Mathematics - 2012-12-02 (page 3 of 7)

@Argon, that's cool! I like term by term integration of power series.

Not that hard to come up with

6:59 PM

@N3buchadnezzar Not really, you are right

@Argon hmm.. nice. could be better if i had chat jax in mobile...

@Charlie HAHAHAHA!

@N3buchadnezzar, seems to be just a curious definition of $\sinh x$, geometric series formula, and something involving . . . integration by parts? Not sure on the last two lines.

two pieces, I should say.

@Argon damn it...

@Argon, $\Gamma$ is magical.

7:01 PM

@Charlie What?

@Limitless 'tis

Chatjax thing.aaron do you know fourier series?

@Limitless $$ \int_0^{\infty} \frac{x^{t-1}}{e^x - 1}\,\mathrm{d}x = \Gamma(t)\zeta(t) \qquad t \in \mathbb{N}$$

@Charlie I've done some, but not well :)

Hmm...

@N3buchadnezzar, content yet? XD

7:03 PM

What about your bridge, @argon?

@Limitless Never ;) I needed to make sure it was right.. =)

@Charlie My teacher said he will test them this week. He needs to mark everything first

@Argon bleh..

@Charlie Yep

A

7:05 PM

@N3buchadnezzar, want to help me out and show me how to apply that in the last two pieces? That is, $$2\sum \int_0^{\infty}x^{\alpha-1}e^{-(2k+1)x}dx=2\sum\frac{\Gamma(\alpha)}{(2k+1)^{\alpha}}=2\Gamma(\alpha)\lambda( \alpha)?$$

A

R

O

@Limitless What does the integral become?

N

Use the $\Gamma$ function

YAY!

7:07 PM

Yay

@Limitless Use the substitution $t = (2k + 1)x$ then you have the gamma function

DOOF. I didn't know that was the very definition. !!! Thanks.

@N3buchadnezzar, yep. And you just have to fix that other part into the expression, the $\frac{1}{(2k+1)^{\alpha}}$.

A quick question

Shoot

suppose I know that a certain quotient space's equivalence relation has to include certain pairs

how do i go about proving that it is spanned by them and nothing else?

7:11 PM

@AlexeiAverchenko, I'm going to leave that to someone far more competent than me.

Ditto

@Argon the pokemon?

Hahahaha!

I never really did watch that show...

i'm trying to see if it's easier to prove that the CW topology on a subcomplex conicides with its subspace topology than doing it by induction

@Argon, was $\Gamma(n)$ originally defined $(n-1)!$ for $\{n: n\in \mathbb{Z}\wedge n >1\}$ and then extended to all $\mathbb{R}$ via the integral representation? Or was it the other way around?

7:13 PM

it's neat that a CW complex is a quotient of its cells' disks

@Limitless Other way around, I think

Factorials were first

I think

@Argon hahaha

with the quotient mapping being the mediating morphism of the characteristic mappings of these disks

so i wondered maybe if i can describe the equivalence relation directly i can prove that the two topologies coincide by simply noting that the equivalence relation for the subcomplex is simply birestriction of the relation for the whole complex

@Charlie What's a good song?

it would be a pretty good proof, except that i've hit a brick wall trying to prove that the equivalence relation is of the certain form

7:16 PM

If you haven't already, this is interesting. Seems very many greats had different ideas about it, @Argon.

@Argon do you like gorillaz?

@Limitless I've seen them, awesome

KaliMa

can anyone shed some light? math.stackexchange.com/questions/249317/…

@Charlie Never heard of them. I do know the the Monkees, however :)

@Argon nice band. listen to a song called "clint eastwood"

7:18 PM

@Charlie Ok

Peter Sheldrick

what does this have to do with upvoting my stuff

@Argon :)

@Charlie Funny

Peter Sheldrick

@Argon, go to piratebay.se, download a torrent for everything gorillaz ever did, and then host said torrent on a powerful server with lots of bandwidth

?

i wrote Argon

Now

Peter Sheldrick

7:23 PM

lies and slander

@PeterSheldrick I will do it if you buy me a powerful server

@Argon i like their song