@robjohn If I post an answer and after getting some up-votes I make it community-wiki. In that case will I loose rep? Will I loose rep if someone down-votes that answer after I make it community wiki?
@Committingtoachallenge, I hope so. We've learned two new verb tenses in French this week, and along with those are tons of irregular verbs. Ugh. I will have a lot of studying to do for that.
French is my first language, but I'm in an accelerated version right now that's combining two semesters into a single semester. It's definitely nontrivial. Probably the hardest non-math course I've taken here at UGA.
@WillHunting, I should. I'm currently sitting on Rosetta Stone myself. I've worked through the first two CD's in the past, but it's been a while. Now that I've gotten a LOT of the theory behind my belt, it would be nice to start over in a more immersive environment to build my working vocabulary up.
@KajHansen Many people think Rosetta Stone is a waste of money. Also, try Routledge's Colloquial French and Colloquial French 2, and Teach Yourself's Complete French and Perfect Your French, these four.
@Integrator I just linked to it, but since I used the link to your post, it came out as a post snippet. Now that I've added more text, it shows up as a link
Keep at it @beginner. Don't forget to give yourself lots of homework. That's how you truly learn math, imo. It's never enough to just read (a trap I fall into all the time).
Yes! That's a great idea @beginner! One of the things Ted had us do on the first day of class was to show that the medians of a triangle always intersect at a single point using vectors.
any one knows where to search for this: I need to find out relation between zeros of polynomial p1 and p2 on their own, vs. zeros of the combined polynomial (1-z)p1+z*p2 where z is a number between 0 and 1. Actually just knowing the relation between p1,p2 and the combined one p1+p2 will help. Not sure if there is a theory on this I should look for
by trial and error, I found that if p1 and p2 are stable polynomials, then (1-z)p1+z p2 is always stable as well. but need to proof this.
$$F(\theta)=\sin(\theta)\int_{-l}^{l} e^{-ikz\cos \theta} h(z)\,dz$$
We know that $F(\theta)$ is defined on $0\le \theta \le \pi$ and $h(z)$ is defined on $|z|\le l$
What is the period of $F(\theta)$?
Is $F(\theta)$ even or odd?
What if we change the dummy variable $z$ in the integral? does it a...
$$F(\theta)=\sin(\theta)\int_{-l}^{l} e^{-ikz\cos \theta} h(z)\,dz$$
We know that $F(\theta)$ is defined on $0\le \theta \le \pi$ and $h(z)$ is defined on $|z|\le l$ and $z$ is real in this case, but it might be complex too!
That is all we know, how can we find the $h(z)$ by having knowledge of $...
@MikeMiller I am trying to find an example of covering maps $p : A \to B$ and $q : B \to C$ with $A, B, C$ path connected such that $r = q \circ p$ is not a covering map. This can be done for non path connected spaces, say, $S^1 \times \Bbb N \to S^1 \times Bbb N \to S^1$, the latter map being projection and the former map just wraps each circle $n$ time around another, $n$ goes to $\infty$.
Maybe $r$ is always a covering space when $A, B, C$ are locally path connected but I am unsure.
What we know is that $\pi_1(p)(A, a_0)$ is inside $\pi_1(B, b_0)$ and $\pi_1(q)(B, b_0)$ is inside $\pi_1(C, c_0)$
Stay up for 2-3 days straight and then go to bed at a reasonable hour on the 3rd day. I speak from experience from when I took differential geometry, 2nd semester abstract algebra, and mathematical methods in physics all at once.
Since you haven't studied inverse limits, here's something that might surprise you : $Gal(\overline{\Bbb Q}/\Bbb Q)$ is the inverse limit of $Gal(L/K)$s, $L$ ranging over all finite extensions of $K$. The morphisms are restriction homomorphisms, but in that case you need to order the extensions of $K$ properly. That, I think, is done by use of the Krull topology which I haven't studied.
But to do that you need to "order the field extensions". For example, both $\Bbb Q(\sqrt{2})$ and $\Bbb Q(\sqrt{3})$ are extensions of $\Bbb Q$, but there is obviously no inclusion.
