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6:00 AM
It's not, right? It sends $b$ to $aba^{-1}$. So it sends one vertex over the fiber to two steps further in the fiber.
Not sure if that makes sense. I don't see purely from the covering space why it's not normal, but let me think
 
and also there's the answer below to witness how messed up math terminology is
lol the wrong answer got accepted
 
It's just one headache problem after another on this HW
"Show that
$$Y_n=\frac{\sum_{k=1}^n X_k}{\sum_{k=1}^n X_k^2 + \sum_{k=1}^n X_k^3}$$
converges in probability as $n\to \infty$, and determine the limit."
 
What do we know about the $X_k$
 
Oh yeah, forgot that part
Let $X_1,X_2,...$ be independent $\text{Uniform}(-1,1)$ distributed random variables.
 
use the python method
 
6:05 AM
Divide bottom and top by $n$, and strong law of large numbers?
 
oh lol
 
Ah, yeah, I do need to practice the python method
 
why didn't I think of that
 
I guess you'd want to know the distributions of $X_k^2$ and $X_k^3$
 
6:05 AM
No they're messed up
 
it isn't that bad
 
What's the expected value of $X_k^{2,3}$ at least
 
I mean you can write down a pdf
 
that's trivial @AkivaWeinberger
just use LOTUS
 
Well $X_k^3$ is 0 clearly
LOTUS?
 
6:06 AM
second central moment is 1/3
 
Makes sense
 
In probability theory and statistics, the law of the unconscious statistician (LOTUS) is a theorem used to calculate the expected value of a function g(X) of a random variable X when one knows the probability distribution of X but one does not know the distribution of g(X). The form of the law can depend on the form in which one states the probability distribution of the random variable X. If it is a discrete distribution and one knows its probability mass function ƒX (but not ƒg(X)), then the expected value of g(X) is E ⁡ [ g ( X...
 
I know the pdf of $X_k^2$ is just $\frac{1}{2\sqrt{x}}$
 
$E[g(X)] = \int g(x) f_X(x) \ \mathrm dx$
 
Wait do we get $\dfrac{0}{\frac13+0}$? That feels wrong
Or $\dfrac{0}{\frac n3+0}$ at least
 
6:07 AM
The second thing you wrote is nonsense
 
$E[X^2] = \frac12 \int_{-1}^1 x^2 \ \mathrm dx = \frac13$
 
Also no expectation doesn't go inside division
So the first thing is also nonsense
 
but the point is that sum X^k / n -> E[X^k]
so the first thing shouldn't be nonsense
 
Yeah I thought Akiva was computing expectation of Y_n
 
Wait are we trying to find $E(Y_n)$ or its pdf
 
6:09 AM
(check bounded variance)
no, we're finding Y_n itself
 
Yeah he was
 
it is a degenerate r.v.
 
So it's nonsense :3
 
Yn -> 0/(1/3+0)
 
Right.
 
6:09 AM
That's what I just wrote
 
Huh, simplifies to just 0
(You guys figured that out faster than me)
 
@LeakyNun The orbit of the action of $a$ by conjugation are geodesic lines, if you think of Cayley(F_2) embedded in $\Bbb H^2$, I believe.
 
on which cover?
 
For O_O_O I think the orbits of the subgroup corresponding to it acting on Cayley(F_2) are all points contained in separate geodesic lines
 
can we restrict to the cover and not the universal cover?
 
6:18 AM
Yeah I am not at all sure how to see if off from the cover. That's such a good question
Ashamed I never thought about this before.
 
what's the group of deck transformations?
why do I feel like it's Z
(shouldn't it be F_infty)
surely I have the wrong picture in my head
 
The subgroup corresponding to the cover is F_infty, right?
 
Alright, one more problem on this HW
 
The group of deck transformations does seem to be Z
 
I thought the group of deck transformations is supposed to be the subgroup
 
6:22 AM
No, the subgroup is the fundamental group of the covering space!
 
then what is the deck transformations?
 
$p : C \to X$, $\pi_1(C)$ injects into $\pi_1(X)$ by $p$
 
I need some context. In the question "Show that $Y_n$, suitably normalized, converges in distribution as $n\to \infty$, and determine the limit." what is meant by "suitably normalized?"
(Same $Y_n$ as before)
 
@LeakyNun In general it's $N_{\pi_1(X)}(p_* \pi_1(C))/p_*\pi_1(C)$ I think
 
@Rithaniel see central limit theorem
 
6:23 AM
Where $N_G(H)$ is the normalizer
 
that's strange
 
Danke
 
I don't think that works
because in general you have non-trivial deck transformations
even for the universal cover
but your quotient is trivial
 
$N_G(H) = G$ if $H$ is the trivial subgroup
So $N_G(H)/H = G$
 
oh I misinterpreted
interesting
 
6:26 AM
For a normal subgroup, $N_G(H)/H = G/H$, so the group of deck transformations of a normal cover is in fact $\pi_1(X)/p_* \pi_1(C)$
Let's see. In our case $H = \langle a^n b a^{-n} \rangle$
 
so $p$ is normal iff the normalizer is the whole group
 
yeah
I think for $H$ that, $N_G(H)/H$ is $\langle a \rangle = \Bbb Z$
Which should check out
 
what's the map from the group of deck transformations to that group?
 
@LeakyNun Hm, now I am confused.
isn't it normal?
 
lmao
 
6:35 AM
The deck transformation group of O_O_O -> S^1 v S^1, like you said, acts by shifting everything to the left say.
That is indeed transitive on the preimage of the basepoint
 
then maybe "deck transformation acts transitively on fibres" corresponds to something else
 
But why isn't it normal? Where did you get that from?
 
I don't know
so _O_O_O_ is normal you say?
 
I just checked Hatcher. It is.
 
So, does it sound right that the r.v. I mentioned before would converge to $N(0,3)$ in distribution after appropriate normalization?
 
6:39 AM
@BalarkaSen then what does normality of cover correspond to?
@Rithaniel not sure
 
I am not sure I understand. It checks out with everything right? Regular graph, deck transformation group is acting transitively, so on
What falls apart?
 
maybe maths is broken
@Rithaniel I don't see immediately why, but it might just turn out so
 
Hello, I wonder if someone could explain this top answer to me: math.stackexchange.com/questions/3039219/…
I've been following the links too, and does this not mean that <x,y> = xtAy = dotproduct between x and y?
 
the "standard" dot product is xt y
without A
 
Oh, so <x, y> is ?
thank you btw
I've been reading up and I come to this notation everywhere but I don't understand it
 
6:42 AM
depends on what the inner product is
 
hmm sorry
 
it's just the notation for a generic inner product
 
and that isn't the dot product?
 
an inner product on R^n is any function f: R^n x R^n -> R satisfying four properties
given such function, the inner product f(v,w) is denoted by <v,w>
 
oh
so what does <x, y> = xtAy
mean?
just that we define the inner product by multiplication with A?
 
6:44 AM
$A$ is a positive-definite symmetric matrix.
 
Yes, and symmetric I think
 
You define the inner product by $\langle x,y\rangle = x\cdot Ay$ with $\cdot$ being the usual dot product.
 
@BalarkaSen does Hatcher say that a cover is normal iff the corresponding subgroup is normal?
 
I tried decomposing x and y in to basis vector combinations, but then when I multiply it out I see that for example in R3 e1 e2 and e3 have to be orthogonal and 1
 
hi, a @Balarka and @Leaky
 
6:45 AM
that means A is just Identity
 
@Ted hi
 
ah sorry you said symmetric
 
@oliver: The identity gives you the standard dot product. But in general you can have $\sum a_{ij}x_iy_j$.
 
@TedShifrin thank you
 
@LeakyNun That is true.
Hi @TedShifrin
 
6:47 AM
what on earth @BalarkaSen
 
so this expression, all it means is that A is positive definite and symmetric?
 
To get the definition of an inner product you need that, yes, @oliver.
 
So $H = \langle a^n b a^{-n} \rangle$ should be normal in $F_2$.
 
@TedShifrin
 
@BalarkaSen so, if Hatcher says that the cover is normal and normal cover corresponds to normal subgroups, the only thing this can mean is that... [sniped]
 
6:50 AM
@TedShifrin would you have a good link for the inner product definition?
Sorry I'm being a pain
 
No, I don't have a link. It's certainly in Artin's Algebra book.
Or any number of linear algebra books.
 
Maybe we should try to see why $baba^{-1}b^{-1}$ is in $H$
 
wait no
 
I see $baba$ and think $is\:\: you$
 
what breaks down is $H = \langle a^n b a^{-n} \rangle$ @BalarkaSen
 
6:52 AM
I've looked in Friedbergs and Dummies as well as lem.ma but I feel like that I just now understood I'll check out Artin's thank you
 
Certainly in Friedberg/Insel/Spence.
 
@LeakyNun I think that is indeed the subgroup corresponding to the cover. Why isn't it believable that $H$ is not normal in $F_2$?
 
because "they all start with $a^k$ and end with $a^m$"
 
Yeah it's in there, I just didn't understand it, I think I spent about 4 hours on it yesterday
Both Friedbergs
that book usually makes everything very clear
 
$baba^{-1}b^{-1}$ is clearly in $H$ because $b \in H$, $aba^{-1} \in H$, so that was a dumb choice.
 
6:53 AM
oh what
 
just this time it didn't make click
 
lmao
 
@LeakyNun It's clear that all conjugates of $b$ in $F_2$ are in $H$.
 
this is surprising
what a roller coaster
 
Cool, right?
 
6:55 AM
yes
it is the kernel of the "count the number of $a$'s" homomorphism
 
Yeah
 
@loch hey
 
hi
 
do you want an AT exercise (don't scroll up :P)
 
no
 
6:56 AM
lol
 
lol
 
don't you TA AT
 
It's a strange thing. You bash your head against a particular kind of problem for a few hours and give yourself a headache because you're not used to it. Then turn around and look at a different sort of problem, which some might say is more difficult, but you relax because you're familiar with it.
 
yes
 
lmao thats a good one
 
6:56 AM
lol
 
ATGC bikhes
It stands for Allen Hatcher is a Topological Giant, C*nts
 
lol
nice
@BalarkaSen so what's the model of:
2 hours ago, by Leaky Nun
why not take $F = \Bbb Q$, $K = \Bbb Q(2^{1/4})$, $f = x^2-2$, $L = K$, $E = \Bbb Q(\sqrt2)$, $a = 2^{1/2}$, $b = -2^{1/2}$.
2 hours ago, by Leaky Nun
then $E \to E$ cannot be extended to $K \to L$
 
Quotient $Cayley(F_2)$ by $\langle a \rangle$ and then further quotient by $\langle b^n a b^{-n} \rangle$ say
The first step you get a non-normal cover, the second step you get a normal cover
 
how about no
 
So $p : Z \to Y \to X$
$Z \to X$ is non-normal
$Y \to X$ is normal
So passing to larger cover, but set of symmetries is being restricted down
Fun how the two examples we thought about gave us the example we needed
 
7:00 AM
c'est la vie
so X = _O_O_O_?
 
X = 8
Y = O_O_O_O
 
oh ok
 
Z = O<(mess)
 
what is "set of symmeries"?
the point is that Z has no non-trivial deck transformations?
(at least I can't come up with one in my head)
 
Yeah, the subgroup corresponding to Z is $\Bbb Z* 0$ in $\Bbb Z * \Bbb Z$. Normalizer of that is $\Bbb Z * 0$ itself, right?
So deck transformation group is $N_G(H)/H = (\Bbb Z * 0)/(\Bbb Z * 0) = 0$
 
7:07 AM
the point is that by collapsing the other parts to circles, we obtain new symmetries by mapping our original circle to one of the new circles
 
$\langle a \rangle$ is self-normalizing in $F_2 = \langle a, b \rangle$.
Yeah
 
nice picture
I'll store that in my head
 
I'll store your example in my head
 
lmao
 
This was a great exchange, we should do this more often
 
7:09 AM
I concur
 
The only problem is contravariance in this whole analogy; you add stuff to get field extensions, you collapse stuff to get covering spaces
I have to psychologically rewire myself for this
 
it's just like going from A to Spec(A)
 
Yeah lol
I have to write down stuff before asking new questions/thinking about newer things, I will forget everything otherwise. Let me spend the next few hours transcribing our discussion lol
Talk later
 
see you
 
7:50 AM
thanks guys, I re read some things and thanks to your explanation I do understand this better. I was thinking about inner products rigidly as dot products from geometry instead of just a scalar product
 
 
3 hours later…
11:16 AM
Hi everybody, is there a standard word for Riemannian manifolds whose sectional curvature is constant as one varies the 2-planes at a given point, but not necessarily constant when moving between points?
 
12:06 PM
@Danu They are best known as space forms. ;)
 
12:35 PM
I wish to ask when they combined the summation $d_1|a$ and $d_2|b$ into a single summation of $d | ab$ (third to second last step), did they not lose several summand terms? For example, suppose $a$ and $b$ were even numbers. Then in the first summation $d_1=2,d_2=1$ and $d_1=1,d_2=2$ were two separate summands. However, in the new summation, $d=2$ and thus there is only one summand.
Could someone please clarify?
 
@MikeMiller Cool, thanks! Didn't know about that.
@MikeMiller I'm dealing with a curvature tensor with a certain decomposition, and one piece is a function times the Kulkarni-Nomizu product of some symmetric bilinear form with itself, so I'm trying to understand what it means for the curvature that that piece is of the type I asked about, though the full curvature tensor is not.
 
1:00 PM
@Danu I certainly have nothing to say
You can follow the conventions of Schur's lemma and call it pointwise constant
 
Yeah, that's the name I came up with myself too, but I thought it'd be confusing when I asked in chat haha
I really confused a fellow PhD student for about 20 seconds with that
 
Just say what it means at the start of conversation
I would have known what you meant it's the obvious name
 
So how's things at Columbia?
you're an assistant professor there, I saw! :D
 
They're fine, finishing teaching a course now
Fancy name for postdoc
They classify us that way mainly for the benefits schemes
 
Oh, okay
Is that positive for you, or negative?
 
1:07 PM
No, I mean the faculty benefits are better than postdoc benefits. Mainly healthcare
 
Okay, great
 
But to be clear it's neutral for me, since I didn't have a choice, I just got the job I got ;)
 
Heh, yeah, that's one way of looking at it.
 
Hope things are well with you too --- sorry to make the convo one-sided, but I've got to finish writing up final exams
 
Anyways, glad to hear you're doing well
Oh no, no worries :-)
 
1:10 PM
You can get in touch with me elsewhere for a longer conversation if you want
 
I'll get back to work... Polishing away!
 
 
1 hour later…
2:19 PM
@GaurangTandon $a$ and $b$ are assumed to be coprime so you can't have two even numbers
that's why $1$ can only occur once in the sum over the divisors of $ab$
(since $a, b$ share only $1$ as a divisor)
 
Hi people! Servus @ÍgjøgnumMeg!
 
Grüß dich @Rudi :)
 
Wia gäds da?
 
Jo scho guat, Masterstudium isch schwer hahaha
wie gohhhts da?
 
Jo mia aa. Bin scho wieder aufm Flughafen, Vortragsreise ...
 
2:25 PM
Hoi, wo gohts ane?
 
Schwer is guad - dann lernst wos! Helsinki
;-)
 
Oo nice :) Viel spaß.. :D
 
Merci! Wos hosdn füa Vorlesungen?
 
Und ja i lean scho viel, i wead nögschte woche mitam Prof readn übr Masterarbeit :D
Modulformen und Algebraische Zahlentheorie
und ein Topologie Seminar
 
Cool!!
 
2:28 PM
yeeeee
 
Und Masterarbeit wos mogst do mochn
?
 
Irgendwas richtung Iwasawa Theorie
 
Pfff nie gehört ...
 
aba da Prof wird ma wahrscheinlich an thema geaba
also des hot mit dr algebraischen zahlentheorie zum tua
hehe
 
p-adische Zahlen auch, oder?
 
2:30 PM
genau :) Die unendlichen Körpertürme die zu betrachten sind heißen $\Bbb Z_p$-Erweiterungen
(also Erweiterungen mit Galoisgruppe $\Bbb Z_p$
 
und ged da auch was mit perfektoide Räume?
 
öööh wahrscheinlich aber ich kenn mich da ned aus lol
 
OK! Sonst? Wie iss leben in HEI?
 
also auf jeden Fall in der klassischen Theorie geht's darum dass man die asymptotische Größe des $p$-Anteils der Idealklassengruppe von Kreisteilungskörpern irgendwie untersuchen kann
Scho nice hier muass i säga :) bin aber richtig busy da ganze zitt hahaha
 
Ferien kemman aa no, oda?
 
2:35 PM
jo bald glob i :) I hob gad usgfunda dass as am 6en Februar a Klausur git :(
 
Do hosd ja no a bissl :-)
Gibts guate Kneippn in HEI?
 
i war no ned fort in Heidelberg :D
i wohn eher so richtung mannheim
 
Ah, OK. Da iss billiger und einfacher zum Wohnen, oder?
 
Are you already thinking about your master thesis?! @ÍgjøgnumMeg
 
@Alessandro I'm just thinking about getting a topic from the prof so I can make sure I'm taking the right courses and stuff lol
@Rudi ja :D
@Alessandro and what texts I'm gonna be focussing on for the next couple of years
 
2:51 PM
Makes sense, it seemed a bit early to already have a topic and everything
 
@ÍgjøgnumMeg I glaab am Rhein konn ma super Radlfahrn.
 
Do you have news from Mathei by the way?
 
Yeah it's too early to have an actual topic, but I know I wanna be doing something Iwasawa theoretic at least
Also, nobody has seen him
 
Ri Koblenz wird des richitig schee
 
@ÍgjøgnumMeg say Modulformen und Algebraische Zahlentheorie in your dialect
 
2:52 PM
but apparently this is exactly what he did a couple of months ago; just withdrew completely for a while to "regroup"
@Leaky Uhh.. Modulforma und algebraische Zahlatheorie
ned so geil..
hahaha
@Rudi ja isch scho nice da zum radla :)
 
@ÍgjøgnumMeg I have no idea what that is, but as long as you enjoy it!
 
aber grad a kle kalt, muass mal wiadr Zwift runterlada
@Alessandro it's just arithmetic of infinite extensions of number fields
or infinite towers
 
@Rudi_Birnbaum you can go all the way from Koblenz to Bonn by bike, there's biking trails all along the river iirc
Coincidentally I wrote that message while crossing the Rhein
@ÍgjøgnumMeg just :P
 
hehe, well anyway I only know some of the basic constructions in the theory atm
but it's pretty cool
 
I see
So that falls under algebraic number theory?
 
2:56 PM
yeah :P
 
Phew
I was worried you were going to become a crazy analytic number theorist with all those modular forms
 
In mathematics, “co” is a prefix to turn something you don’t really know into something you really don’t know.
 
@Alessandro well, I might end up doing some kind of weird p-adic analysis
 
A thing that can assume different values should be called a nstant
 
a party that gives a shit about poor people should be called the nservatives
 
2:58 PM
Did you figure out that functional analysis thing by the way?
 
Not really, my exercise partner kinda did that exercise and she'll explain it to me
lol
or rather, i'll see it in the problem session
 
What is the question?
 
The question is really to show that the first $d$ Poincaré series of weight $k$ form a basis for the space of weight $k$ cusp forms, $S_k$, where $d := \dim_{\Bbb C}(S_k)$
so essentially to show that these guys are linearly independent
 
3:05 PM
Well you lost me :D
 
bleh it's not like I know what I'm doing anyway
lol
 
3:23 PM
Can someone name a topic on cryptography/cryptanalysis on which an undergrad can work?
 
Suppose a group has elements x, y, z that satisfy the equations

x y = y² x
y z = z² y
z x = x² z

Show that x, y and z must all be equal to the identity.
 
@AlessandroCodenotti cool!
@AlessandroCodenotti And Koblenz and sourrounding has some wine culture which is also cool
@ÍgjøgnumMeg Ich fahr no ind Uni 2x10 km, aber ned alle Tage.
 
@Rudi ja für mi sinds ca 2x20km
 
@ÍgjøgnumMeg Im Sommer ideal, im Winter kanns zach sein. Ich hab eine Variante über die Salzach die ist schöner und es sind 18 km.
 
ja im Sommer wirds geil :D aber grad jetzt sinds morgens so -5 grad
da hab i eher kein bock hahaha
 
user12692
3:34 PM
I may miss something, but this answer seems too simple to be true:
 
user12692
0
A: Dirichlet kernel inequality

Umberto P.All you need to know is that $3 < \pi < 4$ and $e > 2$. As $n \ge 1$ you have $1 + \dfrac 1{2n} \le \dfrac 32$ so that $\dfrac 2\pi \left( 1 + \dfrac 1{2n} \le \dfrac 32 \right) \le \dfrac 3 \pi < 1$, and $\ln \pi < \log_2 4 = 2$.

 
@ÍgjøgnumMeg ja genau. Ich hab mir ein Brompton gekauft, dann fahr ich wenns wetter scheise ist mit dem Bus+Brommie das ist cool
 
user12692
Estimates in this one are more commonly known (could be simplified though):
 
user12692
4
A: About the order of the $L^1$ norm of the Dirichlet kernel.

Pedro Tamaroff Claim If $L_n=\displaystyle\frac{1}{\pi}\int_0^{\pi}|D_n(t)|dt$, we have $L_n=\dfrac{4}{\pi^2}\log n+O(1)$. Proof We begin by dividing the interval $[0,\pi]$ into the subintervals where $\sin\left(n+\dfrac 1 2\right)$ keeps its sign. Since $\sin t/2$ is always non-negative in said domain, we...

 
Nice :D ich hab kein geld dafür lol, der Radweg ist hier ein bisschen langweilig auch
alles so flach
es ist eine SEHR einfache Fahrt lol
vielleicht kauf ich mir einfach ein Fixie
lol
 
3:46 PM
@ÍgjøgnumMeg aidos
 
Ja die sind praktisch (und schaun besser aus als ein Klapprad ...)
 
Irgendwelche Mädchen hier?
XD
@Jack pretty advanced...
 
@Abhas warum fragst du das?
@Rudi :D naja ich mag mein Rennrad einfach
bin ein bisschen out of practice im moment
 
@ÍgjøgnumMeg Das war nur ein Witz Kumpel ...
 
3:51 PM
@ÍgjøgnumMeg Rennrad fahren? Das ist einfach...
 
@AbhasKumarSinha meinst Du ohne Stützräder?
 
@Rudi_Birnbaum Na sicher
 
thumb up
 
thumb up
@oliver What happened to your Profile Pic?
It looks error (blank image to me)
404 probably....
 
4:11 PM
@Ultradark
 
@AkivaWeinberger Hint? Can you solve this by just playing around with algebraic manipulations?
 
@F.White I don't actually know the answer
This isn't my puzzle
 
Is it actually true though?
Like definitely true?
I have tried it since you posted it without success
After exhausting low degree manipulations, I think a more sophisticated approach might be necessary
We can reformulate it as: Let $G=(x,y,z)/(xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2})$. Prove that $G=\{1\}$
And considering the subgroup generated by $x$ in here and take quotients or something
@AkivaWeinberger Where did you get the puzzle btw?
 
@F.White Someone posted it on Twitter
 
Oh it is true, it's in Lang
And it does follow by lots of algebraic manipulations
Okay, on a different note, can someone help me with probably a basic number theory problem. Please don't just give the answer, but maybe tell me how to approach it.

I want to find all primes $p$ such that $-x^2+py^2=z^2$ has a solution $(x,y,z)\in\Bbb Q^3$.
In other words, for which prime $p$ does the quaternion algebra $(-1,p)$ split over $\Bbb Q$
 
 
2 hours later…
6:08 PM
Help in Suppose $Γ$ derives $A→¬A$ and $¬A→A$ then $Γ$ derives $A$ and $¬A$.Also how to prove : $∀α, α ∈ ∆$ iff $¬α∉ ∆$ where $∆$ is maximal consistent set.
-1
Q: Is following true in Hilbert System?

MikeSuppose $Γ$ derives $A→¬A$ and $¬A→A$ then $Γ$ derives $A$ and $¬A$. Also how to prove : $∀α, α ∈ ∆$ iff $¬α∉ ∆$ where $∆$ is maximal consistent set.

 
Let's start with the second one, what does it mean for $\Delta$ to be maximal consistent?
 
Hi, I have some doubts about the implicit function theorem; here is my question, that after a day has received a relatively little number of views (imo): math.stackexchange.com/q/3464216/719906 I post it here, so hopefully some of you erudite and interested in this topic may give me a help. Thanks in advance!
 
6:27 PM
I choose a random permutation on 65 elements. What's the expected number of elements that lie on a 27-cycle?
 

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