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4:00 PM
@WillHunting +1!!!!! TU!!!! Sweet comment!!!!
 
Could one have a dev to look at the database whether the voting patterns are sufficiently suspicious?
 
@DanielFischer They are suspicious, but like I said, they can be considered "genuine", whatever that means.
What is there to say I cannot upvote a whole bunch of another's answers because I like them?
 
@WillHunting Maybe. But if the proportion of their votes on each others answers is high enough, that could still be judged inappropriate, even though it doesn't trigger the automatic script.
 
@DanielFischer Ah, I see. In other words, the script should be extended using Hahn-Banach, lol.
I am sorry I gave the wrong movie title yesterday. The movie you should watch is "If I Stay".
 
4:17 PM
@WillHunting And this is your account number?
 
howdy Jasper, @DanielF
 
howdy
 
Hi @Ted.
 
Sup.
 
ah, hi @skull, @Khallil
 
4:27 PM
Inf.
 
liminf
 
@Khallil hello
limsup
 
on break between office hours 1 and office hours 2 & 3 before I go to the cardiologist
 
Hey, @Balarka and @Ted.
 
@Khallil Hey Khalil
 
4:28 PM
Good luck at the cardiologist, @Ted.
 
thanks, @Khallil
 
Hey, @rehband. ^_^
 
guten Tag, @rehband
 
@TedShifrin Guten Tag!
 
@Ted I hope it's only a routine check, and nothing acute.
 
4:29 PM
well, I dunno ... I have had two major heart surgeries, but a symptom has me worried
 
@rehband night is "nikht" in german, right?
 
Nacht
 
or is it "nisht"?
@TedShifrin Ah
 
Ted is right
Do you speak German, Ted?
 
You have stopped ignoring me, @TedShifrin?
 
4:30 PM
Ich hab' deutsch 3 Jahrelang studiert ...
 
Wow
 
If you stay on good behavior, @Balarka
 
Bevor du Mathe studiert hast??
 
@TedShifrin I will do so.
 
please do
 
4:31 PM
nein, nein, auf der Universität
 
Hello, by the way. Long time.
 
Achso, in Deutschland?
 
nein, nein, bei MIT
 
Haha ok
Sehr gut
 
Aber DanielF hat mich korrigiert ... also red'ich nicht so viel auf deutsch :P Ich hab' zu viel vergessen!
 
4:33 PM
Irgendein Deutsch was du gelernt hast hast du vergessen?? Ich bin schockiert!
 
@skull: To whom was that directed? :)
schockiert ... ja, sicher.
 
@TedShifrin Balarka
 
Hahaha'
 
@ParthKohli What do you mean by account number?
 
@WillHunting No, your credit card number and pass
:P
 
4:34 PM
haha @Hippa
 
@Hippalectryon It's actually my mum's card, I don't have one. I am just a jobless lunatic.
 
-____-
Damn you adults :P
 
méchant @Hippa !
 
@TedShifrin Parlez vous francais aussi?
C'est fou
 
Oui, je me suis spécialisé en français, aussi bien qu'aux maths ...
Some of us in the US can speak several languages ... it's not just the Europeans.
 
4:37 PM
Tu es une légende
 
plus programming languages, do they count?
 
I am thinking of getting "French in 30 days" to work through in November and "German in 30 days" to work through in December...
 
Do it, Jasper.
Stop deliberating.
Just do it.
 
@DanielF ... I suppose I'll quit answering homework questions for people with no rep. Someone asked a cute diff geo question, which I've answered, to no response from the OP in several days.
 
^_^
 
4:38 PM
damn. Filled my mouth with caramel now I can't move my jaw
 
@TedShifrin Why don't you try answering my question ? :P
 
Jasper: I doubt you can do much in 30 days. Better to solidify one language before you move on to the next.
 
@TedShifrin OK.
 
I don't see how to do it, @Hippa. I thought robjohn was working on it.
 
@ParthKohli Are you still here?
 
4:39 PM
Cool @Alizter ... It'll keep you quiet for a day.
 
What's the question, @Hippa?
 
@TedShifrin Did you even look at the in between questions ?
 
@TedShifrin I can still type :)
 
2
Q: Limit of the sum of $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t$

HippalectryonLet $f$ be a continuous, decreasing function, with $\displaystyle\lim_{x\rightarrow\infty}f(x)=0$. Let $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t,\displaystyle x>0$. Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x)$, suppose that $\displaystyle\lim_{x\rightarrow0} xf(x)=A...

 
Some, @Hippa. I'm quite busy these days.
 
4:40 PM
Ok :)
 
@TedShifrin Have you seen the ring-theoretic version of the problem Mike gave me/
 
NO.
 
Are you interested/can spare some time?
 
Not really ...
 
eh, OK then.
 
4:45 PM
@robjohn i just want to know if a solution wich is a minimum is bounded ?
 
@Vrouvrou I don't see a reason why, without some other knowledge.
 
hi, @robjohn
 
@TedShifrin hey there... how goes?
 
@TedShifrin You have any teach yourself packages to recommend to me for French and German?
 
so far so good ... we'll see in a few hours
 
4:49 PM
@robjohn Are you still working on my problem ? (or should I ask smone else/set a bounty N)
 
Nope, Jasper, sorry.
 
@robjohn i have also that $u(0)=u(+\infty)=0$
 
@WillHunting Hi.
I was asking you if you'd kept count of the number of accounts...
 
@ParthKohli I forgot your email address. Could you send it to my new email, jasperloy at outlook dot com?
 
@robjohn ?
 
4:57 PM
@ParthKohli Also, what is Sawarnik's last name and email address? You can send it to me too?
@ParthKohli Nope, about 10.
 
@Will It's Sawarnik Kaushal.
 
@BalarkaSen His email is? I forgot, he told me before.
 
I forgot it.
 
Hehe, never mind, I will ask him again next time.
 
hi :)
 
5:03 PM
Hi Rachel.
 
@WillHunting Hi.
 
@math101 Have you graduated?
 
Rach!
 
lmao nooo
 
@WillHunting revenge?
 
5:04 PM
In touch with Bah?
 
@BalarkaSen What do you mean?
 
@ParthKohli Not really
 
Me neither.
 
@Will slowly walks away whistling
 
@WillHunting Got a question Why are the eigen vectors correponding to different eigen values orthogonal to each other?
 
5:05 PM
@BalarkaSen Ha, I have no idea what you mean.
@math101 Sorry, I forgot all my math, lol.
 
@WillHunting click the ping.
it's a link.
 
@math101 Are they really? Isn't this only true for normal operators?
 
@WillHunting I feel the same way... been idling for too long
@rehband idk I was suprised tooo
 
@math101 Are you still interested in going into finance?
 
@math101 I mean, a vector space need not even have a scalar product
 
5:09 PM
@WillHunting Yes its still in the plan
 
In mathematics, particularly linear algebra and functional analysis, the spectral theorem is any of a number of results about linear operators or about matrices. In broad terms the spectral theorem provides conditions under which an operator or a matrix can be diagonalized (that is, represented as a diagonal matrix in some basis). This concept of diagonalization is relatively straightforward for operators on finite-dimensional spaces, but requires some modification for operators on infinite-dimensional spaces. In general, the spectral theorem identifies a class of linear operators that can ...
 
@math101 Cool. It is still in my plan to get well and go to grad school, but I don't know exactly when.
 
@rehband kkkk lemme look at that
 
@math101 A vector space endomorphism is self adjoint $\iff$ There exists an orthonormal basis of eigenvectors of this endomorphism
 
Heya @Vibhav
 
5:12 PM
hi @BalarkaSen
Im facing some trouble with $\sum_{k=1}^{\infty}\frac{2^nn!}{n^n}$
 
@rehband lol didnt process that :P Lemme delve into reading abt this topic
 
:D
 
@VibhavPant $2^n n!/n^n$ goes to $0$ very fast so I'd expect convergence.
how about Stirling?
 
@BalarkaSen but my book tells me that convergence implies the n'th term goes to 0
 
yes, it does.
so?
 
5:16 PM
@Hippalectryon I am still working on it, I had some downtime last night
 
not the opposite
 
Thanks @robjohn
 
I tried the ratio test on it
 
@VibhavPant no.
not the other way around.
take $\sum_{k = 1}^\infty 1/k$
 
5:17 PM
@VibhavPant I meant that $2^n n!/n^n$ decreases to $0$ "very fast", so one would "expect" convergence
@VibhavPant Are you familiar with Stirling's formula?
 
I see
yep
 
apply it.
 
So do I use use $\left(\frac{n}{e}\right)^n$ for $n!$?
 
that's an underestimation.
it's $O(\sqrt{n}) \cdot (n/e)^n$
 
so $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$?
yeah
 
5:22 PM
yep
you'll get something like $O(n \cdot (2/e)^n)$ and $2/e < 1$
 
However, I dont know how to manipulate $O$ notation
 
well, then, take $\sqrt{2\pi} \cdot n \cdot (2/e)^n$
 
This can be solved with the ratio test as well I think
 
202
A: Are there real-life relations which are symmetric and reflexive but not transitive?

amWhy $\quad\quad x\;$ has slept with $\;y$ ${}{}{}{}{}$

This stupid answer should be downvoted because it is not even clear what "x has slept with y" means.
 
Well
The n-th root should converge to $\frac{2}{e}$ I think
 
5:26 PM
@rehband what should?
oh that sum
 
Nono
 
Yet it got over 200 upvotes. People, this is ridiculous.
 
you are using root test then?
 
@BalarkaSen 1 sec
 
@WillHunting Sex sells, you know.
 
5:28 PM
@DanielFischer It just got another upvote, sad...
 
$$\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \lim_{n\to\infty} \frac{n^n}{(n+1)^n} = 1/e$$
So applying root test to the problem above should yield $2/e$
 
Here's what I tried: $a_n = \frac{2^nn!}{n^n} \sim \frac{2^n\sqrt{2\pi n}(n/e)^n}{n^n} = \frac{2^n\sqrt{2\pi n}}{e^n}$ which goes to zero as $n \to \infty$
 
@DanielFischer only to those who are willing to pay for it :-)
 
Since $\frac{2}{e}<1$
 
@VibhavPant that isn't sufficient for convergence
 
5:31 PM
oh yes
:|
 
@VibhavPant Take the n-th root and see what the limit is. If it is smaller than 1 we have (absolute) convergence
 
thunks @rehband to hell with your root tests
=P
 
@BalarkaSen No silly approximations needed! Only algebraic manipulations
=)
 
stirling is not an approximation
it's asymptotics.
 
Stirling is not an approximation -- It's a lifestyle
 
5:33 PM
the nth root $\sim (2\pi n)^{1/2n}\cdot2/e$
 
@DanielFischer I hate my enemy so much that there should be a feature for me to not see any of her posts on the site.
 
@WillHunting Don't look. It's called ignore that which bugs you.
 
@VibhavPant Indeed. Do you know what this converges to?
 
@rehband I think 0, but Im sure its less than 1
 
$\lim_{n \to \infty} n^{1/n}$ is what you need to calculate @VibhavPant
 
5:35 PM
$1$?
 
yes
 
@IceBoy She emailed me to ask me to upvote her in secret, which is clearly cheating. She left stupid comments on my post and made stupid edits on them just to get my attention in the hope that I will upvote her, and then scolded me when I told her to stop. She also plagiarised other people's answers and then accused them of plagiarism. I could go on and on about how evil she is.
 
@VibhavPant $\sqrt[n]{n^k} \to 1$ as $n \to \infty$
 
So we are left with the terms $(2\pi)^{1/2n}\cdot2/e$, right?
 
no $1/n$
 
5:37 PM
@WillHunting If you need additional flags, just tell us :)
 
@WillHunting I remember all that stuff, but like I said if it bothers you don't pay any attention to it.
 
@VibhavPant We are left with $$\sqrt[n]{ \sqrt{2 \pi n} } \cdot 2/e$$
 
$\lim_{n \to \infty} c^{1/n} = 1$ hence we have $2/e$ @VibhavPant
 
@Hippalectryon let's not get into the same "gang up on" mentality that these people are into :-)
@WillHunting It is just some person on the internet
 
@IceBoy I just wanna know who it is :)
 
5:40 PM
why care?
 
@IceBoy a person infected by severe form of RH
 
will it help you in math?
 
@Hippalectryon It is the user with the 5th highest rep.
 
@DanielFischer It only took me like 2 weeks to realize that the logarithm of the gamma function and the log gamma function have different principal branches. Suddenly things make so much more sense. Someone should have told me sooner. :)
 
@BalarkaSen RH?
 
5:42 PM
Reputation Hysteria
@RandomVariable eh, what's the difference?
 
@rehband @BalarkaSen thanks
 
@BalarkaSen well said +1
 
I think I have said enough today. I will try not to mention my enemy again in this chat.
 
@WillHunting It was enough, thanks
 
5:44 PM
An open problem in neuroscience is to determine why so much users in MSE are affected by RH.
 
aka Reputation Obsession
look what it did to Arturo M
it only leads to "burn out"
 
@RandomVariable Weird.
 
Is there anyone in this chat who successfully learned a foreign language all by himself?
 
Who can tell me briefly a nice series expansion of $$\arctan\left(\frac{\sin(2x)}{1-\cos(2x)}\right)$$?
 
@Chris'ssis What's $\arctan$'s expansion ?
 
5:47 PM
@Hippalectryon I have something else in mind.
 
@Chris'ssis Expansion around what point ?
$x=0$ ?
 
@RandomVariable wat
 
@Chris'ssis $\arctan\left(\frac{\sin(2x)}{1-\cos(2x)}\right)=_{x\rightarrow0}\frac{\pi}{2} - x + \mathcal{O}\left(x^{6}\right)$
 
@Alizter the series expansion has different branch structures.
it's news to me though.
 
@Hippalectryon Something in Ramanujan's fashion I'd prefer ...
 
5:50 PM
@BalarkaSen
 
@Chris'ssis Uh ? mine is kinda simple -__-
 
@Alizter Are you familiar with branch cut/points?
 
@BalarkaSen yes yes. it is suprising thats all
I successfully taught differentiation to 14 year olds today.
 
yes, it's surprising.
 
rather than having them wait 3 years for good math
 
5:52 PM
@Alizter excellent. i was teaching them calc a few weeks ago in here.
 
@BalarkaSen in person
to about 10 of them
 
in person
my classmates, actually
 
in here?
 
oof 10
@Alizter no, no, in my school
 
ah
 
5:53 PM
@BalarkaSen Stirling's approximation only applies to the latter.
 
@RandomVariable That must make sense. Steepest descent works only for very well-behaved functions, no?
 
Or we may write it as $\arctan(\cot(x))$
 
@RandomVariable It's rarely nice what you get by applying a branch of the logarithm to the values of a function, unless the entire image lies in the domain of the branch. So one takes a simply-connected part of the domain on the function where it is non-zero and holomorphic, then you get a nicer function. Or one looks at the entire Riemann surface of $\log f$.
 
@Chris'ssis Uh cot :/
I never use $\cot$
 
@Hippalectryon Are you against this function?
 
5:56 PM
@Chris'ssis We don't use it a lot in France that's all
It's not that I don't like it
 
lol they don't do a lot of trig in france, you know?
 
I just answered 2 lhf.
 
"ENS students who have sat through courses on differential and algebraic geometry (read by respected mathematicians) turned out be acquainted neither with the Riemann surface of an elliptic curve y2 = x3 + ax + b nor, in fact, with the topological classification of surfaces (not even mentioning elliptic integrals of first kind and the group property of an elliptic curve, that is, the Euler-Abel addition theorem). They were only taught Hodge structures and Jacobi varieties!"
 
How would you explain colour to somebody who sees in black and white?
 
5:57 PM
I don't believe that true...?
@Alizter One can't.
 
use colorful language
 
Does he have a taste?
 
or any other sensory perceptions?
 
@BalarkaSen Those who know the classification of surfaces may not know the classification of curves topologically, lol.
 
Hm I was under the impression that classification of surfaces usually meant classification of topological manifolds?
Not sure though, never studied them.
 
6:04 PM
@BalarkaSen That is what I meant.
@BalarkaSen That is why we need Lee's Topological Manifolds.
 
@DanielFischer I was struggling to understand why those results in that paper about contour integration were correct even though the author erroneously asserted that the log gamma function has no branch points. Now it makes sense.
 
@Hippalectryon one of the series representations $$\sum_{n=1}^{\infty} \frac{\sin(2 n x)}{n}=\arctan\left(\frac{\sin(2x)}{1-\cos(2x)}\right)$$
 
@RandomVariable Well. It has branch points. One just can place a simple branch cut through them.
 
@Chris'ssis $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} \sin{\left (2 n x \right )}=\sum_{n=1}^{\infty} \frac{2 \tan{\left (\frac{2 n}{2} x \right )}}{n \left(\tan^{2}{\left (\frac{2 n}{2} x \right )} + 1\right)}$
 
@Hippalectryon Do you imagine myself thinking of the obvious way you get that? No, I don't. ;)
 
6:13 PM
@Chris'ssis Who knows :)
 
@Hippalectryon With my identity above one can do pretty nice things (btw).
 
@Chris'ssis With most of the things you post here, one can do nice things :)
 
@DanielFischer I've been debating whether or not I should send an email to the author. I probably won't.
 
@Hippalectryon 10x 10x 10x :-)
 
10x ?
Oooh
I got it :)
 
6:21 PM
@Hippalectryon Thanks=10x
:D
 
ten x?
 
Tanks
 
Tan x
 
@Hippalectryon I have a proof, I just need to juggle some estimates to make things look nicer. Now you've edited the question so that it is much longer. It seems harder to read.
 
@robjohn Oh i'll revert that then
 
6:28 PM
@Hippalectryon I think the rest is best left to the offsite description. I was able to answer the question from the information given
 
Great :D
 
@Hippalectryon Find the closed form of $$\sum_{n=1}^{\infty}\frac{\operatorname{\displaystyle Ci\left(\frac{3}{4}\zeta(2) \space n\right)}}{n^2}$$
@robjohn are you already done with that question?
 
@Chris'ssis That's my 60th screenshot of the things you post there :D
 
@Hippalectryon What will you do with the screenshots? Make them posters and put them on your walls?
 
Make a book muhahahahahaha
kidding
@Chris'ssis Oh good idea :D
 
6:35 PM
hahahaha :-)
@Hippalectryon You can do anything you want to since I made them public.
 
You could be a contributing author :D
 
@Chris'ssis Let's scare 16 y.o. kids with those sums :D
 
wait for Halloween
trick or treat?
 
IT WAS A TREAT TRICK :O
 
6:43 PM
@Hippalectryon @r9m
@Hippalectryon let's see if it's for 16 y old kids ...
 
Upvote my answers at 0 votes please. =)
@Chris'ssis Done.
 
@WillHunting Thanks :-)
 
r9m
@Chris'ssis :) !!
2
 
@r9m Good job (+1 star) :-)
 
r9m
^_^
 
6:53 PM
@IceBoy creepy
 
@Chris'ssis just edited and undeleted my previous hint
@Hippalectryon I've submitted my answer. Let me know if you have any questions.
 
@robjohn Who downvoted it -__-
 
@robjohn Great! (+1)
 
@robjohn How do you get $(3)$ ?
 
The only problem is to somehow know how to make those initial assumptions.
 
r9m
6:58 PM
@robjohn downvoters should really leave a comment .. sometimes people downvote and never look back at that post again :( disgusting !! and (+1) gr8 answer !! :D
 
@Chris'ssis I tried to do something with bounded variation, but it did not work. The current proof was difficult because of the interplay among $\epsilon, x_\epsilon, x,$ and $n$
 
@Hippalectryon @robjohn is over all MSE users in my opinion. Hard to beat.
 

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