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00:10
Soon I'll also compute $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n+1} H_{2n+2}}{(2n+1)(2n+2)}$$
Actually I'm curious to see if I can get rid of the hypergeometric function or any polylogarithm function.
I'll post it to myself as "Compute without pen and paper: $t_1=1$ and $t_n=n(1+t_{n-1})$. Find $\displaystyle \prod_{n=1}^\infty \left(1+\frac{1}{t_n}\right)$" and go to sleep. This one is very enjoyable.
@Winther) An observation regarding your comment to that 'Prime Root Matrix' question: googling "prime root matrix" yields 14 results, and googling "prime root matrix" -"alwyn scott turner" (the author of that question) eliminates all of the the relevant ones. so I think it's pretty fair to characterize that as a crackpot question
OK, OUT FOR NOW
 
2 hours later…
02:27
@r9m Are you feeling better?
 
2 hours later…
r9m
r9m
04:49
@RandomVariable yes ! :) Thanks for asking :-)
04:59
@r9m I have a bit of a cold. Or it might just be allergies. It's hard to say at the moment.
r9m
r9m
@RandomVariable oh ! take care then ! There is a outbreak of cough and cold in our hostel ! a lot of guys are getting it .. I got it from someone else .. and now my roommate has got it frm me :( terrible thing
@Semiclassical, if you hover over the downvote arrow you may decide to vote after all ;)
@AntonioVargas: eh, I'll stick with my initial reaction for now
@AntonioVargas: I'm much more liberal with up-votes than down-votes in general, and i'd rather not react out of pique
@Semiclassical Understandable.
@AntonioVargas: Though I really do hope someone offers an answer in the spirit of Winther's comment to that question---theta functions were my initial reaction upon seeing that question, but i don't know much about approximating them or their antiderivatives
05:15
@Semiclassical Totally, me neither. O.L. gave an awesome answer on this question of mine using them. They're far outside of my day-to-day tools though so I haven't really gotten a chance to play around with them.
@r9m There was only one time in my life when I got really sick from a cold/flu virus. And that's when I developed pneumonia. I missed 2 weeks of school.
@AntonioVargas: That is a really slick answer. I know elliptic integrals far more than elliptic functions / theta functions, which is frustrating since I know the latter is very useful in computing elliptic integrals; i just can never quite keep track of it well enough
@AntonioVargas for example
@Semiclassical Wow, that's nuts. It's like the answer would just fall out of your head onto the paper if you know enough!
Along with some other things, perhaps.
05:31
How to write that vertical bar when I want to say that the antiderivative goes from a to b, for example?
@LucasZanella, try \left. stuff \right|_a^b
@AntonioVargas I tried but the bar is too litle
tiny
sorry
$$\left. e^{2 x^2} \right|_0^1$$
Oh I see
05:33
The bar gets tiny
Maybe $$-\cot t - t \Bigr|_{\arcsin(1/3)}^{\arcsin(2/3)}$$
how's that?
Exactly! Thank you so much @AntonioVargas
@LucasZanella No problem.
06:32
@AlexanderGruber do you know pdes?
@usukidoll Ehhh... not really but try me
(this will depend entirely on what level of PDE we're talking about)
@Alexander I am stuck... how does the $\frac{-2}{5} $ go away?
http://math.stackexchange.com/questions/931921/find-the-particular-solution-of-u-x2u-y-4u-exy-satisfying-the-following-s
@usukidoll ok lemme read through
heya @Alex
@Alex save me D:
how's things
dying in pde land :) sarcasm
@MikeMiller they're alright. working like a madman
that sounds like a good state
06:53
it ain't bad.
@AntonioVargas do you know pdes
@usukidoll Not really. I've taken a couple classes on them though. Why do you ask?
I am super stuck... like how does the $\frac{-2}{5}$ disappear when if I am iwrt z I have to get that fraction otherwise my integration would be wrong. http://math.stackexchange.com/questions/931921/find-the-particular-solution-of-u-x2u-y-4u-exy-satisfying-the-following-s

@AntonioVargas
@usukidoll Sorry, it's way too late for me to read all of that. I don't think I can help :)
:( you suck
07:07
How kind.
there needs to be more pde people..then I wouldn't be saying those words
 
1 hour later…
08:26
Do you have an example of some solved PDE question avaible for you, @usu?
yeah but it had a different side condition @studentmath
and the damn book doesn't go into detail that much
@studentmath do you know pdes
I don't know any DE, so I can't help you.
^ not u
I know, I am just saying.
But I intend to learn about them some time in the future.
good luck
it is hell
08:29
Currently, I know how to solve separable variables, first order linear ODE and second order linear ODE homogeneous with constant coefficients, lol.
dat is ez]
That is all I need to know to read my differential geometry book.
Hi @robjohn you should be sleeping.
@robjohn do you know pdes?
@usukidoll Which area of math are you most interested in?
odes r my fav
08:43
@usukidoll You should read Gerald Teschl's book then. It is the best ODE book I have come across.
I had a crummy ode book , yet I still passed with an A-
I never did ODE in uni, only in high school, lol.
 
2 hours later…
10:30
@Chris'ssis shhhh don't tell ;)
Greetings
@Hippalectryon can you do Alitzer's problem without pen and paper?
$t_1=1$ and $t_n=n(1+t_{n-1})$. Find $\displaystyle \prod_{n=1}^\infty \left(1+\frac{1}{t_n}\right)$
10:50
@Chris'ssis Have you latexed your book already?
@WillHunting No, not yet.
11:00
Are you tired of Dubstep that sounds... well... a few iterations short of a workable numerical solution?

Maybe you just need to try something analytic instead ;)
https://soundcloud.com/full-synthetic/inured-feat-humming-bird-militia
11:22
@usu once I am back from the trip with the dog, if you still need help, I will go over my books again and see if I can help
Or if you could give the example problem and the solution, it will certainly remind me, and maybe also get you to understand the new problem better
12:00
@DanielFischer I'm looking for rationals in $[\log(n), \log(n+1)]$ with given denominator.
Any suggestions?
@BalarkaSen With given denominator, say $d$, you can look at $\lfloor d\log (n+1)\rfloor$ and $\lceil d\log n\rceil$.
yes.
but how does that supposed to help?
Any integer between these two, provided the latter is not smaller than the former, gives a rational in that interval.
uh, $n$ and $d$ are not known actually. i am trying for some diophantine results, to clear up what i mean.
$\lceil d\log n\rceil \leqslant n \leqslant \lfloor d\log (n+1)\rfloor \implies \log n \leqslant \frac{n}{d}\leqslant \log (n+1)$.
Okay, so not a given denominator.
Continued fraction?
12:06
maybe. i don't know of any good diophantine approximation results on logs
Me neither. What is it that you actually want?
well, if you want to know, the problem we (me and a friend of mine) are looking at is to find growth results on $c(n)$, the minimal integer such that there is an integer $x$ such that $2^x$ sits in between $(n+1)^{c(n)}$ and $n^{c(n)}$. all we have is that $c(n) \ll n/2$ (vinogradov notation).
now i am interested in looking at the $x$s. how many could be there for a given $c = c(n)$?
You mean "for how many $n$ does a given $c$ work"?
"how many integers could be there in $[c\log_2(n+1), c\log_2(n)]$"
@DanielFischer that problem is already solved.
there are, of course, infinitely many such $n$s, and we have a closed form
yep
it's not hard to invert $c(n) = c$, btw, for large $n$s
There can be roughly $\left\lceil \frac{c}{n\log 2}\right\rceil$ integers in such an interval. More typical, I guess, would be the floor, and while $n$ is not much larger than $c$, you may need to take higher order terms of the logarithm into account.
12:23
oh by the way our bound was $c(n) \ll n \log(2)/2$, not $\ll n/2$
@robjohn here is my work to that alternating harmonic series
@Chris'ssis It's not hard to do by generating functions, btw
oh i think you have showed me that before
right right
no polylogarithms
right
@BalarkaSen Yeah, that was my point.
is there a high-school method to evaluate it though?
@BalarkaSen Yeah. I get the expansion of $\log(\cos(x))$ by elementary tools.
12:32
fourier analysis is not high-school!
@BalarkaSen No Fourier ... I can get that expansion by high school tools.
ah. post it here.
@BalarkaSen Do you want me to teach you a very brilliant way?
sure why not
Is there an easy way of seeing the series expansion of $\log(1+x+x^2)$ ?
12:38
:17697254
@rehband taylor for $\log(1+t)$?
@BalarkaSen use the result above derived by telescoping sums.
@BalarkaSen I know that $\log(1+t) = t - t/2 + t/3 - ...$
oh noes, i am not going to try your problem. too hard for me, @Chris'ssis
@BalarkaSen That allows you to get the expansion of $\log(\cos(x))$ (elementarily)...
12:40
What would be the easiest/quickest way of showing $\lim_{x\to0} \frac{ \log(1+x+x^2) - \log(1+x)}{x^2} = 1$ ?
i don't see how but i'll take your word, @Chris'ssis
@BalarkaSen It's very easy, I let you think of it.
i will not think of it. too hard for me =P
@rehband partial factorization and then L'Hopital
or a few terms of taylor would do
12:44
@BalarkaSen Okay thanks
$\log(1+x+x^2) = x - x^2/x + O(x^3)$. $\log(1+x+x^2) - log(1+x) = x^2 + O(x^3)$
$$\frac{\log(1+x+x^2) - \log(1+x)}{x^2} = \frac{x^2 + O(x^3)}{x^2} = 1 + O(x)$$.
As $x \to 0$, the error tends $0$
so the answer's 1
so actually no L'Hopital needed
@BalarkaSen Hmm, $x^2/x$ ?
x^2/x?
oh
i mean $x^2/2$
Right
But isn't it still wrong? :P
why?
oh darn
$\log(1+x+x^2) = x + x^2/2 + O(x^3)$
12:49
Yes
Now I agree :)
anyways that's what i had in mind so my calculations aren't wrong
Tyty
@rehband to calculate taylor of $\log(1+x+x^2)$ : $$\log \left ( \frac{1-x^3}{1-x}\right ) = \log(1-x^3) - \log(1-x)$$
Is it okay if I write something similar to what you said in an answer on MSE?
it took me some time, but was elementary
@rehband sure
12:51
@BalarkaSen Thanks!
@BalarkaSen Have you seen the problem I received from Alitzer?
$t_1=1$ and $t_n=n(1+t_{n-1})$. Find $\displaystyle \prod_{n=1}^\infty \left(1+\frac{1}{t_n}\right)$
@Chris'ssis interesting, but too hard for me
@BalarkaSen It's very easy, honestly.
i decided that calculus is too hard for me, so i won't think about it much
@Chris'ssis that's nice
it's too hard for me though
@BalarkaSen You'll have a lot of fun playing with it. Just give it a try.
12:58
nah
i won't, trust me
i will stick to my algebra.
@Chris'ssis Interesting
@rehband Yeah. Try to finish it without pen and paper.
$\prod_{n=1}^\infty \left(1+\frac{1}{t_n}\right) = \prod_{n=1}^\infty \left(\frac{t_{n+1}}{nt_n}\right)$
Take logs?
@rehband no
Huy
Huy
@rehband: $t_{n+1} = (n+1) (1 + t_n) \neq n(1+t_n)$.
13:04
@Huy True :P
$$\prod_{n=1}^N \left(\frac{t_{n+1}}{(n+1)t_n}\right)=\frac{t_{N+1}}{(N+1)!}$$
Then divide $$t_n=n(1+t_{n-1})$$ by $n!$. The rest is a baby job.
2
mmhmm
that was fast O_O
@Chris'ssis :-O
13:09
@anon can we use abc-type arguments to find a certain $k_0$ such that there are only finitely many $k$-tuple with elements in $[c\log_2(n), c\log_2(n+1))$ for some $k > k_0$
that's too difficult for me while I'm doing my combinatorics take-home last minute, unfortunately
@anon aha, no problem. have fun with the test.
@rehband $$\underbrace{\frac{t_n}{n!}}_{\displaystyle a_n}-\underbrace{\frac{t_{n-1}}{(n-1)!}}_{\displaystyle a_{n-1}}=\frac{1}{(n-1)!}$$ and then sum over $n=2$ to $N+1$
even though none is listening, corrections : $k$-tuple consists of integers, for any $k > k_0$
and $c$ is given
@Chris'ssis Ah okay, i see
13:15
@rehband I'll add this to my book.
@Chris'ssis You should add everything you can think of to your book.
@Chris'ssis Weeeee :) The day you release your book will be like xmas for me
@WillHunting But there is a lot of stuff ... I need to make a selection :-)
@Chris'ssis Hehe, why not make it a multivolume work? =)
@rehband lol :-)))
13:17
gah we need a number theory room in MO. all they've got is some homotopy room which seemingly consists of super-nerds and their jibberish formulas.
@WillHunting I might leave the world of mathematics after that ...
(I saw many things that I don't like)
@Chris'ssis what do you mean?
@BalarkaSen Hard to summarize in a few words. I prefer not to talk about it more, not now.
@Chris'ssis Just like Grigori Perelman.
well, Perelman is incomparable
13:21
Dude, Chris's Sis might be more brilliant than Perelman.
he solved a legendary open problem, man!
So what? It's silly to even compare two people.
i am not comparing talents. i am comparing works =P
Same thing.
not really
ok, less trivia time
13:24
How do I add like a star next to something which I wanna refer to later?
in MSE editor?
Huy
Huy
@BalarkaSen: I'm sure you know a lot about free vector spaces?
heh, @Huy? what makes you think that?
Huy
Huy
No idea, just a wild guess, because you're so smart.
you've got a wrong idea about me =P
Huy
Huy
Too bad.
13:26
For instance, to have a paper with the most simple solution ever to the Au-Yeung series rejected it seems to me pretty crazy ... I wouldn't be surprised to see the proof there appears in some book written by I don't know what author.
@Huy =P
you'll find my knowledge limited in number theory and abstract algebra
Huy
Huy
I thought free vector spaces were part of abstract algebra.
i think mostly of ring/field theory while i say abstract algebra =P
it's not that i don't even know what a free vector space is, i just don't know "a lot"
Huy
Huy
@BalarkaSen: I'm looking at different definitions of the tensor product because I need those for my work. And one of them defines it as the quotient space of the free vector space of the cartesian product w.r.t. some subspace.
Do you know anything about that definition?
i forgot my modules, so no.
i am afraid not
13:31
I only remember algebra-precalculus now.
I only remember arithmetic lol
@Chris'ssis One wonders what kind of article they don't reject.
@Chris'ssis You are better than that, those people aren't fit to judge you. I find what you do extremely inspiring!
@rehband Thank you.
@WillHunting Yeah, maybe.
Oh @Huy, found an excellent solution of that $x^y + y^x > 1$ ineq
Bernoulli's inequality tells us that $(1+x/y)^{1/x} > 1 + 1/y$, i.e., $\frac{x+y}{x} > 1/y^x$, i.e., $\frac{x}{x+y} < y^x$. Similarly, $\frac{y}{x+y} < x^y$. Add up and QED.
Fun isn't it? Someone on the forum came up with it.
13:44
@BalarkaSen I think there is a typo.
er, yeah right
i meant $\frac{x+y}{y} > 1/y^x$
and $\frac{x}{x+y} < y^x$
Never mind, the idea was clear. I was just nitpicking.
no, no, that's fine. it's good to be nitpicky =P
i need to leave bu-byes
14:34
Hey, @Chantry.
Hey, @Ice.
Hey, everyone!
Hi pal @Khallil
Hi @Khallil :)
What have y'all been getting up to today?
Just woke up. I'm supposed to be meeting someone to work on a project in -5 minutes. Just getting the energy to leave the door.
You should've left 5 minutes ago, @Chantry?
=P
14:38
Yep lol
To be fair, he didn't exactly give me short notice.
Are you going to be solving the Riemann hypothesis, @Chantry?
much*
Maybe if I get a half hour after lunch.
Anyway, off to class. Enjoy your day Khallil.
@Chris'ssis I haven't looked at that series yet, so I will look at your solution later.
14:48
You too, @Chantry. ^_^
@robjohn do you agree with this?
I don't think I've "read" an entire book on mathematics period
(with the possible exception of the princeton companion)
I prefer to think of all of the books I have access to as one giant book that I can dip into whenever I want

it seems a bit arbitrary to me to divide mathematics up into books
@robjohn OK
@IceBoy I've read sections of math books, but I don't think I've read one cover to cover. I usually know what is in sections that I haven't read and if I want to know more about those topics, I will go back and use them as references. I would rather get the ideas and work out the details myself than read the details from the book.
@robjohn I will try to read my 12 holy books from cover to cover next year!
@WillHunting I wouldn't think you would do anything else :-)
14:55
It's weird to ask this in math room, but any of you can recommend me a good english grammar book?
CGEL
OK, too thick for me. I am looking at something under 500 pages.
@WillHunting I don't think I've looked at an English grammar book for over 35 years.
amazon.com clearly has a preference of american over british dictionaries, lol
15:32
@Chris'ssis Heya
The internet is a great resource, @Will.
Whenever I'm unsure, I just search it on Google.
@Khallil Real men read real books.
The internet is the greatest book in history, @Will.
@Khallil It is also full of shit.
One man's trash is another man's treasure, @Will.
15:37
but turning trash into treasure consumes a lot more energy than is needed
@Chris'ssis Lol that problem is obvious isn't it ?
Is it well known that $t_n=\lfloor e*n!-1\rfloor$
@Hippalectryon Obvious is the most dangerous word in mathematics
@rehband I thought it was trivial :)
@Hippalectryon Trivial is the second most dangerous word in mathematics
third please?
15:40
Easy is the third most.
:D
Are u talking about that infinite product, hippo?
it is easy to show that this is trivially obvious
15:47
That user is posting stupid comments again, lol.
don't read what you find stupid
I think she does that so that the other user will upvote her.
I think she is extremely crafty.
it is only someone on the internet
@Khallil who ?
16:00
Finally I have 700 points. 300 more to reach 1000.
Is it possible to hand MSE points over to other people?
I'd like to donate all of my points to Jasper.
You can award a bounty to his answer, I think. Nah, don't do that.
What's the difference between $\varnothing$ and $\emptyset$?
There has to be a reason why the two exist separately, without a \var prefix.
(Like $\phi$ and $\varphi$)
@Khallil Nice, I was wondering how to make that second phi
Does anyone here know why Mr. Mhenni is suspended?
16:09
@Anastasiya-Romanova No.
Wait a second. @Anastasiya, did you change your username recently?
I feel so confused. It's as if I've seen that avatar on someone else's profile.
@Khallil V-Moy
I think it was Valentina's.
YES! Suspicions confirmed!
@Khallil Yes. I change it but you can easily notice that I'm V-Moy
Mr. @robjohn, may I know why Mr. Mhenni's account is suspended? What kind of rule violations he did so that I can avoid it?
Have you ever seen The O.C, @Anastasiya?
16:17
@Anastasiya-Romanova It is confidential.
@Khallil What is O.C?
@WillHunting At least he can give me a slight explanation
It's The O.C. but don't worry. I can tell you haven't heard of it, @Anastasiya.
Shinpai shinaide.
@Khallil Yes, what does that mean? Explain please...
@Anastasiya-Romanova Hi there
@Hippalectryon Hi too
16:22
math.stackexchange.com/a/933837/150347 Rep seekers, bad answers
@robjohn Could you add in your $\LaTeX$ in chat link a Katex renderer ? It's usually faster
It looks terrible, @Hippa.
@Khallil ?
$\int_0^1\iint_0^1\iiint_0^1\iiint_0^1\int_0^1$
Uh
Heh KaTeX
@robjohn forget that
Nobody beats LaTeX
@Hippalectryon RH?
16:33
What's RH ?
Rampant Hippalectryon ?
Can't even spell his own name correctly
sneer
Hypalaictréon
@Hippalectryon Reputation Hysteria
Many users in MSE are affected and infected by RH.
@Hippalectryon So, you introduced some values of the sequence in OEIS and realized that there is an alternative way? My way is very easy. :-)
16:36
Heh good catch @Chris'ssis
hahah
@Chris'ssis Forget what I said, I realized it didn't lead to any solution
(does it ?)
@Hippalectryon Take a look at my suggestion in the right panel. :D
No -__-
Nevar
I gotta find
lolll OK :-)
you gotta get a spellcheck, @Hippa
16:38
@BalarkaSen A spellwhat ? Mai englich is véri gud
you need to eeemprove your eeenglish, frenchman
@Chris'ssis While I'm searching, have you ever worked on derivating integers ?
anyone here familiar with the Jacobian in integration?
I'm looking for great results one can get from int's derivatives
@Hippalectryon derivating integers?
what's that supposed to mean?
16:42
@Hippalectryon What is that?
Uh well basically
@Anastasiya-Romanova Did Cleo answer my question you asked? (let me guess - No)
We define $D(n)$ for $n\in\mathbb{N}$ by:
$D(n)=1$ if n is prime
ah, right
$D(1)=0$
16:44
$D(uv) = uD(v) + vD(u)$
Yep
That's the three rules
integral derivative, heard of that
you just ringed the bell
I found them while reading that article
16:44
I found them in OEIS =P
@Chris'ssis As you can see, she didn't give a **** for a kid like me, lol
tsk tsk language, kid
@BalarkaSen Censored! I learn from 9Gag
@Anastasiya-Romanova You mean she did not give a fuck? I see.
cool article, @Hippa
16:46
@WillHunting No comment...
although I guess @Chris'ssis isn't interested. it's just number theory ={
wait, wait. i know Edward Barbeau!
he wrote a paper on the SCESS identity.
search "Barbeau + Seraj + Sum of cubes equal to square of sum"
Okay guys, I have to sleep now since there's nothing special today on MSE except Mr. Mhenni suspension.
Wait, before I sleep. @Chris'ssis could you please help me to derive $\dfrac{\ln(1-x)}{1+x}$ or $\dfrac{\ln(1+x)}{1-x}$ into sum of series involving harmonic number?
I asked you because you're good at this subject
@Anastasiya-Romanova What do you think? I can?
@Chris'ssis I don't know, I have no idea. I only know that $\displaystyle-\frac{\ln(1-x)}{1-x}=\sum_{k=1}^\infty H_kz^k$ and $\displaystyle\frac{\ln(1+x)}{1+x}=\sum_{k=1}^\infty(-1)^{k+1}H_kz^k$
@Anastasiya-Romanova $$\int_0^1 x^{n-1} \log(1-x) \ dx =-\frac{H_n}{n}$$ and $$\int_0^1 x^{2n-1} \log(1+x) \ dx =\frac{H_{2n}-H_{n}}{2n}$$

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