Mathematics - 2014-12-06 (page 3 of 6)

Swapnil Tripathi

10:00

bot? @kaj

Integrator

Do you know these reviewers?

Kaj Hansen

I know DonAntonio. Not MegaNoob though.

Integrator

@KajHansen That bot is stupid!

Swapnil Tripathi

I accepted that edit! @Integrator

Kaj Hansen

I agree! LOL. I've seen some ridiculous stuff it produces to try and "trick" you.

Venus

10:03

@robjohn What is wrong with user: Olivier Oloa? Why his reputation below 10K? As far as I remembered, he already reach more than 13K or 15K. Also his answer to my question, As far as I remembered too, his answer got more than 40 upvotes but it gets 22 upvotes. He also got a silver badge Guru.

Integrator

@SwapnilTripathi What?

Swapnil Tripathi

@Venus Talk about consistency.

Integrator

@Venus See this math.stackexchange.com/users/118798/olivier-oloa?tab=reputation

Venus

@SwapnilTripathi Consistency of what?

@Integrator I don't know we can see his history rep

Integrator

@venus it says he lost 10599 rep due to voting reversal!

Swapnil Tripathi

10:09

That's a shocker! @Integrator And yes, I was one of those who reviewed the edit suggested by meganoob

usukidoll

yo

Integrator

@SwapnilTripathi But the review page is not showing your name!

usukidoll

anyone know pdes?

Swapnil Tripathi

@Integrator I don't understand. Even the edit description was different when I accepted.

@usukidoll: Sorry, I don't

usukidoll

ugh

r9m

10:12

@DanielFischer hee ! :D Finally a new pic (avatar) :-)

usukidoll

@DanielFischer know pdes?

Daniel Fischer

@usukidoll Enough to stay away from them when possible.

Venus

@Integrator -10599. That was a huge amount of reputation.

@DanielFischer Are you already healthy?

Daniel Fischer

@r9m After Jasper got so many stars, I couldn't ignore the request, could I?

usukidoll

@DanielFischer do you know integral equations..that uses convolution theorem and inversion theorem

r9m

10:14

@DanielFischer :D

Daniel Fischer

@Venus Not entirely, still coughing.

Venus

@DanielFischer Is there someone who treats you? Are you living alone?

Daniel Fischer

@Venus No need for treatment, it's just an ordinary infection of the kind one has every year or every other year.

Integrator

0

Q: Serial up-voting reversed?

Today in chat somebody mentioned that a user was had rep much above $10$k and today he is having somewhere around $7$k. I checked his reputation history, and it said that he lost almost $10$k rep and the reason being Serial up-voting reversed Link to the reputation history of Olivier Oloa I...

2

GBeau

Putnam is in 5 hours

RIP

My goal is 1 point :D

Venus

10:20

@DanielFischer I see. It's winter now in Europe, you should maintain your health. Get well soon

usukidoll

putnam?

Swapnil Tripathi

@Integrator math.stackexchange.com/review/suggested-edits/314016

Kaj Hansen

I'm taking the putnam too @GBeau

GBeau

@KajHansen Good luck~

Kaj Hansen

I'm hoping I can get ~10 pts.

GBeau

10:22

This is my first time; I didn't study as much as I wish I could have, but I spent some time every week this semester

Integrator

@SwapnilTripathi can you please review this

Kaj Hansen

I haven't prepared at all.

Swapnil Tripathi

Good luck. @Kaj @Gbeau

Venus

@Integrator The closing words should be "could someone here explain this to me? Thanks."

Kaj Hansen

Rough week coming up. I guess I'll get 3 hours of sleep and head over to the math building.

GBeau

10:23

that was actually my plan~

Integrator

@Venus you can edit that! English is my $10$th langauge :(

GBeau

maybe I'll try to catch you tomorrow or something and ask how it went

Venus

@Integrator Me either

Swapnil Tripathi

@Integrator I'm having a problem with my meta.math.SE account. I can log in using my phone but not on my laptop. Any ideas?

Integrator

@Venus edited!

Venus

10:24

@Integrator I just saw it

Integrator

@Venus Thanks!

Venus

Hope it get a good response

@Integrator Upvoted & Favorited

Integrator

@SwapnilTripathi Just after noob edited that somebody noticed that it was a blunder, but before it gets second review! I edited it!

robjohn

@Venus I believe that this page and this page (on Dec 2) explain as much as can be said.

Integrator

@Venus Does this happen regularly?

@robjohn I saw that but! How can you lose $10000$, because somewhere it says that it takes 24 hours for system to revert the serial-voting, excuse my English please!

Swapnil Tripathi

10:27

@Integrator Yes, I understand. And any help on my problem?

Venus

@r9m We have an alternative answer by @robjohn. Have a look. (+1) robjohn

robjohn

@Integrator I bet it says it may take up to 24 hours since the script is run once per day.

Balarka Sen

@KajHansen!

Integrator

@robjohn and you can get 200 rep per day (due to up-votes) right? I hope you get my point!

robjohn

@Integrator Have you looked at my rep? On some days I do, but definitely not all.

Swapnil Tripathi

10:31

@Integrator I have had >200 once or twice. I don't know why they say 200 is the "daily maximum"

usukidoll

@robjohn do u know pdes

Swapnil Tripathi

Is it a fairly simple one? ;) @usukidoll

usukidoll

@SwapnilTripathi it has something to do with fourier transform and heat equation

I am following an example from a book and I have let this thing be q(x,t). Then I take the fourier transform of the heat equation simple enough but what is the fourier transform of that particular thing

robjohn

@usukidoll somewhat

usukidoll

2. Solve the problem \\
D.E $u_t=ku_{xx}+\frac{1}{\sqrt{2kt}}e^{\frac{-x^2}{4kt}}, -\infty < x< \infty, t >0$\\
I.C $u(x,0)=0$\\
@robjohn I have to use fourier transform but I don't know what it is for the 1 over all that stuff

Swapnil Tripathi

10:34

I just know what a fourier transform is. I have never applied it. Why don't you post your ques and see. :) @usukidoll

usukidoll

but I need to know what the fourier transform of that 1 over fdfslfsdlaf;jdfslkfjsdf

@SwapnilTripathi nah... too quiet at this hour

quick question is there any other way to solve this besides fourier transform?

because I used D'alemberts for the wave.. why make things complicated with the transform if I can use just taht?

that?

Integrator

@SwapnilTripathi 200 only for up-votes you can earn even 300000 with accepted answers!

Swapnil Tripathi

Haha. Over to you @robjohn

GBeau

please no game theory on this year's Putnam pls pls pls

Swapnil Tripathi

@Integrator: Oh, it makes sense.

GBeau

10:36

finger crossing

Integrator

@robjohn 200 is only for up-votes you can get more than that iff some of your answers were accepted on that particular day!

robjohn

@Integrator or bounties

GBeau

not that I mind game theory, but the problem last year was insane

Integrator

@robjohn and bounties :)

usukidoll

~_~ pdes are insane how's that

T_T

Integrator

10:38

@robjohn And did you noticed that the reputation history says that he lost 45 rep because his answers were un-accepted, They're not

usukidoll

if I could just finish those 2 sections both which had one problem done already then I can do 7.2 and 6.4 tomorrow after I study for my japanese oral test

robjohn

@usukidoll are you trying to take the FT in $t$ as well? I think you only need to take it in $x$.

usukidoll

@robjohn no... I just want to know how this thing works... so I need the FT in x?

@robjohn $u_t=ku_{xx}+\frac{1}{\sqrt{2kt}}e^{\frac{-x^2}{4kt}}, -\infty < x< \infty, t >0$ so do I take the FT for the e part or for the heat equation? if I shift $ku_{xx}$ over I get $u_t-ku_{xx}= \frac{1}{\sqrt{2kt}}e^{\frac{-x^2}{4kt}}$

hmmmmm :/

robjohn

@Integrator it probably means that the 15 points were removed, not that the answers were unaccepted.

usukidoll

do I just take the fourier transform of $e^{-x^2}$?

r9m

10:47

@Venus yes I have seen it ^_^ (and +1 ed of course)

usukidoll

ughhhhhh I'm pulling my hair off of this thing geez

Integrator

@robjohn If I'm not mistaken then it means that just like you the software also cannot mark answers as un-accepted

robjohn

@usukidoll You have to take the FT of the whole equation. I think you get something like $$\hat{u}_t=-4\pi^2k\xi^2\hat{u}+\sqrt{2\pi}e^{-4\pi^2kt\xi^2}$$

usukidoll

@robjohn ah and then what happens next? Actually I've only did the FT of the heat equation by itself with out that extra jazz on the side. because there was an example that had the heat equation + q(x,t) and since the entire equation was being FT'ed I did the same thing, but couldn't figure out the FT of q(x,t).

robjohn

@usukidoll next, I assume you use that $\hat{u}(\xi,0)=0$ and solve the equation in $t$ for each $\xi$. It looks to be a standard integrating factor situation

usukidoll

10:57

yes because that's the Initial equation

so ... hmm.. wait does the t =0 in this case?

so we need the reverse product rule

this would be much easier without the funky symboles

Integrator

@Venus Fairly good response!

usukidoll

what if...
$\hat{u}_te^{-4\pi^2kt\xi^2}=-4\pi^2\xi^2\hat{u}e^{-4\pi^2kt\xi^2}+\sqrt{2\pi}e^{-4\pi^2kt\xi^2}$

yay wishful thinking

hmm I know in odes I multiply the integrating factor throughout the equation and then use reverse product rule... after....that...then I integrate

$e^{-4\pi^2kt\xi^2}$

$-4\pi \xi^2 u e^{-4\pi^2kt\xi^2}$ @robjohn

Venus

@Integrator You know Arthur. He is very a strict mod. ^^

usukidoll

how to even ......

GBeau

somebody suggest a topic to try to look over before the putnam in 4 hours :3

usukidoll

11:12

is the integrating factor $e^{-4\pi^2kt\xi^2}$ @robjohn ?

robjohn

@usukidoll I think the integrating factor would be $e^{4\pi^2k\xi^2t}$

usukidoll

@robjohn why is the k gone?

robjohn

@usukidoll oops... I dropped the $k$. Let me fix things

usukidoll

ok

LIke this?
This is the FT equation multiplied by the integrating factor
$\hat{u}_t=-4\pi^2\xi^2\hat{u}[e^{4\pi^2\xi^2tk}]+\sqrt{2\pi}e^{-4\pi^2kt\xi^2}[e^{4\pi^2\xi^2tk}]$\\

Committing to a challenge

Did we @BalarkaSen, thanks for saying so

I just woke up after 10 hours of sleeping, and it is 9:30PM

usukidoll

11:21

$\hat{u}_t=-4\pi^2\xi^2\hat{u}[e^{4\pi^2\xi^2tk}]+\sqrt{2\pi}$\\

robjohn

@usukidoll you didn't multiply the $\hat{u}_t$

usukidoll

oops

$[e^{4\pi^2\xi^2tk}]\hat{u}_t=-4\pi^2\xi^2\hat{u}[e^{4\pi^2\xi^2tk}]+\sqrt{2\pi}e^{-4\pi^2kt\xi^2}[e^{4\pi^2\xi^2tk}]$\\
$[e^{4\pi^2\xi^2tk}]\hat{u}_t=-4\pi^2\xi^2\hat{u}[e^{4\pi^2\xi^2tk}]+\sqrt{2\pi}]$\\

Integrator

@Venus Better not to say anything!

usukidoll

late night XD that's where the errors live

and then we take the antiderivative /integration of what the heck? the RHS is too hard to do that! should we use inversion and convolution theorem @robjohn

Venus

@Integrator Freedom of speech ^^

@N3buchadnezzar It turns out, it has a nice closed-form

Committing to a challenge

11:27

@Venus What strictness did Arthur display to you? The closing of your 'love in SE' question?

robjohn

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{4\pi^2k\xi^2t}\hat{u}\right)=\sqrt{2\pi‌}$$ so $\hat{u}=\sqrt{2\pi}\,t\,e^{-4\pi^2k\xi^2t}$

usukidoll

that was fast! How did you do that @robjohn ? oh wait reverse product rule.. and then we take the antiderivative of $ \sqrt{2 \pi}$ right?

Committing to a challenge

Raff(Behaviour), made a clone: math.stackexchange.com/users/179835/passing-by?tab=answers

usukidoll

@robjohn so there is a substitution for the u hat?.. oh yeah that makes sense...if I put it back together the e's cancel

@robjohn so do I take the antiderivative of the rhs or ...use inverse fourier transform...or convolution...or inversion theorem?

robjohn

@usukidoll take the inverse FT to get $u$ back

usukidoll

11:43

which part gets the inverse FT? @robjohn

ah I need a $g( \xi)$

robjohn

@usukidoll Take the IFT in $\xi$

usukidoll

$g(\xi )e^{i \xi x}\frac{1}{\sqrt(2 \pi)}$

$\sqrt{2\pi}\,t\,e^{-4\pi^2k\xi^2t}e^{i \xi x} \frac{1}{\sqrt{2 \pi}}$

$t\,e^{-4\pi^2k\xi^2t+i \xi x} $

o-o

it's @Ted!

Ted Shifrin

Howdy.

usukidoll

@Ted I'm still trying to figure out fourier transforms and heat equation... ugh the wave equation is much easier

Ted Shifrin

Parabolic is subtle ... But I haven't thought about this in almost 30 years ...

usukidoll

11:53

the inverse fourier transform in $\xi?$

robjohn

@usukidoll I get $$\sqrt{\frac{t}{2k}}\,e^{-\frac{x^2}{4kt}}$$ now to plug it in and see if it works

usukidoll

arghhh that's exactly what it is....but I just don't know the steps to it :(

I know it's going to work, but how did you get that @robjohn

robjohn

@usukidoll I took the IFT of $\sqrt{2\pi}\,t\,e^{-4\pi^2k\xi^2t}$

usukidoll

but how does the IFT process work? @robjohn$

don't you need the $g(\xi)$ and the $e^{-i \xi x}$

Daniel Fischer

Morning @TedShifrin, you're up early.

robjohn

11:59

@usukidoll $$\hat{f}(\xi)=\int f(x)\,e^{-2\pi ix\xi}\,\mathrm{d}x$$ $$f(x)=\int \hat{f}(\xi)\,e^{2\pi ix\xi}\,\mathrm{d}\xi$$

usukidoll

ok but what is the f hat xi? @robjohn

robjohn

@usukidoll just use the $\hat{u}$ we computed above

usukidoll

@robjohn but I need to figure out who my xi is before I continue

so would the f xi hat be the u hat?

robjohn

$\sqrt{2\pi}\,t\,e^{-4\pi^2k\xi^2t}$

usukidoll

$\hat{f}(\xi)=\int\sqrt{2\pi}\,t\,e^{-4\pi^2k\xi^2t}\,e^{-2\pi ix\xi}\,\mathrm{d}x$

robjohn

12:05

@usukidoll I used the fact that the FT of $e^{-\pi x^2}$ is itself

and scaled things

usukidoll

?

?!!

what I need right now is some sleep...I'm starting to lose my mind lol

night everyone

Venus

@robjohn Does mse or other se sites have a chief of moderators?

Will Hunting

@Venus No, but they should make me king of the world.

Integrator

@WillHunting You betray me ? :P

user130018

12:27

Hi @JasperLoy

robjohn

@Venus No.

Committing to a challenge

@Kaj Yeah I am loving all the music, thanks so much!

Funny that Ted's starred comment was in regards to Huy and Venus, and it precedes Venus's one by three days.

@Will So UserX is gone, I believe you were there when it happened right?

Integrator

12:50

@robjohn I've edited my question on meta could you please take a look

Integrator

13:01

@Venus Do you know about this about our strictest mod

Will Hunting

@Committingtoachallenge Yes.

@user130018 Hi.

user130018

@JasperLoy How are you

Will Hunting

@user130018 I have been thinking of how to save myself from my mental problems and how to save others from some other problems. I hope I achieve both over the course of 2015.

GBeau

nervous for Putnam

user112495

How can you go about evaluating the following integral:

$\int_0^{\infty} \frac{x}{e^x-1} dx$

I've tried a few methods but I can't get any of them to work. I don't want the method to assume the solution to the Basel problem or anything about the gamma function

Will Hunting

13:08

I see that both @alizter and @DanielFischer have changed their pictures, lol.

user130018

@JasperLoy it's almost time for you to start doing math..a few more weeks

Venus

@Committingtoachallenge Why is that funny?

@Integrator Eh, he knows that comment? Well, I'm glad he takes that as a random praise ^^

Will Hunting

@user130018 Yes, I hope to finish my 12 holy books next year. If that is too much, at least 6 of them, enough for the GRE.

Venus

@Integrator No. I didn't know him. I've never had a conversation with him

Will Hunting

I just had a great shit.

Venus

13:11

@WillHunting We are king of our own

user112495

Or can it even be solved in a relatively simple way?

Will Hunting

@Venus Yes, but in this sick world, we can still be punished by our parents and by the law for doing harmless things.

Venus

@WillHunting I've never been punished by my parents

But I've been punished by a padre in my church

Will Hunting

@Venus Are you going to become an engineer after graduating?

Venus

@WillHunting I dunno

Alizter

13:15

@WillHunting Because we look fabulous.

Will Hunting

@Alizter I won't put up my pic. I might even delete my account for good one day.

Alizter

@WillHunting Why?

Will Hunting

@Alizter It's a long story. Anyway, I think I should forget about Sarah, lol.

Alizter

We should have a voice over IP for this chat.

Will Hunting

@Venus Is the orphange you grew up in a religious organisation?

Venus

13:17

@Alizter Is the one in your ava you?

@WillHunting Yup

Alizter

@Venus The one in my ava is me

Will Hunting

@Alizter When I am gone from SE, you can always email me. That is why I keep sharing emails with people. I will respond if I am alive, not in hospital and not in prison.

I might end up in hospital if my mental illness worsens.

I might end up in prison for trying to help certain people in my country.

Venus

@Alizter I want to praise you but I don't want to offense anyone else here

@WillHunting Why would you be in prison?

Alizter

@WillHunting No you won't. You are taking GRE no?

Will Hunting

@Venus What I am about to do is a secret. I hope to change certain laws in my country, to cut the long story short.

Venus

13:21

@WillHunting Why would you end up in prison for trying to help people?

Will Hunting

@Alizter I will, when the time is ripe. But I am not sure exactly when. I definitely want to leave my country and go to grad school and become a mathematician.

Alizter

@Venus Praise away who cares? It does say general discussion

Will Hunting

@Venus Maybe not in prison, but there could be other punishments.

@Venus What I consider right, others can consider wrong. What I consider wrong, others can consider right.

@Venus In Iran, they still execute gays. But I am not in Iran.

@Alizter Good observation, lol.

Venus

@Alizter Nah. I don't wanna get a trouble for that but I believe you know what exactly my praise to you

Alizter

If we have a skype group we won't have any trouble

Will Hunting

13:24

@Venus Do you think Ali is handsome?

Venus

Is Arthur trying to enter this chat room? I just saw his ava shows up but where he's now?

@WillHunting In Russia gay people will go to jail

Will Hunting

@venus Your English is terrible, lol.

@Venus Really? I did not know that.

Venus

@WillHunting What do you thing? ^^

Will Hunting

@Venus Hmm, I think @ali is very cute, lol.

Venus

@WillHunting It should be: "In Russia gay people will be put in a jail"

Is that correct?

Will Hunting

13:29

@Venus I was talking about another sentence, not that one, lol.

Venus

@WillHunting This one: "Is Arthur trying to enter this chat room? I just saw his ava shows up but where he's now?"

Integrator

@Venus I've deleted my question as it is concerned about some specific user!

Will Hunting

@Venus Yes. You should write 'I just saw his ava but where is he now?'

Venus

Maybe : "Is Arthur trying to enter this chat room? I just saw his ava showed up but where is he now?"

Will Hunting

@Venus Seems OK too.

Venus

13:32

@WillHunting Wait!? Are you Arthur?

Will Hunting

@Venus No, I am JL.

Venus

Who is JL?

Will Hunting

JL = Will Hunting

JL is my real name.

Integrator

@venus Jasper Loy!

Venus

What does it stand for?

Oh Jasper

@WillHunting Do you have a Chinese name?

Will Hunting

13:35

@Venus Yes, I do. But most people in this chat do not know what it is.

Venus

So that a secret I presume?

Will Hunting

Well, sort of.

My location is also sort of secret for now, though many people know it.

Venus

I know where you are

Will Hunting

You are right beside me, lol.

@venus So when are you getting married?

Venus

We are separated by sea

@WillHunting I dunno. 25 maybe

Will Hunting

13:45

@venus Have you considered switching to doing math?

Venus

@WillHunting No, I'm not good at math

user112495

Does anyone know how I can evaluate this integral:

$\displaystyle\int_0^{\infty} \dfrac{x}{e^x-1} dx$ without using the gamma function (or the Basel problem).

Venus

@user112495 Let me try

Committing to a challenge

His name obviously isn't Jasper Loy, or he can prove me wrong and say that it is. I always remember him saying it is JL and never confirming Jasper Loy.

Anonymous

14:03

@WillHunting SOrry Srry

Anonymous

@WillHunting I spoke a lot of shit

Anonymous

@WillHunting I beg your pardon

Venus

Yes, we can. Use substitution $t=e^{-x}$, then we have
\begin{align}
\int_{0}^\infty \frac{x}{e^x-1} \, \mathrm dx &= \int_{0}^\infty \frac{x\,e^{-x}}{1-e^{-x}} \, \mathrm dx\\
&=-\int_0^1\frac{\ln t}{1-t}\, \mathrm dt\\
&=-\int_0^1\sum_{k=1}^\infty t^{k-1}\,\ln t\, \mathrm dt\\
&=-\sum_{k=1}^\infty\int_0^1 t^{k-1}\,\ln t\, \mathrm dt\\
&=-\sum_{k=1}^\infty\frac{\partial}{\partial k}\int_0^1 t^{k-1}\, \mathrm dt\\
\end{align}
Can you take it from here?

Anonymous

@Venus Are you marrying @Huy?

user112495

@Venus Sorry, could you explain how you get from the second to last line to the last one?

@Venus Sorry, I see it now!

Venus

14:09

@AshwinGokhale Why do you think so? I already have a BF

Huy

@Ashwin: Why do you think so? I already have a GF

Anonymous

That's called symmetry LOL

Committing to a challenge

@Ashwin Are you the same Ashwin?

Anonymous

lol ya

Committing to a challenge

@Ashwin Are you Mathwonk?

Anonymous

14:11

same old dumbass :D

Anonymous

@Committingtoachallenge idk who Mathwonk is

Integrator

@Venus I hope this will receive fairly good response.

Committing to a challenge

@Ashwin Oh I thought you were mathwonk aha

mathwonk.wordpress.com/2014/12/05/what-is-this

user112495

@Venus Wait, but doesn't this only work if you assume that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}$ which I said I don't want to do?

Integrator

Oh, @Venus You're popular now!

user112495

14:13

@Venus As my whole reason for asking this is to try and find another (albeit possibly much longer) way of finding the sum of this series using the gamma function.

Anonymous

@Committingtoachallenge How you doin?

Anonymous

@Committingtoachallenge I know how to spell Riemann

Committing to a challenge

@Ashwin Oh ahaha, sorry I didn't mean to imply anything, I just hadn't talked to this user prior. I am fine thanks

Alizter

@Venus Can I be your best friend?

Committing to a challenge

I woke up at 9:30PM to start the day, so that is somthing(it is 12:14am now)

Venus

14:15

@user112495 I don't get it you. you said, "without using the gamma function (or the Basel problem)". So I give you other way. It easy to evaluate the last series

Alizter

Assuming thats what BF is

Anonymous

@Committingtoachallenge Do you have permanent jet lag?lol

Venus

@Integrator Upvoted

@Integrator Why so?

Committing to a challenge

@Ashwin I normally get up at 4am each day, and go to sleep at 9pm. A few days ago I slept for 11 hours, then again the next day(not normal for me). I think I am sick or something.

Anonymous

@Committingtoachallenge That's your caffeine!

user112495

14:16

@Venus But the last series is $\sum_{n=1}^{\infty} \frac{1}{n^2}$ isn't it? I don't want to assume a value for this series as my reasoning for evaluating this integral is so that I can evaluate this series.

Venus

@Alizter Are you sure? Well, I don't mind if you want it. I love making friends ^^

Anonymous

@Venus Have you done group theory?

Committing to a challenge

@Ashwin I actually never checked the caffeine quantity of my current coffees, I am going to do that now

Alizter

@Venus We can be friends. You are it. runs away

Integrator

@Venus I see something on the sidebar ;)

Alizter

14:18

We should have a skype group

Anonymous

@Committingtoachallenge Whst are you studying now?

Committing to a challenge

I have been having 2000mg of caffeine a day

Integrator

@Committingtoachallenge $\star$ for "Oh fuck"

Anonymous

@Alizter Why?

Venus

248

Q: Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$

As I have heard people did not trust Euler when he first discovered the formula $$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us? This is being...

sequences-and-series complex-analysis fourier-analysis big-list faq

Alizter

14:19

@Ashwin Because then we can talk. I am too lazy to type.

Committing to a challenge

I have been having 20 coffees a day WTF

I have to go

Anonymous

bye @Committingtoachallenge

Venus

@Ashwin No, I'm not doing math

Anonymous

@Venus ?

Anonymous

@Venus It's dangerous to go for engineering!

Venus

14:21

@Alizter What did you mean by "You are it. runs away."? Sorry, I'm not a native

Alizter

@Venus Like tag.

"Tag you are it"/

and the game proceeds by avoiding the person who is 'it'

Venus

@Integrator What did you see?

Anonymous

Anybody here a gamer?

Integrator

@Venus $\Large\color{red}{❤}$

@Ashwin Me! I love Mathematical games!

Anonymous

Cool!

Venus

14:23

@Alizter Sorry, I still don't get it

Alizter

Tag (game)

Tag (also known as it, tip you're it or tig (in regions of Britain), and many other names) is a playground game that involves one or more players chasing other players in an attempt to "tag" or touch them, usually with their hands. There are many variations; most forms have no teams, scores, or equipment. == Basic rules == A group of players (two or more) decide who is going to be "it", often using a counting-out game such as eeny, meeny, miny, moe. The player selected to be "it" then chases the others, attempting to get close enough to "tag" one of them (touching them with a hand) while the others...

Now it sounds weird.

user112495

@Venus I know there are other ways to evaluate the sum. I'm currently writing an essay on the gamma function, and I thought it would be cool if I could use the link between the gamma function and the zeta function to evaluate that sum. But doing so involves evaluating that integral (which I obviously can't assume the sum is pi^2/6 for). I think I probably won't bother with it though as it doesn't seem like there is a simple enough solution to be worth putting in.

@Venus Thanks for your help though.

Venus

@Integrator I ask everyone here on chatroom how to make that shape using LaTeX & only Huy gave response

Integrator

@Venus Oh!

Venus

@user112495 Are you saying that you wanna evaluate the integral using the gamma function?

Anonymous

14:26

I have a new mental issue:i get sad if I go online :(

Venus

@Alizter Oh, I know this game. I get it now, haha

user112495

@Venus I want to use the fact that $\zeta(2) = \dfrac{1}{\Gamma(2)}\displaystyle\int_0^{\infty} \dfrac{x}{e^x-1} dx$ to show that $\zeta(2) = \frac{{\pi}^2}{6}$

Integrator

@Ashwin I see you haven't started participating on MSE!

Anonymous

Nope

Integrator

@Ashwin use @Integrator

teadawg1337

14:29

Can anyone spare some Tums? My stomach is very volatile today...

Anonymous

@Integrator NOOOOOOOOOOOOOOOOOOOOOOOOOO

Anonymous

@Integrator Heard that?

user112495

@venus But i'm starting to think there isn't a particularly nice way of solving that integral without assuming what I want to prove.

Integrator

@Ashwin Heard what?

Venus

Ted said this to me & Huy, "This is not supposed to be a dating room ... and this verges on that ... " :D

It has lots of stars

Maybe I should refrain myself from entering chat room ^^

Anonymous

14:34

Anyone reading Putnam and beyond?

Committing to a challenge

@Venus I dislike people making statements like this.

Anonymous

@Committingtoachallenge Even I feel like not coming here

Committing to a challenge

Can my "oh ****" get unstarred before I get banned :P.

@Ashwin I mean people saying they won't come, because $X,Y,Z$

@Ashwin Make a blog for me to follow if you do leave :)

Anonymous

@Committingtoachallenge Even I feel so

Anonymous

@Committingtoachallenge That would e great

Committing to a challenge

14:36

Sorry I mean to say people saying they will leave because person $A,B$ did $X,Y,Z$

Huy

I never realised how much better Sumatra PDF is than other PDF readers for working with LaTeX.

Good for me that I decided to try out something new once in a while.

Venus

@Committingtoachallenge Why so? It's not a bad word

Anonymous

@Committingtoachallenge Were you serious when you said that you take 2000mg os caffeine?

Committing to a challenge

@Ashwin I thought I did by mistake after doing some checking, but I am fine

Anonymous

@Committingtoachallenge 600mg+ is dangerous

Committing to a challenge

14:38

@Ashwin My coffee counter seems to be correct forunately. Google has instant coffee listed at 3,100mg per 100g, which freaked me out

Anonymous

@Committingtoachallenge What book are you on now?

Venus

@Committingtoachallenge I never says "Oh ****" here

Committing to a challenge

@Ashwin Still just working on Axler,Cohn and Zorich

@Venus I have put you on ignore now

Anonymous

@Committingtoachallenge I am doing Apostol,Burkill,Carter,Artin,Strang,...

Venus

@Committingtoachallenge Be my guest. Feel free ^^

Committing to a challenge

14:39

@Ashwin All simultaneously?

Anonymous

@Committingtoachallenge Yes

Anonymous

@Committingtoachallenge Can you clear my doubt?It's one SE

Committing to a challenge

@Ashwin Wow that is a bunch to do a once xD. I was thinking about starting Rudin

@Ashwin What is the doubt?

Anonymous

@Committingtoachallenge I feel like talking to only you and nobody else.Can we have a separate chat room?

Anonymous

@Committingtoachallenge How do you deduce that the symmetry of a frieze pattern is always one among seven?

Anonymous

14:43

@Committingtoachallenge I am not quite getting it,though it might be simple to everyone else.

Transcript for

$Mathematics$ Mathematics