« first day (1443 days earlier)      last day (3580 days later) » 

12:13 AM
hey y'all.
 
12:34 AM
@AlexanderGruber Yo!
 
@Shisui what's up?
 
@AlexanderGruber What is up?
:O
 
@Shisui $\mathbf{\hat{k}}$?
 
Vectors!
How are you?
 
pretty good
just gave my students a nightmare-mode level exam and a bunch of them got perfects
 
12:39 AM
Is it late where you are?
(Where I am it's 1:39 AM)
 
@Shisui i think it's around 9pm.
where are you?
you should pin yourself!
 
1:29 AM
@AlexanderGruber The UK :-)
@AlexanderGruber I think I already have … I'll check again to be sure.
 
2:10 AM
Students getting perfect @Alexander is a sign of a good teacher :-)
 
 
3 hours later…
5:12 AM
I have linear Algebra problem,
let $A$ is real symmetric positive definite matrix, show that: there exsit complex matrix $B$,and postive integer $m$ such
$A=B^m$
I can prove when $B$ is real matrix
But for complex I can't
 
5:35 AM
oh,I know ,this is fast solution[math.stackexchange.com/questions/869653/…
 
@skullpatrol i like to think so. i am not an easy grader.
 
5:57 AM
@AlexanderGruber Hallo?
NOOOOOOOOOOOOOO
 
6:28 AM
Mathematicians should never perform a calculation in their life time.
ID
 
@EnjoysMath how is your complex analysis?
 
^_^
My complex anal lysis?
 
:(
 
It's shitty but what's your quest?
 
18
Q: entire 1-1 function

PeteyCan we prove that given an entire function $f$ that is also one to one then $f$ must be linear? Thanks for any help.

I don't get why the function is n+1 to 1
 
6:30 AM
What's entire?
What?
 
It's in the first answer.
Entire means analytic everywhere, I think.
 
Oh, $(n+1)$-to-$1$ just means there are exactly $n+1$ inputs that map to each existing output
What does a linear function $f: \Bbb{C} \to \Bbb{C}$ look like? Is it only the $f(z) = a z$ for some $a \in \Bbb{C}$?
 
I know what it means, but why is that?
 
What? Is this relevant?
 
6:37 AM
Greetings
 
Hey!
 
@chinamath In my research I found a very nice integral
 
Can you explain the n+1 to 1 thing in the first answer here math.stackexchange.com/questions/29758/entire-1-1-function
@Chris'ssis
 
@Chris'ssis,which integral?
 
@chinamath Find the closed form $$\int_0^{\pi/4} \frac{\log(1-x^2)}{1+x^2} \ dx$$
 
6:40 AM
maybe I see it
 
@robjohn have you ever met the integral above?
@Anthony If you have questions related to integrals, series and limits, I can ... :-)
 
lol
 
@r9m don't miss the integral above, it's impressive! ;)
 
r9m
@Chris'ssis :-) okay .. simple substitutions won't kill it ?! :o
 
@r9m :D
 
6:49 AM
\int_{0}^{1}\ln{(1+\tan{x})}dx
 
@r9m Will simple substitutions kill it? Teach me your way. :-)
@chinamath What is that integral?
 
is your integral
 
@chinamath No, it's not.
 
.......
$ln{(1-x^2)}=\ln{(1+x)}+\ln{(1-x)}$
and let $x=\tan{t}$
 
r9m
shows standard computation time exceeded :| .. as usual :P ... @Chris'ssis .. does it have a nice closed form .. I mean free of Li and stuff like that ?
 
7:04 AM
@r9m Sure, it has a very nice closed form. :D
 
which closed form?
 
@chinamath If I tell it now all the beauty of it will disappear. It's just a candy as @r9m uses to say ... :-)
 
..................
 
@chinamath I mean it is too easy for you if I tell you the closed form.
 
yestedy My integral you solve it?
 
7:06 AM
@chinamath Sure.
 
post it?
 
@chinamath Once you use the substitution $x=\operatorname{sech}(t)$ things become easy and clear ...
 
note this two lnln
Not a one ln
 
@chinamath Did you note what I wrote above? From that point, the integral is for kids ...
 
I think you can't solve it, only let $x=\operatorname{sech}(t)$
 
7:09 AM
@chinamath I know, but the last part is easy ...
 
No,follow is not easy
can you post full solution to seen ?
 
@chinamath No. I don't have things in latex, but on paper.
 
you can send JPG
I'm sure you can't solve it
 
@chinamath sorry?
 
because this integral is very hard in china
 
7:12 AM
@chinamath Yeah, sure. I can't evaluate your integrals.
 
....
 
@chinamath I just wrote you the thing you wanted to read.
 
so I think you can't solve it my intergral.because this integral is open problem to sometimes year ago
 
@chinamath Do you understand that this integrals is for kids?? How to be an open problem? Come on ... :-)
 
only to let $x=\operatorname{sech}(t)$ ,follow is very ugly too
because some people reaserch the integral in china
there can't solve it. there are famous ,like Wei Zhi Sun
he can't solve it a year ago
 
7:15 AM
@chinamath I repet, that integral is very easy and boring from that point.
 
No!
 
@chinamath You'll lastly get something in terms of the derivative of the Riemann Zeta function ...
@chinamath Things are clear, obvious, very easy ....
 
........
I post it,if you can solve it,you post your solution?
 
@chinamath I don't need a solution, I can do it mentally.
 
r9m
jaw drops ... @Chris'ssis you did that mentally ?? :O !!! ...
 
7:19 AM
so you can't solve problem.and you only ask question
 
I'm dubious too @Chris'ssis
 
@chinamath You know, I'm going to ignore you.
@r9m well, it was an exaggeration ... :-) I only wanted to emphasize that the problem is easy.
 
r9m
long sigh of relief ^^' .. I started to wonder if if were Cleo :P @Chris'ssis
 
@r9m Can you confirm if that integral is easy or not?
 
r9m
@Chris'ssis I've got no idea ... I couldn't do even with pen or paper .. ^^'
 
7:23 AM
@r9m I'll show you my solution when I put things on paper (in Latex I mean).
 
r9m
I'm not good with evaluating Integrals .. unless its already fits into a pattern I recognise :|
@Chris'ssis Thank you ^_^ ... I'm eager as ever :D
 
and in china @tian27546 can't solve it a year ago.
and at last is solve by other people
 
r9m
@chinamath tian27546 is math110 ?
 
yes!
if you fell easy,please post your follow solution...
 
@chinamath are you a student ?
 
7:27 AM
He is a famous teacher
 
@chinamath what about you ?
 
I'm student
 
r9m
@chinamath ah !! .. he's good :D .. no doubt about that !! :D
 
@chinamath how old are you ?
 
I am a graduate student
 
7:29 AM
I see
 
I think I want to introduce you to the math room.that this room more and more people in it
I think I want to introduce someone to the math room.that this room more and more people in it
because I think lot of people can't know this room
 
When I was just a beginner in this area I came up with this proof (that is related to the main question)
It's long since that moment, I didn't even check the proof again. Now, I'd probably tackle the problem in a different way.
I have to go! Have fun!
@robjohn please delete my posts above when you have time.
 
7:57 AM
and I post it,I don't know why have three people closed it?
 
 
1 hour later…
9:12 AM
Isaac Newton: If I have seen further it is by standing on the shoulders of giants.

I seem to be standing only on the shoulders of the above-average.
 
@r9m
 
@FredKline "If I have seen further than others, it is because I have been surrounded by midgets" - Murray Gell-Mann
 
9:28 AM
Hi, soft off topic: Given a sequence $x_n$ defined as follows: $x_1 = 1, x_{n+1}= \sqrt (x_n + 3)$, why does one need to show that $\lim x_n $ exists before finding the limit? If I can show $\lim x_n = (1+\sqrt 13)/2, doesn't that imply that the limit exists (since it's equal to a real number)?
 
9:47 AM
@TheSubstitute yes inherently if you prove by the definition of a limit that it is $L$ then that is a prove that it exists and is equal to $L$. Proving existence firsthand w/o finding the actual limit allows one to use other methods to find the actual limit. For instance for your $x_n$ support I prove that the limit exists (i.e. $x_n$ converges to a real number) then I can say $L = \sqrt{L+3} \implies L^2 - L - 3 = 0$ and then we can solve that quadratic and pick the appropriate solution.
 
9:59 AM
A new tag called was created. No tag-wiki, I am not sure for what kind of question it is intended. Should it be removed?
 
@TheSubstitute Soft off topic?
 
@MartinSleziak no
 
r9m
@Sawarnik kya baat hai imp ? :P
 
@r9m how is your back feeling?
 
r9m
@skullpatrol more or less okay ... I can move now :D .. thanks for asking :-)
 
10:07 AM
:-)
@MartinSleziak in my opinion, it would be like removing the percentage tag :-)
 
r9m
@Sawarnik sorry .. I didn't see that :|
 
@r9m great :D
 
in English Language & Usage, 1 hour ago, by Johan Larsson
why do you delete stuff all the time? Deleted messages always makes me think I missed something interesting :)
in English Language & Usage, 1 hour ago, by skullpatrol
that's exactly why I do it :D
 
r9m
@Sawarnik ?! >8(
 
$\Huge \text{>8(}$
 
10:19 AM
@r9m well it was something liek this ... [.. chris'sis ko khush rakhne ke liye bahut cream dalne lage ho na? ;)] ... now plz dont get angry!
 
r9m
@Sawarnik not exactly :| ... its not about the cream but identifying the eye of the storm :P
 
Also @MartinSleziak what better place to discuss the impossibility of division by zero in a physical sense :-)
 
@skullpatrol ohai
 
r9m
@skull are you involved in any sports activity ? :-)
 
10:34 AM
@r9m not seriously, how about you?
 
r9m
@skullpatrol neither me ... day before yesterday I visited a boxing gym out of whim :P
 
heavy weight boxing is my favourite sport to watch
 
r9m
oh .. !! it sure is interesting !! :D
 
yep, the russian Wladimir Klitschko is about to set a world record for most title defenses by a heavy weight
Klitschko is the longest reigning IBF, WBO & IBO heavyweight champion in history with the most title defenses for those organizations. Overall, he is the second longest reigning Heavyweight Champion of all time with the second most successful grand total title defenses with 21 (including his "super" title recognition).
 
r9m
its an insane sport :| ... I think boxers are not exactly sane people :P
 
10:42 AM
Klitschko is tied with Muhammad Ali for second on the all-time list for most heavyweight title fights with 25, behind only Joe Louis with 27
 
11:04 AM
@r9m my net is bad now .. ooh i see.
 
Is Wikipedia not working for some of you as well ?
 
r9m
its working like rain for me :D
 
r9m
@G.T.R ya .. totally worked :D
 
even en.wikipedia.org is down for me...
damn my ISP must be in serious trouble then :/
Hello @math110, I heard you're a teacher
 
11:12 AM
@G.T.R,Hello
yes
 
are you teaching university in China ?
 
yes.
 
@math110 Hello!
 
hmmm
 
Chinese people are very intelligent generally.
 
r9m
11:16 AM
@Chris'ssis 'sup ? :D
 
$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_n}{n} \left(1-\frac{1}{3}+\frac{1}{5}-\cdots +\frac{(-1)^{n+1} }{2n-1}\right)$$
 
Hello,everyone,I found this nice solution:[math.stackexchange.com/questions/390640/…
 
@r9m Creating some new stuff. :D
 
@DanielFischer have you read Vom Kriege by Clausewitz ?
 
r9m
@Chris'ssis Nice !! :D .. @Sawarnik claims ... I cream you a lot to make you happy :P LOL
 
11:22 AM
@r9m Do you have a question for me?
@r9m Wat? You sure?
 
@r9m lolllll :-)))
 
r9m
@Sawarnik I'll have to think for a while ... something that you can't google out :P
 
@r9m hmm .. lol .. you won't be able to :P
 
Now, I'm going to create something very special ...
Till then, there is a nother interesting piece to try $$\sum_{n=1}^{\infty} \frac{H_n}{n^2} \left(1-\frac{1}{3}+\frac{1}{5}-\cdots +\frac{(-1)^{n+1} }{2n-1}\right)$$
 
@r9m Meanwhile you can enjoy this translation ..
 
11:31 AM
@G.T.R No.
 
r9m
@Sawarnik I can read hindi ... you brat :|
 
The masterpiece is coming ...
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left(C-\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\left(1‌​-\frac{1}{3}+\frac{1}{5}-\cdots +\frac{(-1)^{k-1}}{2k-1}\right)\right)$$
 
@r9m -_- i know ... i was pointing to the quality of translation .. which nowhere conyes the original meaning.
btw i thought i was a rat not a brat :D
$I=\frac{a}{b+c}
+ \frac{b}{c+a} + \frac{c} {a+b}$
Any solutions for natural $I$, when $a,b,c \in \mathbb N$? @r9m
 
r9m
11:55 AM
@Sawarnik you know the lower bound for that right ? .. now use rearrangement to find a strict upper bound :-)
 
@Chris'ssis The upper limit of $\pi/4$ seems odd. I can see that there might be a closed form for an upper limit of $1$.
 
@robjohn I'll recheck that. I think I initially got a closed form that contained $\displaystyle \arctan\left(\frac{\pi}{4}\right)$.
 
r9m
$101$ answers !! YaY !! :D
 
12:34 PM
Mma gets $$\frac{1}{8} \left(-4 i \left(\text{Li}_2\left(\left(-\frac{1}{8}-\frac{i}{8}\right)
(-4+\pi )\right)-\text{Li}_2\left(\left(-\frac{1}{8}+\frac{i}{8}\right) (-4+\pi
)\right)+\text{Li}_2\left(\frac{4-4 i}{4+\pi }\right)-\text{Li}_2\left(\frac{4+4
i}{4+\pi }\right)\right)+4 \log \left(\frac{64}{(\pi -4)^2 (4+\pi )}\right) \tan
^{-1}\left(\frac{4}{\pi }\right)+\pi \left(\log (8)-4 \tanh ^{-1}\left(\frac{\pi
}{4}\right)\right)+8 \tan ^{-1}\left(\frac{\pi }{4}\right) \tanh
^{-1}\left(\frac{\pi }{8+\pi }\right)\right)$$
 
hmmm, this answer looks ugly.
 
indeed
The integral for $[0,1]$ is $\frac\pi4\log(2)-C$ where $C$ is Catalan's Constant
that seems right to me
 
r9m
1:06 PM
@Chris'ssis have you seen this one ? :D $$\lim\limits_{n \to \infty} \dfrac{1}{\log n} \sum\limits_{k=1}^n \dfrac{1}{k}\tan \frac{\pi k}{2n+1}$$ :D
:16634166 idk :-)
 
@r9m Yeah, I have something like that. I need to take a look at my paper.
 
r9m
@Chris'ssis okay :-)
 
Huy
1:35 PM
What would be some applications of solving the Dirichlet problem for laymen?
 
@r9m Done without pen and paper ...
$$\lim\limits_{n \to \infty} \dfrac{1}{\log n} \sum\limits_{k=1}^n \dfrac{1}{k}\tan \frac{\pi k}{2n+1}=\frac{1}{2}$$
Nice question. This is related to one of my questions that bothered me in the past.
 
r9m
@Chris'ssis oh !! WOPORPAP !! fantastic :D
How did you do it ? :-) hints plz :)
 
@r9m You need to use a well-known product identity ... :D
 
r9m
which is ?! :o ... come on ... at this rate .. you are teasing me :P
 
Hullo
@Chris'ssis Given the main answer to the interview riddle, which seems totally correct, i'm starting to wonder whether the answer they gave you is the right one ...
 
1:41 PM
@r9m LOLL I missed that $1/k$.
 
Hi there. How to solve it $\displaystyle\int_0^{\infty} x^2 e^{-2x^2}\, dx$ without Gama function. Maybe to observe some parametric integral?
 
@Hippalectryon Maybe they didn't want me there ...
 
@Chris'ssis the latest point is $(36)$
 
@Hippalectryon Yeah, my bad ...
 
r9m
@Chris'ssis okay .. thanks :D .. I figured i'd be 35 :-) .. but that is with products :-) .. I have sums :)
 
1:45 PM
@r9m That identity is simply amazing!!! I saw many professors failing to prove that.
 
@Chris'ssis How do you prove it ?
 
r9m
@Chris'ssis seems you like to put professors in uncomfortable situations :P :D
@Chris'ssis hmm ... these seem related !! :D
 
@Chris'ssis Btw, any idea in how many month your book will be published ?
 
9
Q: Prove the trigonometric identity $(35)$

Chris's sisProve that \begin{equation} \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)= \left\{ \begin{aligned} \sqrt{n} \space \space \text{for $n$ odd}\\ \\ \ 1 \space \space \text{for $n$ even}\\ \end{aligned} \right. \end{equation} I foun...

 
Thanks
 
r9m
1:52 PM
@Chris'ssis oh !! nice :-)
 
@Hippalectryon I cannot offer these details at the moment since it depends on many factors. I hope this will happen soon.
@r9m It's a marvellous identity.
 
r9m
@Chris'ssis yes .. make it happen as soon as possible :D ... we are waiting like the hungry hatchlings :P
 
@Hippalectryon I have an offer from a mathematician to publish the book along with him.
 
@Chris'ssis Ooh i see
Da book :C
 
r9m
@Chris'ssis is he going to put his work alongside with yours in that book ?!
@Hippalectryon lol
 
1:55 PM
@r9m Yeah.
 
r9m
@Chris'ssis so .. there will be 3 authors of the book ? :-)
 
@r9m It's hard to publish a book when you have no math background ...
At any rate, I might publish a book every week since I create a lot of stuff :D.
(theoretically I mean)
 
Or, publish a huge book :D
 
@Hippalectryon :D
 
r9m
@Chris'ssis :D ... you can do that practically I think :P
 
2:00 PM
:D
 
How's the lap top temperature @Hippalectryon?
 
@skullpatrol I'm not at home, and therefore not on my computer
 
2:26 PM
@r9m Yeah, that's true. However, one may come up with a very nice limit using that product.
(I wrongly read that identity)
 
r9m
@Chris'ssis okay :-)
 
@Chris'ssis Publish what book?
i.e., in which topic?
 
r9m
@BalarkaSen 300 spartan integrals and series :P
 
Haha
Great to hear that, @Chris'ssis. Congrats.
 
@BalarkaSen A book containing very nice integrals, series and limits (and solutions, of course).
@BalarkaSen Thanks.
 
2:34 PM
@r9m Algebrain' hard.
$xy = yx$ everywhere.
 
@r9m :D
 
r9m
@BalarkaSen good :D
 
@Chris'ssis Do let me know when it will publish.
@r9m Let's do some group theory, alge-bro.
 
@BalarkaSen OK. I'll do it along with a mathematician that made me this offer.
 
@Chris'ssis Do you mind if I ask who is the mathematician you are referring to?
Apparently, Romania is filled with mathematicians.
 
2:36 PM
@BalarkaSen I can only tell you that I'll do this with a pretty known mathematician.
 
@Chris'ssis I was asking about the name. But if you don't want to tell, that's fine.
 
@BalarkaSen That's why I said "I can only tell you ...".
 
Talking of mathematicians, I met with V. K. Murty a month earlier.
Great man. Knows a great deal of arithmetic geometry.
He and his brother M. R. Murty has kind of an international reputation.
Both are, of course, number theorists.
 
50
Trolling

Proposed Q&A site for this site is for any part of Trolling. For those who want to prevent trolling, design systems that discourage trolling, or for people who want to successfully troll others.

Currently in definition.

 
2:52 PM
@r9m @robjohn this seems very nice $$\int_0^{\infty} \left(\frac{\tanh(x)}{x^3}-\frac{1}{x^2\cosh^2(x)}\right) \ dx=\frac{7\zeta(3)}{\pi^2}$$
3
 
r9m
@robjohn I only partially understood the contents of the link that Steven Stadnicki linked to in the comment :| ... but getting the asymptotics for the expression with $x$ shouldn't be too difficult ... imho :-)
@Chris'ssis monstaaa!!! :O
 
@Chris'ssis Very cool!
 
@BalarkaSen Yeah, it is. :D
 
@Chris'ssis The result is especially amazing.
 
r9m
2:56 PM
@robjohn XD LOL .. what is that ? :P
 
trolls
 
trolls
Jinx.
 
r9m
oh ! so thats how them trolls luk like :D ... nice :P
 
@r9m That was what I mistakenly posted to you a while ago. I have a proof of their formula that is more understandable.
$$\sum_{k=0}^n\binom{n}{k}^px^k \sim\left[p^{-\frac12}\left(1+x^{\frac1p}\right)^{p-1}x^{\frac{1-p}{2p}}\right] (2\pi n)^{\frac{1-p}2}\left[\left(1+x^{\frac1p}\right)^p\right]^n$$
 
r9m
@robjohn OH .. !! post it plz .. when you have time !! :D
 
2:59 PM
@blue OK OK OK. OK everywhere.
 
@robjohn can you help me please
 

« first day (1443 days earlier)      last day (3580 days later) »