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1:01 PM
@TheArtist forming a reduction formula and using it is more time consuming than the IBP approach though
 
@UserX $I_n=\int_0^1 e^x(1-x)^n dx$

$I_n= [(1-x)^n e^x ]_0^1 + n\int_0^1 e^x (1-x)^{n-1}$

$I_n=-1+nI_{n-1}$

$I_4=-1+4(-1+3(-1+2(-1+1I_1)))$

You only have to find $I_1$ which is elementary, and plug that value you get :

$I_4=-12+24(-2+e)$
@UserX I don't know :) it takes less than 5 mins for me to do that :) this function is easy for reduction :)
Infact most of such functions or trig function that aren't too complex with high powers is much better done in reduction I think :)
 
If $b_1,\dots,b_n\in V$ form basis of vector space $V$, does that mean that there is no linearly independent vector $v\in V$ of $b_1,\dots,b_n$, right?
 
@UserX using it again and again? It's jus arithmetic after you form it...
@Sush how is $v$ linearly independant? Can you tell me it's properties?
 
@TheArtist, ok I got it!
 
@Sush :)
@Anastasiya-Romanova your very cute :p because of you someone else is not coming online
 
1:16 PM
Why are columns with no pivots in reduced echelon form of a matrix are linearly dependent on the columns with pivots?
 
@Sush write out the equations, it's clearly the way it is
 
@TheArtist, I can see that for particular matrix. How to generalize it?
 
@Sush because pivot columns have only that term in 1 equation (say you have four equations) so there is no way of expressing other variables in terms of a tat variable
 
@Sush Sus-hi!
 
I got a weird question
 
1:21 PM
Weirdd.
 
This Saturday I'm participating in a math contest exam, and the last question is always euclidian geometry
 
:D
 
@UserX what is it called? :)
 
But I haven't taken euclidian geometry for about 2 years and only remember trivial theorems that help me in complex numbers and functions
 
@Sawarnik how are you ? Still angry? :p
 
1:22 PM
@TheArtist Thales
 
@TheArtist Yes.
@UserX So what is the question?
 
@Sawarnik comeon bro :) chill :)
 
Greetings noble people around the world!!!
 
Can I say "Without loss of generality, assume that "$blabla$ is in a coordinate system" and use analytic geometry?
 
@UserX Hello Chris's sister
 
1:23 PM
$$\LARGE{\text{I'm terribly happy!!! :-)}}$$
 
@UserX Why not?
@Chris'ssis Hmm.
 
@Chris'ssis $$\LARGE{\text{That is wondrous!!!!}}$$
 
Is it possible to prove every result from euclidian geometry through analytic geometry?
 
@Chris'ssis greetings to the awesome one who asks questions on math stack, and when people answer using paragraphs and paragraphs of working....at the end the one who answers it using the least amount of steps and space :p
 
Also, to what extent can I assume stuff to assign them to coordinates?
 
1:25 PM
@TheArtist :D I computed 2 hardcore integrals by elementary tools only!!!
 
Chris'ssis which ones?
 
@Chris'ssis you are too awesome :)
 
@UserX and the version with minus in denominator
 
I'm still fighting with the $\int_0^{\infty} \frac{x}{\sqrt{e^x-1}} dx$ for about 2 days now
High-school methods only
 
1:27 PM
@UserX As long as it preserves generality.
Ok gtg..
 
@TheArtist :D
 
@Sawarnik Cya friend
 
@TheArtist, thanks for reply, but if you can elaborate it little, please!
@Sawarnik, hi.
 
@Chris'ssis is there a trick to evaluate $\int_0^{\infty} \tan^{-1} (\sqrt{e^x-1}) dx$?
 
1:39 PM
@user130018 Are you here?
 
@UserX Can you use $\sqrt{e^x-1}=\tan(y)$?
 
@TheArtist, oh, got it!
thank you so much.
 
@Sush I know you would :) after looking carefully for sometime ...UR welcome
 
:D
 
@UserX WAIT .... that integral does not converge ...
 
1:42 PM
Agreed
 
@Chris'ssis what integral book do you advice for high school finishers ????
 
Not convergent on $[0,\infty)$
 
@Chris'ssis to become really good I mean :)
 
Damnn
I assumed convergence
 
@TheArtist Have you seen this one? springer.com/mathematics/analysis/book/978-1-4614-6761-8 Try to get that and study it in small details again and again. Besides that, on MSE there are lots of qurstions from where you can learn a lot of very precious things.
 
1:44 PM
@UserX do you know to integrate $\int \frac{ln(x^2+1)}{x(x^2+1)}dx$
@Chris'ssis thank you :) i will check it out :)
 
@TheArtist I also recommend you this one
It was published this year, but it's not bad to study the very old books too. In fact it's a very good idea.
These books have very nice questions with very detailed solutions such that you can learn a lot.
 
@TheArtist Find it's antiderivative? No single idea. Integrate over positives? Probably yes, I think I've seen it before
 
@Sush After a long time!
 
@Yes!
 
How is life now?
 
1:49 PM
@JasperLoy Why seek out 130018 but not me </3
 
@UserX If you know that then that's a hint for you to solve the 2 day problem you had :)
 
@Committingtoachallenge Currently, I have a closer relationship with him than you.
 
@JasperLoy Oh, I thought he was new, when did he start coming here?
 
@Committingtoachallenge He has changed his usernames many times.
 
@JasperLoy What was the precursor name?
 
1:50 PM
@TheArtist does it use high-school methods?
 
@Committingtoachallenge Can't recall.
 
Probably Bart?
 
@UserX I got really attracted to your unsolvable integral :) note $x$ is not the $x$ in your question, it's $u$ , I did a subsitution
 
@JasperLoy I will win your heart
 
@Committingtoachallenge Both of them had tough childhood, I think.
 
1:51 PM
@Committingtoachallenge Focus on your girlfriend instead.
 
Otherwise I could try poisseaux series and piecewise integration I think. I'm restricted to high-school methods.
 
@UserX Yep :) bro im a high school finsiher too :)
 
I had a very tough childhood as well if it wins points, I was out on my own at 15
Now his heart is mine
 
@TheArtist Anyway, I gotta leave to eat and go attend an optics and a math tutoring class, so if you have any "eureka" moments or hints, tag me and I'll see them later.
 
@UserX yes it's high school :), subsitutions ,maybe by parts , etc etc
@Chris'ssis appreciate it :) Thank you :)
@UserX goodbye :)
 
1:54 PM
@TheArtist Cya artsy
 
@Committingtoachallenge im not going :p
 
@TheArtist Believe in yourself entirely, work extremely hard, be persistent, never give up, and then you reach the sky. ;)
 
@Committingtoachallenge im not going :p
 
@TheArtist Hey artsy
 
@Chris'ssis I will get the first book first ;) must not get more than one at one time :p I tend to then check both books at curiosity and not finish none :/
 
1:56 PM
@TheArtist OK
 
@Committingtoachallenge hey :D
@Chris'ssis soemtimes it's hard to believe in yourself :p
@Chris'ssis thanks for the advice :)
@Chris'ssis and for the book ;) :)
@Committingtoachallenge you know I stalked some profiles today of top users, there was this guy on the top list...he has put that he will deactivate it :p is there anyway to get hold of his account? :p why deelete wen you can give off to others :/
 
@TheArtist Haha yeah you can take it, just run a rainbow table hash hack for a few days
 
@TheArtist, I think I couldn't understand! $$
\begin{matrix}
0 & 0 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0\\
\end{matrix}
$$ is echelon matrix and third column has no pivot, but it is not linear dependent on first two columns!
 
@Committingtoachallenge nono I don't want to steal.. If they have collected 70k reputation, why just delete it. Give it off to someone else who wants :p
 
2:01 PM
@TheArtist Ask for it all in bounties xD
 
@Committingtoachallenge why not give the account with the badges :p if they don't want....and v can share it amoung ourselves by bounties
 
They usually come back
 
@Sush the first two columns are dependant on the third , the third column contains free variables
@Sush if you think I didn't get your question properly :) please rephrase it :) because im not 100% sure what you mean either
@Committingtoachallenge but he's finding a way to deactivate it :p , if one can contact him and tell okay give me the password we will deactivate and then change the pro pic and name :p
 
@TheArtist, ok!
 
2:05 PM
@Committingtoachallenge he won't be able to find his account, so that's like deactivating :p
@Sush , it's always better to see by writing the equations after you get the rref
 
He wouldn't possibly fall for that xD. Anyway I must go now, three questions to go and I can sleep(12:06AM) here now
 
@Alizter Ok +1
 
@Committingtoachallenge oh ok :) goodnyt
 
@TheArtist nyt
 
Best of luck friends
 
2:07 PM
@TheArtist, what is rref?
 
Row reduced echelon form
 
Hi @Commit
 
@Sush il be glad to help if you can make it more clearer.....RREF means the short form Ya committing asnswered it :)
 
@Sush How is life?
 
@Sawarnik im not going anywhere , why are you telling night to me ? :p
 
2:08 PM
@Sawarnik, very pleasent.
 
@Sush oh why? :)
 
@Sawarnik, I don't know!
 
@Sawarnik that was mean :p
 
@Sawarnik, (removed)
 
@Sawarnik ok goodnyt :) im going :)
 
2:12 PM
@TheArtist yes it was.
@Sush Ok! You didn't tell me the name of your college :P :|
 
@Sawarnik are you going to tell me goodnight or what? :p
@Sawarnik chill bro relax...I was just joking that day :) sorry bro :) I apologize :) from that time I was being nice Ryt?
 
ryt :)
 
Ok gtg @Sawarnik goodnyt :) bye cya
 
A new very nice question I just posted!!!
1
Q: Evaluating by real methods $\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$

Chris's sisI'm sure you guys can briefly get the result by some methods of complex analysis, but now I'm only interested in real analysis methods of proving the result. What would you propose for that? $$\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$$$$=\frac{\pi^6 \sqrt{2}}{768}+\frac{5 \sqrt{2}\pi^4}{64}\op...

 
Bye.
 
2:17 PM
@TheArtist partial fractions would be the first step
 
@Chris'ssis Can you check this limit? $$\lim_{n\to 0}\left(\frac{1+(-1)^n}{n+1}\right)^{1/n}=e^{-1}$$
 
@Alizter that notation is horrible
 
@Chris'ssis How about now?
 
@Alizter The limit doesn't smell well ... are you sure it is correct?
 
@Chris'ssis That is why I am asking you/
My idea was to first start with turning it into a limit to infinity
then I can split that into $e^{-1}$ and $(1+(-1)^{1/n})^n$
which the latter I compute to be 1
as $n\to \infty$ of course
 
2:23 PM
$$\lim_{n\to 0}\left(\frac{1}{n+1}\right)^{1/n}=e^{-1}$$
 
hmm
oh
there is a factor of a half floating around that I forgot to put in
Which means the geometric mean of $[-1,1]$ is the same as the geometric mean of $[0, 1]$
wat
 
a floating half is better than a hole in the graph :P
 
Hey @Alizter!
 
@Chris'ssis can you suggest any book which you have or read or consider good for the types of integration you posted just before or for the related topic
i have no idea about Li, Ei, and other special functions, which I am interested in
 
2:39 PM
@Aditya Major part of the integrals and series I posted you won't find in books (especially the last ones), but for a start you might check this one springer.com/mathematics/analysis/book/978-1-4614-6761-8
2
and this one
 
why is the first two columns are dependent on the third of matrix $\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\\ \end{matrix}$?
 
@Aditya there you have lots of very nice such questions with very nice detalied solutions. In general the questions I post come from my research.
 
next year i'm going to learn bessel functions
 
Hi @JasperLoy
 
@JasperLoy Oh hey
 
2:50 PM
@IceBoy nope, never read that.
for introductory elementary-level survey on mathematics, i'd actually refer to Courant-Robbins.
@Alizter
 
@BalarkaSen
 
@Alizter To the NT chat.
 
dan ananana daaan
 
3:09 PM
I also added a supplimentary question that is similar to the one of Ron Gordon here
In the beginning he says "This integral is deviously difficult.". However, the question on main seems way more advanced ...
 
@robjohn partial fractions ? For?
 
1 hour ago, by The Artist
@UserX do you know to integrate $\int \frac{ln(x^2+1)}{x(x^2+1)}dx$
 
@TheArtist $\frac1{x(x^2+1)}=\frac1x-\frac{x}{x^2+1}$
 
@robjohn ohhh yes now I got what Robbin is referring to :)
@robjohn how does one enable latex on chat ? :/ I have to go to "ask a question" on math stack and paste codes here to view them
 
@TheArtist when you hover over someone else's comment in chat, do you see the bent arrow at the right side?
 
3:16 PM
@TheArtist here
 
@robjohn yes I can see :)
@robjohn when I click that then the box gets framed sort of
 
@TheArtist If you click on the arrow on one of those lines, it will make your comment in response to that one... it will put an arrow to the left of your comment which can be used to find out to which comment yours refers. Furthermore, you don't need to type the user's name; it is put there for you.
 
@IceBoy my net is too slow Ryt now for links :/
@robjohn yes got that :) how to see latex? Any idea? :)
 
3 mins ago, by Ice Boy
@TheArtist here
 
@TheArtist ChatJax installation <- That link is not going to slow you down.
 
3:20 PM
:D
 
@robjohn do you think this is correct? :)
@IceBoy how are you :)
@robjohn thank you :) :)
 
@TheArtist It's correct.
 
@TheArtist It is...
@TheArtist the links make it way easier to know what is going on during multiple conversations.
 
@robjohn no I mean the question had $log (x^2+1) $ on the numerator, so you have left those out as 1?
 
@TheArtist No, I just rewrote part of the integrand. It should help, if there is a nice closed form
 
3:30 PM
@robjohn oh yes :) should see if I can integrate now :)
 
@TheArtist I believe the integral is $-\frac14\log(1+x^2)^2-\frac12\mathrm{Li}_2(-x^2)$
 
@robjohn how did you do this so fast? :o
@robjohn ohhh what is $Li$ I don't thihk I can do this :/ since I haven't learnt what that is :/
 
@TheArtist It is a special function, which means that your integral cannot be written with the usual functions $$\mathrm{Li}_{\color{#C00000}{2}}(x) =\sum_{k=1}^\infty\frac{x^k}{k^{\color{#C00000}{2}}}$$
 
@robjohn oh ok :/
@robjohn thanks :)
@robjohn do u think math exams is a good test to check the math knowledge?
 
@TheArtist It can check whether you know the essentials from a given course.
 
3:39 PM
@robjohn OHHH the 2 is related by that :) thanks for highlighting
 
@TheArtist yeah, you can replace the $2$ with other numbers to get other functions.
 
@robjohn yes but competition for perfect scores is bad right?
@robjohn Yep :)
 
@TheArtist I guess so, but are you using them to test your knowledge, or are you talking about the grades from a course reflecting general math ability?
 
@robjohn Could you read this?
I don't understand why the question was closed, it seems perfectly reasonable for the site, according to the rules given in the help center.
 
@robjohn I know its good for a teacher to check if students know the material....but using it as a measurement of your knowledge and intelligence and math ability is wrong right?
 
3:44 PM
@robjohn did you see my last question on main? It's too nice ...
 
@Vÿska I added a comment. I agree that it would be better asked on another site.
 
so weird
this room doesn't let me leave from my PC being off
 
@ZachSaucier it will after a while or if there is some reason for it to check if you are still there (someone trying to ping you for instance)
 
but it hasn't for the past couple of days
I'm a regular in several chats but this one always shows my avatar greyed out when I come back
 
@ZachSaucier perhaps no one has tried to talk to you and so the room has assumed you are still here.
 
3:49 PM
the rest have me at the front of the que because I'm rejoining
maybe so
 
@Chris'ssis Hello !
 
@Hippalectryon Hi
 
@Chris'ssis Have you ever dreamed of an interesting integral, then solved it after waking up ? :D
 
@Hippalectryon Often (just the part with dreaming integrals). However, there is something interesting in my dreams, there I'm like a super human, the way I talk, the way I move, the way I solve questions, my creativity is beyond measure, but then I wake up ... (and I lose all)
 
@Chris'ssis That's alcohol, not sleep >.>
xD
 
3:55 PM
@Hippalectryon lolllllllll
 
@Chris'ssis Anyway, I had one of those weird dreams this morning
No integrals, though. Recursive functions.
 
@Hippalectryon I don't drink alcohol at all, I don't smoke either. ;)
 
@Chris'ssis Me neither, I'm underage anyway :)
Water and fruit juice is way enough
And cheaper
 
cheap is good :D
 
@Hippalectryon Ah, I was about to forget my water. I go to the store ... (back in 20 min or so)
 
3:58 PM
:-)
 
Has any algorithm that can calculate the closed form of all the roots of a polynomial, failing only if there is no closed form, been found?
 
@Chris'ssis I will take a look. I liked the question yesterday. I think I would have gotten the answer had I gotten home a bit earlier.
 
@Chris'ssis You buy water?
@Hippalectryon Great. Tell to UserX to follow that :D
 
@robjohn Yeah, it's a very nice question.
 
@Sawarnik :D
 
4:00 PM
@Sawarnik Yeap, it's cleaner the one I buy than the one I have at home.
 
@Hippalectryon Not really .. its dreams :D :D
@Chris'ssis Why don't you install a filter?
 
@Sawarnik I don't rely on filters either. I like the water brought from mountains. :-)
 
And how do you trust that water in the store is from mountains?
 
@Sawarnik It's a known producer here, often checked, verified.
 
@Sawarnik He personally knows one of the mining dwarves :D
 
4:04 PM
I'm away.
 
Everybody knows that water is pumped in the moutains by dwarves
 
@Chris'ssis I don't like questions where the answer has a bunch of special functions in it. $\mathrm{Li}_n$ is not as bad as some, but it still leaves a bad taste.
@Chris'ssis I spent what time I had before I left in trying to write out what you were asking. There are only a few ways I know to approach things like that complicated sum of quotients.
 
@robjohn Do you remember that sum I showed you? Did you solve it?
 
@Alizter which sum?
 
Oct 21 at 21:00, by Alizter
@Chris'ssis $$\sum_{\substack{k_1,\ \cdots,\ k_n\\ k_1\ne\ \cdots\ \ne k_n}}\frac{1}{m^{k_1+\cdots+k_n}}$$
where $k_i \in \Bbb N_0$
uber symmetric
@robjohn Do you remember?
 
4:24 PM
Back.
@robjohn Trust me, it's not as bad as it seems to be. My research shows that the most terrible things seems to be a piece of cake when approaching them properly.
 
I love this site :D
2
 
@TheArtist why?
 
@Alizter why not?
 
@Alizter why not :D
 
@Alizter Yes... It involves counting how many ways you can sum $n$ distinct positive integers to a given number.
 
4:34 PM
@IceBoy exactly
Name any site better than this ;)
 
The library
 
@IceBoy website *
 
WorldCat
 
@IceBoy :p
 
:D
 
4:35 PM
This is the best :D
 
@robjohn and then those weight the geometric series?
 
@Alizter yes.
 
4:49 PM
@robjohn @Chris'ssis What title should I give to the question ? I'm out of ideas.. gyazo.com/8faf06515c6b86b1a47f89b4827cbca9
 
@Hippalectryon A question from the dreams realm
@robjohn see @Anastasiya-Romanova, this is the way to go
 
@IceBoy :c
 
@Hippalectryon just kidding pal c:
we all have dreams...
 
@IceBoy xD
@Chris'ssis Do you really think that could work as a title ?
 
@Hippalectryon Sure, 100%. You say the truth! :-))))))))
 
4:56 PM
I hope I'm not asking too many questions in one question though :/
People have told me this on some previous questions
 
@Chris'ssis yes. That is almost exactly the way I was approaching it. I guess I will stop. Too many interruptions these days. :-(
@Chris'ssis I was solving the recurrence $a_n=6a_{n-1}-a_{n-2}$ which has the characteristic roots $3\pm\sqrt8$
 
@Hippalectryon upvoted :)
 
@TheArtist Thanks :)
 
Out for most of the rest of the day... BBML
 
later
 
5:05 PM
@Hippalectryon UR welcome :)
@robjohn BBML???
 
be back much later
 
Be back, my lady
:c
ok i'm out
 
@IceBoy ok :D
 
@robjohn OK
 
Damn. Got to learn basic number theory&all HS euclidian geometry in 1 day and 3 hours
Just learnt that the contest I'll participate in has exercises even from stuff we don't get taught at school but are in the book
Tips?
 
5:13 PM
read the book :P
 
300 pages lol
 
speed read the book
 
There is no single chance I'll be able to remember more than 5 theorems this way
 
which book?
 
School book
 
5:19 PM
@UserX open up the books and read the theory. understand them thoroughly. read the theorems, memorize them but not as facts but as problems. exercises. then sleep on them. you'll figure why they are true.
 
@UserX Which contest?
 
@BalarkaSen Time!
 
@Chris'ssis Thanks, I will look through it in a sec!
 
@Sawarnik is enough
 
5:20 PM
@BalarkaSen sleeping on them is the more likely scenario
 
@rehband learn, learn, learn ... ;)
 
@Chris'ssis how would you cram an insane amount of material in a day?
 
@UserX Just speed read.
It works.
 
no that works if you are digesting stuff
math should be understood, not gulped
 
But if circumstances are not such...
 
5:23 PM
@UserX Hard to say ... I'm used to attend an insane amount of material every day, but people that do not do this often might have a problem ...
 
@Sawarnik i have done it, kiddo.
all of fundamental galois thy in 2 days
 
Ah. Hmm. Hmmmmm.
 
Step 1. Get off the internet and start studying.
7
 
Which contest, btw?
 
@robjohn I also have in mind an idea to compute this one by real analysis (in a very nice way)
 
5:32 PM
Bye people, bye world for a day. If I got questions I'll ask here. I found notes on the contest material that are about 100 pages.
 
@Chris'ssis Mike didn't like it :) math.stackexchange.com/questions/998535/…
 
Apparently, I gotta learn discrete math too
 
Good luck :D
 
@Hippalectryon lolll
 
@UserX Bye :D
 
5:35 PM
I'm afraid that one day there would be no challange in terms of integrals, series and limits ... or too rare :-(
 
@Hippalectryon such nonsense
 
@Chris'ssis then your challenge will be to teach others :)
2
 
@Chris'ssis So true! I learn all day every day...but these days mostly Lebesgue theory and topology :P
 
@IceBoy That sounds terribly nice. :-)
 
@BalarkaSen -__-
 
5:44 PM
@rehband Good ;)
 
@Chris'ssis I just saw u recommended this book springer.com/physics/…
Never heard of that one before, gonna see if i can find it at the library
 
@Chris'ssis Of course, I own that one
 
6:24 PM
Ah, yeah, there one may use the Binet's log formula ...
 
6:49 PM
The Physics chatroom is called "The h Bar"
 
Hello, i try to ask a question but i can not find the formatting help to write prosper math symbols.
 
hmmmm
This answer here seems a bit too complicated ...
5
Q: A closed form for the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$

Olivier OloaI have found a closed form for the following new series involving non-linear harmonic numbers. Proposition. $$\sum_{n=1}^{\infty} \dfrac{H_n^2-(\gamma + \ln n)^2}{n} = \dfrac{5}{3}\zeta(3)-\dfrac{2}{3}\gamma^3-2\gamma \gamma_{1}-\gamma_{2} $$ where \begin{align} & H_{n}: =\sum_{k=1}...

 
ah,.... $\sum etc... is it LaTeX ?
 
@Johannes here
 
@Ice Boy thx
 
6:58 PM
:D
 

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