Mathematics - 2014-09-23 (page 2 of 3)

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3:00 PM

@MatsGranvik Note that $$x^5 + x = \frac{x}{1/(1+x^4)} = \frac{x}{ 1 - x^4 + x^8 - x^{12} + x^{16} - x^{20} + x^{24} - x^{28} + \cdots}$$

I will do a numerical check.

Try inverting that.

@MatsGranvik No, no, use that^ as an input

It's of the form x/(1 + blah*x + blah*x^2 + ...) like you wanted

ok

{1, 0, 0, 0, -1, 0, 0, 0, 5, 0, 0, 0, -35, 0, 0, 0, 285}

Yes, that works.

Congratulations.

Thanks.

3:05 PM

Funny though. I am interested in seeing your technique.

It is not entirely new.

Oh?

There is something called the INVERT transform, written with large letters in the OEIS. This algorithm is a generalization of the invert transform, applied infinitely many times.

Link?

oeis.org/transforms.html

3:08 PM

Heya @anon

hi

nvm about my previous question, it was silly.

i was confused by the notations

interesting @Mats. I am going to implement that in my machine, I think.

@Balarka!

@Studentmath!

@BalarkaSen!

@Studentmath!

3:16 PM

@Nick!

@Nick!

@anon!

ho hum.

@BalarkaSen: I have finally digested what you said to me about integration constants. Thank you for that. I realized stuff after about 93 problems.

For a relative max/min at $x_0$ of a function $f: \mathbb{D} \to \mathbb{R}$, $f'(x_0)$ AND a change of sign of $f'(x)$ at $x_0$ is necessary and sufficient, right?

3:17 PM

@Huy: Yes

It seems like I won't be able to advance my analysis in my uni-studies before graduating anymore. It's stuck on calculus+multi.var calculus, without real epsilon-delta proofs (and soon set-point topology but that doesn't count in). Any suggested books one can read to advance up?

Hopefully not too boring, yet rigirious and stuff.

@Nick: do you by any change remember if there is another sufficient and necessary condition? I remember the second derivative being strictly positive/negative being sufficient, but not necessary.

@Huy: sorry, I only know enough monotonicity to graph basic functions. I don't know any textbook points.

horrible internet connection

@BalarkaSen: get Idea, Sirji

3:21 PM

haha

Ermagard Ermagard Ermagrad, i forgot. I have a huge boolean algebra test tommorow and I don't know potatoes about K-Maps!

What do I do. What do I do?

Solve questions with solutions about K-Maps

@Huy $f$ must be differentiable at $x_0$ for one.

@BalarkaSen: Thank you very much for your information.

@Chris'ssis Cleo is TF ??

3:24 PM

@Huy =P

glad to be of no help

@TheGame I was inclined to believe so.

@Studentmath: I accidently burnt my textbook. Any good links!

@Chris'ssis Why ?

@Nick you accidently burnt your textbook?

Ah, @TheGame

3:26 PM

5 hours ago, by Chris's sis

@robjohn I have an indirect proof that Cleo is Tunk-Fey ..

What book, @Studentmath?

@Studentmath: In a huge dramatic gas cylinder explosion.. which is irrelevant, right now. Because I need to know how to use K-Maps in half an hour and all that google is giving me is gibberish.

@Balarka suggest me books!

@Studentmath That is my job, lol.

@TheGame I think that one was a proof done in a hurry. However, I think Cleo might be Tunk-Fey, or at least someone close to him, her.

3:27 PM

@Nick a huge dramatic cylinder explosion is never irrelevant. But don't you have some course-website with tests? I will give it a google-search try.

@Willhunting @Balarka I need to advance up my analysis. I had calculus and multi.var calculus, not epsilon-delta proofs. I skimmed through Rudin's honour's calculus too. Anyhow, rather something not too boring as I will have to force myself to find time to read and practice it, also something rigirious =.

@Chris'ssis But, why do you think so ?

@TheGame by the way, did you see this answer? I got the real generating function of $$\sum_{n=1}^\infty \frac{H_nx^n}{n^3}$$ The one in the proof is wrong. math.stackexchange.com/questions/909228/…

@Studentmath Then just read the whole of Rudin thoroughly.

@Chris'ssis no. I didn't

Which one of Rudin?

3:29 PM

@Studentmath: It was a year ago, burnt down a relative's house. Charred up a lot of my novels and textbooks. No one got hurt. Except my relative who lost a large chunk of his house and property.

@Nick hrmpf kvvayusenanagar.edu.in/userfiles/file/BOOLEAN.pdf got 5 questions on k-maps. Not sure if they are relelvant level-wise.

@Chris'ssis Hence your suspicion ?

@Nick eecg.utoronto.ca/~jayar/ece241_06F/solved4.pdf This looks much better

@Student Rudin.

Principles of mathematical analysis

Thanks @Will @Balarka

3:32 PM

@Studentmath Anytime you need a book recomm, just ask me!

@Studentmath: Holy crap. You're batman! Thank you very very very much!

@Studentmath jasper will only give you hipster books.

@Nick glad to be of help :P

Don't forget to thank me when you win the Fields medal, lol.

@Will @Balarka Rudin's hipster?

3:34 PM

@Studentmath I don't know what he means by hipster, so no comments.

oh by the way @Studentmath @Nick.

dis

@Studentmath: He'll also give you nice educational books. Actually, he won't give it to you. You have to sneak into house, take it and leave an IOU and hope he won't notice until you finish reading it.

@Balarka I wonder if searching Grrrh will have similliar result

@TheGame You should note that Tunk-Fey is a specialist in handling the polylogarithm. On the other hand, I noted that Cleo answered a lot of questions where handling the polylogarithm, generalized harmonic numbers wisely you get great results.

@Chris'ssis They made the same mistake, right ?

@Chris'ssis Why don't you post the correct answer ?

3:36 PM

@BalarkaSen: I bet you can't find the last time I said misbehaving in this room

@Nick You just did

you already said it, @Nick

The n-1 th time, genius.

@TheGame Not really. I wrongly noted something. Only Tunk-Fey made a mistake.

@Nick The n-1th time what ?

3:38 PM

OK, I need to leave now. my internet connection is too disturbing.

@TheGame: I liked Hippa better

brb

@Nick But now you're gonna lose every day :)

@Chris'ssis Doesn't he say that his answer matches Cleo's ?

@TheGame: The n-1 th time I said the m word, if you have just witnessed the nth iteration.

@Nick Aww you didn't fall for it :P

3:40 PM

@TheGame: I paused and I thought.

@TheGame Initially he wrote $$\color{purple}{\frac{7\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2} 8 \zeta(3) + \operatorname{Li}_4\left(\frac12\right)}$$ (if I'm not wrong)

Now, let me go screw with Karnaugh.

@Chris'ssis Indeed

@TheGame I think he had a great luck to answer that question ...

Hehe

3:42 PM

@TheGame Do you know why? Look at that relation in $(2)$. The value that remains out due to a wrong application of the variable change is then recovered by I don't know what means.

Nope. I'm not good enough :)

Let's compare with the old one

@TheGame the generating function there is wrong, that's clear.

@Chris'ssis $(3)$ is the same

@TheGame Yeap, it's wrong.

Wait

Let me find where the difference comes

What I'm saying is that up to (3), the proof is the same for different results

3:48 PM

@TheGame Hold on a second to explain you something ...

@Chris'ssis The only difference I see is the computation for $x=1/2$

That nice difference, that is $\pi^4/120$, you only get for $x=1/2$ since there an interesting thing that happens and one can make use of that invoked reflection formula.

@TheGame The problem appear from $(2)$. The rest of the work is not safe.

@TheGame For instance, if you check $(3)$ for $x=1/3$, you get different numerical results.

@Chris'ssis Where exactly is the problem ?

What line in (2)

(I mean the right side is different from the left side)

What line is the first wrong one ?

3:53 PM

$$\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\
&=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2
\end{align}$$

@TheGame I refer at that variable change.

What are the rules for variable changes on indefinite integrals ?

If there were bounds, they would change too

@TheGame If you like this way, then you can use it too in your problems. :-)

Uh ? What do you mean ?

@TheGame Did you read the comments below (referring to that answer)? Just take a look there.

@Chris'ssis Ok. In the meantime, can you answer my question about the var change ?

'@ SuperAbound I don't have a rigorous explanation but I think the constant of integration will adjust the final result.'

WAAAT ???

That's not maths -__- that's cooking

3:58 PM

@TheGame that was my point ...

Even I, who do not know much about this stuff, saw that -__-

@Chris'ssis I believe you have found a good way to get the solution ?

@TheGame Sure, I have the correct generating function.

@Chris'ssis How do you do it ? (or, is it secret ?)

@TheGame kind of ...

Ah :/

For the book ?

4:01 PM

:D

@TheGame You know what? I'm sure many here would say I just lie.

@Chris'ssis Btw that might be a weird question, but do you have any idea on the book's price ? For instance, @Ted 's book are really expensive because of the publishers

@Chris'ssis Nono you've already solved way harder stuff

That is beautiful

@TheGame I need to use one more formula to make it more beautiful.

(working on it)

@Alizter Hullo !

You just lost

@Chris'ssis where does Tunk-Fey's computation go wrong?

14 mins ago, by Chris's sis

$$\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\
&=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2
\end{align}$$

4:07 PM

hi @hippa

@robjohn As I said above, from $(2)$ on.

@TheGame maybe my math is criticized sometimes, but with my math I get great results. And soon I get this one

@Chris'ssis By who ?

$$\sum_{n=1}^\infty \frac{H_nx^n}{n^4}$$

@TheGame There are always some. :-)

@TheGame Oooooo, I should ask a question on main.

What one ?

~~maybe I can answer it!~~ uh no

@TheGame $$\sum_{n=1}^\infty \frac{H_n}{2^n n^4}$$

4:18 PM

Haha

They'll just think it's a ripoff of the previous

They'll make use of the already known generating function (divide by $x$ and integrate again):-))))))))))

Hehe

Then ask $\displaystyle\sum_{n=1}^\infty \frac{H_n}{3^n n^4}$ and by using $x=1/3$ they'll be wrong :D

My life is complete. I have gotten the 16 books that I need to study. 12 math, 1 LaTeX, 1 English, 1 French, 1 German.

@WillHunting Then it's not complete, you still have to study them

@WillHunting Burn the German one

Since I only mentioned my 12 holy books, let me now mention all of my 16 holy books...

4:23 PM

$12=16$. Q.E.D.

Spare us!

"Since I only mentioned" only?

Best. Comment. Ever.

Marsden and Weinstein: Calculus I, Calculus II, Calculus III
Cohn: Classic Algebra, Basic Algebra, Further Algebra
Rudin: Mathematical Analysis, Real and Complex Analysis, Functional Analysis
Lee: Topological Manifolds, Smooth Manifolds, Riemannian Manifolds
Kopka and Daly: Guide to LaTeX
Living Language English Complete Edition
Living Language French Complete Edition
Living Language German Complete Edition

Someone now please star my 16 holy books. They will prepare me for the GRE general paper, the GRE math paper, the PhD qualifying exams, and the foreign language exams.

Why would one star them ?

I don't have any of those exams in my country

@TheGame Because it would be a good set to use for anyone intending to do a math PhD in the US or Canada?

I won't star anymore.

4:27 PM

@WillHunting We don't need to star them

The chat log will be there if we need them

@TheGame OK. =(

And, who buys a book on LaTeX -__-

I have taken a few years to come up with this list, lol.

Now anyone who reads the transcript only has to follow it, lol.

Hey @ParthKohli

@BalarkaSen
http://math.stackexchange.com/questions/584866/algorithm-for-reversion-of-power-series

I put the reversion algorithm there.

@WillHunting Good evening.

4:36 PM

@ParthKohli Please admire my list of 16 holy books above, lol.

@TheGame I don't refer to myself now, but I only wanna say one needs to be highly skillful to deal with those sums, otherwise you have no chance.

I work for some hours on it.

@Chris'ssis I know :)

4:53 PM

afk

Go jogging, I need to free my mind from all thoughts. :-)

bbl

5:11 PM

back

Does anyone know what I can look into for possible research topics that ties together probability and some aspect of gambling like poker?

Hey @nablablah

@user60887 blackjack theory

5:36 PM

@nablablah gyazo.com/0ff2ac07a6dec209123a59b12421226b :C

6:10 PM

@TheGame there is something to fix in my generating function formula. Not a big problem.

Done

@Chris'ssis Show me :D

@TheGame Let's choose $x=1/3$, or do you want another value?

For any $x$ :D

The general one

(1/72)*(4*Pi^4 - 12*I*Pi*Log[3]^3 + 3*Pi^2*(Log[2]^2 + 10*Log[3]^2) +
3*(5*Log[2]^4 - 5*Log[3]^4 +
6*Log[2]^2*(PolyLog[2, 1/4] - 2*PolyLog[2, 2/3]) +
24*Log[3]*PolyLog[3, 1/3] +
6*Log[2]*(PolyLog[3, 1/4] + 4*PolyLog[3, 2/3]) +
24*(PolyLog[4, -(1/2)] - PolyLog[4, 2/3] - 2*PolyLog[4, 3]) -
3*Log[839808]*Zeta[3]))

OMG

6:20 PM

@Chris'ssis Uh ugly in mathematica :)

For $x=1/4$?

(7*Pi^4)/360 + (1/8)*Log[4/3]^4 - Log[4/3]^3*Log[2] -
Log[4]^4/24 + (1/2)*Log[3]^2*PolyLog[2, -(1/3)] + (1/2)*Log[4]^2*
PolyLog[2, 1/4] +
(1/2)*PolyLog[2, 1/4]^2 + (1/4)*
Log[4/3]^2*(Log[4]^2 - 2*PolyLog[2, 3/4]) - (1/2)*
PolyLog[2, 3/4]^2 +
(1/6)*Pi^2*(Log[4/3]*Log[4] + Log[4]^2 + PolyLog[2, 3/4]) +
Log[3]*PolyLog[3, -(1/3)] + Log[4/3]*PolyLog[3, 1/4] +
Log[3]*PolyLog[3, 1/4] +
Log[3]*PolyLog[3, 3/4] + PolyLog[4, -(1/3)] + PolyLog[4, 1/4] -
PolyLog[4, 3/4] - PolyLog[4, 4] - Log[4]*Zeta[3]

Oh, let me try another one ...

I mean, it looks so much better in LaTeX :D

It looks cool in mathematica.

marcelolpjunior

Hi

I will do a search on my university course in mathematics, there are 110 students in the course, and how do I scale my sample?

The research seeks to explain the socioeconomic reality of college

Hi all! A quick stat question: If $\mathbf{e}\sim N(0,\sigma I_n)$ what is the expected value of the product $(\mathbf{u}\cdot\mathbf{e})(\mathbf{v}\cdot\mathbf{e})$, where $\mathbf{u}$, $\mathbf{v}$ are $n$-dimensional vectors (not random). Please help! Thanks a lot!

6:29 PM

Where is @BalarkaSen to see the polylog and the polygamma at the same table?

$$\sum_{n=1}^{\infty} (-1)^n \frac{H_{2n}}{n^3}=$$

(1/960)*(-((711*Pi^4)/4) +
10*Pi^3*(128*ArcTan[2 + I] - 29*I*Log[2]) + 530*Pi^2*Log[2]^2 -
10*(6*Log[2]^4 - I*PolyGamma[3, 1/4] + I*PolyGamma[3, 3/4] -
192*(Log[4]*(PolyLog[3, 1/2 - I/2] + PolyLog[3, 1 - I]) +
4*(PolyLog[4, 1/2 - I/2] - PolyLog[4, 1 - I]))) +
120*I*Pi*(3*Log[2]^3 -
48*(PolyLog[3, 1/2 - I/2] + PolyLog[3, 1 - I]) + 35*Zeta[3]))

That is disgustingly beautiful :D

6:44 PM

I have a question on topology

@TheGame Some time ago you showed me this math.stackexchange.com/questions/765198/… amd told me I'm not there ...

Can someone help me ?

Yeah @Chris'ssis

@TheGame Yeah, I'm not, and I don't wanna ever be there. ;) (this doesn't mean I cannot do nice things)

@Carpediem Ask your question first :)

@Chris'ssis Why ?

6:46 PM

@TheGame Ok.. wait

I have two topologies

$\mathcal{T}_3=$ the topology having as basis all open rays $(- \infty, a)$ $\mathcal{T}_4=$ the topology having as basis all open intervals $(a,b)$ and all one-point sets $\{c\}$, such that $c \in \mathbb{Q}$

I am having trouble finding the closure of $A=(2,\sqrt{7})$ in each of these topologies

I think in $\mathcal{T}_4, \bar{A}=(2,\sqrt{7}]$

Ah not my field sorry. Maybe someone else in the chat knows.

How do you approach and understand this type of question?

user image

I do not get how you can get the gradient (delta f) from multiplying 19 (max) by the unit vector of the first. What makes the max special?

@Carpediem You're right for $\mathcal{T}_4$. You should be able to say why you're right, though. For $\mathcal{T}_3$, consider $x < 2$ and $x > \sqrt{7}$. Which has neighbourhoods not intersecting the interval?

@MLM The gradient always points in the direction of steepest growth. So if the directional derivative is largest in direction $v$, then the gradient is a multiple of $v$.

7:01 PM

@DanielFischer ahh this "gradient always points in the direction of steepest growth" really cleared it up

Thank you

@Chris'ssis So ?

@TheGame I only do math for fun, not for being told I'm a guru or something like that. ;)

@Chris'ssis That doesnt mean you have to refuse being considered good :)

worshsip the guru @Chris'ssis

3

@TheGame lolll :D

@DanielFischer Do you remember how one shows that any positive polynomial in $\mathbb{R}[X]$ is the sum of two polynomials squared of $\mathbb{R}[X]$ ?

If it's positive then it can be expressd $P=a\displaystyle\prod_{k=0}^p(x^2-2\Re(b)+|b|^2)$

I think there's some Laplace Identity to convert $A^2+B^2$ into something useful, but I can't remember for sure

@Chris'ssis , do you know ?

7:17 PM

@TheGame: I recall using $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ for that proof.

@Huy Exactly. I was trying to remember it. :-)

Didn't know it had a name though. :)

@TheGame see @Huy's answer

@Huy There was a nice question given on some local contest here that used the identity above.

@Chris'ssis: I don't really like the typical kinds of contests, but what question was it?

@Huy I have it here on some paper. Let me find it.

7:21 PM

But here I don't have anything in the form $(a^2-b^2)(c^2+d^2)$ do I ?

@TheGame: I think we expressed the polynomial in a different way.

@Huy I have $P=a\displaystyle\prod_{k=0}^{2p}(x-b_k)$, where all the roots $b_k$ have a conjugate root

@TheGame: That's not the factorisation into irreducibles yet, right?

Hence my expression above (it was supposed to be $b_k$ not $b$)

It's just the expression using roots

I think you can express it with irreducibles and then you can apply the identity.

7:24 PM

This question seems interesting to me. I can come up with ${}_2F_1$-based expressions for $E_2$ that contain three ${}_2F_1$s, but there might be something simpler.

PS: Apparently it's Fibonacci's identity. en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

@Huy The above form is already with irreductible polynomials

They all have degree $1$

@Huy

Prove that for any $m,n\in \mathbb{N}$, we have $x, y \in \mathbb{N}$ such that

$$(m^4-m^2+1)(n^4-n^2+1)(m^4+3m^2+1)(n^4+3n^2+1)=x^2+y^2$$

@Chris'ssis Each factor can be written as a sum of two squares. Euler takes over.

Euler wins

7:32 PM

@ccorn Or we can use Lagrange's identity ...

Ye even easier

Lagrange 2 - 1 Euler

It doesn't help with my problem though :)

Tunk-Fey = Cleo. That explains a lot.

@DanielFischer

Oh Balarka sen-pai is there

@Chris'ssis Beh easy.

7:36 PM

Still not sure how to find in $\mathcal{t}_4$

@BalarkaSen It was given on some math contest, and some kids were really in trouble with it.

Haha, that's because they never Eulered.

@BalarkaSen Help me on my problem ? I' just missing one step

what problem?

24 mins ago, by The Game

@DanielFischer Do you remember how one shows that any positive polynomial in $\mathbb{R}[X]$ is the sum of two polynomials squared of $\mathbb{R}[X]$ ?

21 mins ago, by Huy

@TheGame: I recall using $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ for that proof.

18 mins ago, by The Game

But here I don't have anything in the form $(a^2-b^2)(c^2+d^2)$ do I ?

There

7:40 PM

Yes, that's a famous problem, IIRC.

I know

But i can't find that last step

I don't think only that identity is involved in the proof.

Let me recall it.

27 mins ago, by The Game

If it's positive then it can be expressd $P=a\displaystyle\prod_{k=0}^p(x^2-2\Re(b)+|b|^2)$

Wait a sec, it's false as you have stated it @Hippa

Supposed to be $b_k$

7:43 PM

$-1 \in \Bbb R[x]$ cannot be written as sum of squares, for example.

And ?

I didn't say that only positives one could

I just wanna show that all positives ones do

Positive

By that I mean $P(X)\geq0,\forall X$

Ah you mentioned positive polynomials.

Oh wait

I just realized It can have real roots too

I have no idea what you are talking about (O_o)/

So $P=a\displaystyle\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)$

7:45 PM

Unfortunately I have to go. Can't think about that yet.

$a_k$ are the real roots

$b_k$ the complex roots, conjugated

Search for Hilberts 17th problem in one variable, maybe?

@BalarkaSen Mh ok

Anyways, got to go

I'll ask on main

7:46 PM

I managed to get 12/19 in a test I fell asleep in in maths

still highest in the class

cool

i got 3 less than the highest this time.

bleh

Good evening!

@BalarkaSen tests never mean anything

no mathematician solved an important problem under pressure

i know

$\pm$ a few pressures

7:48 PM

i was actually not under pressure though LOL

i had a fever and were on the teachers room in a comfortable and nice and peaceful atmosphere

@BalarkaSen Would you like a hard elementary algebra problem?

i won't like anything now since i got to go.

byes.

user image

4

see if anyone can solve this

@Alizter by elementary algebra do you mean the algebra taught in elementary school?

@IceBoy pre uni

7:56 PM

icic

@Alizter That's just from MHB

@BalarkaSen I got it from somewhere else

these problems usually circulate

Aha. Give me the link.

It'd be helpful for the internet spy thingy I am doing right now.

@Alizter: "no mathematician solved an important problem under pressure" What about Dantzig?

20 mins ago, by Alizter

$\pm$ a few pressures

8:09 PM

I see. So you withdraw your previous statement.

that's just some modus ponens and modus tollens reasoning infants are born with

@TheGame

Wow

@TheGame If you kill the imaginary part, you get nicer forms.

@Alizter there seem to be either no solutions or infinite. is the smallest wanted or something?

8:41 PM

@cxseven There definitely is a solution

@Alizter if a,b,c,d,e,f solves the equation then ak,bk^2,ck^4,dk,ek^2,fk^4 also does

@cxseven The answer is 64 if you must know

and also they are all equal

Can't you just hammer the algebra out, @Alizter?

@Alizter 64*k^14 is another solution for abcdef

@Khallil yes. That was the answer

8:46 PM

Square, group the appropriate terms, square, rinse and repeat.

a von Neumann answer

I'll save it to my desktop and try it when I get some time. Thanks for the problem, @Alizter. ^_^

i think my statement is verifiable without even doing any of that

@cxseven not my problem. Literally I did not create it.

then whoever did has a problem

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