|Tarantulafudge||TRWBW: I'm confused with your answer|
TRWBW: how do I get it in y=a*x^2 form
|TRWBW||Tarantulafudge: just saying that once you solved this first example, you can solve any other example by rotating and tranlating it to be in the form y=a*x^2|
|Tarantulafudge||TRWBW: oh okay|
|Adlai||what's the directrix again?|
|Dovetailed||Hey, don't suppose anyone knows a free piece of statistical software which can be used to compute robust correllation matrices?|
|demute__|||Steve|: isn't it possible to do a newline if i use displaymath?|
|Tarantulafudge||I'm so lost -.-|
|TRWBW||Tarantulafudge: or start with wikipedia article on conic section even|
||Steve|||demute__: Why don't you tell me what you are really trying to do and I can offer a suggestion for how to go about that.|
If q(x) is a polynomial over some field and we define the derivative q'(x) in the obvious way without using limits, is there some simple way to consider the gcd(q(x), q'(x))?
I wrote out the first iteration of the algorithm, but I don't think it's going to get any simpler.
|Kasadkad||Isn't it just q(x) with the multiplicity of each zero reduced by 1?|
||Steve|||I have no idea.|
What makes you think that?
|TRWBW|||Steve|: you can show that (qp)'=qp'+q'p|
|Steve|: so if you have p=(x-r)^kq, then (x-r)^(k-1) divides p'
|Steve|: ignoring things like k=0 in the field
Do I need to do anything like look at it in the splitting field of the polynomials so that I have linear factors?
|Tarantulafudge||Okay I read some of the wikipedia stuff but it really didn't help much|
I'm not sure how to find the coordinates of the vertex because my book has an example but in y=ax2+bx+c form
|demute__|||Steve| using eqnarray instead of displaymath solved the problem.|
||Steve|||Don't use eqnarray.|
It puts ugly spaces in the equation.
Use align (as I suggested above).
|TRWBW|||Steve|: the big idea here is if r|p, then p=r*s, p'=r'*s+r*s'. so r|r'*s, which has smaller degree. which is impossible. so gcd(q,q')=1|
|Steve|: impossible by induction
|powerfox||I have z/ (z^2 + z +1) dz If I'm integrating it by parts I have a very strong integral...|
Maybe I should do something else?
||Steve|||Don't do it by parts.|
u = z^2 + z + 1, du/2 = z dz.
*grumble, grumble* Stupid calculus. *grumble grumble*
|powerfox||Thanks. I got it :) You are saving my stydy task, folks! Thanks!|
|Cale||gcd(f,f') is a good way to notice multiple roots of f.|
||Steve|||TRWBW: how did you get gcd(q,q') = 1?|