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masfenix
masfenix
12:00 AM
@KevinDriscoll Thanks. What is the intuition behind Laplace Transforms? What do they do to a function? I don't understand why the condition of decay (ie, function must be of exponential order) is needed.
 
Kevin Driscoll
Kevin Driscoll
@masfenix Well if it grows faster than exponentially then the transform doesn't exist for any $s$
 
Pedro Tamaroff
Pedro Tamaroff
@KevinDriscoll YO YO
 
Kevin Driscoll
Kevin Driscoll
The intuition though is that the Laplace domain is one in which differentiation by $t$ is mapped ot multiplication by $s$ @masfenix
What it does to functions though..... I can't say for sure. It's more clear with Fourier and Mellin. I don't work much with Laplace @masfenix
 
masfenix
masfenix
Oh, I think I get it now. The decay condition is needed for the integral to converge. If the integral dosn't converge then obviously we don't have a Laplace transform.
 
Kevin Driscoll
Kevin Driscoll
Indeed.
@PedroTamaroff Yo! Good to see you.
 
Pedro Tamaroff
Pedro Tamaroff
12:06 AM
@KevinDriscoll What's up?
 
Kevin Driscoll
Kevin Driscoll
@Pedro Working on my PhD thesis proposal. And crying on the inside at how few jobs are available in my field.
 
Pedro Tamaroff
Pedro Tamaroff
@KevinDriscoll Ah, cannot be less than those for a mathematician, for sure!
 
Kevin Driscoll
Kevin Driscoll
@Pedro You'd be surprised. The problem is physics is more segmented. Whether you do theory or experiment you're automatically disqualified from the other's job postings.
I think there are also fewer math PhDs. But I know its also pretty bad over there. From my cursory estimates and personal experience I'd say its about even.
 
Pedro Tamaroff
Pedro Tamaroff
@KevinDriscoll Hehe, I'll tell you when I graduate.
 
Kevin Driscoll
Kevin Driscoll
@pedro I'm hoping to not wait until I'm ready to graduate to really be looking for things
 
Pedro Tamaroff
Pedro Tamaroff
12:22 AM
@KevinDriscoll But you're already a PhD student right?
@AlexanderGruber HAI.
 
Alexander Gruber
Alexander Gruber
@PedroTamaroff Bon soir.
 
Kevin Driscoll
Kevin Driscoll
@Pedro Yeah, so a good bit farther along. Seems like a lot of people don't start looking for jobs until they graduate tho.
 
Pedro Tamaroff
Pedro Tamaroff
@AlexanderGruber What have you been up to?
 
Alexander Gruber
Alexander Gruber
@Kevin oh you're a physics PhD student?
@PedroTamaroff torturing my business calc students :)
 
Kevin Driscoll
Kevin Driscoll
@AlexanderGruber Yup. Cold atom theory.
 
Alexander Gruber
Alexander Gruber
12:24 AM
@KevinDriscoll sounds fun. i like physics, did my undergrad in it
i'm not quite awake yet.
 
Kevin Driscoll
Kevin Driscoll
@AlexanderGruber Did you also do math?
 
Alexander Gruber
Alexander Gruber
@KevinDriscoll yeah, doubled up
 
Kevin Driscoll
Kevin Driscoll
AH okay. I wish I had done math sometimes.
 
Alexander Gruber
Alexander Gruber
@KevinDriscoll did you during undergrad?
@PedroTamaroff yourself?
 
Kevin Driscoll
Kevin Driscoll
Nope. I was a physics/philosophy double @alexander
 
Alexander Gruber
Alexander Gruber
12:27 AM
@KevinDriscoll Ahhh interesting.
So you can ponder the ethics of relativity
 
Kevin Driscoll
Kevin Driscoll
Hahahaha. Well it all depends on you frame of reference really.......
2
 
Pedro Tamaroff
Pedro Tamaroff
@AlexanderGruber Algebrain' pretty hard.
I failed my combinatorics test as I expected. =P
 
Kevin Driscoll
Kevin Driscoll
@pedro you failed a test on combinatorics? Howd that happen?
 
Pedro Tamaroff
Pedro Tamaroff
@KevinDriscoll A series of bad choices and mistakes during the exam.
 
Kevin Driscoll
Kevin Driscoll
@pedro What topics did it cover?
 
Pedro Tamaroff
Pedro Tamaroff
12:38 AM
Elementary Counting, Generating Functions, Graph Theory, Incidence Algebras of posets.
 
Kevin Driscoll
Kevin Driscoll
@pedro I probably would fail that test too. Generating functions are aparticular weak point for me. Though we use a generalized version in Quantum Field Theory
 
 
5 hours later…
G.T.R
G.T.R
5:25 AM
@Hippalectryon $\mathbb Q$ Lacks Bolzano-Weierstrass property
 
 
1 hour later…
Balarka Sen
Balarka Sen
6:33 AM
@mick I can't talk about magic floopskintunks and say that they make sense but I can't explain, right?
First show me how you define exp(X) in R[X] modulo X^3 - 1. You are not even sure whether or not it makes sense, i.e., is a polynomial.
Hey @Alexander.
 
Alexander Gruber
Alexander Gruber
hi
real empty in here tonight
 
Balarka Sen
Balarka Sen
empty?
 
David Kirby
David Kirby
∅
 
Balarka Sen
Balarka Sen
Heh.
I couldn't find any stuff in Isaacs, by the way @Alexander
 
David Kirby
David Kirby
@Balarka Sen, I've been on a quintic trip for the past few days thanks to you. Teehee
 
Balarka Sen
Balarka Sen
6:42 AM
@DavidKirby Oh? Tell me about it.
@Alexander I totally not approve the weird man mustache.
 
Alexander Gruber
Alexander Gruber
@BalarkaSen hahaha
i grew it out specifically for my ID picture
i don't normally have facial hair
 
David Kirby
David Kirby
@Balarka Sen, well, I've been working a lot with Harmonic Series lately and have come up with notebooks full of interesting series. After talking about quintic equations with you I started trying to solve various series with a quintic equation on the right hand side, as well as trying to rewrite different types of series in quintic form and the results of have been really interesting.. Roots galore, which I guess is to be expected...
 
Balarka Sen
Balarka Sen
@Alexander To be a mathematician, you need to look like this
bad pun
@DavidKirby Examples?
 
David Kirby
David Kirby
@Balarka Sen, sure, gimme a sec and I'll type in an example
Ok, here's one -- $$H_{-x-\frac{3}{4}}-H_{-x-\frac{1}{4}}=x^5-x-Z$$
Solving for Z gives a really interesting result :
$$Z\to \frac{1024 H_{-x-\frac{1}{4}}-1024 H_{-x-\frac{3}{4}}-1024 \left(-x-\frac{1}{4}\right)^5-1280 \left(-x-\frac{1}{4}\right)^4-640 \left(-x-\frac{1}{4}\right)^3-160 \left(-x-\frac{1}{4}\right)^2+1004 \left(-x-\frac{1}{4}\right)+255}{1024}$$
 
Balarka Sen
Balarka Sen
Whoa.
That's odd.
 
David Kirby
David Kirby
6:56 AM
yeah now check this :
$$H_{-x-\frac{3}{4}}-H_{-x-\frac{1}{4}}=\frac{-1024 H_{-x-\frac{1}{4}}+1024 H_{-x-\frac{3}{4}}+1024 \left(-x-\frac{1}{4}\right)^5+1280 \left(-x-\frac{1}{4}\right)^4+640 \left(-x-\frac{1}{4}\right)^3+160 \left(-x-\frac{1}{4}\right)^2-1004 \left(-x-\frac{1}{4}\right)-255}{1024}$$
solving for x :
$$\{\{x\to -1\},\{x\to 0\},\{x\to -i\},\{x\to i\},\{x\to 1\}\}$$
 
Balarka Sen
Balarka Sen
Ah, so you have a solvable quintic up there.
Let me understand what you are trying to say,
$x^5 - x - \tau = \mathcal{H}_{-x-3/4} - \mathcal{H}_{-x-1/4}$
When $\tau$ is the humongous thing above, right?
 
David Kirby
David Kirby
Correct, ... my thinking was that this is the expected behavior though, right?
 
Balarka Sen
Balarka Sen
So shouldn't the harmonic number terms cancel out, leaving you some algebraic identity?
 
David Kirby
David Kirby
Yup, it reduces to : $$-H_{-x-\frac{1}{4}}+H_{-x-\frac{3}{4}}-x^5+x$$
 
Balarka Sen
Balarka Sen
How do you find it interesting?
 
David Kirby
David Kirby
7:06 AM
Well, it just hit me by surprise, and it's a relationship that I only understand in a one-sided sense.. I think I need to study up on Galois groups (and cyclic permutations especially) .. Not sure where to begin
 
Balarka Sen
Balarka Sen
How do you even get that identity? Seems weird to me.
I never imagined a connection of quintics with harmonic numbers.
Could harmonic numbers then be inverted to solve a quintic?
 
David Kirby
David Kirby
Heh, neither did I.. It was the weirdest thing, tried it on a whim after talking to you the other night.
 
Balarka Sen
Balarka Sen
Like, asking to solve $\mathcal{H}_x = \tau$ for $x$.
 
David Kirby
David Kirby
@Balarka Sen, I was wondering that, but hadn't figured out how to go about it yet
 
Balarka Sen
Balarka Sen
How is $\tau$ related to $x$ in the above quintic identity?
$$x^5 - x - \tau = \mathcal{H}_{-x-3/4} - \mathcal{H}_{-x-1/4}$$
@David
 
David Kirby
David Kirby
7:16 AM
like this? $$\tau =H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$
uhm, err
and solving for $\tau$ gives you that big equation
and the big equation reduces back down to $$\tau =H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$
it's got my brain all blended for sure heh
 
Balarka Sen
Balarka Sen
Ah. But that's not weird at all, is it?
We just have another algebraic form for $x^5 - x$
And the solutions you get for $x$ are really the solutions of $x^5 - x = 0$
 
David Kirby
David Kirby
Yup, -- so that definitely makes sense.. Ok, so I'm still a little baffled by the solution after substitution :
$$H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}=H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$
$$\{\{x\to -1\},\{x\to 0\},\{x\to -i\},\{x\to i\},\{x\to 1\}\}$$
so instead of the 5 roots of the quintic, you get that
 
Balarka Sen
Balarka Sen
Yes, the roots of $x(x^4 - 1) = 0$, the simplest kind of quintic you can think of.
 
David Kirby
David Kirby
AH
Hah, ok it all becomes clear to me now
... I have fooled myself with slight-of-hand
 
Balarka Sen
Balarka Sen
7:34 AM
I am trying to prove that a quintic cannot be solved in terms of polygamma and polylogarithms, based on Murty-Saradha-Gun conjecture about their algebraic independence.
Essentially tracing steps of T. Chow, who proved it for exponential and logarithms.
(assuming Schanuel's conjecture)
 
David Kirby
David Kirby
If so, it could be a boon to transcendental number theory
 
Balarka Sen
Balarka Sen
Probably. I've never heard of any result regarding the fact that polylogarithmically-closed of polygamma-closed fields only contain algebraic with solvable galois group.
 
Studentmath
Studentmath
7:53 AM
David, I have a probability question I think you will enjoy
I am enjoying it, at least
 
Studentmath
Studentmath
8:10 AM
Hrm, unsure..
 
David Kirby
David Kirby
@Studentmath, .... cuuuurrrious
 
Studentmath
Studentmath
You have a box with 6 coins, numbered 1 to 6. Each has a probability of 1/i (where i is its number) to land Head up. You have a fair dice, with 6 sides.
You throw the dice once, pick up the coin according to the number that it landed on, and throw it until you land Head up.
Let X be the RV representing the result of the dice, and Y be the RV representing the number of times you throw the coin until (including) it lands Head up.

E[Y], Var(Y), and $P(Y\ge 5$| X is even)
It's really nice, you get to toy with few known probabilities but in a nice controlled and adjusted enviroment
 
Chris's sis
Chris's sis
8:29 AM
Greetings
@robjohn Is there a known form of $\zeta''(2)$ expressed in terms of known constants?
@robjohn I wonder what is the best place to publish such a discovery (just in case).
 
robjohn
robjohn
@Chris'ssis I think so... let me look
 
robjohn
robjohn
8:47 AM
@Chris'ssis Heh... I entered Sum[(Log[k]/k)^2, {k, 1, Infinity}] into Mathematica and got a useless result :-)
 
David Kirby
David Kirby
@Studentmath, interesting.. i gotta jet right now, but i'll be back later on this morning and will tinker with that a bit more
 
Chris's sis
Chris's sis
@robjohn I have to go now. Pls let me know if you find something about it.
 
Balarka Sen
Balarka Sen
9:35 AM
@Chris'ssis Unlikely.
 
Chris's sis
Chris's sis
@BalarkaSen A friend of mine said he discovered such a thing ... (I need to see to believe)
 
Balarka Sen
Balarka Sen
$\zeta'(2)$ has a closed form though. Interesting question.
@Chris'ssis Try arxiv for a preprint, as an initial publication.
 
Chris's sis
Chris's sis
@BalarkaSen If someone like me discovered such things, most probably would be accussed by intellectual property theft or something like that. I can guess many would claim it's their discovery (since I have no math background).
 
Balarka Sen
Balarka Sen
@Chris'ssis Your works are original, no? So why would anyone accuse you of thieving?
 
Chris's sis
Chris's sis
@BalarkaSen Hypothetically speaking ...
 
Balarka Sen
Balarka Sen
9:41 AM
You're being silly.
Nowadays, it doesn't matter if you don't have a math background.
People publish there work and none will claim it's theirs.
 
Chris's sis
Chris's sis
@BalarkaSen Maybe ...
 
Balarka Sen
Balarka Sen
If your works are original and of general interest, publish them!
It's far better to publish research works than publishing a contest book.
@robjohn
Given any formal power series $f_1,...,f_n$ in $t\Bbb C[[t]]$ which are linearly independent over $\Bbb Q$, then the field extension $\Bbb C(t,f_1,...,f_n,\exp(f_1),...,\exp(f_n))$ has transcendence degree at least $n$ over $\Bbb C(t)$. [Ax-Schanuel theorem]
What the heck is $t\Bbb C[[t]]$?
 
robjohn
robjohn
@BalarkaSen I assume complex power series without constant term
 
Balarka Sen
Balarka Sen
@robjohn $\Bbb C[[t]]$ is the ring of formal power series, but I don't understand what's the $t$ sitting up in the front.
 
robjohn
robjohn
@BalarkaSen it means multiply by $t$ to get rid of the constant term
 
Balarka Sen
Balarka Sen
9:51 AM
Weird notation for that.
@robjohn OK, thanks. That makes sense.
But wait a minute. The aren't $f_i$s linearly dependent over $\Bbb C(t)$? All having a factor of $t$?
 
N3buchadnezzar
N3buchadnezzar
Heyall
 
Balarka Sen
Balarka Sen
Ahoy.
 
N3buchadnezzar
N3buchadnezzar
How can I use that $\sqrt{1-t} \leq 1-t/2$ holds for $t \in (0,1)$ to prove that
$$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$$?
 
Balarka Sen
Balarka Sen
Yuck.
Back to Ax-Schanuel : Doesn't the linear independence of $f_i$s over $\Bbb C(z)$ mean we are excluding the $n$-tuple candidate $\{f_i\}$ outright?
Plus, excluding all $n$-tuples with multiple $f_i$s in it?
 
Daniel Fischer
Daniel Fischer
1
Q: Using $\sqrt{1-t}\leq 1-\frac t2$ to show that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

AleksanderI have a problem that tells me to use that $\sqrt{1-t}\leq 1-\frac t2$ for $t\in(0,1)$ to show by induction that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$ So far I have shown that $\frac{1\cdot3\cdot5\cdots(2n-1)}{1\cdot2\cdot3\cdots n}\leq 2^n$ from a previou...

induction
 
N3buchadnezzar
N3buchadnezzar
10:06 AM
Cheers
 
Balarka Sen
Balarka Sen
Should I jettison the dead weight then (i.e., @N3buchadnezzar)?
 
N3buchadnezzar
N3buchadnezzar
?
 
Balarka Sen
Balarka Sen
@N3buchadnezzar We are in a pirate ship. Didn't I say "Ahoy"?
 
Chris's sis
Chris's sis
@N3buchadnezzar I think you wanna see my proof ...
 
N3buchadnezzar
N3buchadnezzar
@BalarkaSen yarr
 
Balarka Sen
Balarka Sen
10:09 AM
That sum looks familiar.
Note that the identity below $(3)$ is weirdly resembles the Chudnovsky brother's formula, @Chris'ssis @N3buchadnezzar
These are called Ramanujan-Sato series. Who knew?
So rigorous, much results, very identities, wow.
 
Studentmath
Studentmath
There has to be a simpler way to do this.. sigh
Any probability lads around? Nothing too complicated, just trying to figure out a shorter way to do something
 
Balarka Sen
Balarka Sen
10:29 AM
@Chris'ssis Wiki says Ramanujan knew something to compute these series we don't. Interesting, isn't it?
 
Chris's sis
Chris's sis
@BalarkaSen Indeed. (I hope to be like Ramanujan one day)
 
robjohn
robjohn
@Chris'ssis There are claims that he had alien help :-)
4
 
Balarka Sen
Balarka Sen
I reckon Ramanujan must've thought about modular forms and the surrounding algebraic theories long before we did.
 
Chris's sis
Chris's sis
@robjohn lol :-)
 
Balarka Sen
Balarka Sen
As I always say : Ramanujan is the father of modular forms. He knew much than we think he did.
Perhaps he created them exactly with the same rigor we introduce these days.
 
N3buchadnezzar
N3buchadnezzar
10:31 AM
@DanielFischer That's funny, the exact same guy asked the same question on a different site. (Norwegian). Seems the problem with your explenation (which I found exellent) is that he seems to miss the crucial fact that $\prod_{k=1}^n k =\prod_{k=2}^n k = n!$
 
Daniel Fischer
Daniel Fischer
Hmm. How?
 
Balarka Sen
Balarka Sen
@Chris'ssis Well, it's never late to be like a classical hero in mathematics.
 
Chris's sis
Chris's sis
True
 
N3buchadnezzar
N3buchadnezzar
@DanielFischer Well it seems he does not understand why the product equals $1/\sqrt{n}$
 
Balarka Sen
Balarka Sen
I sure do hope to be like Erdos someday. But alas, I don't think I can ever be like him.
 
N3buchadnezzar
N3buchadnezzar
10:34 AM
I hope to be like Galois one day
 
Daniel Fischer
Daniel Fischer
@N3buchadnezzar Yes. I wonder how that is possible.
 
Balarka Sen
Balarka Sen
@N3buchadnezzar Whoa.
He was another guy like Ramanujan.
 
N3buchadnezzar
N3buchadnezzar
Swimming in p***y, and death by manly duel + mathematician.
 
Daniel Fischer
Daniel Fischer
@N3buchadnezzar Pah, better be like Ian Curtis. Or the Lizard King.
 
Balarka Sen
Balarka Sen
@N3buchadnezzar Ah.
@DanielFischer Lizard? Curtis Connors?
 
Daniel Fischer
Daniel Fischer
10:37 AM
@BalarkaSen James Douglas Morrison
 
Balarka Sen
Balarka Sen
Don't know who that is.
K, I gotta go.
 
skullpatrol
skullpatrol
later pal
 
Daniel Fischer
Daniel Fischer
 
N3buchadnezzar
N3buchadnezzar
@DanielFischer There was an idea of making a nirvana guitar hero game, where the controller would be the shape of a shot gun..
 
skullpatrol
skullpatrol
..not exactly sympathetic imo
 
Daniel Fischer
Daniel Fischer
10:45 AM
@N3buchadnezzar That could be considered a little in bad taste.
 
N3buchadnezzar
N3buchadnezzar
Yeah, although I would not say no to more nirvana
 
skullpatrol
skullpatrol
sometimes suicide is the only way to relieve the pain.
 
Chris's sis
Chris's sis
@robjohn the solution to the integral involving the dilogarithm
 
robjohn
robjohn
@Chris'ssis why are you writing $i^2$ instead of $-1$?
 
Chris's sis
Chris's sis
@robjohn it looks nice that way ( the negative result annoys me)
 
skullpatrol
skullpatrol
10:52 AM
@Chris'ssis Why? Haven't you accepted negative numbers yet :-)
 
Chris's sis
Chris's sis
@skullpatrol :-))))))))
 
robjohn
robjohn
@skullpatrol you've heard of the double negative, this is the double imaginary.
 
skullpatrol
skullpatrol
:D
 
N3buchadnezzar
N3buchadnezzar
I do not believe in negative numbers
 
Studentmath
Studentmath
@N3b what about non-integers?
 
robjohn
robjohn
10:54 AM
@DanielFischer I feel so old.
 
Daniel Fischer
Daniel Fischer
@robjohn Why that?
 
robjohn
robjohn
@DanielFischer people here don't know about The Doors...
 
N3buchadnezzar
N3buchadnezzar
._.
 
skullpatrol
skullpatrol
@N3buchadnezzar Do you believe in zero as a number?
 
N3buchadnezzar
N3buchadnezzar
nah
 
Daniel Fischer
Daniel Fischer
10:57 AM
@robjohn That's not age, that's lack of education. Like if they didn't know about Marlowe (both).
 
N3buchadnezzar
N3buchadnezzar
< 3 Riders on the Storm
 
robjohn
robjohn
@DanielFischer It has to do with when people grew up and what music was playing at the time.
 
Studentmath
Studentmath
I grew up with hip hop nods
 
Daniel Fischer
Daniel Fischer
@robjohn Good music is timeless.
 
Studentmath
Studentmath
There has to be an easier way to do Var(Y), I can't believe I actually have to do all these $Cov(Y_i, Y_j)$. I'd like to think they are all zero, but I can't find a good explaination. So I guess they aren't. Which means, I am stuck.
 
N3buchadnezzar
N3buchadnezzar
11:00 AM
@DanielFischer Cream?
 
Studentmath
Studentmath
They are exclusive, so they can't be independent..
 
Daniel Fischer
Daniel Fischer
@N3buchadnezzar Not so much my thing. Good, but not with the necessary je ne sais quoi.
 
Chris's sis
Chris's sis
 
N3buchadnezzar
N3buchadnezzar
Creedence?
 
skullpatrol
skullpatrol
When people say "grew up with;" that means, to me, the popular music when they were 13 to 19 years old.
 
Daniel Fischer
Daniel Fischer
11:02 AM
@N3buchadnezzar Definitely.
 
N3buchadnezzar
N3buchadnezzar
=D
 
robjohn
robjohn
@skullpatrol That's about right, but for me, I think it was until 26 (I listened to and still really enjoy music from when I was in grad school and a couple of years after)
 
N3buchadnezzar
N3buchadnezzar
 
skullpatrol
skullpatrol
Indeed, music is a very personal taste...
...like the music of the spheres :D
 
Hippalectryon
Hippalectryon
Hello there !
 
skullpatrol
skullpatrol
11:07 AM
Hi!
 
Chris's sis
Chris's sis
@robjohn in my proof there is a text that I forgot to remove ... "By letting $x\mapsto -x$ in $X$, we get"
 
Hippalectryon
Hippalectryon
http://math.stackexchange.com/questions/397097/uniform-convergence-of-sum-n-1-infty-frac-sinn-x-sinn2-xnx2/398085#398085 why do we have $\begin{align}
\sum_{n = 1}^N{\sin{(nx)}\sin{(n^2 x)}\over n + x^2} & = \sum_{n = 1}^N {1\over n+x^2}(S_n - S_{n-1}) \\
\end{align}$ ?
($S_n = S_n(x) = \sum\limits_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$)
 
robjohn
robjohn
It may have something to do with $\sin(a)\sin(b)=\frac12(\cos(a-b)-\cos(a+b))$
 
Chris's sis
Chris's sis
@Hippalectryon Isn't it related to the summation by parts?
 
Hippalectryon
Hippalectryon
@Chris'ssis it is
 
Chris's sis
Chris's sis
11:14 AM
@Hippalectryon OK :-)
 
Hippalectryon
Hippalectryon
@Chris'ssis but it comes after that part
 
Balarka Sen
Balarka Sen
Music is noise.
 
robjohn
robjohn
@BalarkaSen how long ago did you sell your soul? ;-)
8
 
Balarka Sen
Balarka Sen
a couple of years ago, i believe.
 
Hippalectryon
Hippalectryon
Don't worry you can get a cheap new one on ebay :3
 
Balarka Sen
Balarka Sen
11:17 AM
wait, what?
@Hippalectryon this is star-worthy
 
Hippalectryon
Hippalectryon
:)
'Mathematician looking for a soul - cheap - good quality - science resistant - contact me anytime'
 
robjohn
robjohn
@Hippalectryon no denomination specified?
 
Balarka Sen
Balarka Sen
Who starred?
You fool.
 
Parth Kohli
Parth Kohli
I did.
 
Balarka Sen
Balarka Sen
@ParthKohli I know you did.
 
Parth Kohli
Parth Kohli
11:20 AM
@BalarkaSen only robjohn's.
 
skullpatrol
skullpatrol
@robjohn But a zero denominator is not allowed ;-)
 
Balarka Sen
Balarka Sen
Wait, 3 stars?
 
Hippalectryon
Hippalectryon
@robjohn I'm not used to doing this - i already stole my 30 classmate's souls :D
 
Balarka Sen
Balarka Sen
@Hippalectryon And now you're selling them cheap?
 
robjohn
robjohn
@BalarkaSen gotta keep the product moving
 
Balarka Sen
Balarka Sen
11:22 AM
Just don't make a horcrux outta those, that's all.
 
Hippalectryon
Hippalectryon
@BalarkaSen Exactly, i need money for an ice cream :D
Once i have my ice cream, i multiply it into 2 ice creams thanks to the axiom of choice
And therefore i have infinitely many ice creams
 
Balarka Sen
Balarka Sen
No, it's Banach Tarski.
 
Hippalectryon
Hippalectryon
Isn't it based on the axiom of choice ?
 
Balarka Sen
Balarka Sen
yeah, but it's nonobvious.
 
Hippalectryon
Hippalectryon
Hey i'm not telling how i multiply ice creams to everyone :P
 
Balarka Sen
Balarka Sen
11:23 AM
Haha
 
skullpatrol
skullpatrol
If I take your ice cream away; is that negative ice cream?
 
Hippalectryon
Hippalectryon
@skullpatrol :O Don't ! nuuu my ice cream - hey wait, i'll just need to buy something to take the absolute value of ice creams :D (Unless it creates negative entropy duh)
 
Balarka Sen
Balarka Sen
You should bye a comathematcian, @Hippa, if you're interested in turning cotheorems into ffee.
 
Hippalectryon
Hippalectryon
@BalarkaSen HTCPCP is enough for me :)
 
Balarka Sen
Balarka Sen
?
 
Hippalectryon
Hippalectryon
11:27 AM
Hyper Text Coffee Pot Control Protocol
The Hyper Text Coffee Pot Control Protocol (HTCPCP) is a facetious communications protocol for controlling, monitoring, and diagnosing coffee pots. It is specified in RFC 2324, published on 1 April 1998 as an April Fools' Day RFC, as part of an April Fools prank. An extension is published as RFC 7168 on 1 April 2014 to support brewing teas, which is also an April Fools' Day RFC. Protocol RFC 2324 was written by Larry Masinter, who describes it as a satire, saying "This has a serious purpose – it identifies many of the ways in which HTTP has been extended inappropriately." The wordi...
Is the Banach Tarski's 'paradox' related to Cantor's bijection between lines and areas/volumes ?
 
Balarka Sen
Balarka Sen
Q : Why did Bourbaki stopped writing books?
A : Lang was only one person.
 
Hippalectryon
Hippalectryon
what is $||A||$ for $A\in\mathbb{R}^n$ ?
 
Balarka Sen
Balarka Sen
@Hippalectryon What is $||\cdots||$?
 
Hippalectryon
Hippalectryon
11:42 AM
Is it $\sqrt{\sum\limits_{k\in A}k^2}$ ?
@BalarkaSen The norm i think ... not sure
 
Balarka Sen
Balarka Sen
Maybe. Euclidean norm, you mean?
 
Hippalectryon
Hippalectryon
In a correction of an exercise I have (with some conditions before)
A is a square matrice and $X\in\mathbb{R}^n$
 
Balarka Sen
Balarka Sen
Is X also a matrix?
ah.
Then it's Euclidean norm.
 
Hippalectryon
Hippalectryon
That is .. ?
 
Balarka Sen
Balarka Sen
square root of sum of squares blah
 
Hippalectryon
Hippalectryon
11:47 AM
Ok ty
 
usukidoll
usukidoll
 
Hippalectryon
Hippalectryon
@usukidoll lol ?
 
usukidoll
usukidoll
what kind of question is that?
does it even have a question?
 
JMK
JMK
@usukidoll looks like the OP must learn some typesetting
 
Chris's sis
Chris's sis
@robjohn have you tried this one? I plan to compute it now. $$\sum_{n=1}^{\infty} (-1)^{n+1 }\frac{1}{n} \left( 1+ \frac{1}{2^2}+ \cdots+\frac{ 1}{ n^2}\right)= \zeta(3)-\frac{\log(2)}{2}\zeta(2)$$
 
usukidoll
usukidoll
11:54 AM
I actually went apes*** on someone because the op was taking an exam and posting his test paper on here
it was in another language
anyway, anyone on here know partial differential equations?
 
JMK
JMK
@usukidoll That's an irony, I was about to ask the same thing
@usukidoll What is the problem?
 
usukidoll
usukidoll
well I was just practicing but I think my outline book is making errors yo
 
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