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2:32 AM
How do you find the sum of the first n perfect 5th powers
 
2:43 AM
@WilliamYi Use Faulhaber's formula.
 
 
1 hour later…
3:51 AM
0
Q: Which of the following are reduced modulo residue systems modulo 18?

usukidollQuestion: Which of the following are reduced modulo residue systems modulo 18? $a. 1,5,25,125,625,3125$ $b. 5, 11, 17, 23, 29, 35$ $c. 1, 25, 49, 121, 169, 289$ $d. 1, 5, 7, 11, 13, 17$ Attempt: Our mod n is mod 18, so n = 18. Now, we need to remove the integers that aren't prime to n. Now, ...

 
 
3 hours later…
6:40 AM
Does the minimal polynomial of a vector space endomorphism always divide any other annihilating polynomial?
 
7:09 AM
@Alizter Noo.
@IceBoy lol :D
 
7:33 AM
Greetings
 
7:59 AM
Can we find a nice alternative for expressing

$$\int_0^{\infty} \frac{x}{\cosh^s(x)} \ dx=\frac{2^s}{s^2}{_3F_2}\left(\begin{array}c\large\tfrac{s}{2},\tfrac{s}{2}, s \\ \large1+\tfrac{s}{2},1+\tfrac{s}{2}\end{array}\middle|\,-1\right)$$?
@DanielFischer what do you think? Might we possibly avoid the use of the hypergeometric function?
@robjohn have you seen this one?
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)}\left(H_k-\sum_{i=1}^{\infty} \frac{1}{k+i+1}\right)$$
I saw it somewhere, but I don't remember now exactly where.
@robjohn However I think something is wrong with this question ...
(I mean it seems it diverges)
 
8:43 AM
hi
 
9:36 AM
hi
 
@IceBoy My appt with the therapist is tmr.
 
@WillHunting great, have you thought and written down what you want to talk about?
 
Wait ...
 
@IceBoy No, I can just go there and talk tmr. I am not sure if it would help, but I want to try a couple of sessions to see how it goes first.
 
Ok @WillHunting the first session should give you a good idea of what to expect. Be sure to be honest and as open a you can :-)
They are trying to help you.
May I suggest you go over what you have worked on over the past 7 years trying to heal yourself @WillHunting and try to find out things that you haven't tried or even thought about.
 
10:05 AM
Some lessons to do for a kid. Back later on.
 
10:25 AM
Later pal.
 
11:03 AM
@Chris'ssis Umm... $\displaystyle\sum_{i=1}^{\infty} \frac{1}{k+i+1}=\infty$
 
11:49 AM
$$\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$$
Anyone familiar with a growth result of that^?
I believe it's of number theoretic importance. Jacobi's two square theorem?
@ParthKohli!
 
12:09 PM
@BalarkaSen Can't you look at it as if it was a Riemann sum?
 
You know me, @Chris'ssis, I am trying a number theoretic approach =P.
 
@BalarkaSen Hi.
 
Herro.
 
How're you?
 
Not particularly bad.
 
12:14 PM
When $n$ large enough $$\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)$$
Yeap
 
OK, what about an asymptotic for the tan integral?
 
@BalarkaSen I like this approximation, it looks awesome.
 
I want an asymptotic for $\text{Ti}_2(n)$ though.
But anyway I guess I can derive one from Li asymptotics. Thanks, @Chris'ssis.
 
@BalarkaSen Yes, the inverse tangent integral can be expressed in terms of dilogarithm.
 
yep.
 
12:21 PM
@BalarkaSen You can simply make use of the relation $(9)$
Thus, when $n$ large enough, we have that $$\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2} \log(n)$$
 
Interesting. Thanks, @Chris'ssis
You want to replace $\approx$ by $\sim$, btw.
 
@BalarkaSen It depends on the meaning. $\sim$ usually means a very good (sharp) approximation.
 
No, @Chris'ssis
 
(if I'm not wrong)
 
$f(x) \sim g(x)$ means $\lim_{x \to \infty} f(x)/g(x) = 1$, which is what happens here.
i.e., the relative error goes to $0$ as $x \to \infty$.
 
12:26 PM
@BalarkaSen Just look at these notations here
or here
 
Sad use of $\approx$, @Chris'ssis.
 
In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. It is a very powerful approximation, leading to accurate results for even small values of n. It is named after James Stirling. The formula as typically used in applications is The next term in the O(ln(n)) is (1/2)ln(2πn); a more precise variant of the formula is therefore Being an asymptotic formula, Stirling's approximation has the property that Sometimes, bounds for rather than asymptotics are required: one has, for all so for all the ratio is always between  A019727 and  A001113. ...
 
They are all $\sim$
Yes, wiki used correct notations, @Chris'ssis
 
@BalarkaSen If you look a bit downward you also see the use of $\approx$
 
well, it's not incorrect, but $\approx$ is a very vague notation. nobody knows what it means.
for example, why can't we say $1 \approx 10$? =P
 
12:30 PM
@BalarkaSen lol, of course not :-)
 
why, @Chris'ssis? what's your definition of $\approx$? =P
(btw, we CAN. google "Fermi's approximation" =P =P =P)
 
@BalarkaSen In terms of asymptotics, I think it's good the way you defined it. That $\sim$ is just a very sharp approximation.
 
well, what do you mean by "sharp approximation"? $\approx$ just doesn't make sense mathematically.
 
(actually, you simply have more terms for that asymptotic, not like $H_n\approx \log(n)$)
@BalarkaSen I'd say that $$H_n \sim \log(n)+\gamma+\frac{1}{2n}$$
 
@BalarkaSen Think of how much the edges add as you increase $n$ by $1$: $$2\sum_{k=1}^n\frac1{k^2+n^2} =\frac2n\sum_{k=1}^n\frac1{(k/n)^2+1}\frac1n \to\frac2n\int_0^1\frac{\mathrm{d}x}{x^2+1}=\frac\pi{2n}$$
 
12:40 PM
This might be a nice limit to be tackled by Cesaro-Stolz theorem $$\lim_{n\to\infty} \frac{1}{\log(n)}\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$$
 
@Chris'ssis using the estimate I gave above, it would follow that way
 
@robjohn Precisely. I just noticed that.
 
@Chris'ssis The interesting question is what is $$\sum_{j,k=1}^n\frac1{j^2+k^2}-\frac\pi2\log(n)$$
 
@robjohn hmmm, indeed
 
is it finite?
 
12:45 PM
@robjohn I think so.
 
@Chris'ssis I do, too, but it is not immediately evident.
@Chris'ssis $$2\sum_{k=1}^n\frac1{k^2+n^2}-\frac\pi{2n}=O\left(\frac1{n^2}\right)$$
So I think the difference I asked about is finite.
 
@robjohn it might be $-\pi^2/12$
 
@Chris'ssis really? do you have a reason?
 
@robjohn Well, initially for some finite value I got $-0.824076$, but then, increasing the number of terms, the situation changes ...
 
For $n=1000$, I get $-0.824076366923426977$
 
12:59 PM
What if $n$ is over $9000$?
 
@Khallil I am computing $n=10000$ now
 
^_^
 
@robjohn Better give up, it requires a lot of time.
 
@Khallil HAHAHAHA
 
@robjohn N[1/Log[2000] Sum[1/(i^2 + j^2), {i, 1, 2000}, {j, 1, 2000}], 10]
$1.462260712$
 
1:11 PM
I'll soon try a number theoretic approach, @robjohn.
Essentially it's equivalent to counting number of solutions to $x^2 + y^2 = n$, which Jacobi counted using theta functions.
Hello by the way @Khallil
 
Heyo all!
Anyone here a moderator or have at least 10k and want to gander at summat for me? ♪
 
@robjohn is both, @GraceNote
 
We've got this deleted question here, closed as a proper duplicate of this not-actually-as-ancient-as-I-expected question here, and as far as I can tell, the lone answer to the former isn't amongst the answers of the latter. And maybe could fit there. But I'm also not mathy enough to know proper that it's of enough intrigue to preserve.
 
1:31 PM
@Chris'ssis $n=10000$ gives $-0.825682642054696606$
 
@robjohn wow, what kind of computer do you have?
@robjohn However, I think Mathematica is wrong. It gives another answer is you use $N$ in front.
 
@Chris'ssis It's an old laptop
I used f[n_] := NSum[1/(j^2 + k^2), {j, 1, n}, {k, 1, n}, WorkingPrecision -> 20, NSumTerms -> 10000] - Pi/2 Log[n]
 
1:44 PM
@Chris'ssis robjohn is working with the standard error, whereas you are computing the relative error. =)
 
@BalarkaSen My Mathematica gives me some crazy results.
@robjohn D*mn it, that limit is not that easy to compute, but it's very nice! :-)
 
The few people who downvoted my meta question must be X, Y, Z et al, lol.
 
@BalarkaSen we work on a more advanced limit that derives from your question.
 
@Chris'ssis That's because you are using different expression.
@Chris'ssis I can see that.
 
Sorry for not replying sooner, @Balarka. I was afk for a while. Hello!
 
1:49 PM
I need some sweets before continuing my work. Go to the store. (back in 20 min)
 
Robjohn is computing $\sum_{i, j = 1}^n \frac1{i^2+j^2} - \pi/2 \log(n)$ and you are computing $1/\log(n) \cdot \sum_{i, j= 1}^n \frac1{i^2+j^2}$ is that right?
I believe both are $O(1)$.
@Khallil Heya!
 
@BalarkaSen $$1/\log(n) \cdot \sum_{i, j= 1}^n \frac1{i^2+j^2}=\frac{\pi}{2}$$ that I just computed it above by using the approximation I showed you.
 
You mean $\pi/2 + o(1)$
 
@BalarkaSen $$\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2} \log(n)$$
 
Yes, I got that.
 
1:52 PM
@BalarkaSen Hence, that limit is simply $\pi/2$
 
Yep.
 
OK
Away
 
I'll wait for more an opening to re-posit then.
 
2:09 PM
@Khallil Hey.
 
Hey, @Sawarnik!
How've you been?
 
@V-Moy Hello :)
@Khallil In the middle of semester exams :|
 
Oh, cool. Good luck!
22 hours ago, by Khallil
When considering an or statement in math like '... if $m>0$ or $0>M$', do you consider both cases at the same time?
Any ideas, @Sawarnik?
 
1 min coming...
@Khallil Back. I don't think so but not sure :|
 
@GraceNote Sorry, I was out for a while. I will take a look.
 
2:18 PM
No rush, take your time.
 
@robjohn it might be the catalan's constant. I need to approach things from scratch.
 
@Chris'ssis It appears to be approximately $-0.825861+1.785/n$
 
2:36 PM
@robjohn hmmm, I have some new ideas ...
 
@Khallil What do you mean?
 
Don't worry about it @Balarka. I got it. If you're really curious, I was looking at part b from this question.
I was supposed to consider the two cases separately for each bullet point using the fact that $M>m$, which I derived earlier in the question.
 
With regards to the bounty notice on this question...really?
 
-34. Record.
 
what makes it especially weird to me is that the +100 bounty is derived entirely from that user's association bonus, which leaves them now at rep=1
 
2:51 PM
@BalarkaSen Sure?
@Semiclassical Oh!
 
@Semiclassical "I'm not using multiple accounts" that's like stealing all the cookies from the jar on your kitchen and say "it wasn't me" with chocolate sticking to your lips.
 
@BalarkaSen: pretty much, especially when their way of 'opening the jar' (i.e. the bounty) involves no actual rep from this site.
 
Mr. @robjohn: Would you be so kind as to do something for me? May I know Ms. Cleo's email? I know it's confidential info but at least you as a moderator could send her a message & ask her whether she permits you share this info to me or not. Please... ≥Ö‿Ö≤
 
@BalarkaSen: Even if it's not a sockpuppet, though, it still strikes me as a completely incorrect use of the bounty system; "why does this question have so many downvotes" is a question for meta or chat.
 
well, it's not (yet) illegal.
 
2:56 PM
@robjohn I'm thinking to ask it on main, it would be fun seeing more solutions to it. Latsly I thought of using $\coth(z)$ but still there are some problems ...
 
maybe there should be a rule. the association rep should be solely for the purpose of posting comments/editing questions/etc in the site
@Anastasiya-Romanova mods can't "send message" as far as I know.
can they?
 
@Parth Science went awesome!
 
@balarkasen: i'd think the simplest solution would just be to bump up the 'can award bounties' privilege a bit past 75 (say 125 or so) so they can't convert association bonus immediately to bounty
 
what line, @Sawarnik?
@Semiclassical that's plausible.
 
@BalarkaSen They know all of users' email, I think it can be done
 
3:02 PM
¯\(O_o)/¯
 
thinking of starting a blog so i have a place to put random math stuff. any suggestions for a good platform?
 
@Semiclassical i don't blog, but wordpress is well-known.
and it supports latex, i think. for example, Tao did some black magic to make it work.
 
thx, i'll take a look
 
@Semiclassical You may soon get bored.
Wordpress is the best option I know though.
 
@Sawarnik do you have experience with blogs?
 
3:09 PM
@Sawarnik: lol, probably. mostly i just want a place to refer people if i want to ramble on about something that wouldn't be in the MSE scope.
 
@BalarkaSen Not actually, but the idea of having a blog is attractive at first, then soon you get bored.
 
how can you say that if you never blogged, @Sawarnik?
 
I just know ... not for all people though.
 
my 'probably' is based on the fact that I do get bored pretty easily:P
 
@Sawarnik you just know?
 
3:11 PM
@Sawarnik Good job, boy.
 
@ParthKohli English is the last one, finally :D
 
@Sawarnik I admit that too.
 
@BalarkaSen Yeah.
 
@Sawarnik how come?
 
@Parth Why did you remove your age? :O
 
3:13 PM
personally i never blogged as i know that i am not good at expository writing but i do know a lot of people who are excellent at writing out blog or blog-like stuff
 
@BalarkaSen Speaking for just myself, I don't really get motivated to write blogs.
 
yeah. it's a bit different than writing up MSE solutions (at least, depending on how wordy your answers are)
 
Yeup.
 
MSE answers itself are as big as blogposts though =P
 
In MSE, reputation, competition things like that are factors.
 
3:16 PM
@ParthKohli I don't get motivated by MSE reputations either =P
 
@BalarkaSen I have no idea how you aren't. You can easily answer the majority of questions on main.
...Or can you?
 
@ParthKohli Which novel do you have to study?
 
I don't think so, @Parth.
 
@BalarkaSen Once upon a time, I used to be but not now.
 
@Sawarnik The Story of My Life by Helen Keller.
 
3:18 PM
@ParthKohli Oo, someone said it was Anne Frank's diary?
 
@Sawarnik We have two options, dude. You do too.
You have a choice between Three Men in a Boat and Gulliver's Travels.
 
@ParthKohli :O
 
I think I like forums better than Q&A sites.
 
Choice? Never heard of that! I have to study Three Men in a Boat.
And I hate that one.
 
And doing math better than helping people =P
 
3:21 PM
Choice :O Which one did you take?
 
Three Men in a Boat.
 
Wasn't it kind of nonsense?
 
@Sawarnik I don't even know of any of the novels you guys are talking about except Gulliver's Travel.
aren't those all super-silly novels?
 
@BalarkaSen Y ouk nowa bout Gulliver's Travels? :O
 
@robjohn By using $\coth(z)$, one may understand why the limit is so close to $-\pi^2/12$, but it cannot give you the limit.
 
3:23 PM
@Sawarnik mmhmm
 
@BalarkaSen Yup. And its all something .. ahem CBSE related.
@Parth Why didn't you take Anne Frank?
 
@Sawarnik Our teacher recommended Helen Keller.
 
@Sawarnik sure it is
 
Actually, Sawarnik, I've seen that in most of the schools - 99% of them - the choice is not actually given to an individual. The choice is given to the teachers who get to decide the novel they'd be teaching and testing.
 
such silly. much english. so boring.
 
3:26 PM
@ParthKohli Yeup.
 
@BalarkaSen Nice maymay.
 
@ParthKohli Did you like Three Men in a Boat?
 
@Sawarnik did they all drown?
 
@BalarkaSen Unfortunately not.
 
^true.
lol
 
3:29 PM
@Sawarnik I actually did.
 
@ParthKohli Waat? Hoowww???
 
@Sawarnik If you begin to appreciate even a little bit of literature, you would too.
 
-_-
<big> -_- </big>
 
fuck, meant to edit, oh well... should have just used the keyboard shortcut
 
I do appreciate literature, @Parth, but not high-school ones.
 
3:36 PM
@BalarkaSen There's no "high-school" literature.
 
You know what I mean
 
If you mean that you don't like how it's tested, then that's a different thing. But the beauty of the thing remains as it is.
 
<huge> -_- <\huge>
 
@BalarkaSen you know what's great about that search query that got starred?
 
i don't, @Mick
 
3:38 PM
If you remove the restrictions for Ted only, and for this room only, the only other message is your link to the query itself
 
haha
oh. the uniqueness just got void.
sorry, @Mick
 
lol, no worries bro
 
lol
 
Here is a marvellous question to upvote
@robjohn I posted the question on main.
 
3:57 PM
@Chris'ssis Oh
 
5
Q: Computing $\lim\limits_{n\to\infty} \Big(\sum\limits_{i = 1}^n \sum\limits_{j = 1}^n \frac1{i^2+j^2}-\frac{\pi}{2} \log(n)\Big)$.

Chris's sisIn the chatroom we discussed about the asymptotic of $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2...

@robjohn Did you see why that limit is so closed to $-\pi^2/12$?
 
@Chris'ssis What am I missing? It appears to diverge at a similar pace with $-n^2$
 
What is wrong with $-1=(-1)^{1/3}=(-1)^{2/6}=((-1)^2)^{1/6}=1^{1/6}=1$?@robjohn
 
@Sush: Branches
 
4:13 PM
@Nick, what do you mean by branches?
 
@Chris'ssis Marvelllllous
 
 
@Hippalectryon :-) Did you upvote it?
 
@Sush: Exactly what you think I mean.
@Chris'ssis: Is it not against some precious policy to ask for upvotes?
 
@Chris'ssis Of course
 
4:18 PM
@Nick, do you mean that 2/6 should first be converted to 1/3?
 
@Nick Is it against some policy to recommend you to support a marvellous question?
 
@Chris'ssis: I don't know. Hopefully not.
 
@Nick I don't think so. I can recommend something, but I do not oblige anyone to proceed like that. The purpose is to attract more people to work on it.
@Nick by the way, did you upvote it? :-)
 
I'm a teenager, it's in my nature to oppose hypocritical authority. Sorry for the misunderstanding.
@Chris'ssis: You can't make me.... but I did ...lol
 
@Nick Yee another teenager :D
 
4:24 PM
:D
 
Yee 63 wpm :D play.typeracer.com
@Chris'ssis I really am though -___-
 
@Chris'ssis: I just wrote a little C++ program to evaluate that using a lot of dirty for loops. It crashed after approximating it to "-0.2617929333". .. Is that close to being right?
 
@Nick Well that's strange, I've noticed for most teenagers it's actually in their nature to emotionally distort situations where they would otherwise feel uncomfortable admitting that they are actually learning something that they were ignorant / misinformed about previously.
 
@Nick The limit is around $-\pi^2/12 \approx -0.8224670334241132$
@Hippalectryon Do you refer at the speed of typing? I'm pretty clumsy at this chapter.
 
@Chris'ssis Yep
 
4:28 PM
@Chris'ssis: mmh, I should have used Python.
 
@Chris'ssis I only type with two fingers so it's not as fast as it should be
I've never learnt to type 'properly'
 
@Hippalectryon: Half of the people who knew how to type properly are dead.
 
Meh so poor :/
I make 0€/year
@Nick Python yaay
 
@MickLH: I have a brain. Don't tell me to replace it with a computer.
@Hippalectryon: What does that graph imply?
 
4:32 PM
That's fine, just expect to be laughed away from my engineering firm.
 
@Nick It's somthing that's been running on my comp for xx second (look at the X axis)
 
@robjohn I'm shocked by the beauty of this question, it's simply too nice ... here the art comes in place ...
 
I feel it's morally wrong to let a human do arithmetic
And extremely silly
 
@MickLH: ... You run a firm!!?
 
So go ahead, reject mathematica. Just make sure to tell me any bridges you've helped engineer.
 
4:34 PM
@MickLH Mathematica is 99999999$. Kind of.
So expensive
 
lol it's ok, just get the free trial
 
5 mins ago, by MickLH
or, if you're poor: http://maxima.sourceforge.net/
 
it will convince you to shell out the cash
 
@Hippalectryon: ^
 
@MickLH I don't have any money ._.
 
4:36 PM
@MickLH: Me neither and I want mathematica
 
@Nick By broadband is not good enough xD
That's everal gigs isn't it ?
 
well then it will convince you to download a program that makes the other program think you shelled out the cash cough cough wink wink
 
@MickLH I alreday have a cracked mathematica
But it won't open
 
Actually I had a similar issue once
 
A month ago it stopped working
Crashes at launch
 
4:37 PM
@MickLH: Should I try to keep the pirates at bay for Mathematica?
 
Not sure how it was resolved, for a while I just used the command line interface
 
My other pb is that i only have 4BG ram
Too few for big computations
 
I have Nokia Asha cellphone! It can at most do 3 digit multiplication.
 
@Nick You're not talking about T igers P ossibly B ored are you ?? wink wink
 
Well that's a bit tougher, when I find myself running into the memory limit, I usually think harder about the math and figure out a way to get what I want without large computation
But sometimes that's not actually possible, in those situations I just use C++ and libGMP
I want to use the big data STL implementation, but I've not had a real necessity for it to date
Maybe because I have 20 GB of RAM to work with
 
4:41 PM
@MickLH 20GB WOW
 
... and I still had to set up a swap file
 
@MickLH My graph alone takes me 500 Mb or RAM. It's Python based.
 
damnit! that thing in the corner got my hopes up for this one having a legend
lol, what does the graph represent?
 
Some online currency stock exchange on a website
I wanna analyse that :)
 
ah, playing with neural networks?
 
4:43 PM
That thing
I'm getting the info with some python program and cookies libs
Then storing them into text files
And updating the graph every 5 secs
 
that's fairly gangster, sir
 
Uh -__-
Nothing illegal there
 
Hehe who said anything about illegal?
 
But i'm pretty sure one can find interesting patterns
 
The biggest gangsters are the legal ones
 
4:45 PM
And make MONI
moniiiii
Well, virtual money -__-
But still, toying with currency exchanges can be fun.
 
If a cubic has a distinct root and a repeated root, would you say it has 3 roots or 2 roots? Or would you say that one of it's roots is of multiplicity 2?
 
@Hippalectryon: MON$i$ is indeed imaginary.
 
@Nick -___- Nuu
 
aw, I was hoping that went the other way, to "undead". I'd have found my motivation to study or work today, if it had turned out that there were "undead" imaginary numbers
 
Y u shatter my dreams @Nick ;)
@Khallil I'd say 2 roots
One of which has multiplicity 2
 
4:48 PM
Gotcha.
Thanks, @Hippa!
 
@Hippalectryon: Because nobody said 'hi' to me today.
 
No wonder why :P
 
@Khallil: I'd say 3 of which two are equal but it doesn't really matter. Nothing really matters.
 
4:53 PM
@Hippalectryon: I need atleast one greeting per day inorder to get high enough to get up the next morning.
 
I appreciate this computer age so deeply, I would never have found how deep my love for mathematics really goes without such excessive computational force at my fingertips
 
That's it. I'll recycle!
Nov 13 '13 at 17:05, by Charlie
@Nick hi little nick
Ah, a greeting at last
 
hi @Nick
 
Bye! :D
I'm leaving
actually
Good night!
 
lol goodnight
 

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