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23:00
yeah
You don't install it, click the button
yeah got it now
eh this seems too simple to be worth 1/7 of our total score
it looks like what i was doing was the number of ways to select something, not the coefficient
It's really amazing computing this one without pen and paper in one line proof $$\int_0^1 x^m \log^n (x) \ dx, \space m,n \ge1, \space m,n \in \mathbb{N}$$
@Chris'ssis Yeah that one was always fun
@Committingtoachallenge did you ever think of computing $$\lim_{x\to 0} x^m \log^n (x)$$?
23:10
For arbitrary $m,n\geq 1,m,n\in\Bbb N$?
@Committingtoachallenge Yeap.
It's pretty cool.
What methods do you use for that one?
Hi @Committingtoachallenge @Chris'ssis
@ᴇʏᴇs Hi
23:12
You should change your name to something that I can type :)
I thought my midterms were next week but they begin tomorrow
@Committingtoachallenge Eyez is only 4 characters
But I can't tag you without clicking your name :\
Normally I just type a few letters like @Chr
@Chris'ssis Does this limit to $0$?
@Chris'ssis When $m\geq n$?
@Chr is 4 characters if you don't include the Shift button which makes it 5
But I can't tag you without clicking your name @Eyes Doesn't tag youy
@Committingtoachallenge I won't tell you since it's nice to do it alone.
23:15
Only @ᴇʏᴇs tags you, and I have to grab the mouse
@Chris'ssis :'(
@Chris'ssis Okay, but I will have to do it next week since I have so much worrkkk
You can do @ + Tab button which is still 3 including Shift button
Oh wow yes
That is good thankz
@ᴇʏᴇs
23:20
Quick question: $\forall n \geq 1 \; \sqrt{n^2-1} \notin \mathbb{Q}$?, I tried the standard $\sqrt{2}$-esque proof, but I don't know if I did it correctly.
You just have to show that $n^2-1$ is not a perfect square (i.e., square of an integer). Then if you know that $\sqrt{c}$ is either an integer or irrational, you're done. And you can prove this last bit by following the same idea as the $\sqrt{2}$ proof.
@MikeMiller and $n^2 - 1 = r^2$ implies a contradiction?
because $r^2 = (n-1)(n+1)$, and this would only hold if $n-1 = r = n+1$ (by the fundamental theorem of arithmetic), from which $-1=1$?
23:46
Why does it hold that $\cos{(kx- \omega t)}=\cos{ k \left( x+\frac{2\pi}{k})- \omega t\right)}$ ?
Going back to binomial stuff how do we go about finding the coefficient when the inside of (x+y) has coefficients, such as (x^3+y)^5 (random made up numbers)?
also, how do you do exponents in latex?
i haven't messed with this in about a year
Hi @meer2kat
It's ^
@ᴇʏᴇs yeah it didn't change it
@ᴇʏᴇs and how are you?
@meer2kat You need to surround your LaTeX code with dollar symbols
hi mr eyeglasses
23:48
like $\ x^3 $
@TedShifrin @ThomasAndrews @DanielFischer Do you have an idea why it holds that $\cos{(kx- \omega t)}=\cos{ k \left( x+\frac{2\pi}{k})- \omega t\right)}$ ?
Hi @TedShifrin
@evinda: You need $\omega/k$ in there.
@meer2kat One more dollar sign at the end to complete it
@meer2kat You have ChatJax running in here, right
Because $kx-wt+2\pi = k(x+\frac{2\pi}{k})-wt$. @evinda
23:49
@ᴇʏᴇs yep
Oh ... Or a parenthesis in front of the $k$.
hi @Thomas
@meer2kat Okay, you had it before
i quit
I quit years ago, @meer2
23:51
@ThomasAndrews So because of the fact that we know the following? $\cos(x+2 \pi)= \cos x$ ?
@ᴇʏᴇs this is my new favorite video
Yes, @evinda
Nice, thank you @ThomasAndrews @TedShifrin :)
@TedShifrin i think ^ is common enough that people will get what i'm sayin, ya know? i'm not a programmer for a reason. me and code don't get along
I'm not a programmer, either ... but I've managed to typeset 4 books in LaTeX. :P
23:53
@meer2kat $\$x\textrm{^}2\$$ should work otherwise your ChatJax must be not working or something
Can't even escape a carrot
mr eyeglasses, perhaps @meer2 doesn't have ChatJax loaded for here.
you want celery instead, mr eyeglasses? :D
@TedShifrin no i have it on
@ᴇʏᴇs that showed up fun. bolded and with red things
no backslash in front of the caret ...
$x^3$
@meer2kat That's weird, I saw your $x^2$ working fine earlier, so you should've seen it too
23:55
HA GOT IT
okay now i have to figure out this binomial crap again
Reviews just came in, @Ted. Let's see...
coefficient of $x^6y^12$ of $(x^2+y^3)^7$ is 0, yeah? it's not possible
nah, the coefficient is $\binom{7}{3}$

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