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5:13 PM
@Lord_Farin Lordy, lord.
 
@Lord_Farin I think there is a problem with this definition, your $\partial_X V$ always belongs to $TM$ (if X and V are vector fields), but in the case of surfaces in $\mathbb{R}^n$ it isn't true in general and one defines projections of this on $TM$ and on $NM$ (first is connection $\nabla_X Y$, second is second quadratic form of surface)
 
5:26 PM
Hi folks
 
@RajeshD Hi
Are you from Hyderabad?
 
yes
I think I have spoken to you once before
 
I think so too. :-)
What do you do? (If you don't mind me asking that is.)
I am starting grad school at TIFR CAM this year.
 
hah I am an engineer, I am developing speech processing algorithms for speech analytics applications in a call center
 
@RajeshD Ahh, I see. I am an electronics engineer too in Bachelors.
 
5:30 PM
so u liked signal processing?
 
Yes, I liked signal processing.
 
good to know... TIFR hyd or Mumbai?
 
TIFR Bangalore actually.
Center for Applied Math.
 
ok
 
TIFR Hyd just has physics I think.
 
5:32 PM
ok
 
@Nimza @Peter: Sorry, was called away for dinner.
 
@Lord_Farin I've found a solution!
 
Please tell me all about it!
 
Since $Y = Y^i \frac{\partial \vec r}{\partial u^i}$ then $\frac{\partial Y}{\partial u^j} = \frac{\partial Y^i}{\partial u^j} \frac{\partial \vec r}{\partial u^i} + Y^i \frac{\partial^2 \vec r}{\partial u_i \partial u_j}$
 
Oh yes, of course.
 
5:41 PM
And $\partial_X Y$ is just a sum $X^j \frac{\partial Y}{\partial u^j}$, where $X = X^j \frac{\partial \vec r}{\partial u^j}$
hehehe)
 
Huh? But $Y$ is a vector field, and you're treating it as a function.
 
Hi :)
Can anyone tell me how I can calculate the probability that a sample is outside of the coincidence-range?
I have already calculated this range...
 
@Lord_Farin aha, and I can do the same with matrix-valued function $Y$, where do you see a problem?
$X^j$ is a scalar, $\frac{\partial Y}{\partial u^k}$ is a vector
 
So the latter is actually $\sum_i \dfrac{\partial Y^i}{\partial u^k} \dfrac{\partial \vec r}{\partial u^i}$.
 
@robjohn With spectrum in a sector.
 
5:49 PM
@Lord_Farin hm, no, $Y$ has $n$ components, but surface is $k$-dimensional (so there are only $k$ variables $u^j$)
 
@Nimza Dude.
 
@PeterTamaroff a?
 
@Nimza: $\dim M = \dim T_pM$, all $p$.
 
@Lord_Farin right, $\dim T_p M + \dim N_p M = n$
 
A monotone function discontinuous at a countable dense subset of the reals.
Let $E$ be dense and countable in $(a,b)$, say $E=\{x_n:n=1,2,\dots\}$
Let $c_n> 0$ be such that $\sum c_n$ exists.
Set $$f(x)=\sum_{n\in\Bbb N \; ; \;x_n<x} c_n$$
 
5:52 PM
@Lord_Farin our surface is submanifold in $\mathbb{R}^n$ and for each point $p$ $T_p \mathbb{R}^n = T_p M \oplus N_p M$, sorry, I'm called to eat
 
@Peter: You want $c_n > 0$ (strict).
 
@Lord_Farin Sorry?
 
Take $c_n = 0$, all $n$; no discontinuities.
 
@Lord_Farin Yes, yes.
 
5:53 PM
@Nimza Ok, enjoy!
 
thanks!
 
@Lord_Farin We show that $f(x+h)-f(x)\to 0$ simply using that $[x,x+h)\to\{x\}$, yes?
 
Gay Lord.
 
While if we chose $x_k$, we sum only on $k$ and $f(x_k^+)-f(x_k^{-})=c_k$
 
@PeterTamaroff Yes.
 
5:54 PM
@Lord_Farin I did that.
Pretty interesting construction.
Never encountered such a thing.
Can we write the sum in terms of indicator functions, instead?
Or Iverson brackets?
 
@Peter: $f = \sum_n c_n \chi_{\{x \ge x_n\}}$ (actually, I now see that your $f$ is left- rather than right-continuous).
 
Iverson brackets, of course.
 
@JonasTeuwen I can't comprehend the origin of this childishness... 't Makes me sad.
 
@Lord_Farin Wait, don't we want the set to be $$E\cap(a,x)$$?
OK, maybe that doesn't work with the indices well.
The conditions is $n\in\Bbb N$ and $x_n<x$. So we sum through the indices $n$ for which $x_n<x$ is true. I think $[x_n<x]$ works nicer, @JonasTeuwen, yes.
 
Hi, I'm trying to prove an inequation using complete induction. I managed to break down the given problem to proving that: 1/2 \leq sum 1/k, from k=2^n +1 to 2^(n+1). Unfortunately I'm kind of stuck here. Do you have any advice?
 
6:01 PM
Oh yes I meant $\chi_{\{\uparrow x_n\}} = [\uparrow x_n]$.
 
@feaDawn $$\frac 1 2 \leq \sum_{k={2^n+1}}^{2^{n+1}}\frac 1 k $$?
 
exactly
 
@feaDawn OK. For $n=1$ we have $$\frac{1}{3} + \frac{1}{4} \geqslant \frac{1}{4} + \frac{1}{4}$$ yes?
 
@feaDawn: $\dfrac 12 = \sum\limits_{k=1}^{2^n} \dfrac1{2^{n+1}}$,
 
Try first to see the idea of this.
 
6:03 PM
My guest seemed to find my stuff 'intimidating'
Sad? Why not be gay instead.
 
@JonasTeuwen Your maths?
 
Stop being sad, start being gay!
2
 
@JonasTeuwen Which stuff? ;-)
 
@Lord_Farin k=2^n+1 to 2^(n+1) for greater parallelism
 
Some of my results - or their proofs -.
 
6:04 PM
@anon I didn't want to give away everything :).
 
@JonasTeuwen Ahh. That is a compliment I will say.
 
$$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \geqslant \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}=1/2$$
@feaDawn And so on..
You have $2^n$ terms at step $n$.
Each term before $1/2^{n+1}$ is greater than it.
So you have $2^n$ terms greater than or equal to $2^{n+1}$, which accounts to $$\geq 2^n/2^{n+1}=1/2$$ see?
No induction needed.
Nice way of showing $H_n$ diverges, by the way.
 
Wow, first of all thank you very much for your help! I guess I need some time to think about it.
 
$$\eqalign{
& \frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}} \cr
& \geqslant \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} = \frac{1}{2} \cr} $$
 
@Lord_Farin I'm back, all is clear with it?
 
6:14 PM
I still think that our definitions agree.
At least for proper vector fields on $M$.
 
it uses induction
 
@Lord_Farin but in your definition $\partial_X Y$ is an element of $T_p M$ if $X$ is a vector, $Y$ is a vector field, since it is a linear combination of $\frac{\partial}{\partial u_i}$
 
@caveman OK, yes. My point was first we must develop the idea, then the proof.
 
@Nimza Yes, naturally. But I was also under the impression that $Y$ was $M \to TM$, while you seem to have $Y: M \to \Bbb R^n$.
 
@PeterTamaroff I think I finally got it! Your explanation was great - thank you very much - you really helped me out!
 
6:20 PM
@Lord_Farin So I think that what you're speaking is $\nabla_X Y$ in the case of surfaces in $\mathbb{R}^n$. It is projection of my thing on $TM$
 
Equivalently, I'm assuming that $Y$ is zero on the $n-m$ coordinates that are perpendicular to $M$.
 
@Lord_Farin yes, it is always like this since $Y$ is tangent to $M$
 
But if $Y$ is tangent to $M$ then so will $\partial_X Y$ be.
 
So you think that $P_{TM}(\partial_X Y) = \partial_X Y$ is we use my definition?
then our definitions agree, if it is like that; here $P_{TM}$ is the orthogonal projector
 
@feaDawn I am glad to read that.
 
6:27 PM
For if $Y = \sum\limits_{i=1}^m Y^i \dfrac{\partial}{\partial u_i}$ (i.e. with $Y^i$, $i>m$ all zero), then: $$\partial_X Y = \sum_{i=1}^m X^i \sum_{j=1}^m\frac{\partial Y^j}{\partial u_i} \frac{\partial}{\partial u_j}$$
Where the inner sum is what I take you mean by $\dfrac{\partial Y}{\partial u_j}$.
(Sorry for mixing all those $i,j$ up; it should be correct now.)
 
So your vectors are $m$-dimensional?
 
Well, their $n-m$ remaining components are zero (if we take $\Bbb R^n = TM \oplus (TM)^\perp$ with the first $m$ components spanning $TM$).
IOW I am avoiding at this point to deal with the action of vector fields on submanifolds.
 
Okay, enjoy))
 
@Nimza: What we went through is very reminiscent of my first encounter with these things; I recall I grasped it only one week before the exam (so after practically a whole semester).
 
@Lord_Farin heh! your course was about general smooth manifolds or about embedded in $\mathbb{R}^n$? I'm not sure, but maybe there is a difference in notion of this thing $\partial_X Y$ for these two types of manifolds (because for embedded ones we have such cute thing as normal space e.t.c.). but covariant derivative will agree
 
6:39 PM
The basic theory was developed in full generality. Also submanifolds were done in full generality (but admittedly I've forgotten most of that). It influences how vector fields may be defined (i.e. in "my" setting we always have $Y: M \to TM$ while in "yours" $TM$ is the ambient $\Bbb R^n$) which likely was the cause of most of the confusion.
 
yes, I think so too
 
But lengthy confusion is often, once it is resolved, a great source for deeper understanding of the subject :). Good luck with the remainder of your course; you might be off better with someone else (trying to) help you with further questions :).
 
Heh, great thanks for help and interesting talk, good luck with group actions)
 
Thanks. Should you want a good though dense source, be sure to check out the course notes I used!
(Updated to link to newer version.)
 
Hi all , I need the definition of trivial element in field. Didn't find it in Google. Who knows ?
 
6:42 PM
Oh - those notes I do know.
 
@Ashot: Presumably it's the additive identity, but to assess that properly a bit more context would be nice.
 
I need to find all trivial elements of field F_13
 
@Ash, what's a trivial element?
 
I am asking that :)
 
it doesn't mean anything
 
6:46 PM
maybe a element which generates the field, what is it called ?
 
A generator? 8-).
 
ok, something is wrong with question, thank you
 
perhaps the reason you're looking for a definition is that it appears in some context. why not explain the context instead of keeping others in the dark?
 
a generator
 
for each n and prime p I need to find a polynomial which solutions over field F_q are trivial , but I dont know what is trival element
 
6:53 PM
Has anybody heard about characterisation of completely-monotone functions in terms of Mellin transform?
 
I'm calling it a day; goodbye all!
 
bye
 
@JonasTeuwen Ah, then it is not so surprising :-)
 
7:08 PM
Hey Rob
 
@FedericaMaggioni Your gravatar seems broken.
 
7:56 PM
@robjohn But still took like 10 years to prove 8-).
Actually, the equivalence with properties of associated wave equations and square functions and such were quite hard to tackle.
 
@PeterTamaroff Those sound interesting I have very much wanted to learn about the first one, unfortunately I am quite lacking in many other subjects which I think I would need.
@robjohn The only 'slightly' related thing I can say I have done in the past was I used hurrwits fourier expansion for bernoulli polynomials
$$\text{B}'_{2r}(x)=\frac{-r!}{(2\pi i)^r}\sum_{k=-\infty}^\infty'\frac{e^{2\pi i x}}{k^r}$$
And properties of dirichlet convolutions one can show identitys for arithmetic functions like,
$$\frac{(-1)^{r+1}2(2r+1)!}{(2\pi)^{2r+1}}\sum_{n=1}^\infty\frac{(f*1)(n)\sin(2 \pi n x)}{n^{2r+1}}=\sum_{n=1}^\infty\frac{f(n)}{n^{2r+1}}\text{B}'_{2r+1}(nx)$$
$$\frac{(-1)^{r+1}2(2r)!}{(2\pi)^{2r}}\sum_{n=1}^\infty\frac{(f*1)(n)\cos(2 \pi n x)}{n^{2r}}=\sum_{n=1}^\i
 
@Ethan What are you talking about?
 
I suppose any function with multiplictive fourier coeiffients would yeild the same results though
 
@Ethan, do you mean {px}
 
@PeterTamaroff I meant to ping robjohn my bad
@caveman yes
 
8:02 PM
@Ethan, I can't beleive you have a proof of this
 
@caveman do you want to see it?
 
@Ethan, look at the big about Vinogradov en.wikipedia.org/wiki/Equidistribution_theorem
 
@caveman let me show you
@caveman $$\sum_{n\leq x}(f(n)*g(n))\beta(n)=\sum_{n\leq x}f(n)\sum_{k\leq \frac{x}{n}}g(k)\beta(kn)$$
Keep in mind for integer functions f(n) g(n) and B(n) respectively
Let x tend to infinity in both expressions
Supistute f(n)=1, g(n)= $\Lambda(n)$, the vonmangoldt function
Now let $B(n)$ be $\sin(2\pi n)$
Keep in mind the fourier series for the first bernoulli polynomial
The right hand side, should come out to $\Lambda(n)/n$ times $B_1(x)$
 
I love that Dirichlet theorem
so cool
 
@caveman Yes it was the first thing I used to study it lol, I could prove the cases for a=4, a=3
But this stops working
Its based apon the fact you can write indicator functions for congruences as a linear combination of fractional parts
It stops working at around modulo 5 though so..
 
8:09 PM
Could someone point me to the latex rendering bookmark, I had formatted my system recently
 
wow Ethan
you are really scary
this is some advanced stuff
 
A little bit creepy.
@RajeshD Perhaps it has already been done so.
 
I know, I am asking for the link
 
There is a bloody link.
 
give me
 
8:12 PM
No.
 
someone talking about Dirichlety theorem, may i know in words what you are talking about?
 
Dirichlet theorem?
 
lol
 
@RajeshD I have been trying to write my own proof with out characters for about a month now lol, the best I can show is that atleast $\frac{\phi(a)}{2}$ of the residues in any reduced residue class modulo a, contain infinitely many primes
 
8:15 PM
It is remarkbly easyier to obtain results on semi primes then primes
 
@Ethan, you should read about sieving
 
Using convolution identities like $\mu(n)*ln(n)^2=\Lambda(n)\ln(n)+\Lambda(n)*\Lambda(n)$
The key here is although the mobius function is a pain in the ass to work with
the double logarithm factor trumps this
In selbergs elementry proof of the pnt, it is precisely this technique he uses
The right hand side can be restated in terms of semi primes with some additional factors that vanish when summed right
 
Sigh : I am not able to add a bookmark in chromee!
 
You can obtain things like $$x\ln(x)+O(x)=\sum_{pq\leq x}\ln(p)\ln(q)$$ with only elementry algebraic techniques
this is infact a modification of selbergs formula, in which $p$ can equal $q$, that is , it also sums prime squares, not just distinct primes
 
When I right click on it I do not get the add as bookmark option @Jonas
 
8:25 PM
Is it possible to do LaTeX searches in MSE?
 
oh you were talking about number theory
 
8:42 PM
@robjohn here?
If anybody knows what I'm doing wrong trying to create a spoiler in this deleted answer, tell me!
 
Can someone help me with this?
0
Q: Weight function

Carpediema) Find $P(D)$ whose weight functions is $\omega(t)=100e^{3t}\sin t$ and find the exponential solutions for this operator $P(D)$ to $$P(D)y=e^{3t}$$ We have $W(s)=\mathcal{L}\{w(t)\}= 100\mathcal{L}\{e^{3t}\sin t\}=100\frac{1}{(s-3)^2+1}=\frac{100}{s^2-6s+10} $ Does that mean that $P(D)=100(D^2...

 
@anon I have no idea! =D
 
@PeterTamaroff Can you help me ?
 
@Carpediem I am leaving now, so I cannot.
 
9:24 PM
Oh.... Saturday should be called "sad-urday"
Or If I follow jonas' advice, "saturgay"
 
Better than sad-urectum.
 
Don't you think @anon ?
@JonasTeuwen yes :-/
 
9:43 PM
8-).
 
can someone help me with number thoery question?
 
sure. I remember you from the mainsite.
 
@anon Can you help me ?
 
If I could have I would have.
 
if $p\equiv1(3)$ then exist integers x,y $p = x^2 + x^y + y^2$. @anon
i think this has something to do with minkowski thm
 
9:55 PM
x^y? is that written correctly?
 
sorry xy
 
my immediate impression is that it's related to the Eisenstein integers
 
ok it is a 2 part question, I made a thread about part a here: math.stackexchange.com/questions/360417/… this is part b, perhaps it helps
Eisenstein integers, I've never met these before
?
 
"Natural primes that are congruent to 0 or 1 modulo 3 are not Eisenstein primes: they admit nontrivial factorizations in Z[ω]" (Wikipedia). This fact is a direct restatement of what you seek to prove, so there's probably a way to translate a proof of this into a more elementary proof in your problem.
 
that seems a bit tricky. I don't know what a natural/Eisentein prime is or Z[omega]. maybe I could ask you something a bit easier to start off with?
can i please send you a pdf
and if you can pick any question that seems easy that we can work on?
 
10:09 PM
maybe you should try to work on your homework before I do :)
 
its not
you can check the date
it is in preparation for my exam
the deadline passed
 
Alex Bartel, where have I seen that name before..
 
he's the lecturer
 
hmm, I have his webpage in my bookmark archive
 
10:12 PM
#2 is pretty easy for me and I've already written about it before
 
would it be possible to work on a q with you
ok
i will read it now and get back to you
 
#3 I know as well.
ah, geometry of numbers. something I probably should know about but don't.
 
that link you sent looks really difficult :(
 
#1 is really interesting too, I should learn about it for my own p-adic course
@MathproofP. I can digest it for you in chat
do you have the latex bookmark up and running?
 
what is latex bookmark
thanks that would be great
 
10:17 PM
See "LaTeX support for chat" link pinned on the starboard.
 
i can copy the latex from what you write in lyx?
i dont mind
i cant see latex on the chat board
 
did you even go to the link? the bookmark, once you click it, runs automatically all the while you're in chat
 
i know drag and drop
bookmark no
the link you mentioned goes to a forum page
 
@MathproofP. the link goes to a specific answer that explains everything
what internet browser do you use, and do you have the bookmarks bar/folder visible?
 
i use firefox
yes i have it
 
10:23 PM
then drag the "start MathJax" blue link from the webpage I linked to your bookmarks bar, then click it while you're on this tab
 
ok done
$\in$
 
does it work $\LaTeX$ ?
 
it works
 
alright, fire away
 
ok so for question one then
can we go through it
I will just read it one sec
 
10:26 PM
I'm not familiar with continued fractions. Do you mean #2?
 
so it is a continued fraction
ok yeah 1 sec let me read it and tell u my thoughts
so if $a\inZ_p$ does this mean that it is a set $a+pk$
or is this about p-adic integers
in which case we need to show that sequence they gave, perhaps the series with that sequence converges to 0
p adically
?
 
"it is in the set $\{a+pk:a,k\in{\bf Z}\}$" would correspond to $x\in a+p{\bf Z}$. The notation $x\in A$ always means that $x$ is an element of $A$. And the notation ${\bf Z}_p$ in the context of $p$-adics stands for the $p$-adic integers.
@MathproofP. "perhaps the sequence with that series converges to 0" nope
 
ok how about finding what it converges to?
if I put n=3
then i need to work out the padic valuation right?
 
They converge to roots of unity (which we'll show in a bit) that have complicated expansions that you don't need to worry about.
Whereas sometimes it is easier to prove something converges when you know what it converges to, this will not be one of those times.
First note that if $x$ is a unit (equivalently, has absolute value $1$), then so does any power of $x$, e.g. $x^{p^n}$.
Since $|x|=1\implies |x^a|=|x|^a=1^a=1$. Follow? (I mean the $p$-adic absolute value. I am suppressing the designator because we should both know I'm talking about the $p$-adic one.)
 
Let $p(D)=D^2+aD+b$

The weight function is given by $\displaystyle W(s)=\frac{1}{s^2+as+b}$
Let $w(t)$ be the inverse laplace transform of $W(s)$

If I find an expression for $\displaystyle W(s)=\frac{\lambda}{s^2+as+b}$, what will be $p(D)$ ?
 
10:34 PM
@Carpediem The makeup of the room hasn't changed since the last time you spammed your question, and you're interrupting us. I realize you're desperate or excited or whatever, but please have some control.
 
yes i understand |x| means $|x|_p$
 
I didn't realize you were having some other discussion.
 
how did you pull out that a from |x|
 
You can go discuss it in a private room though..
 
@MathproofP. Since $|xy|=|x||y|$, we can prove by induction that $|x^n|=|x|^n$. does that make sense?
 
10:36 PM
yes that great from rules of padic absolute value
$|x|^{n-1}|x|$ etc..
 
to show a sequence converges to zero, we can show that the difference between consecutive terms approaches zero. that is, we want to show $x^{p^{n+1}}-x^{p^n}\to0$ in the $p$-adics as $n\to\infty$
 
yes cauchy seq. implies conv.
 
now, $x^{p^{n+1}}-x^{p^n}=x^{p^n}(x^{p^{n+1}-p^n}-1)=x^{p^n}(x^{p^n(p-1)}-1)$. since $|x^{p^n}|$ is bounded, it therefore suffices to show that $|x^{p^n(p-1)}-1$ converges to zero
 
but why to zero?
 
sorry, I meant "to show a sequence converges to something in an ultrametric space, we can show the difference between consecutive terms approaches zero"
note that this is false in the archimedean world. just look at the sequence (1,1+1/2,1+1/2+1/3,...) in the reals for instance (these are the partial sums of the harmonic series).
 
10:42 PM
yes that is harmonic seq, which is cauchy i think? but divergent?
 
correct
 
what is $x^{p^n}$ bounded by?
 
the reason it holds in the nonarchimedian world is: suppose $\{a_n\}$ is such that $a_{n+1}-a_n\to0$. pick $\epsilon>0$ arbitrary. by hypothesis, there is an $N$ for which $|a_{n+1}-a_n|<\epsilon$ for all $n>N$. hence given any $n>m>N$, we have $$|a_n-a_m|=|(a_n-a_{n-1})+\cdots+(a_{m+1}-a_m)|\le \max_{m\le i<n} |a_{i+1}-a_i|<\epsilon,$$ (by the ultrametric inequality) fulfilling the conditions of a cauchy sequence.
 
@anon did you solve your problem?
 
@MathproofP. every $p$-adic integer has $p$-adic absolute value at most $1$.
@robjohn nope
don't really care anymore though
 
10:45 PM
@anon okay. Sorry I was gone.
 
so, once we finish this stage, the next stage will be to show that $x^{p^n(p-1)}-1$ converges to $0$ in the $p$-adics
anything you want me to clarify at this point?
 
yes can i have just 2 minutes to brisk over and ask a question
 
of course
 
1: you used the ultrametric inequality inductively right? 2: what is $x^{P^n}$ bounded by
I'm quite happy with the rest so far
 
10:52 PM
@MathproofP. like I said earlier, $|z|\le 1$ for every $p$-adic integer $z\in{\bf Z}_p$. that applies to $z=x^{p^n}$ just as well.
yes, the ultrametric inequality is generalized (via induction) by $| u_1+\cdots+u_l|\le\max |u_i|$
in fact by taking limits it applies to infinite sums (you can take limits with nonstrict inequalities in the reals)
 
ok but why is z a p-adic integer in that case?
 
what are you talking about? I was using z to denote an arbitrary $p$-adic integer. should I have used a different letter or something?
or are you asking why $x\in{\bf Z}_p\implies x^{p^n}\in{\bf Z}_p$?
the latter is because ${\bf Z}_p$ is a ring, hence closed under multiplication and addition, hence closed under arbitrary positive integer powers.
 
ow i see ok, yes i was wondering about the latter.
so we just do x.x.x...... $p^n$ times
 
mmhmm
 
and have closure under mult.
 

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