6:00 PM
how can i find solutions to 4a^2(b^2 + c^2) = 5b^2c^2 ?

@Argon Lets say you have an integral J, and some integral K

ok

Then sometimes it can be easier to solve the system \begin{align} \alpha K + \beta J & = d_1 \\ \delta K + \epsilon J & = d_j \end{align}

@N3buchadnezzar I've seen stuff like that before

To find $J$ and $K$.
A prime example is ofcourse $$K = \int \frac{\cos x}{\cos x + \sin x} \mathrm{d} x$$

6:03 PM
What would $J$ be?

@Argon Take a guess ;)
Often much of the cleverness is choosing a wise $J$..

Hmmmm
@N3buchadnezzar Yep, would the top become $\sin$ to get $\int dx$?

@robjohn do not spoil it!

Give me another then:)

6:06 PM
@Argon Indeed!
@Argon Well you still have to set up the system ;)

Well, $K+J = x+C$
and...

aaannnndddd?

$K-J=-1+2K$
That's not right

$K-J$ is easy to integrate btw

Is it not $2K-1$?

6:12 PM
@N3buchadnezzar If you see the trick...

@robjohn Trick?

$$\frac{2\cos x}{\sin x+\cos x}-1$$

@N3buchadnezzar that $\mathrm{d}(\cos(x)+\sin(x))=(\cos(x)-\sin(x))\mathrm{d}x$

@robjohn Substitution is no magic trick.. =)

$\log |\sin x+\cos x|$
Oh noes

6:15 PM
@Argon Yay!

$K+J = x+C$
$K-J=\log |\sin x+\cos x|+D$
$-2J = \log |\sin x+\cos x|-x+C$

@N3buchadnezzar another approach is $\cos(x)+\sin(x)=\sqrt{2}\cos(x-\pi/4)$

And then add them to find $K$, easy peasy.
@robjohn Yeah, or Weierstrass substitution

Hi everybody!

I just wanted to show of the nifty $J$, $K$ technique.

6:18 PM
@N3buchadnezzar That probably complicates this integral, but it is possible.

$\frac{x + \log |\sin x +\cos x|}{2}$

@Argon very good.

@Charlie Hi Marilia!

Hides the big box labeled secret stuff not for charlie
@robjohn It is a fairly easy integral, does not get that messy.
@Argon Very nice

Cool trick!

6:20 PM
@Argon hi aaron!

@N3buchadnezzar $\boxed{NFC}$

:)

I am not supposed what it contains, but it is for christmas and makes unicorn sounds.
@Argon Yeah, loads more integrals can be solved this way

@Argon So are you saying that I have been summing fish?

6:22 PM
@robjohn Poisson summation?
:) Yep!

@Argon yeah

$$K = \sin( \log x), \quad J = \cos( \log x )$$

@jayesh hi

@Argon Sorry for posting that one, it involves some more techniques...
$$K + J = \alpha \\ K - J = \beta$$
Dont use subs

6:27 PM
Ok
$x(\sin \log x +\cos \log x) - \int \cos \log x - \sin \log x \, dx = K-J$

How are you @argon @n3buchadnnezar
?

@Charlie Fine, as always

@Argon wrong

hmmm

@Argon hmm...

6:31 PM
What did you do to try to solve J + K?

@Charlie Hmmmm....
@N3buchadnezzar Yes. A futile attempt

What is different between the K integral, and the J + K integral?

@Argon hmmm hmmm

@Charlie hm hmhm hm

$\begin{bmatrix} H & m \\ m & H \end{bmatrix}$

6:32 PM
@N3buchadnezzar I noticed that $d/dx \sin \log x +\cos \log x = \frac{\cos \log x -\sin \log x}{x}$

@Argon hhhhhhhhmmmmmmm

I said no substitutions ;)

@N3buchadnezzar That's not a substitution. I used by parts

Ah, parts is correct.
Try parts on just $\sin ( \log x )$ or $\cos ( \log x)$
Should yield some absurd cancellations ^^

$\int \sin \log x\, dx = x\sin \log x - \int \cos\log x\, dx$
No?

6:35 PM
@N3buchadnezzar that sounds funny

@Charlie He sounds like he is ordering food

@Argon Indeed

I see

Exactly @argon

anyone good with diophantines

6:36 PM
@Charlie Whyyyy?

$K+J=x\sin\log$

@Argon Yes =)

Woopie!

@N3b no substitutions

The $\sin \log x$ and $\cos \log x$ are some of my favourite integrals, so many techniques and tricks!

6:38 PM
$$\int \cos \log x\, dx = x \cos \log x +\int \sin \log x \, dx \implies J-K = x\cos \log x$$

@Argon Almost

@N3buchadnezzar Did I flip them?

@Argon Should be right now =)

Ok
$J+K = x\sin\log x$
$J-K = x\cos\log x$
$J = \frac{x(\sin\log x + \cos \log x)}{2}$
$K = \frac{x(\sin\log x - \cos \log x)}{2}$

Indeed
Cool integrals?

6:43 PM
Yep

I love that cancelation by parts trick
I performed parts once, and that solved my entire integral? Well okai..

Neat

I know several integrals that is almost impossible to solve without using it

Sorta reminds me of $\int_0^\pi\log \sin x\, dx$
You know it?

@Argon I gave you a link for it

6:48 PM
@N3buchadnezzar Wait... you did?
1 sec

Bye guys, so long.

@Charlie So soon?

Oh, right!

@Argon I'm not doing anything around, so...

6:50 PM
@Charlie Well....hm..
@Charlie Do you like integrals?

@Argon well..hmm what?

@Charlie Thinking

Yes i like

@Charlie Nonelementary?

Yeah cool

6:53 PM
@Charlie I will get you a good one

@Argon nice

@Argon!!!!

@Limitless Heyyy!

It would appear that I am possibly hitting my rep cap.
NOOOO.

@Limitless You mean YAY!

6:54 PM
The site says I have gained 185 rep today. I don't know how it's calculating that, @Argon.

Have yourself a merry little hanukkah let heart be light...

@Charlie :)
Almost!

@Argon OH. I see. Different time setting!

@Argon ;)

@Charlie $$\int_0^\infty \frac{x^{\alpha-1}}{\sinh x}\, dx=2\, \Gamma(\alpha)\, \lambda(\alpha)$$?

6:56 PM
What is $\lambda(\alpha)$?

$$\lambda (a) = \sum_{k=0}^\infty \frac{1}{(2k+1)^a}$$

@Argon, that certainly "simplifies" the problem . . .

$$\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}}=2e^{-x}\frac{1}{1-e^{-2x}}=2e^{-x}\sum_{k=0}^\infty e^{-2kx}=2\sum_{k=0}^\infty e^{-(2k+1)x}$$
$$\frac{x^{\alpha-1}}{\sinh x}=2\sum_{k=0}^\infty x^{\alpha-1}\, e^{-(2k+1)x}$$
$$\int_0^\infty \frac{x^{\alpha-1}}{\sinh x}\, dx=2\sum_{k=0}^\infty \int_0^\infty x^{\alpha-1}\, e^{-(2k+1)x}\, dx=2\sum_{k=0}^\infty \frac{\Gamma(\alpha)}{(2k+1)^\alpha}=2\,\Gamma(\alpha)\,\lambda(\alpha)$$
(I didn't come up with this)

@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

@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

7:05 PM
@N3buchadnezzar, want to help me out and show me how to apply that in the last two pieces? That is, $$2\sum \int_0^{\infty}x^{\alpha-1}e^{-(2k+1)x}dx=2\sum\frac{\Gamma(\alpha)}{(2k+1)^{\alpha}}=2\Gamma(\alpha)\lambda( \alpha)?$$

O
@Limitless What does the integral become?

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

@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

what does this have to do with upvoting my stuff

@Argon :)

@Charlie Funny

@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

7:23 PM
lies and slander

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

@Argon i like their song

@Limitless I like your "L."

@Argon, thank you! I decided to do something creative.

@Limitless It looks like a Laplace transform L
@Charlie How is Fourier coming along?

7:26 PM
@Argon, it is close. It's $\mathfrak{L}$ whereas Laplace transform L is $\mathcal{L}$.

@Argon i will start to study it

@Charlie You can figure out what $\zeta(2)$ is!
:)

:)

@Argon hilarious

7:31 PM
@Charlie They are so funny

Yup!
@Argon do you like anything else but integrals?

@Charlie Sums :)

What else?

euler-maclaurin formula?

@Charlie Classical numismatics

7:35 PM
Hmm

Ancient history

Cool

You?

Other then math

7:37 PM
orange juice?

@PeterSheldrick What?

i like orange juice...

Physics, neoroscience, art...

@PeterSheldrick Good. Me too

Me too

7:39 PM
Yay!

Yay!

O.J.

...

Simpson

Haha
I like him in naked gun

7:43 PM
?

O j simpson

Oh, hahaha

Do you like this movie?

@Charlie Nope
Golden gun

@Argon :(

7:46 PM
@Charlie Man with the golden gun
James Bond
M

OH NO

I AM RUNNING OUT OF PAPER IN MY ART BOOK
NOOOOO

7:46 PM
OH NOES!
L

I have 26 pages! 26!
Well
26 sides of pages . . .
So 13 pages in total.

Yay!

7:49 PM
giving up
thanks anyway oldjohn, alexei, etc
i think it may be too hard for me

@Argon i like bagels

@Charlie Me too. And lox

@Argon :)

Starring?

Accidentally

7:52 PM
@Charlie It's Hebrew at 1:52, btw!
E(=
Buckteeth

I know Aaron
Such a nice language. what does he say?

Yay!

Hey gang

@Jordan hey!

I can't figure out how to do this, but how do you do $16^{\frac{-3}{4}}$ in your head

7:57 PM
@Charlie Hahah! Mahamoud, my freind howz it going?
@Jordan Hie!

Ah!

@Jordan $$\frac{1}{16^{3/4}} = \frac{1}{(16^{1/4})^3}$$
$2^4 = 16$
$\frac{1}{2^3} = \frac{1}{8} \qquad \blacksquare$