it they become a circle of cusps. So it's $Z \times S^1$ where $Z$ is the zig-zag
Which has a portion of the form $Z \times I$
I haven't checked the quantitative criterion but pretty sure it'll hold somehow
i.e., width of the $I$
 
On a separate topic
I feel like this should be easy but I'm not sure how to do it:
If I have a ratio $p:q$ ($p,q\in\Bbb R$) I can find good integer approximations by doing continued fractions
and cutting it off at some point
If I have a ratio $p:q:r$, how do I find good integer approximations?
 
Speaking of continued fractions, cool fact: if you take an unknot in $S^3$ and do a $p/q$-surgery on it, that's the same thing as a Hopf linkage where you do integer surgery on each component, which turn out to be the coefficients of the continued fraction
 
What if the continued fraction has more than two coefficients
 
Hopf linkage will have more than 2 components.
 
4:24 AM
But then you can't know what order they're in
Hopf linkages are completely symmetric, aren't they?
 
Maybe I'm misusing the term Hopf linkage. I mean a sequence of unknots $1, \cdots, n$ such that $i$ and $i+1$ are linked by a Hopf link
Nothing about $n$ and $1$
 
Ah, OK
I was think like preimages of points in the Hopf map $S^3\to S^2$
in which case every pair is linked, and permuting them corresponds to permitting points on the sphere
 
Yeah, not a complete graph of Hopf links
 
Got it
You can reverse the order, though, no?
Like, surgery with slope $[a,b,c]$ (continued fraction notation) would equal $[c,b,a]$
 
Yeah I think that part you can never tell. So if you have rational surgery $r$, it's either $r$ or $1/r$ you can read off from there.
Those give the same manifolds also.
 
4:28 AM
But those aren't reciprocals
 
Oh, erm
Yeah OK, here's the simplest example. Take $L(pq + 1, p)$, that's the same as $L(pq + 1, q)$.
Lens space
$(pq + 1)/p = q + 1/p$, so it'll be a Hopf link with $p$ and $q$ surgeries
But switch $q$ and $p$ and you get the same thing.
 
So a+1/(b+1/c) would give you the same as c+1/(b+1/a)
 
Yeah thats sort of weird maybe.
It is what it is
By the way did you know if you take a triangle of Hopf links (three unknots, every pair is a Hopf link, no further Borromean structure), the knot complement of that covers the knot complement of the trefoil knot
I am going off on way too many tangents
 
0
Q: About definition of continuity (Let $U\subset\mathbb{R}^n$ be an open subset containing a neighborhood of $\mathbf{a}\in\mathbb{R}^n$)

tchappy haThe following definition is in "Multivariable Mathematics" by Theodore Shifrin. Definition Let $U\subset\mathbb{R}^n$ be an open subset containing a neighborhood of $\mathbf{a}\in\mathbb{R}^n$, and let $\mathbf{f}:U\to\mathbb{R}^m$. We say $\mathbf{f}$ is continuous at $\mathbf{a}$ if $$\lim_{\m...

 
I think it's a 6-sheeted cover.
 
4:33 AM
Oh, bizarre
I bet there's a simple picture if only I can see how
 
Should be obvious from taking three 1/1 circles on the torus, and try to make it into the toral embedding of the trefoil somehow
three 1/1 circles on the torus is that triangle of Hopf links
Take the embedding of the triangle of Hopf links on $T^2$, and then act by $S_3 = \langle a, b \rangle$ where $a$ is a transposition and $b$ is a $3$-cycle, the action of $b$ being rotating by $2\pi/3$ along the axis that goes through the donut-hole of the torus and $\pi$ along the center circle of the torus, simultaneously.
Something like this
 
Are you thinking $6_1^3$ or $6_3^3$ (chart)
I think $6_3^3$
 
Yeah
$6^{3}_3$
What's the meaning of this notation btw lol
$6$ is crossings
 
$a_c^b$ a=number of crossings (minimally), b=number of components
c=where it is in the list
 
Oh ok
 
4:42 AM
@BalarkaSen Wait what does a do
 
Not sure, but maybe it should just rotate by $\pi$ along the center circle of the torus.
 
You can extend a transposition to a map on the torus?
Hm I'll think about it
 
Just a fixed-point free involution of the torus. Not many of these.
 
I feel like a transposition's gotta at least reverse the orientation of the torus though
(which probably forces a fixed point? which is troubling)
 
Just rotate a circle counterclockwise by $\pi$. That's a transposition of a circle without any fixed points. Thicken the circle.
 
4:45 AM
Aren't we trying to map the link onto itself
 
You can do the same thing where the axis of rotation, instead of being the one which goes through the donut hole, is inside the solid torus -- namely the center circle.
 
The three 1/1 circles
 
Yeah
I guess it should be $\pi$ rotation in the meridian coordinate and $\pi$ rotation in the longitude coordinate
Erm, no. Maybe?
I suck at these geometric computations. Topology so flabby, geometry so rigid.
 
@BalarkaSen Nah that's not gonna affect the circles at all
 
That sounds right
 
4:48 AM
Think in terms of the universal cover tiling. It's a bunch of diagonal lines, and you're translating horizontally and vertically
 
Translating horizontally by how much
The spacing between the diagonal lines is $2\pi/3$
 
Doesn't matter, you're not gonna swap two of them and leave the third fixed
 
@AkivaWeinberger Wait I'm not sure I understand. $(\pi, \pi)$ rotation does act freely on the complement of the links, why does it need to permute the circles? The 3-cycle already does that
 
I thought the point was $S_3$ was acting on the circles
 
Shouldn't just one of the elements permuting the circles suffice to get a trefoil cover
 
4:55 AM
Uh, sure, I guess
I thought you were leading this in a different direction
 
Nah
 
In any case if the goal is to get the complement in $S^3$ to map onto the complement in $S^3$,
we're gonna need to expand this map from just the torus to the whole 3-sphere, with no fixed points
 
Sounds like a job for quaternions
 
I guess this actually does work
Right, was about to say that
Is there a subset of $S^3\subset\Bbb H$ that's isomorphic to $S_3$?
 
$\mathrm{O}(2)$ embeds in $\mathrm{SO}(3)$. You make the reflection embed as a rotation
$S_3$ embeds in $\mathrm{O}(2)$
Dihedral
 
4:58 AM
That sounds right
Yeah
 
I guess we need an embedding is $\mathrm{SU}(2)$ though
Double cover of $\mathrm{SO}(3)$
 
Oh OK so here's how you think about it
Isometries of $S^2$ can permute three points on the equator
Now take the preimage in the Hopf map
Well, no, we still get fixed points we don't want
 
Geometry too hard for my flexible brain bro
 
Sphere. Rotate?! Ow
(sorry)
(I'm slightly sleep deprived)
 
what is sleep
 
5:05 AM
I guess I forgot who I'm talking to
We're actually exactly 12hrs off now I think
'cause it's 10pm here
Wait, no, 12hr30min
 
10:36 AM here, but that's probably because of DST
Antipode!
 
I forgot India's time zone is halfsies
 
Is it antipoday or antipodee
Which one sounds right
Trick question. Neither. Truth hurts
Tooth hurts
 
An-TIP-o-dee I thought
Nope
AN-ti-pode, says Google
 
Yup
 
5:09 AM
Oh, Wikipedia says both
but lists the sane one first
 
Who in the world says an-TIP-o-dee
 
Merriam–Webster says one AN-ti-pode but many an-TIP-o-dees
 
counterintuitive
im gonna head out, see ya!
 
One day I should finish playing Antichamber
@BalarkaSen See ya
 
stream it
ill watch
 
5:13 AM
Or as they say in Spanish, silla
 
 
1 hour later…
6:36 AM
So here's a question
For what value of $c$ does the spiral $r=e^{c\theta}$ have the property that, the points with vertical tangent are on the same y-coordinate as a point with horizontal tangent?
(Its approximate form is 0.27441…)
 
6:55 AM
The same but with $r=e^{-\theta / \tan\theta}$ and $r=e^{\theta\tan\theta}$ graphed also
(plus zoomed in a bit)
 
 
8 hours later…
2:48 PM
i am reading a proof of the "root" of semi definite hermetic endomorphisim. and in the proof he says, we will choose a base of eigen vectors. Why does this base exist? what attriute of the Semi definite hermetic endomorphism makes it possible for the existence of such base of Eigenvektoren
nvm found the solutiion
the matrix is n x n, so if we want to use spectral theorem, we need to show that the complex conjugate is equal to the matrix it self, since positive definit, this implies that the eigenvalues are real.
how do you continue
 
3:30 PM
Given $B(x)$ then doing $B'(x)$ do you lose some information about $B(x)$? That is to say the properties $B(x)$ has, may not necessarily translate to that of $B'(x).$ Clearly in differential equations if you differentiate the terms it's a completely different equation.
you're operating on $B(x)$ with the differential operator
 
4:06 PM
@MadSpaces Continue what? It's just the spectral theorem. Period. Semidefinite has nothing to do with it.
 
Spectral theory states a real matrix.
 
Nope.
 
well, it is atleast in my script. a real symmetrical matrix.
we have another theorem, that if an endomorphism normal, then we can get the same results as spectral
But the said endomorphism is not normal
neither am i
 
4:28 PM
question to sequences usually i see its written something like $ 0 \rightarrow V \stackrel{\phi}{\rightarrow} V_2 $ what is meant by the first arrow? is it the zero function? or any function ? usually nothing is written above it
 
What is the definition of normal?
The only linear map on $0$ is the $0$ map.
 
function o adjungate of the function = adjungate of function o function
Ie sumlitaniously diagonalisable
 
So is a hermitian matrix normal?
 
not as far as i know
Because $AA^* \neq A^*A$
Oh wait
it is
Since its the same. $A= A^*$
oh jeeeeeeeeeeeeeeze
I am dumbbbbbbbbbb :(
Love you teddy :3
 
It's worth figuring out which sorts of matrices you might encounter are automatically normal.
 
4:38 PM
wow i dont know how i missed this? So apparently i just did the calcs and hermitic, skew hermitic, unitary , are all normal.
 
Right.
 
 
2 hours later…
6:33 PM
@TedShifrin It isn't PC to call them "normal" anymore. I believe that the correct terminology is "neurotypical".
 
Across all fields?
Subgroups, vectors, topological spaces, probability/statistics …
 
is there a name for matrix whose entries are real numbers between 0 and 1 and whose sum is 1?
the sum of all the entries is 1
 
if you're talking row or column sums, 'stochastic' matrix might be it. en.wikipedia.org/wiki/Stochastic_matrix
 
@geocalc33 I've heard of Stochastic Matrices, where each row, or each column, sums to $1$, but not one where all the elements sum to $1$
 
if you mean the sum of all the entries in the matrix, dunno.
rob and i are two lobes of the same brain sometimes.
 
6:44 PM
I'm assuming there's a good reason not to consider what I considered
 
keeps silent
 
and to consider the stochastic thing
 
Is there a good reason to consider it? In other words, why are you considering it?
 
that would be a good subject for a book. kind of an anti-math book. reasons not to consider a mathematical object or property.
 
haha
 
6:46 PM
the index of the book would be one line: "WE TOLD YOU NOT TO CONSIDER THESE THINGS"
 
by writing the book you'd force people to consider them
 
there'd be a big elephant on the cover
 
I feel like there's a lot of good material for such a book.
 
you could really troll people with it. i'd put p-adic numbers in there. and the henstock integral.
 
@leslietownes Yay!
But everyone should spend more time thinking about the p-adics
 
6:51 PM
non-separable hilbert spaces. do we need them?
 
no throw them to the wolves
 
You can’t have p-adics without q-adics. Fair time.
 
@TedShifrin If we let in the q-adics, then we have to let in the d-adics and b-adics, and I don't think that we should do that.
On the other hand, once you start getting really into the p-adics, you end up in the adeles, where your really rolling in the deep...
 
I think we should put the p-adics in a box and put them in the adic, with adele
 
we should chain the p-adics to the radiator
i don't really understand the difference between puns and the stuff i say
 
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