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5:00 PM
Is this factorization correct????
 
it's algebra. if you're not sure, write it out.
 
@r9m @Chris'ssis Please help :p
@Semiclassical nono im looking for a pattern
 
how is that relevant? if you want to check if your factorization is correct, multiply it out and check.
 
@Semiclassical my next question was going to be , what about z^7-1 , z^9-1 ,z^8-1 ???
 
sure, but none of that helps if said factorization isn't actually right. that's why i said to look at it
 
r9m
5:03 PM
@TheArtist $z^6 - 1 = (z^2)^3 - 1 = (z^2-1)(z^4+z^2+1)$
 
@TheArtist No
 
@r9m thank you very much, how did you go from middle to last step?
@Semiclassical Im looking for an easy way for factorization :)
 
r9m
@TheArtist don't you know how $z^n - 1$ factorizes ? .. you can always write $z^n - 1 = (z-1)(z^{n-1}+z^{n-2}+\cdots+1)$
 
@r9m no not always. That's why im asking :/ what's this called?
 
do you remember what a finite geometric series sums to? factorizing $z^n-1$ is just that in reverse
 
5:06 PM
@r9m for some familiar powers yes, but that's because I have a memory about them...not generally
 
r9m
@TheArtist yes @semi says it right !! think about the geometric progression !! :)
 
@r9m the general formula you said was what I thought was right. Until something unusual happened
 
r9m
@TheArtist unusual happened ? what happened that made you feel that way ?
 
@r9m in your general formula the powers in your second bracket goes as n-1 , n-2 , n-3 after each term.....but look at how you did to z^6-1 , the power goes as 4,2, which is n-2 , n-4
 
r9m
@TheArtist then you should try to multiply out the RHS of your expression and check if what you wrote was correct :)
 
5:12 PM
bounty plug!
1
Q: Eigenvalues of tridiagonal matrix

Tobias Hurthon page 13 of the paper here there is a proof in theorem 4 that all eigenvalues of this tridiagonal matrix, which has strictly positive entries down the subdiagonals, are simple. Unfortunately, I don't get the argument. Apparently, it is almost immediate to the editor that $ker(J-\lambda I)$ must...

 
$z^4 + z^2 + 1$ but shouldn't it be $z^4+z^3+z^2+z+1$
 
write $w=z^2$. then you're factorizing $w^3-1$
 
@r9m I know your right, that's the problem , why doenst this general formula work ???
 
the pattern would be correct if you had pulled out a factor of $z-1$ from $z^6-1$. that gives $z^6-1=(z-1)(z^5+z^4+z^3+z^2+z+1)$
if you instead pull out $z^2-1$, though, then you're not using the same pattern as before
 
@Semiclassical ohhh ok :D so what's the pattern when you pull $z^2-1$ ? It's n-2 , n-4????
@Semiclassical Thank you so much :)
 
5:15 PM
eh, just do what i did with $w=z^2$. then $w^3-1=(w-1)(w^2+w+1)$
and then replacing $w=z^2$ gives you the correct second factor
 
$z^z$? :)
 
psh, you and your eye for things which are actually correct @MikeMiller
 
@Semiclassical ohhhh got it :D Thank you :D
 
@r9m see now in the new answer I received the last amazing limit related to Catalan's constant - math.stackexchange.com/questions/996384/…
I was planning to compute it soon. :-)
 
@przemo: it's a pity the notation is such a burden with the integrals in your two simplex-problems, i rather like these high-dimensional integrals.
 
r9m
5:19 PM
@Chris'ssis hoo !! Nice !!
 
@r9m That limit is crazy nice! :D
 
@Semiclassical what about $z^n+1$ ? :)
 
r9m
@Chris'ssis oh !! $(19)$ there is by Oloa !! Cool :D
 
eh, those are a tad more annouying
simplest way I know is to make a substitution such that $w^n = -z^n$
which means taking $w=e^{i \pi /n}z$
 
@r9m Yeah, it's nice. Oloa is very creative too. :-)
 
5:23 PM
once it's in terms of $w$, though, you've just got $1-w^n$ which again factorizes nicely
or, if you've got $n$ odd, then $w=-z$ will also work
it's really only $n$ even that things become tedious, since there's no roots on the real axis in that case
 
r9m
@Semiclassical but that need not be a factorization in $R[z]$ :) n could be a power of $2$
 
psh, you and your polynomial rings :)
 
@Semiclassical Thank you very much :D
 
but, yes, my comments above aren't relevant in the ring context
 
how's the nervousness now? @TheArtist
 
5:43 PM
@Chris'ssis well, after a bit of manipulation, I've shown that my answer is the same as Felix Marin's.
 
@robjohn Yeah, I saw. The last series is crazy hardcore.
 
@Chris'ssis judging from the difficulty of the $A(k,n)$ evaluation, I am not too hopeful.
 
Hi all.
 
hi there
 
@Chris'ssis I believe that I can get the value to a high precision.
 
5:47 PM
@robjohn OK. Also a closed form would be amazing. :-)
 
@Chris'ssis Of course, the closed form is the target, but it is going to be quite a difficult feat.
 
@robjohn OK
 
I can't find any question about coupon's collector problem
but without replacement.
Or perhaps I'm over complicating and I just need to use the hypergeometric distribution? My problem is something like:
There are N balls. K of them are white. I draw until all white are drawn. What's the expected number of draws?
 
6:08 PM
A NEW VERY NICE QUESTION
0
Q: A limit evaluating to $2 K$ (Catalan's constant)

Chris's sisExperimentally I discovered the limit below that says that $$\lim_{n\to\infty} \int_0^{\pi/2} \frac{1}{\displaystyle \cos\left(\frac{x}{2}\right)\left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^2}\right)\right)\cdots \left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^{2n+1}}\right)\r...

 
@Sawarnik No it is a username. My real name is Ali.
 
@robjohn do you know to answer my last question?
 
@Chris'ssis Not off-hand... I am looking at it.
 
@robjohn It's hard to write it in a better form and look nicer then. (there is a cyclic arrangement)
@robjohn after posting this question I don't expect anyone to upvote the question (better say) ... :-))))))
I should not post questions anymore and let people focus on the question that has a bounty attached.
 
r9m
6:30 PM
@Chris'ssis Hoh !! that crazy integral again !! :D Awesome !!
 
@r9m HAHAHAHAHAHA, niceeeeeeeeeee :-)
 
@r9m haha, awesome. Still need to see that full movie
 
r9m
@ZachSaucier you've seen it ? ;)
 
only parts
 
@Chris'ssis Is this the integral?
$$
{\Large\int}_0^{\pi/2}\operatorname*{\Large\sum}_{j=1}^{2n+1}\left[\cos\left(\frac{x}{2^j}\right)\prod_{\substack{k=1\\k\ne j}}^{2n+1}\left(\cos\left(\frac{x}{2^j}\right)-\cos\left(\frac{x}{2^k}\right)\right)\right]^{-1}\mathrm{d}x
$$
 
6:37 PM
@r9m And you had some work?
 
r9m
@Sawarnik its done now .. had to tex my assignment :)
 
O.
I m 15 now :'(
 
@robjohn Yeah, it seems OK. My notations often fails, thanks! Can I post it on main?
 
r9m
@Sawarnik good good .. not bad !!
 
@Sawarnik why the tear?
 
6:39 PM
Being 14 was so nice :) ..
 
I doubt much changed in 1 day...
 
@Chris'ssis sure
 
r9m
@Chris'ssis That dance is Awesome !! :P I like it !!
 
@robjohn thanks!
 
@ZachSaucier I mean there were things assoicated with being 14 :(
 
6:41 PM
such as...? 15 is still quite young
 
Well yes.
:)
 
Guys, by the previous question of mine, the penultimate one, we've possibly found another amazing way of computing $$\lim_{n\to \infty} \frac{1}{(4n+1)^{2n}}\prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}=e^{2 K/\pi-1/2}$$ Isn't really this amazing?
 
@Sawarnik heck, had I started programming at 13 (like you started math on here) instead of 17 I would be much better than I am now xD be glad you have a good start
 
All we need is to combine my work with Winther's work here
@Chris'ssis Hehe, didn't think about that. Yes that is indeed a quicker way to show the equivalence:) — Winther 9 mins ago
 
@r9m So how is she 23? Tell me!
@ZachSaucier Ok :)
 
r9m
6:45 PM
@Sawarnik I have no proof ! lol .. thats why you should ask her
 
Oh, I placed that really in the wrong place :-)
 
@r9m So any reason to doubt what she has written on her profile? :P
 
@robjohn have you ever seen that limit above?
 
r9m
@Sawarnik her Brilliant profile says 23 .. and that seems more acceptable :P
 
@Chris'ssis That one does look familiar...
 
6:47 PM
@r9m Will you come to FB?
 
r9m
@Sawarnik I'm on fb (online)
 
Good.
 
@robjohn it's familiar to me too, but I didn't attend it so far I think. Well, I have at least a proof now. :-)
 
Hi @Huy
 
Huy
Good evening, @IceBoy.
 
6:55 PM
@Huy I would like to apologise for my comments last time.
 
@robjohn actually 2 proofs, one coming from a paper.
 
Huy
@IceBoy: Which ones?
 
Someone like geometry here? ^^
 
@Huy I was too blunt about saying you should leave...
 
Huy
@IceBoy: How come you want to apologise?
 
6:57 PM
for being overly blunt
 
Huy
@IceBoy: Why, not what for?
 
why was I blunt?
 
Huy
@IceBoy: No, why do you want to apologise?
 
I shouldn't have said it like that
 
Huy
Ok.
 
6:59 PM
:-)
 
@Chris'ssis I went in to edit the limit into the second formula, but it was already there...
 
@robjohn OK
 
@Chris'ssis I am using this
$$\lim_{r\to1}\,(r-1)\sum_{j=1}^{2n+1}\int_0^{\pi/2} \prod_{k=1}^{2n+1}\left[r\cos\left(\frac{x}{2^j}\right) -\cos\left(\frac{x}{2^k}\right)\right]^{-1}\mathrm{d}x$$
 
Huy
@MarcGato: Depends on what kind of geometry.
 
I would to plot some figures, do you know how can I do this online?
 
7:03 PM
@MarcGato what type of figures?
 
@Huy basics geometry: circle, triangle, square. I am going to post my question
 
@robjohn hmmm
 
@ZachSaucier Triangle, circle, pentagon.
 
in a grid format? just the shapes?
 
yep :)
 
r9m
7:05 PM
@MarcGato have you tried Geogebra online ?
 
@r9m I will try now :p. Thanks
 
@robjohn above when I said I have 2 proofs I was referring to the other limit ...
 
@Chris'ssis I understand that.
 
@robjohn $$\lim_{n\to \infty} \frac{1}{(4n+1)^{2n}}\prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}=e^{2 K/\pi-1/2}$$
@robjohn OK
 
@Chris'ssis you said you already had proofs for that, so I was working on the other.
 
7:08 PM
@Huy Hi.
 
@r9m How can I adjust the "size"? In particular to draw an equilateral triangle.
 
@robjohn Ah, OK. Still, I need to find an elementary way to this one which I don't have, I mean the one I just posted above now.
 
Huy
Hi, @Sawarnik.
 
r9m
@Chris'ssis the $\log \log$ integral (one which felixmarin and robjohn answered) seems to be related to Randemacher's formula :O .. I'm trying to find an online reference :-)
 
@robjohn Clever move though ...
 
7:08 PM
@r9m TELL ME!
 
@r9m Never ever heard of it (Randemacher's formula). Does it have a closed form?
 
@r9m The sums that we are stopped at currently are pretty hard.
 
@Huy Has that being a parent personal attack anger subsided by now? :D
 
Huy
@Sawarnik: I don't follow.
 
@r9m Can you give me some reference?
 
7:10 PM
Sorry, typos.
 
Huy
@Sawarnik: Which one exactly?
 
Ok, leave it.
 
r9m
@Chris'ssis no no no closed form .. the series seems to be related to it (can also be derived from) .. wait I'm looking for an online reference (if I don't find one .. I'll type out the whole thing .. I read it from Apostol's ANT book)
 
@r9m OK, no hurry with that
 
r9m
:18386865 I thought we were having the conversation on fb ! :( why are you dragging this here ?
 
7:14 PM
@r9m Nah, you werent responding that's why.
 
7:32 PM
@r9m Indeed. Thanks. I think I saw this paper before ...
 
@Huy If you are interested, here is the question: math.stackexchange.com/questions/997180/…
 
Huy
@MarcGato: What do you mean by "explode"?
 
@Huy Exactly.
 
The size of the figure will swell ?!
 
r9m
@Chris'ssis googling leads me to a lot of interesting pages .. like this and etc :-) this integral is beginning to look very 'tasty' !! ;)
 
7:37 PM
@r9m :-)
 
r9m
 
@r9m What's it?
I was reading on stock exchanges just now (RBIQ).
 
r9m
@Sawarnik fb monitors all our activities ... and generates suggestions :-)
 
..ok got to go..
 
@r9m how does one even think of that lol(referring to the solution for that parameteic integral in your link)
 
r9m
7:43 PM
@UserX ask Adamchik .. he seems to be at the root of this conspiracy :P
@MarcGato interesting question there :) .. how would you find the radius of the circle circumscribing a regular n-gon in terms of the incircle (circle inscribed in the n-gon) ?
ah .. someone already answered the :P fast guns all around ! :D
 
I posted too many questions on main in the last period of time ... (I think it's enough)
(my fault is I'm very active in general - hopefully I don't have ADHD :-))
 
8:03 PM
I see this question, and the only response I can come with is "and I should be interested because...?"
0
Q: Function which has the following properties

Drew I wanted a function which is strongly positive (sharp gradient) in the ++ and positive in the +, and the same in minus

 
r9m
@Semiclassical I couldn't make heads or tails out if that :O
 
My interpretation is "do my homework for me kthxbai"
 
r9m
@Semiclassical haha !
@Chris'ssis ADHD .. try Amphetamines ! if not .. Modafinil (Puff, the Magic Dragon) :P
 
@r9m :D
@r9m Do you still take Modafinil? Maybe, once you told me "Lord ...". :-))))))))) (an overdose?)
 
8:09 PM
hahahaha :-)
 
@Semiclassical what's that sharp gradient he talks about?
 
Good question. I may recognize the intention of his question (see above) but the question itself is pretty inscrutable as well
 
A google search returned me a PhD thesis on Sharp Gradient bounds( google.gr/…)
I bet someone working on something that specialised can grab a better picture than paint one is msdraw
 
r9m
@Chris'ssis oh ! I'm Lord Drugs then ! :P lol (I've only taken 5 so far .. no apparent harm came if I ate and slept in time)
 
@r9m :-))))
 
8:13 PM
though i'm being a bit careless myself, for i missed the true unifying meaning of his question
"i am lazy"
 
0
Q: integral sulotion over a and t

mojwhat is the solution of this integral:$$\int^1_0 \frac{-2(t+a)+(1-a)}{((t+a)^2+(1-a)^2)^2} dt$$ canyou help me? that is a part solultion of a question which I should to solve it!

 
@r9m Good! Do this job carefully though for your health! :-)
 
That answer... He's getting a downvote if he doesn't manage to get the antiderivative and fully explain it in the 5 minute edit gap
 
i already flagged it as 'not an answer.' i'm not waiting that long
the fact that that poster is nearly to 3k rep is deeply disturbing to me
 
has anyone understood what happens at the end of Naruto's latest chapter ?
 
8:21 PM
@G.T.R Wrong room?
 
given that last comment, it seems appropriate that i just gave an answer to this question (look at the poster)
2
Q: How many distinct ways to climb stairs in 1 or 2 steps at a time? (Fibonacci puzzle)

Naruto UzumakiThere is a very interesting puzzle for fibonacci sequence You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? The answer is fibonacci sequence: F(n) = F(n-1) + F(n-2). The explanat...

 
@Semiclassical I told Sonnhard that he should use \frac instead of $x+y/z$ (no brackets too). And he has answered 300 questions...
 
quantity >> quality
is his approach
 
By the way, did you guys read a LaTeX book somewhere or something to type answers? I just read that brief guide and watched other people's codes, should I do something more for my typesetting?
 
@robjohn are you going to answer my last question?
 
8:29 PM
@Chris'ssis which one are you asking about?
 
8
Q: A limit evaluating to $2 K$ (Catalan's constant)

Chris's sisExperimentally I discovered the limit below that says that $$\lim_{n\to\infty} \int_0^{\pi/2} \frac{1}{\displaystyle \cos\left(\frac{x}{2}\right)\left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^2}\right)\right)\cdots \left(\cos\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2^{2n+1}}\right)\r...

 
@Chris'ssis I am working on it, but I have not come up with anything yet
 
amusing that that's again equivalent to "show that one can also define catalan's constant with this expression"
 
@robjohn Ah, OK.
 
except seemingly much harder
 
8:31 PM
@Chris'ssis Do you have a solution?
 
@robjohn I think I found some (some additional work is required). The problem is that now I try to take an English lesson, I wanna improve my English, so I'll postpone the work on it for some minutes.
 
@Chris'ssis Other than what I posted above and writing the difference of cosines as the product of sines, I have not come up with anything else promising.
 
@robjohn I have a very brilliant idea, but I need to work on it a bit more.
 
@Chris'ssis it's too large to fit into the margins of this chatroom? :>
2
 
@Semiclassical Do you refer to this? $$\lim_{n\to\infty}{\Large\int}_0^{\pi/2}\operatorname*{\Large\sum}_{ j=1}^{2n+1} \left[ \cos\left(\frac{ x}{2^j}\right)\prod_{\substack{ k=1\\k\ne j}}^{ 2n+1}\left( \cos\left(\frac{x}{ 2^j}\right)-\cos\left( \frac{x}{2^k}\right)\right)\right]^{-1}\mathrm{d}x=2 K$$
 
8:36 PM
Sure is large.
 
nah, to the answer you just mentioned you had (i was making a Fermat's last theorem joke, sry)
 
@Semiclassical Ah ... :-)))
Yeah, brilliant often also means very short. ;)
 
quite. there's a quote (by mark twain, i think)
"i didn't have time to write this letter short, so i wrote it long"
 
@Semiclassical To me it sounds like: "I wanna clarify all now, to tell all I have to tell, not only some parts"
 
it's more in the spirit of "if i had time, i would work out how to say everything i need to in as concise a manner as possible."
 
8:41 PM
the quote just emphasizes the fact that it's harder and more time-taking to writing up what you want to tell concisely than writing it up the way you find natural
well @Semiclassical beat me to it
 
@Semiclassical The problem with the literature is that you never know what the author had in mind when he wrote what he wrote. I almost always quarreled with my professor about that. Who knows what the author had really in mind?
 
the quote is more properly rendered as "I didn't have time to write a short letter, so i wrote a long one instead"
 
@Chris'ssis somewhat.
 
human communication would be easier if you didn't have to do it with people :P
 
just stop communicating, like me.
@MikeMiller yes. actually my prof (you can google him up, name's mahan mitra, now a monk =P) is a big fan of gromov. i never actually asked him whether he actually ever met that guy but i should now that i mention it.
in fact he is sold on hyperbolic geometry actually.
 
8:53 PM
am i just being thick with regards to this answer?
 

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