7:00 PM
you can't deny the prize it may never fulfill you it longs to kill you, are you willing to die?
2
THE COLDEST BLOOD RUNS THROUGH MY VEINS YOU KNOW MY NAME
2

user19161
I see this weird thing. When I go to answers then sort by newest in my profile under chrome I see a scrollbar on the right. When I switch the sort to votes and then back to newest the scrollbar disappears.

user19161
That is called over-starring.

That is called $over-over-starring$.

@JasperLoy Oh you! touches left shoulder with back of left hand

@JonasTeuwen I don't wanted to be rude. Sorry if that was the case. I'll keep it in mind

7:06 PM
Okay 8-). Did you solve it?

@JasperLoy I wonder why...

user19161
@leo Actually if you are desperate to find an answer you may do it!

@JonasTeuwen yes, @robjohn helps me :-)

BTW, iwonder why a book like fifty shades of grey is published and people read it...

user19161
@Charlie Well, if I were her age, I would fall in love with her.

7:08 PM
@JasperLoy ...

The silent ping^

I was trying to solve this: Let $T:K\subset\Bbb R^n\to K\$ a map so that $d(Tx,Ty)\leq d(x,y)$. Show that $T$ still have fixed points.

@skullpatrol shhhhh

@skullpatrol ?

7:10 PM
shhh

:)

:-*

user19161
You are all nuts!

@JasperLoy we're all mad here....

Takes one to know one

7:11 PM

And it is false: Take $K$ the unit circle and $T$ the map that sends each point to its antipodal point. $T$ has no fixed points

@Charlie Some of them a long time ago.

@JasperLoy The exercise I'm talking about is from Boothby's book. I want to email him to say that he asking to prove something false

@skullpatrol i like it.it's funny. here, it's not that funny, like it once was... they changed the editor, now it sucks.I like to watch mad tv, on cartoon network

It needs some additional hypothesis

user19161
7:14 PM
@leo Ah OK, I am not sure if he is still alive. I often wonder whether this author or that one is still alive or not!

@Charlie I agree.

@skullpatrol :)

@Charlie :)

@JasperLoy However I can't find his email, at least in the book. I have the second edition. Don't know what is the newest

My advisor called my preprint "impressive" 8-))).

7:17 PM
I would not expect anything else from you.

@JonasTeuwen cool!

@JonasTeuwen GREAT!

@JasperLoy it seems he is
I must go. See you later!

Later!

later

7:24 PM
@skullpatrol what have you been doing?

@Charlie Not much, you?

@skullpatrol studying

user19161
When argon comes I must give him the Friday song...

Of The Cure?

hehe
@JayeshBadwaik Good night!Sleep tight!

user19161
7:35 PM
@Charlie How do you know when he sleeps?

@JasperLoy I know everything
it makes five months that i have MSE account!

user19161
@Charlie So do I. I know what is going on there...

@JasperLoy no you don't
And i would like to thank everyone here!
Such nice and cool people I found here!

user19161
@Charlie They should have an annual interest, like add 10 per cent of rep points every year to our MSE account.

Also a bullshit and asshole tax!

7:45 PM
@JasperLoy That would reward just sitting on our thumbs and doing nothing...
@JonasTeuwen Now that is a better idea. For pain and suffering.

Yes.

@skullpatrol after sitting on a thumb?

so what would Z[2 sqrt(3)] be?
that's not the same as Z[sqrt(3)] is it?

Sometimes I wonder:"how people who i barely know, from distant lands, people that i barely know how looks can be nicer to me than the ones I know?"
I am grateful to all you guys!!!Who answered my questions, had some nice chatting, teach me a lot of things !It's a pleasure to talk to you!
Special thanks to:@JasperLoy,@Jayesh , @Jonas , @Skullpatrol , @PeterTamaroff , @Oldjohn , @EdGorcenski , @robjohn, @anon and everybody else who helped me!THANK YOU VERY MUCH!

7:48 PM
@robjohn Possibly.

what's that from

@Charlie Wait... are you dying or something? 8-(.

@JonasTeuwen No, not at all!

Good.

I needed to say it

7:51 PM
@Charlie Do you feel better now?

@skullpatrol Yes! I really like being here

user19161
I want to show something that a nice bro from TeX made for me...

Show us

user19161

@Charlie You're welcome.
@JasperLoy modest, eh?

7:53 PM
@robjohn :D
@JasperLoy Nice!

@JasperLoy It should wrap and complete a full cycle, or bounce back and forth in a harmonic motion.

user19161
in Elaborating Herbert Voss' short answer, yesterday, by ガベージコレクタ
\documentclass[pstricks]{standalone}
\usepackage{pst-node,multido}

\begin{document}
\multido{\n=-2.0+0.2}{21}{%
\begin{pspicture}(-4,-3)(4,3)
\pnode(\n,0.5){center}
\multido{\nx=0+10,\ny=5+10}{36}{\pswedge*[linecolor=red](center){8}{\nx}{\ny}}
\pscircle[fillstyle=solid,fillcolor=red,linestyle=none,border=2\pslinewidth,bordercolor=white](center){1.75}
\rput(center){\textcolor{white}{\bfseries\LARGE Jasper Loy}}
\end{pspicture}}
\end{document}

user19161
That is the code.

@JasperLoy How do you get it to work?

\usepackage{pst-node,multido}  nice packages

user19161
7:56 PM
@skullpatrol Hmm, I think you need to produce it first and then paste it here as an image. You can't use chat jax for that.

@JasperLoy yeah, you can't load packages with MathJax (as far as I know)

user19161
This bro from TeX is also mysterious like me, he seems to have many secrets too!

user19161
But we shared some of them via email. =)

Mystery, mystery , mystery...

Hello everyone, can anyone help me to show that partial sum(-1)^n/n! converges to same limit as (1-1/n)^n?

8:00 PM
you can sort of see it by expansion, if you think about the binomial coeffiecents tending towards factorials as n gets big

user19161
@jdoe I see you have hidden your true identity. =)

I don't know how to go back to my previous name :(

user19161
@jdoe You should copy to all accounts, log in to all of them for changes to be effected.

user19161
The copy is right at the bottom if you edit your profile, just click when done.

yes but they alternate
like odd members of sum are bigger than 1/n! and evens are smaller
so it is confusing me
oh sorry i got it

8:06 PM
LOL
did you know this , @Jasper ?

user19161
@Charlie Nope, never heard of. I am going to sleep now. Good night M!

@JasperLoy Good night ,J!

why people say :"Happy halloween"?Shouldn't it be "Sad halloween"?

Scary Halloween

8:20 PM
@skullpatrol hehehe
@skullpatrol Inteligence is sexy

Note the tattoo.

@skullpatrol yep!

@Charlie There is so much life still to live Charlie! Please!

@Charlie I think Einstein would be more like this compared to everybody else's intelligence.

@PeterTamaroff Yes, I'm living it!

@skullpatrol hehehe
@Argon Hey, Aaron!!!

@Charlie Hey!

@Argon wassup?many tests coming?

@Charlie Lets see...
Monday
Thursday
2 on Friday
And I probably will get more in between

Geez

8:32 PM
@Argon Did you decide if you are going to rationalize the denominators?

@skullpatrol Sure :)
When neccisary

@Charlie ...but physics and chem are starting to make sense!

@Argon GOOD!

hi @Chris'ssister

8:33 PM
Hi.

@Charlie Its about time, right?

I fail to apply Chebyshev here $\int x^{\frac{1}{2}}(1-x)^{\frac{1}{2}} \ dx$
It's the case with $\frac{m+1}{n}+p \in \mathbb{Z}$
I simply obtain a negative value under the radical sign. (something is wrong I suppose)
I used the substitution $x-1=t^2 x$

why people keep on skullpatroling?

Patrolling Skulls is a wonderful passtime.
4

8:38 PM
@Chris'ssister why $x-1$ and not $1-x$?

@Charlie: I only applied a formula from school.

@Chris'ssister excuse me?

$ax^n+b=t^s x^n$, where $s$ is the denominator of $p$

user19161
@argon I have a song for you. youtube.com/watch?v=kfVsfOSbJY0

@JasperLoy Haaahahaha!!:)
Perfect!

8:42 PM
@Chris'ssister ok

@Charlie: is that formula correct?

@Chris'ssister honestly speaking, i never saw that, as far as i remember... :(

@Charlie: OK. Thank you for looking at my problem.

@JasperLoy you came from the nothing just to show this abhorrent song...
@Chris'ssister you're welcome!let's proceed!

user19161
@Charlie Hehe, not sleeping yet, maybe a while more.

8:45 PM
@JasperLoy yuppie!
:(

user19161

i forgot.... you maybe have nothing to say...

@JasperLoy Another classic :)

user19161
@Argon These two songs make me go wild!!!!!!!!!!!!!!!!!!!!!

@JasperLoy watch this

user19161
8:48 PM
@Charlie WTF!

@JasperLoy LOL
oppa gangnam style

user19161
I just realized that "Friday, Friday..." sounds like "Baby, baby...", LOL.

@JasperLoy only now you realized that?...

user19161
@Charlie Don't you like the two songs?

@JasperLoy hmmm...No.
too annoying

8:54 PM
Heya

I am going to ask a question on the main!

@N3buchadnezzar do it ,sir

But I need someone to help me formulate it in a good way! =D

user19161
@N3buchadnezzar Before that, click on my two songs above!

8:55 PM

:)

Lets say I have a sin(p_n(x)) or cos(p_(x))

I want to show that the solutions of the double derivative always lie exactly between the max and min of the function

8:57 PM
hmmm

user19161
@N3buchadnezzar You mean zeroes of the second derivative?

@JasperLoy are you happy today,Jas, you seem happy!

Where $p_n(x)$ is an arbitary polynomial
@JasperLoy yes!
$$p_n(x) = \sum_{i=0}^n a_i x^i$$

@Charlie He's happy to talk to you.

Charlie - Unicorns!
Argon - Chemicals!

9:01 PM
@Argon oh..I know it!

Jasper - Bieber!

Haaaaa

N3buch - rubik

user19161
I think Benja likes Cody Simpson.

@JasperLoy Who?

9:03 PM
@Argon benja lim

user19161
@N3buchadnezzar Note that in English, zeroes and solutions mean slightly different thing. The zeroes of f(x) are the solutions of f(x)=0.

@Charlie No, I mean the other guy

@Argon i don't know him either

user19161

oh my....
what a teenager thing

9:05 PM
My ears hurt.

Jas per Loy

@JasperLoy Something like this ? math.stackexchange.com/questions/221767/…
Jasper is bold

he's italic

user19161
@N3buchadnezzar Do you mean the max and min of the original function or the second derivative?

@JasperLoy you like RB, this is one gorgeous!

9:10 PM
@JasperLoy Original function

user19161
@N3buchadnezzar Let me edit for you.

=D
Also $A$ and $d$ are real constants

@Argon did you have problems with chemistry?

@Charlie Some. There are two many exceptions that I don't like, and it's not as straightforward or intuitive as math or programming, which I hated.
But stoichieometry seems good so far

user19161
@N3buchadnezzar Actually you are not referring to the zeroes but to the y values of the zeroes right?

9:17 PM
@JasperLoy Yeah

@Argon hmmm i liked chemistry.once i made a "professional affinity" test and the three first results were:Mathematics, chemistry and physics

@Charlie I like physics better then chem.
"professional affinity" :)

user19161
@N3buchadnezzar And what do you mean by exactly between? Do you just mean between in the sense of $A\leq B\leq C$?

@Argon i liked too.but i had a crush on chemistry.

:)

9:19 PM
@JasperLoy All the y-coordinates to the zeros of the double derivatives are always A/2 + d
Half the amplitude, always

user19161
@N3buchadnezzar Ah you mean that single value $A/2+d$?

@robjohn Is mods vote binding for reopening questions too? Or do we need 5 users with sufficient rep? Basically I'd like to know whether this suggestion from meta would work: Voting to close as duplicate without taking the time to check if they are

@JasperLoy yes?

@Argon but i didn't get a real big enthusiam

@MartinSleziak A mod's votes are binding everywhere they are cast.

9:23 PM
@robjohn Hey! =)

Great. When you have time, please have a look at meta, robjohn.
It would be much easier if a mod can to that than finding 5 people, who would vote to reopen.

Nothing much, just wanted to say hi. And that you are doing a great job to help the site =)

user19161
@N3buchadnezzar Edited the body, let me think of a better title.

user19161
@N3buchadnezzar Done editing, check to see if that is what you mean...

9:30 PM
@Argon did you see this?

@Charlie Nope, that's cool!
Too bad I don't live around there

@Argon :D

user19161
I think nebu has been kidnapped...

Oh hi
I was just plotting

@JasperLoy what?

user19161
9:33 PM
4 mins ago, by Jasper Loy
@N3buchadnezzar Done editing, check to see if that is what you mean...

@N3buchadnezzar In a nefarious way or like drawing $f(x)=\cdots$

Both!
I am going to kill the queen using a function that is neither well behaved nor bounded, muhahaha..
@JasperLoy It seems fine =)

user19161
@N3buchadnezzar It seems you have gotten an answer. Also note the skill of the editor. =)

@JasperLoy did he say "thanks"?

user19161
@Charlie Hehe, I would prefer ... , never mind.

9:37 PM
@JasperLoy I said Thanks to you.

@JasperLoy The editor must be an amazing person with loads of friends and be very popular.

user19161
@Charlie There were times I spent half an hour editing a question. It's quite amazing I can pull it off in the end.

@JasperLoy :D

$$e^{-\Omega}=\Omega \iff \log \Omega = -\Omega$$
$$W_0 (1)=\Omega \approx 0.5671$$
Yay

Yay
Too many parties this weekend!
Was invited to 1 today, 3 tommorow and 1 on sunday

10:07 PM
Functions on $\Re^3$ - is this an advanced topic?

10:24 PM
I call for an asshole tax.

Sorry for posting a question here, but...
1

Here's the situation: I have a (fair) die, which I roll successively until I get 5 consecutive ones, in which case I stop rolling. My questions are: What is the probability that I stop after exactly 11 throws? (or $k$ throws in general?) What is the probability that I roll the die at least 9 ti...

Can someone tell me if this is a really easy, obvious question, or if it's tedious?

Sapiosexual?

11:23 PM
I just realized how Gentzen-style deduction systems make an obvious Gangnam style parody

Mmm. When is a $C^\infty$ function analytic again? Surely, if we overload it with a complex variable and if it still is $C^\infty$ then it is analytic. But are those the only ones?
Perhaps there is weird-ass shit with branches that can still be $C^\infty$ on the real line.
This should teach me Not To Do Math When Tipsy.

then again, it's not like that would reduce the time you do math sober

Exactly.
This should teach me to drink less, how about that?

11:41 PM
how would it teach you to drink less??