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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!
 
@N3buchadnezzar >8(
 
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 $\boxed{NFC}$
 
:)
 
@N3buchadnezzar funny
 
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
 
@N3buchadnezzar Hmmmm.....
 
@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
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
 
@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?
 
About math?
 
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
 
A
 
R
 
OH NO
 
Í
 
I AM RUNNING OUT OF PAPER IN MY ART BOOK
NOOOOO
 
7:46 PM
OH NOES!
L
 
I
 
I have 26 pages! 26!
Well
26 sides of pages . . .
So 13 pages in total.
 
A
 
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 $$\frac{1}{16^{3/4}} = \frac{1}{(16^{1/4})^3}$$
$2^4 = 16$
$\frac{1}{2^3} = \frac{1}{8} \qquad \blacksquare$
 

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