The h Bar

7:01 PM

@0celo7 That shit ain't PhD dawg

Hi @DanielSank

I saw your question, but there's something I don't understand. What do you mean by specifying that the noise process shouldn't be stationary?

0celo7

@BernardMeurer it takes a PhD to decipher his shit notation

Bernard Meurer

@0celo7 Fair enough lol

Secret

A 4D game

jayisgames.com/review/alex.php

DanielSank

@MarkMitchison I think I mean that the "coordinate" can drift all over the place.

Secret

7:08 PM

and the mapping of the final level

DanielSank

And actually, I think I may have figured out the answer to the question...

Mark Mitchison

@DanielSank OK, but the statistics are still stationary, right?

Ah ok fair enough

Because it seems like you simply want the autocorrelation function in time

DanielSank

@MarkMitchison I won't say yes or no because I'm not sure quite what the words mean.

rob

3

Q: What is "white light" ? Uniform wavelengths or uniform frequencies ?

Suppose you have a mixture of electromagnetic waves of wavelengths spreaded on the visible spectrum only (from $\lambda_{\text{min}} \sim 400 \, \text{nm}$ to $\lambda_{\text{max}} \sim 700 \, \text{nm}$). At some ideal detector, the light spectral distribution is described by a functional like ...

DanielSank

@MarkMitchison Hmmmm.

Mark Mitchison

7:09 PM

and nthe autocorrelation function is just the inverse fourier transform of $S(\omega)$, right?

rob

Do the comments suggest this question is "too broad", or "primarily opinon-based"? Or is it okay?

DanielSank

@MarkMitchison I'm not sure that's what I want to know.

Mark Mitchison

up to the terms proportional to $\langle v(t)^2\rangle$

0celo7

@BernardMeurer I did a PhD level induction proof today

user116211

You want to do infinitesimal calculus @0celo7? Non-standard analysis?

0celo7

7:12 PM

Why would I want to do that?

user116211

Hmm; I saw your comment in the transcript where John asked you to explain infinitesimals.

0celo7

@BernardMeurer have you done continuous functions in analysis yet?

Bernard Meurer

@0celo7 Yes, but it was Klingon to me

I still don't get the formal definition of continuity at a point

user116211

@0celo7 ohh!

0celo7

Why not?

user116211

7:14 PM

@BernardMeurer You doing Calc I?

Bernard Meurer

@MAFIA36790 Analysis in R

0celo7

He is doing PhD analysis in R

user116211

@0celo7 They are all just based on the fact that $\mathbb N$ is inductive.

0celo7

What are you babbling about

user116211

But I would like to see some monster induction proof.

0celo7

7:15 PM

Your face is inductive

user116211

._.

0celo7

@MAFIA36790 Sard's theorem.

DanielSank

@MarkMitchison Yes....

user116211

okay; checking...

Mark Mitchison

@DanielSank Yes, as in, you do want the autocorrelation function?

user116211

7:17 PM

It's not in Royden :/ Let me google.

user116211

Why should it be in it? I'm dumb.

0celo7

It's in books on diff topology.

user116211

14

A: Show that $f(C)$ has Hausdorff dimension at most zero.

I guess that your source was Wikipedia article on Sard's theorem which states at the end In 1965 Sard further generalized his theorem to state that if $f:M\rightarrow N$ is $C^k$ for $k\geq \max\{n-m+1, 1\}$ and if $A_r\subseteq M$ is the set of points $x\in M$ such that $df_x$ has rank less ...

Bernard Meurer

@0celo7 Can you get me a picture of a non-hausdorff manifold?

user116211

Really? Was it really required to define something like stable subset? It's a trivial definition.

0celo7

7:25 PM

@BernardMeurer no!

You cannot draw such a thing actually

There's a theorem which says so

@MAFIA36790 shall prove it

user116211

0celo7

what's that?

user116211

This is the dangerous warning sign.

user116211

Used by Bourbaki.

user116211

I'm warning not to let me explain such stuffs which I don't even know.

0celo7

7:30 PM

@BernardMeurer how the fuck do I start Python on a mac again

Bernard Meurer

@0celo7 Open terminal and run python

run =
1. Type python
2. Press Enter

0celo7

thx

@MAFIA36790 Hint: $X$ Hausdorff, $Y\subset X$. What must the subspace topology on $Y$ be?

If you are reading Bourbaki this should be trivial.

user116211

I'm not still into PhD topology.

0celo7

This is intro baby topology -.-

user116211

@0celo7 I'm reading Bourbaki Algebra I.

user116211

7:34 PM

Not the Topology one.

0celo7

Maybe you should read Munkres then.

user116211

Yes, I would.

user116211

But I will do analysis first.

0celo7

When are you going to read Wald?

user116211

Now, I'm into Group Theory.

0celo7

7:35 PM

Why?

Group theory is hardly useful.

user116211

@0celo7 Don't know.

user116211

@0celo7 Why?

user116211

It's really interesting.

0celo7

You are not a physicist

user116211

...

0celo7

7:37 PM

If you enjoy ~~abstract~~ useless algebra and set theory

You are a philosopher

user116211

I should complete Jech first before reading Wald.

DanielSank

@MarkMitchison I'm not sure!

user116211

I know they are not related. But yeh, I would do that like that.

user116211

@0celo7 I am not.

DanielSank

I want to know, given an initial value of the process, what is the distribution (or at least variance) of the process at a later time.

I think I can reason as follows:

Call the process $\phi$ and its velocity $\dot{\phi}$.

user116211

7:39 PM

@0celo7 They are not useless T__T

0celo7

@MAFIA36790 Complete Jech?

See you in a few year.

user116211

yes.

user116211

yes; that was my point.

DanielSank

Then we can say $$\langle \phi(t) ^2 \rangle = \left \langle \int_0^t \dot{\phi}(t') dt' \int_0^t \dot{\phi}(t'') dt'' \right \rangle \, . $$

If I know the spectral density of $\dot{\phi}$, I can use the Weiner-Kinchein theorem here to convert $\langle \dot{\phi}(t') \dot{\phi}(t'') \rangle$ to $S_{\dot{\phi}}$.

0celo7

Weiner or Wiener?

DanielSank

7:41 PM

^ Not sure.

Spelling was never my strong point.

0celo7

At least you're better than dmckee.

user116211

@0celo7, Anywyas, I learnt about Fubini's Theorem today while doing some bilinear transformations of random variables. Prof said I will get it in analysis class later.

0celo7

You know more analysis than me, I dunno what a bilinear transformation is.

user116211

No, I don't. I'm a freshman. I'm completing all my works in hands to read your notes and re-start analysis.

0celo7

@MAFIA36790 You should try to prove everything in those notes

I will try to keep them updated

user116211

7:47 PM

sure.

0celo7

Some things are trivial, some are insanely hard

You might need help for some

user116211

I'm thrilled to go with those. Maybe I need a week or so to start.

0celo7

line 5 is very important

user116211

Before that, I have to read Born Wolf, do my works on probability Theory, Linear algebra and Group theory. I have also chapter 3 of Lanczos to complete. I can complete them in a week or so, I expect.

user116211

@0celo7 in the note?

0celo7

7:49 PM

my notes in a week?

user116211

@0celo7 yes.

0celo7

you can prove everything in those notes in a week?

user116211

I will start them after a week.

0celo7

oh

user116211

@0celo7 What? no.

0celo7

7:50 PM

I might update them with some measure theory from @obe 's book.

We shall see.

user116211

very well.

user116211

I'm going to start Lebasgue measure: chapter 2 of Royden very soon after all my due works get completed.

Sir Cumference

8:24 PM

Who here is good with degenerate matter?

I'm really desperate

I asked the TA about it, she said she was still learning it herself...

Why in hell is my professor going so far into this?

rob

@SirCumference Have you asked the professor?

Sir Cumference

@rob He just repeats what he says in class. It's almost like he's avoiding answering the specific questions I'm have.

I ask why something happens, he just repeats what happens.

I ask why this equation works, he just tells me what it does.

I ask him what a graph is about, he just tells me what it looks like.

It's becoming useless at this point. I don't press on after though, since I'm afraid I'll annoy him.

user116211

That's one of the adverse side-effects of listening to the SU anthem on daily basis.

rob

One technique that's sometimes helpful in such cases is to sit outside (or inside) the professor's office while you work on homework for that class. Then when you're stuck, you can clarify just exactly where.

And don't be afraid to clarify your questions. Communication goes two ways.

e.g. for a graph: "yes, that's what it looks like. how would the relationship change to give it some other shape?"

Sir Cumference

@rob Oh yeah, get this — he doesn't give homework. He's strangely lenient considering the college is usually brutally hard.

0celo7

8:38 PM

2

Sir Cumference

My questions are about the material he went over in class and the notes I took.

0celo7

@obe <3

Sir Cumference

@0celo7 Oh my god

rob

@SirCumference Are there exams? term papers? attendance-based pass/fail?

Sir Cumference

He actually got you that?

@rob It's not pass/fail, and there are exams

user116211

8:39 PM

@0celo7 \o/

Sir Cumference

@0celo7 Does obe know you irl?

rob

@SirCumference Well. That's stressful.

0celo7

no

Sir Cumference

@rob Yeah. Add to that the fact that he speaks so damn fast, I can't write down one thing without missing another.

@0celo7 How'd he send it to you?

0celo7

Something called "the mail"

Sir Cumference

8:41 PM

@0celo7 So you told him your address?

0celo7

He already knew it. Kind of weird but whatever

Sir Cumference

@0celo7 I think "kind of" is a bit of an understatement...

0celo7

Time to rearrange the books!

Sir Cumference

So, uh, if anyone knows about degen. gases, that'd be really helpful

user116211

@0celo7 ;))

0celo7

8:43 PM

also I might as well put the library books I have on the shelf too.

heather

@SirCumference, not that I'd be helpful or anything, but out of curiosity, what specifically is the issue with degenerate gasses?

Sir Cumference

@heather I need someone to tell me if my understanding for why degenerate gases form in collapsing stars (which become white dwarves) is correct

0celo7

Ask Chris...oh.

Sir Cumference

@0celo7 ... ;-;

0celo7

Oh, ask Kyle...crap.

You're SoL.

rob

8:45 PM

@SirCumference What's your understanding?

0celo7

@MAFIA36790 I have too many damn books.

user116211

I know.

0celo7

So, topology and geometry books up top, GR and Analysis books on the bottom.

heather

@SirCumference, I am intrigued. What is your understanding? (::frantically googling so I can know what in the universe you're talking about::) =)

0celo7

And this singular algebra book wherever it fits.

Hmm

this isn't gonna work

I need to completely rethink this

Sir Cumference

8:49 PM

@rob Well, here's my reasoning: when a star collapses, it begins compressing towards the center of mass. This lowers its gravitational potential energy more and more, so much so that the atoms will eventually reach their zero-point energy (or the lowest energy configuration each atom can have, since only two electrons can occupy the ZPE).

At this point, the atoms physically cannot get closer to the center of mass (since doing so would require GPE to decrease) and must resist gravitational contraction, releasing a pressure known as degeneracy pressure.

Is this correct?

This is all in my head

0celo7

Proof?

Sir Cumference

@0celo7 Just leave

I'm stressed out as it is

0celo7

:(

rob

Play nice, please.

@SirCumference That's different from the way that I think about it, but it's not wrong.

0celo7

crap

Sir Cumference

8:51 PM

@rob Is it necessarily correct?

0celo7

@MAFIA36790 I need help

user116211

?

0celo7

Dover books are too small

user116211

I know they are.

rob

@SirCumference You're taking this class as a freshman? I think your explanation is fine.

I can elaborate more later, perhaps, but got to go for now.

0celo7

8:52 PM

look at that size difference.

user116211

okay, this is not good.

Sir Cumference

@rob The problem I'm having is that, as the radius gets smaller and smaller, the kinetic energy should become more and more influential to the total energy than the potential energy, so why wouldn't the radius just continue decreasing and rise in kinetic energy?

heather

@SirCumference, well, from my speedy reading, here is my probably incorrect reasoning: when a star collapses, it begins compressing towards the center of the star. This shoves the atoms together, and eventually the atoms are so compressed Pauli's Exclusion Principle comes into play, preventing the atoms from coming closer together.

However, there is still a lot of pressure pushing the atoms together, so the atoms "push back" on each other. This new pressure that is created is called degeneracy pressure. It is this that prevents stars collapsing into white dwarfs from collapsing into a blac…

I don't know how bad that explanation is, thhough.

0celo7

@MAFIA36790 yeah, so large books can't be near Dover books

Sir Cumference

@heather There's a few fallacies. First, atoms don't have definite position.

rob

8:54 PM

@SirCumference Because of the uncertainty principle. But like I said, I'll have to elaborate later.

user116211

hmm.

0celo7

but I use dover-size books as much as my Huge AMS books

Sir Cumference

Quantum weirdness dictates over that realm, so things like a definite position and size of a particle aren't existant.

heather

@SirCumference, hmm, yes. That is a rather large flaw.

user116211

I have Fermi, Seth Werner currently from Dover. They really have peculiar dimensions.

0celo7

8:55 PM

ridiculous

user116211

._.

Sir Cumference

@heather The Pauli exclusion principle says that no two fermions (half-integer spin particles) in a system can occupy all the same quantum states (energy levels, for example, which are discrete energies that the particles can have). What we can have is fermions sharing some quantum states and differing in others.

0celo7

No two particles?

Be precise, dammit.

Sir Cumference

For example, we can have at most two electrons (of opposite spins) for each energy level.

heather

@SirCumference, so this is really a completely flawed explanation.

user116211

8:56 PM

Fermions.

0celo7

Exactly.

Sir Cumference

@MAFIA36790 Crap, yeah

user116211

Not Bosons.

Sir Cumference

Forgot the most important thing to mention

heather

@0celo7, I apologize, that is the sort of thing that bothers me to no end.

Sir Cumference

8:57 PM

@heather So we can have a spin-up electron and a spin-down electron with the same quantum state.

heather

@SirCumference, right.

user116211

Lanczos is long; even though it is from Dover.

0celo7

ok now I have 19 books on my desk

wtf

user116211

WoW!

0celo7

this looks silly

Sir Cumference

8:58 PM

@heather So the atoms don't really push each others at those scales. Usually when you "push" an object in everyday life, the electrons in your atoms are repelling the electrons in the other atoms. So it comes down to the electromagnetic force.

0celo7

@MAFIA36790 that's the ones not on the shelf

probably 25 there

Holy shit I have an addiction

Sir Cumference

With such massive objects like white dwarfs, gravity easily wins over the electromagnetic force.

user116211

WWoWW!!!

user116211

@0celo7 me too.

0celo7

@MAFIA36790 that's not including all the physics books...

Sir Cumference

8:59 PM

Even though gravity is usually the weakest force at the low-mass scale.

heather

@SirCumference, right, okay. And then your explanation comes in.

user116211

@0celo7 Holy crap! You have some monster.

heather

@SirCumference, you know, you say you don't understand it, but you are explaining it rather well.

0celo7

@MAFIA36790 Well my job pays for them

Sir Cumference

@heather Yeah, mine talks about energy. See, all particles have a lowest energy, from which they physically can't get any lower. That's a positive nonzero energy though. Such an energy is called the zero-point energy.

0celo7

9:00 PM

I bought most of them with my own money

user116211

good.

0celo7

So...

heather

@SirCumference, ah, okay.

Sir Cumference

I'm reasoning that, since electrons lose gravitational potential energy as they get closer to the center of mass, they'll eventually reach the zero-point energy.

0celo7

But this is silly. I need to do some rearranging.

Sir Cumference

9:01 PM

I'm not sure if my logic is totally sound in practice.

user116211

Take time.

heather

hmm, wow, that makes sense.

user116211

Anyways, I'm off for now. Night comrades.

Sir Cumference

@heather I might add something else: a degenerate gas is basically a gas, but is pretty special. The electrons in its atoms are all at their lowest possible energies (zero-point energies). To get such a phenomenon, you'll need a bunch of really cold gas (which has lost so much heat energy that the electrons have reached their ZPE), as in Jupiter's case. Or you'll need a dense gas at whatever temperature (I was trying to reason why a degenerate gas would form from density back up there)

Degenerate gases aren't something you see everyday, they exert a pressure (called degeneracy pressure) and are found in lots of astronomical objects. They have really strange properties too.

heather

@SirCumference, wow, you're really good at explaining this. I think I'm starting to (maybe) get it.

Sir Cumference

9:08 PM

@heather That's good to hear. It's prevalent in massive objects like brown dwarves, white dwarves, neutron stars and other things that don't fight against gravitational compression by nuclear fusion. Instead, degeneracy pressure supports them.

0celo7

Ok, I have an arrangement that works.

::looks down, sees two more books::

Welp.

Sir Cumference

I was trying to reason why degeneracy pressure would exist, and I need someone else to tell me if I'm correct.

heather

@SirCumference, how (if at all) would this relate to black holes...? (probably nonsensical, but)

Sir Cumference

@heather It's actually very sensical and a great question

heather

@SirCumference, really? Oh, cool!

Sir Cumference

9:10 PM

Say we have objects like neutron stars. As they get more massive, the star begins to compress slowly and slowly from gravity, becoming gradually smaller. This causes degeneracy pressure to increase, which fights off against that gravity.

0celo7

prooooooooooof

heather

Hmm, that makes sense

Sir Cumference

@0celo7 Go learn it on your own

I'm explaining to someone with no QM background

0celo7

Hmm, makes it sound like you do.

heather

@0celo7, come on, shoo =P

Sir Cumference

9:12 PM

@0celo7 I know what I need to know. I'll learn it when I need to learn it.

0celo7

@heather We were gonna talk about sequences.

heather

@0celo7, okay, I'm ready, I have plenty of time right now =)

Sir Cumference

@heather However, at some point that becomes a problem. See, in general relativity, mass isn't the only source of gravity. Other factors, like energy, momentum, and even pressure are sources of gravity. So eventually when the pressure is high enough, it will actually cause gravity to increase.

Pressure will thus increase, and gravity will thus increase, continuing over and over. This inevitably leads to a black hole.

0celo7

@heather Ok, so we had the definition of $|x|$ as $x$ when $x\ge 0$ and $-x$ when $x<0$.

heather

@SirCumference, yeah, but then how does everything else play in, like degeneracy pressure? I assume it doesn't just disappear. I guess I don't understand how that works...

0celo7

9:15 PM

It's very easy to see that $|x|=0$ iff (if and only if) $x=0$.

heather

@0celo7, hmm, yes, I remember that.

0celo7

Furthermore, we have $|xy|=|x||y|$.

A formal proof takes some time, but it should be obvious.

Sir Cumference

@heather Degeneracy pressure is the pressure that's prevalent in neutron stars and some other objects. It usually fights gravity, but when there's enough pressure, it'll act as a source of gravity and contribute to the collapse into a black hole.

0celo7

Now, the first nontrivial thing is $|x+y|\le |x|+|y|$, called the triangle inequality.

heather

@0celo7, hmm, yeah, didn't we talk about the triangle inequality when talking about metric spaces?

0celo7

9:17 PM

Yes! We will see shortly that the real numbers are a metric space.

So, how would one verify such a thing?

Sir Cumference

@Qmechanic You good with degenerate gases?

0celo7

Do you want to suggest something or should I tell you?

heather

@0celo7, I'm thinking, but I guess I would say you could just draw a triangle, and show it that way, right?

I guess I don't know how to prove it formally.

0celo7

So

I think I got a 99 on that QM test after the curve.

But the grade system looks weird.

So I won't celebrate yet.

@heather This is on $\Bbb R$, what triangle?

We will talk about $\Bbb R^n$ much later, it's legitimately hard to prove it there.

Hard if you don't know the trick I guess

heather

@0celo7, well, is there such a thing as a unit triangle, like the unit circle? Or no?

0celo7

9:24 PM

So the traditional way of proving this is called Fallunterscheidung in German.

No clue what it is in English.

Basically, you suppose $x,y>0$, then prove it in that case

Then $x,y<0$, and prove it in that case

etc.

But there's a much better way that I can hopefully remember :)

You first show $(x+y)^2\le (|x|+|y|)^2$, from which we get the original result by taking the square root.

heather

@0celo7, okay...how do you show that?

0celo7

Well, it's obvious that $x^2=|x|^2$, right?

So expand both sides: $x^2+y^2+2xy\le x^2+y^2+2|x||y|$

Cancel stuff: $xy\le |x||y|=|xy|$

But $a\le |a|$ is obvious.

So it's true!

So we have $|x+y|\le |x|+|y|$.

Now we use a trick: $x+y=x-z+y-z$.

Oops. Forgot a step.

@heather You can check that $|x-y|\le |x|+|y|$ is true too.

heather

right, okay, I think I followed there. maybe. I'm reading through it again.

0celo7

So now $x-y=x-z-(y-z)$, and $|x-y|=|(x-z)-(y-z)|\le |x-z|+|y-z|$.

But this is exactly the triangle inequality for a metric $d(x,y)=|x-y|$.

heather

okay, yeah, that makes sense.

0celo7

9:37 PM

@heather Ok next we need the notion of an upper bound of a set.

heather

@0celo7, okay.

0celo7

Let $A\subset \Bbb R$, then $b\in\Bbb R$ is an upper bound for $A$ if $b\ge a$ for any $a\in A$.

Pretty easy.

If no such $b$ exists, then $A$ is unbounded.

heather

oh, I see. but how could a set be unbounded? Except the empty set, I mean.

0celo7

$\Bbb R$ is unbounded

$A$ is nonempty.

heather

Oh, hmm, yeah. So if it is an infinite set or it is empty, then it is unbounded?

0celo7

9:43 PM

If it's empty, we don't talk about it being bounded or unbounded.

heather

That makes sense.

0celo7

Ok, last thing for today is the least upper bound. A least upper bound for a set, called $\sup A$ (supremum) is a a number such that (i) $\sup A$ is an upper bound for $A$, (ii) if $b$ is an upper bound for $A$, then $\sup A\le b$.

heather

wait, how does that work? how can $\sup A$ be an upper bound for $A$ if it is less than another upper bound $b$?

0celo7

less than or equal to.

It's less than or equal to all other upper bounds.

heather

sure, but how would it ever be possible for $\sup A$ to be less than $b$? I understand how it could be equal to $b$.

0celo7

9:48 PM

Example!

Take $A=[0,1]$.

What do you think $\sup A$ is?

What's the smallest upper bound for $A$?

@heather can you give me any upper bound for $A$?

heather

@0celo7, sorry, had to check something real quick. I think it would be 1, right?

0celo7

The sup?

Yes.

@heather How about $A=(0,1)$?

heather

Wouldn't it still be 1? I'm not sure there's a difference there.

0celo7

currect

but notice in one case that $A$ contains its sup, but not in the other case

ok, Fact: Any bounded subset of $\Bbb R$ has a sup.

Proof. Too hard, you never want to do it.

In many treatments we take it as an axiom.

@heather So just take that for granted, the way you'd prove it is via Dedekind cuts.

heather

@0celo7, okay, perhaps I need to back up a bit. there's a difference between the set [0,1] and (0,1)?

0celo7

9:59 PM

You don't know the difference between open and closed intervals?

heather

Oh, wait, ( means 0 and 1 aren't in the set in this case, but everything between is...oh, okay, I remember now. Sorry, I didn't realize that applied here.

So then, I'd still say 1 for [0, 1]

but for (0, 1) I don't know what I would say.

0celo7

It's 1. I can give a formal proof tomorrow

heather

Well, I guess because 0.9 repeating equals 1, right?

0celo7

But it should be intuitively clear

Oh god

Don't say stuff like that

heather

erm, why not? Isn't that true?

@0celo7, what exactly is wrong with that statement?

heather

10:29 PM

@DanielSank, would it be possible to visit the afternoon of Tues, Nov 1st or morning of Wed, Nov 2nd?

@DanielSank, and just to confirm, what is the address of the lab?

DanielSank

10:44 PM

@heather I believe so.

@heather 6868 Cortona Drive Suite B.

0celo7

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\
\da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\
0 & \ras{h_1} & 0 & \ras{h_2} & E & \ras{h_3} & F & \ras{h_4} & 0 & \ras{h_5} & 0 \\
\end{array}$$

Oh the power.

@ACuriousMind It is here.

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