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00:27
hi @Daminark
00:44
Yo
@Antonios
Dom
Dom
Hi folks, I wonder if anyone could answer a pretty simple math question for me please
I'm pretty sure someone could
Dom
Dom
Given that where 'ax^2 + bx + c = 0', if the discriminant (D = b^2 - 4ac) is less than 0 there are no solutions, equal to 0 and solution is x = -b/2a, and if greater than 0 the solution is either of x = (-b-sqrt(D)) / (2a) and x = (-b+sqrt(D)) / (2a) ---------------- does this hold true for any value (instead of 0), or only where ax^2 + bx + c equals 0?
Well, your problem can be reduced to the following:
$ax^2+bx+c=d\iff ax^2+bx+(c-d)=0$. So, just use the previous criterion on the new equation.
So, basically yes, depending on what you mean.
Dom
Dom
00:51
hmm, thanks, I wonder if that'll work
I need to think about it for a few moments
One way to get a handle on what's going on is to think about the graph of a quadratic
No problem. Feel free to ask more if you need.
Dom
Dom
I'm applying it to a physics equation for projectile trajectory, and trying to refactor the equation to figure out how long it will take to collide with a boundary on the x axis
I figured I could translate the path over to move x to 0, which is what it looks like that would do
it's a parabola which either opens up or down. If it opens up and the x-axis is above/below the vertex of the parabola, then there's 2 / 0 real solutions
Dom
Dom
but it's complicated and my maths are out of practice
00:53
if you change the value you're setting the quadratic equal to, that's basically the same as shifting the x-axis up/down
so if said axis moves across the vertex, you'll change the number of solutions
Dom
Dom
@Semiclassical yeah all that matters is the trajectory if I can reverse whatever changes I make to make the calculation
hey @Daminark what's the tikzcd code for putting two arrows between the same two nodes so they don't overlap
Not sure but do you know if the tikz editor online?
nope, but i figured it out
Y_1\arrow[rr,"f",yshift=.85ex]&&Y_2\arrow[ll, "g"]
this will do it
00:59
the only exotic bit is yshift
Dom
Dom
@Antonios-AlexandrosRobotis Oh yeah it's coming back to me now, thanks
Dom
Dom
that was a pretty basic question :D
better to know than not to know :)
01:15
@Antonios-AlexandrosRobotis Shouldn't you just bend the arrows?
I think it looks better with straight arrows in this case
but that's also an option :P
Hi chat
@Antonios-AlexandrosRobotis Actually that looks really nice
I like Antonios , he is a nice guy, never let me down when it comes to say hi :D
Hi @KasmirKhaan
So transparant Jewel -.-
but hi :)
@Narcissusjewel yeah it can look nice. sometimes bending is better
anyways, any of you folks familiar with rep theory of finite Groups?
@KasmirKhaan Not sure what you mean
01:17
a matter of taste, I guess.
Serre book :D
@KasmirKhaan What about it?
I am doing a little bit right now @KasmirKhaan but I'm only learning.
I need help with some questions from that book
It is so hard -.- and its bad that we did not do linear algebra properly in my uni
01:18
@Narcissusjewel no second , third and forth exercices :D
Dom
Dom
Is there a formula to find the length of the quadratic curve from origin to a specified vector/set of coordinates on the graph?
@Narcissusjewel page 12 !
Dom
Dom
or from an arbitrary point, I chose origin to keep it simple
@KasmirKhaan Well tell me what you've tried for each and I'll help you
Or do so in an hour, since I shouldn't get distracted from my work yet
01:22
@Narcissusjewel well thanks for the offer, keep working on your thing, my main problem is not understandign the defintions well enuf ><
we meet again in 1 hour @Narcissusjewel
which defns @KasmirKhaan
hmm many tbh
character, tensor ,
okay, well what do you understand about characters thus far
Sym^2 and ALT^2
well it is a trace
Dom
Dom
@Antonios-AlexandrosRobotis thanks, it's a lot to consume
01:24
no problem @Dom, but that is the best way
okay so, the basic setup is we have a hom. $\rho: G\to \text{GL}_n(\mathbf{C})$ -- at least in the basic theory.
Right?
Dom
Dom
unfortunately I get stuck on a lot of symbols and I don't use them often enough to memorise or remember them
Dom
Dom
It's hard to even describe a lot of them and I don't think they copy-paste in most cases
that is the defintion of a representation
for each g in G, we associate with it an element of GL(V)
Yep. All we're doing with a character is we're sending $g\mapsto \text{tr}(g|_{V})$.
01:26
bad notation but GL(n,K)
That is, $\chi(g)$ is exactly that trace.
but this trace does not depend on basis?
Trace is invariant under change of basis.
I have some gaps in linear algebra
aha so it is well defined nice
a change of basis amounts to $M\mapsto A M A^{-1}$
01:28
okay say taht we did that for all elements of G
Trace is invariant under cyclic permutations of the argument: $\text{tr}(ABC)=\text{tr}(CAB)=\text{tr}(BCA)$.
since we only work with finite groups
In particular, it is invariant under change of basis.
thanks that is very usefull :D
01:29
but once one have the character of any element
it shouldn't really be a surprise: the eigenvalues of a matrix shouldn't depend on the basis, and the trace gives the sum of the eigenvalues
g in G
by that i meant, g---> a in GL(V)
we get few values
2,-2,0,4 say
what does that say to us?
@Semiclassical Semi the last thing i want to hear is that something is obvious -.- i hate that Word now
well, I think the "big deal" with characters is that representations (in this specific context) are uniquely determined by their characters.
hey, just because it's not surprising in retrospect doesn't mean it's obvious
01:34
I would recommend checking out the section in fulton & harris. It's fairly clear.
@EricSilva [F4, Section 3.2.29]...that can only be Federer
@Semiclassical @Antonios-AlexandrosRobotis thanks guys, i Think I should keep trying on my own and come up with good questions if needed for later :D
okay :) good luck
@Antonios-AlexandrosRobotis what section ?
phd level linear algebra above
01:35
section/lecture 2
hmm is this online ?
i meant is it a video or a book
lecture 2?
the book is divided into "lectures"
but it is in fact a book
haha makes sense =P sorry for being slow :D
01:36
I introduced the ambiguity :P
01:56
yo @Antonios-AlexandrosRobotis do you know little bit about Ext functor ?
for groups
for simplicity
not much, I'm afraid.
Do you know why Ext^i(A,G) = 0 for i >= 2 ?
oh wait I get it
@Antonios-AlexandrosRobotis do you want to know why ?
oh hi what's up?
sure
howdy @Antonios, Karim
No homological algebra for Ted, thanks.
@TedShifrin Hi :D Yeah homological algebra gives you magical things
02:07
how about some fractal homology?
Nope, none of that, either.
@Antonios-AlexandrosRobotis so any free group is project
projective
Look, it isn't that hard---it's just a graded group!
Of course, the grading is $\mathbb{R}$, but that shouldn't be too bad to work with, right?
so once we have Ext^2 it becomes free so it is projective and so the homology vanishes
I mean, it could be worse...
02:08
@Antonios-AlexandrosRobotis
@TedShifrin hahaah do you want to hear funny story ?
@TedShifrin So today I was sitting eating while thinking about some sequence (diagrams). I was saying to myself oh yeah you apply some yoga to the sequence and by chopping off at this end. My wife tells me what yoga and was telling me how mathematicians have weird phrasing
haha
@TedShifrin yeah my jokes don't make sense to me as well :D
lolz
@TedShifrin Meeting with the professor of the algebra course tomorrow [she couldn't make friday]. Trying to think of how to word what I need to say. I'm thinking that maybe I should hold a few short "review" lectures.
@Antonios-AlexandrosRobotis I just discuss ideas with my professors
I don't discuss like rigorous things
02:22
this is for teaching
oh
@Adeek wot
@0celo7 WOOT WOOT turn down to wh!t
haha
that music video always make me laugh
> "I don't know how many of you have ever met Dijkstra, but you probably know that arrogance in computer science is measured in nano-Dijkstras."
Also:
@AkivaWeinberger the witcher 3?
you don't know what the witcher 3 is?
Oh apparently it's a game
with a character named Dijkstra
You know, when someone says "Dijkstra", some obscure video game is not the first thing that comes to mind...
Depends on the person, I'm guessing
In any case, Edsger Wybe Dijkstra (Dutch: [ˈɛtsxər ˈʋibə ˈdɛikstra]) was a computer scientist
The "ij" sounds like the "ij" in @Krijn
Hey y'all. Trying to work problem 2.8 here --http://www.math.cornell.edu/~sjamaar/manifolds/manifold.pdf

It's a bit cumbersome to write out.

In any event, I'm getting dg = f_1(x1,...xn)dx1 + f_2(0, x2, ..., xn)dx2 + ... + f_n(0, 0, ..., 0, xn) dxn. Why is that the same as alpha (as described in the problem)? I'm not taking advantage of the fact that alpha is closed, i.e. d alpha = 0 wherein lies the problem in all likelihood.
02:45
His name is fun to say @Akiva
Who, Dijkstra or Krijn
Edsger Dijkstra
@XanderHenderson obscure?
my god
are you serious?
@XanderHenderson ehh. I’m not sure one can call Witcher 3 ‘obscure’ among video games
you're not sure!?!?
how about "it's literally insane to call it obscure"
I was trying to be diplomatic :p
@0celo7 Wrong crowd
I used to be quite into video games. Long before witcher 3 came out though, I imagine.
@AkivaWeinberger what do mathematicians do if not play video games
@XanderHenderson it sold 10 million copies
As of last March it had sold 10 million copies
02:56
@0celo7 presumably other things :P
@Antonios-AlexandrosRobotis How can a young male in 2018 not play video games
Well not all mathematicians are young males in 2018. But, on the flip side, you can quite simply have other hobbies lol
I don't play video games
wtf
why not?
Actually, most of the friends I had in undergrad who were also math people didn't play video games.
02:57
@0celo7 is that a lot?
TBH I don’t play video games either
is this the twilight zone?
I’m too cheap to pay for it
I recently bought Quantum Conundrum and then realized it's PC only
(I have a Mac)
I don't play video games anymore because it doesn't feel like time well spent for me right now.
02:58
On the other hand it was like two dollars
no differential geometers here?
Because, I do other things? I like listening to podcasts and working out and sleeping. And I don't have time to do all that and school and play video games.
@Antonios-AlexandrosRobotis it's definitely not time well spent, but that isn't the point
That said, I do follow a decent amount of video games news
@JoeShmo what do you need
02:58
Not to cast judgement on other people – I just can't do it anymore.
I used to be really into Minecraft
I dunno, high school is a big time drain for me
Is Witchar 3 anything like Castlevania? 'cause that was a great game...
@0celo7
Trying to work problem 2.8 here --math.cornell.edu/~sjamaar/…

It's a bit cumbersome to write out.

In any event, I'm getting dg = f_1(x1,...xn)dx1 + f_2(0, x2, ..., xn)dx2 + ... + f_n(0, 0, ..., 0, xn) dxn. Why is that the same as alpha (as described in the problem)? I'm not taking advantage of the fact that alpha is closed, i.e. d alpha = 0 wherein lies the problem in all likelihood.
or is it more like Megaman?
In genre, yes
02:59
@XanderHenderson it's like Skyrim
and if you've never heard of that, we're done here
well, then I guess we are done
Skrim sounds like a silly name for a game
holy shit you're serious?
I do like video game music, but for that you just need YouTube and a good sound system
02:59
Megaman makes sense... you plan a MAN who is quite MEGA
@XanderHenderson how old are you?
(Hm. If it's written for the NES, do you need a good sound system?)
Skyrim was meh

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