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10:13 AM
Good morning.
$$
\frac12\zeta(2)^2-\frac12\zeta(4)
$$
No paper or pencil :-)
@Chris'ssis what? that's ridiculous! I did it while staring at it on the screen. (if my answer is correct :-)
Plugging in values, it is $\dfrac{\pi^4}{120}$
I just checked with Mathematica, and it agrees with me :-) Sum[HurwitzZeta[2,n+1]/n^2, {n, 1, Infinity}]
Note that
$$
\begin{align}
\cot^{-1}(n)-\frac1n
&=\tan^{-1}\left(\frac1n\right)-\frac1n\\
&=\int_0^{1/n}\frac{\mathrm{d}x}{1+x^2}-\frac1n\\
&=-\int_0^{1/n}\frac{x^2\,\mathrm{d}x}{1+x^2}\\
&=-\int_0^1\frac{x^2\,\mathrm{d}x}{n(n^2+x^2)}\tag{1}
\end{align}
$$
and partial fractions gives
$$
\frac{x^2}{n(n^2+x^2)}=\mathrm{Re}\left(\frac1n-\frac1{n-ix}\right)\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^{\infty}(\cot^{-1}(n)-1/n)
&=-\mathrm{Re}\left(\int_0^1\sum_{n=1}^\infty\left(\frac1n-\frac1{n-ix}\right)\,\mathrm{d}x\right)\\
 
10:32 AM
@robjohn do you have any advice for working without solutions? The texts I'm using for math right now (Hoffman and Kunze, Spivak) have no solution manuals
 
@Astrum if you are unsure of your answer ask Wolfram Alpha if the problem can be answered that way, or ask here.
 
@robjohn all of the problems are proofs, wolfram isn't any help with that
 
@Astrum then that leaves the second option. Try your hardest first. That way the help here will mean more to you.
 
@robjohn it seems to be helpful in a way, because I don't have a choice but to solve it myself (or ask for help). Math comes much harder to me than physics =/
you guys have been a lot of help so far, really been a great thing for me
 
@Astrum that's what MSE is for.
 
10:35 AM
heres a similar way
wait wait
let me try at it
 
the chat at physics SE is less populated, and their rules on HW help are a lot stricter
 
$$\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{a^2(a+b)^2}=\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{b^2(a+b)^2}=s_1$$
By changing the order of the summations
$$2s_1=\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{a^2(a+b)^2}+\frac{1}{b^2(a+b)^2}$$
 
@robjohn without pen and paper? That's good ...
 
ahh screw it
its 2 am here
 
@Ethan it's 5:40 here, I'm going to sleep soon
 
10:39 AM
@Chris'ssis It seems really pretty easy...
The sum of $\frac1{n^2}\zeta(2)$ is easy $\zeta(2)^2$
The other part is $\frac12(\zeta(2)^2+\zeta(4))$
 
oh no I got it
$$\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2(a+b)^2}=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{b^2(a+b)^2}=A$$
$$2A=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2(a+b)^2}+\frac{1}{b^2(a+b)^2}$$
$$2A=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2b^2}-\frac{1}{ab(a+b)^2}$$
 
@robjohn I also have 3, 4 ways, but nothing without pen and paper.
 
$$2A=\zeta(2)^2-\sum_{n=2}^\infty\frac{1}{n^2}\sum_{k=1}^{n-1}\frac{1}{k(n-k)}$$
alright I don't know what i'm doing anymore
I took adevan to go to sleep but its not working
 
@Chris'ssis I pictured it on the integer grid
 
lol
I did something similar a while back
Define $\beta_s$:
$$\beta_s(x)=\sum_{n=1}^\infty\frac{1}{n^{s}+x^s}$$
$$\sum_{m=1}^\infty\frac{\beta_s(mx)}{m^s}+\sum_{m=1}^\infty \frac{\beta_s(mx^{-1})}{m^s}=\zeta(s)^2$$
$$\beta_2(x)=\frac{\pi}{x}\frac{1}{e^{2\pi x}-1}+\frac{\pi}{2x}-\frac{1}{2x^2}$$
------------------------------------------------------------------------------------------------------------$$\frac{\pi^3}{180}(x^{-2}+x^2+5)-\frac{1}{2}\zeta(3)(x^{-1}+x)=$$
$$\frac{1}{x}\sum_{m=1}^\infty\frac{1}{m^3(e^{2\pi mx}-1)}+\frac{1}{x^{-1}}\sum_{m=1}^\infty\frac{1}{m^3(e^{2\pi mx^{-1}}-1)}$$
 
10:49 AM
@Ethan maybe you like this one. Compute
$$\sum_{k=1}^{\infty} \frac{\zeta(2n)}{n(2n+1)2^{2n}}$$
 
hmm
I am really screwing up my sleep schedule
 
ok guys, I'm goin to sleep, see you all later
 
I should be studying for the SAT
 
@robjohn bye, and thanks for the help earlier
 
$$\begin{align}
A
&=\sum_{a=1}^\infty\sum_{b=1}^\infty\frac1{a^2(a+b)^2}\\
&=\sum_{a=1}^\infty\sum_{b=a+1}^\infty\frac1{a^2b^2}\\
&=\sum_{b=2}^\infty\sum_{a=1}^{b-1}\frac1{a^2b^2}\\
&=\sum_{b=1}^\infty\sum_{a=1}^{b-1}\frac1{a^2b^2}
\end{align}$$
So twice $A$ plus $\zeta(4)$ (the $b=a$ terms) is $\zeta(2)^2$
@Astrum you're welcome
 
10:53 AM
I had alot of interest in these sorts of double sums a while back
can't find my stuff
why are you still up robjohn?
 
@Ethan What do you mean?
 
its like 3 am in the morning
 
@Chris'ssis did you see the answer $-\gamma-\arg(i!)$ ?
 
@robjohn where?
 
@Chris'ssis here
 
10:56 AM
@robjohn Ah, beautiful!!! That's a really nice closed form!
 
@Chris'ssis Nicer than the one with the Gammas and logs
 
@robjohn Yeah, far nicer.
 
I think I remember seeing a question just like that on the main
 
N[-EulerGamma - Arg[I!], 20] gives $-0.27557534443399966272$
 
hello
 
11:03 AM
good night everyone
 
can someone help me on relative homology
???
 
@Chris'ssis: I thought this answer and this answer were kind of neat.
@Ethan Good night!
@Vrouvrou Sorry, above my pay grade...
 
@robjohn Indeed. I like them.
 
@Chris'ssis This morning has been a good one so far :-) good problems
 
@robjohn You seem to be in a very good shape. :-)
 
11:11 AM
@anon can you help ?
 
@Vrouvrou I said above that that is not my area. Sorry.
 
i understand but i say to to @anon
 
@Vrouvrou Oh, geez. I'm sorry. I saw Chris'sis ping to me just above yours and mistook it :-)
 
no problem !^^
 
I wish I had found this triangle earlier: oeis.org/A042977 It is the generalization of the LambertW power series.
Dated at Sat Dec 11 03:00:00 EST 1999.
 
11:26 AM
@MatsGranvik does this triangle bear on what you are doing?
 
@robjohn yes it is exactly what I have been doing. Only that it is expressed through differentiation, and not power series reversion.
 
@MatsGranvik Ah. Reversion is often easier. You can get some nice recursive formulae that way.
 
@robjohn My trouble now seems to be that I expand exp(-x) around Log(z), but tautologically expanding exp(-x) around LambertW(z) seems to work too.
In fact it is identity. Expanding exp(-x) around LambertW(z) gives LambertW(z). How weird. Or true. But Wouter Meeussen differentiates and gets expansion around Log(z).
Through series reversion.
 
@Rasmus
 
The reason must be that Log(z) and ProductLog(z) are close to each other, as I think I have seen in a comment by Charles Greathouse.
 
11:43 AM
@robjohn I think also this one works with some integer grid $$\sum_{n=1}^{\infty}\frac{1}{n}\left(1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{‌​n+1}}{n}-\log(2)\right)$$
@robjohn I finally managed to compute one question without pen and paper (the first one this day)
 
Note to self. Expanding exp(-x) around LambertW(z) gives exact answer while expanding around Log(z) is only approximate.
 
hello
 
@Chris'ssis Is it $\frac12\log(2)^2-\frac12\zeta(2)$?
 
@robjohn No.
 
@Chris'ssis Ah, there is only one sign alternation. Sorry. I was trying just looking at it.. I need to look harder :-)
 
11:52 AM
@robjohn hehe, I like this kind of approach! :D
(brb - I have lunch)
 
@Chris'ssis Is it $\frac12\log(2)^2-\zeta(2)$?
I think that is it
 
@robjohn No.
 
really? Hmmm
 
@robjohn It's a very precious question. I love that.
 
12:23 PM
Ah, $\frac12\log(2)^2$
 
@robjohn No.
 
Hello
 
@Chris'ssis Are you sure? I actually used pencil and paper and $\sum\limits_n(-1)^{n-1}\frac{H_n}{n}$...
 
@PabloRotondo Hi!
 
I will check it over and write it up in LaTeX
was the sign off?
 
12:27 PM
@robjohn Indeed.
 
It would be convenient to write the difference of the sum and the log as an integral I think.
 
Okay
@PabloRotondo I do that in computing $\gamma$
 
@Chris'ssis Is this your problem, or from your students?
 
@PabloRotondo No, this one comes from a professor that challenges me once in a while.
 
The terms are $\frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^{n+1}}{1+x}dx$, right? or something like that
 
12:30 PM
(brb - lunch time)
 
@Chris'ssis Britain?
@Chris'ssis Okay, I think I have computed it correctly, is it $$ \frac{\log^2(2)}{2}$$ ?
 
Anyone know the official definition of a pointed polyhedron?
 
$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1n\frac1k(-1)^k
&=\sum_{k=2}^\infty\sum_{n=1}^{k-1}\frac1n\frac1k(-1)^k\\
&=\sum_{k=1}^\infty\sum_{n=1}^{k-1}\frac1n\frac1k(-1)^k\\
&=\tfrac12\zeta(2)+\sum_{k=1}^\infty\sum_{n=1}^k\frac1n\frac1k(-1)^k\\
&=\tfrac12\zeta(2)+\sum_{k=1}^\infty(-1)^k\frac{H_k}k\\
&=\tfrac12\zeta(2)+\tfrac12\log(2)^2-\tfrac12\zeta(2)\\
&=\tfrac12\log(2)^2
\end{align}
$$
 
@robjohn Nice.
 
@PabloRotondo I had previously computed the sum with the harmonic numbers
 
12:39 PM
@robjohn I see
I did it by noticing $$ \left(1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{‌​n+1}}{n}-\log(2)\right) = (-1)^{n}\int_0^1 \frac{x^{n}}{1+x}dx $$
(integrating the tail of the geometric series)
 
12:55 PM
@PabloRotondo Yes, I use that in equation $(13)$ of this answer to accelerate convergence.
 
1:25 PM
@robjohn Hey, how's it going?
 
@robjohn What about this one? $$\sum^\infty_{n=0}\frac{(n-1)^{1-n}\;\Gamma(n+1, n-1)}{n!}$$
I got it "down" to $$\sum^\infty_{n=0}\left((n-1)e\right)^{1-n}\sum^n_{k=0}\frac{(n-1)^k}{k!}$$
 
@alizter: hi there Al. Um, since when did the gamma function take in two inputs?
 
@Nick It's the incomplete gamma function
 
1:41 PM
oh, you mathematicians with your new-fangled functions. How adorable.
 
In mathematics, the upper and the lower incomplete gamma functions are respectively as follows: : \Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t ,\,\! \qquad \gamma(s,x) = \int_0^x t^{s-1}\,e^{-t}\,{\rm d}t .\,\! Properties In both cases s is a complex parameter, such that the real part of s is positive. By integration by parts we find the recurrence relations :\Gamma(s,x)= (s-1)\Gamma(s-1,x) + x^{s-1} e^{-x} and conversely : \gamma(s,x) =(s-1)\gamma(s-1,x) - x^{s-1} e^{-x} Since the ordinary gamma function is defined as : \Gamma(s) = \int_0^{\infty} t^{s-1}\,e^{-t}\,{\...
 
2:33 PM
@robjohn did you manage to attend that Fibonacci series?
 
@Chris'ssis HI
 
@Alizter Hello
@robjohn btw, if you compute it, no need to post the answer (solution). I'm only interested to see that value you get.
 
$$\sum_{k=0}^n \frac{e^{1-n}(n-1)^{1+k-n}}{k!}$$
 
@Chris'ssis I haven't worked on it yet.
 
@robjohn ok
@robjohn I'm writing up the proof immediately.
 
2:50 PM
@Chris'ssis Found it... I will work on it.
@Chris'ssis Got it... That's cool. The partial sum is $\tan^{-1}(F_{2n})$
 
@robjohn hehe, glad you like it.
 
Ergo, the sum is $\frac\pi2$
 
@robjohn :D
@robjohn I hope your son knows what kind of nice things his father can do!
:-)
I've always wanted to have someone like you around ...
out for a while
 
@Alizter, have you encountered using operators to simplify series?
There's a small chance it may work with your problem.
 
@Alyosha Do go on
 
3:03 PM
I have no idea why, but you can treat operators like numbers if you are summing them so that they form $e^{\text{Operator} ...}$ (see page 3 here 129.81.170.14/~vhm/papers_html/rmt-final.pdf for an example).
 
Note that
$$
\begin{align}
\frac{F_{2n-2}+\frac1{F_{2n-1}}}{1-\frac{F_{2n-2}}{F_{2n-1}}}
&=\frac{F_{2n-2}F_{2n-1}+1}{F_{2n-1}-F_{2n-2}}\\
&=\frac{F_{2n}F_{2n-3}}{F_{2n-3}}\\
&=F_{2n}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^n\tan^{-1}\left(\frac1{F_{2n-1}}\right)=\tan^{-1}(F_{2n})
$$
 
@Alyosha It looks interesting but so far it only looks like it works for special cases.
 
In the example, the operator $E$ is defined such that $E(f(n))=f(n+1)$, and $E^n(f(0))=f(n)$. $f(0)$ is moved out of the integral, so I assume it doesn't matter whether you operate before or after the integral.
There's a chance that your series could be converted into that special case, although with the emergence of the integral it seems unlikely.
It depends whether one can find a nice enough operator.
 
@Alyosha This is beyond me right now :P
 
3:21 PM
This hardly helps, but $\sum_{n=0}^\infty \frac{e^{-n}}{n!}=e^{e^{-1}}$.
 
@Alyosha So the two ways I can turn the function into an integral is by looking at the gamma version in which i steered away from in the first place. The problem is I either get an incomplete gamma function integral or a reciprocal of the gamma function. Both of which I have no idea what to do with.
 
You could first split the integral into $\int_0^\infty e^{-x}x^ndx- \int_{0}^{n-1} e^{-x}x^ndx=n!-\int_{0}^{n-1} e^{-x}x^n$ for integral n.
 
sos440 , user : 9340 , Byrom Schmuland user : 940
 
@Alyosha Lets work on this $$\sum^\infty_{n=0
}\frac{(n-1)^{1-n}\;\Gamma(n+1, n-1)}{n!}$$
 
3:37 PM
$\sum_{n=0}^\infty\frac{(n-1)^{1-n}}{n!}\left(\int_{0}^\infty e^{-x}x^{n}dx-\int_0^{n-1}e^{-x}x^ndx\right)
=\sum_{n=0}^\infty(n-1)^{1-n}-\sum_{n=0}^\infty\frac{(n-1)^{1-n}}{n!}\int_0^{n-1}e^{-x}x^ndx$
As the first integral is simply $n!$.
The rightmost integral is probably doable by IBP, it will form a sum, which may not be nice.
 
4:51 PM
Can anyone help me with my coordinate geometry problem ?
Will anyone be kind enough to help me ?
 
@user2369284 It's best if you just ask. Then, if someone comes along who can answer, they will.
No one knows ahead of time if they can help.
 
5:08 PM
@Vrouvrou people who know homological algebra need help too :D
 
5:28 PM
@EnjoysMath i know have you a question ? if i can help you
 
No, I was just commenting that people who know something that advanced still need helps, lol
I don't know homological algebra yet, but soon
*one day
 
5:46 PM
hey
i do i find the matrix rotation around the vector 1 1 1
i found rodrigue formula, but looks a bit overkill for my situation
 
Rotate around $\hat{z}$ then rotate $\hat{z}$ into $(1,1,1)$.
 
Guys, anyone knows anything about regular language and disproving regularity?
 
6:10 PM
SUP DAWGS
anyone there?
 
yo dawg
i heard u like functions
so we put a function in yo function $f(g(x))$
 
How can I prove that the hypotenuse is the longest side of a right triangle without using Pythagorean theorem?
anyone there?
 
6:28 PM
yea im here but unfortunatly for you, i understood fuck all of my discrete math class
 
:-), i dont follow how this has anything to do with discrete math
 
yeah, exactly what i said
i seriously understood fuck shit
and i got B
 
you are talking about a university class, right?
 
but isnt it realated, i mean, discrete math is about proofs
yea absolutly
 
yeah , but isnt every math class in some way about proofs?
 
6:31 PM
No.
im taking linear algebra right now and we dont do any proof at all
we apply the stuff and understand the context, but no proofs
well maybe one once in a while, but in discrete class it was all about proofs, very similar to your question
 
No proofs? So you are discussing theorems without proving them?
 
yeah kinda
lets say we use a rotation matrix, he shows where it comes from, but no intense proofs about it
 
But I suppose that anyone interested can look up the proofs?
Anyway I dont go to a university. What other classes are you taking?
 
well, im in computer science, so i took about 7-8 programming classes, and some math classes
i work full time , and take 2 classes per semester
so to do my full bachelor degree, its gonna take me 9 years
ive been in it for now 4 years i think
i wish i started university earlier and did it full time, but oh well, i still can do it partial time, its all about perseverance etc.
 
After you complete your Bachelor, do you intend to continue your study?
 
6:42 PM
oh yea absolutly
i want a master degree,
my girlfriend is doing a phd right now
 
Isnt it going to take a very long time?
 
its in developpement régional, which means something like how does a small town growth, business etc
yeah well, you know, as i said, its all about perseverance
 
so you are not a math major?
 
nope, math as always been a struggle for me, but im starting to love it
2-3 years ago i had to take some highschool math class, before taking the college math class, and now im doing the university level ones
i started from the absolute bottom
 
what is your favourite branch of math?
 
6:46 PM
hmm, right now its linear algebra, because the teacher is good, he is very calm, he explains slowly and carefully, and i feel like i understand what we are doing
i did a calculus class, and understood nothing, it was going way to fast, the teacher was young and crazy,
we could use a note sheet for the exam, so i wrote a shitload of stuff on it and relied a lot on it for the exam, thats how i got my not bad grades,
 
yeah but to understand what you are doing seems to be very little, does linear algebra push your buttons?
 
yeah kind of, as i said im starting to love it, im taking this class and a programming languages class right now and i prefer the math one
im getting kind of bored with programming stuff
math is much more challanging for me , and i feel 'stronger' doing that
 
I also did a calculus class in UNI (later was kicked out) and while I understood things pretty well because I have been going through a book on calculus, I felt the proffesor wa putting too much emphasis on doing problems and too little on the underlying theory and proofs.
 
oh yea, maybe he was not a good teacher, you know, theyre not all good
 
I think I am going to learn Haskell, it is useful to know how to program, for testing conjectures and stuff....
 
6:54 PM
i dont know
i would suggest an easy, popular, modern, all around language
 
like?
 
depends what paradigm you want to dig in, youre really into functionnal languages ?
im not so much in functionnal languages, the only one i tryed is F#
 
like find all partitions of 27...
they seem well suited for the kind of thing I am doing
 
yeah thats right
functionnal languages are very well suited for scientific/math stuff in general
 
you write that your are becoming bored of programming, I would venture to say that if you like math, you also will like computer science, dont you think?
in general...
 
7:05 PM
i dont know i dont feel that these 2 are related at all
 
:-D
 
there is indeed a branch of computer science that is realted to math
but most of modern computer science stuff is not about science/math
 
come on computer science is just math in disguise....
 
no it is not
114
Q: Is mathematics necessary for programming?

simsimI happened to debate with a friend during college days whether advanced mathematics is necessary for any veteran programmer. He used to argue fiercely against that. He said that programmers need only basic mathematical knowledge from high school or fresh year college math, no more no less, and th...

 
But do you agree that progress in cs comes through proof and not experiment?
 
7:07 PM
i feel like programming can be done in a naive way, like test, debug, repeat,
and there are a lot of hacks
maybe what im saying relate more to web programming,
 
That maybe possible, but I am talking about computer science, you know? computability, complexity, that kind of stuff, say THAT is not math.
 
no one needs to know of to write a sort algorithm in log n for web programming
yea, but thats my point, the stuff you are saying is very rare around here
 
what do you mean?
 
i know literally 10's of programmers or 100's and no one is in that field,
 
time to end this discussion: CS is practical math, math is well...theoretical
 
7:09 PM
its not that simple
 
in fact CS is a practical field
write applications
thats al you do
*all
 
you dont write code its obvious
 
*Scrtches head
http://math.stackexchange.com/questions/587158/gaussian-quadratures-adaptive-method
 
okay then you have the higher-level theory, but it's all practical in the end. math is just theoretical for its own sake
or i just hate CS (i'm not really too interested in computers atm)'
lol
 
take P=NP, I dont say the question does not have ramifications into real life, but the methods you would use in investigating such a question are very theoretical. I dont think there is anything practical about Razborov´s proofs about possible proof barriers
are you still there Dave?
or what about game theory, what do you think about that?
is anyone there?
 
7:28 PM
@Adam What's the fuss?
 
Hi, Pedro. Wanna chat?
 
@Adam Depends on the topic. =)
 
What interests you?
 
@Adam That's a good question.
 
:-)
 
7:32 PM
Obviously mathematics.
What do you do w.r.t. math?
 
what a relief, I thought you are going to say gardening or something
what is w.r.t. ?
 
"with respect to"
 
I am interested in and all math. Currently I am studying Euclidean geometry at high-school level.
*any and all
and you?
 
@Adam So you're in high school?
 
not exactly, I just happen to read a book that is at that level
 
7:37 PM
@Adam Ah.
 
You study at a university?
 
@Adam Aha. Just returned from a midterm, actually.
 
are you a math major?
 
@Adam I'm from Argentina, not sure what that means. =)
 
well someone studies biology someone studies geography, and someone studies math - he is a math major
 
7:41 PM
Well, then you can say I am a math major. But my career lasts 6 years, IIRC careers are shorter there.
Well, 5 years without introductory courses, if you may.
 
by a career, you mean time it takes to complete the school?
 
Well, yes. But it is a "career".
 
in what year, are you currently?
 
@Adam First year at university, second year of studies. (As I said, one has a year of common introductory courses. In my case, I had common courses with people from my faculty, exact sciences, and engineering)
 
what classes are you taking?
 
7:46 PM
@Adam Analysis II, Linear Algebra and a course on Sequences and Series.
 
I am surprised that there are two courses one on Sequences and Series and one on Analysis.
 
@Adam Well, Analysis II is about integration and differential equations.
By integration I mean line integrals, surface integrals, Stokes, Gauss, Green, &c.
 
multivariable calculus?
what do you learn in linear algebra?
 
@Adam I wouldn't call it that, but yeah, a bit of it.
 
my rating broke 3k today
loads a big bong bowl
you know instead of champaigne
 
7:55 PM
:-D
 
@TedShifrin !
Had my Analysis II midterm today.
 
@Pedro! Anything good?
 
@TedShifrin The first problem was nice, I guess.
 
And all downhill thereafter? :D
 
Heh, what do you mean? =P
 

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