Recent content by abelian jeff

Anyone in pure math hear from Brown yet?

If you don't mind me asking, how were you informed that you were rejected from Cornell? Hang in there, I'm sure you'll start hearing some good news.

Good point, Jason! That's very encouraging!

Does anyone know if schools generally send out all of their first-round acceptances together? According to GradCafe, Cornell sent out at least one acceptance today via email. I applied to Cornell but haven't heard anything yet. Should I take this to mean I've been rejected, or waitlisted at best?

Brandeis isn't admitting any math grad students this semester. They made this decision about a month ago, and they refunded hopefuls' application fees, with the option of deferring their applications until next year at no charge. Whether this is a result of the current economic situation, they...

So, the Joint Math Meetings were awesome. I highly advise any undergraduates who can get funding (or pay their own way) to attend MathFest and JMM. Both are great experiences.

I have an art history minor...does that count?

I went to MathFest two years ago, and it was amazing. I'll be attending the JMM next week, and if I remember, I'll let you know how it was, but my guess is that it, too, will be amazing. (I'm currently a senior undergraduate.)

Hi vertices, The complex contour integral (2) you are considering equals the original real integral (1) plus the integral over a semicircle of radius R. It turns out that, in a lot of cases, the integral over that semi circle goes to zero as R goes to infinity, so you get the equality you...

Great! You're finished! :smile: As a bonus, you can easily adjust the argument to see that, in general, if x is the order of a, and y is the order of b, the order of (ab) is actually lcm(x,y). Of course, when gcd(x,y)=1, we have lcm(x,y)=xy, hence what you just showed.

Hi HF08, I think you're getting too caught up in the notation. My k's, etc. are just letters I'm using to illustrate the main ideas. Look, this is what you have shown: If x is the order of a, and a^k=1, then x divides k. Similarly, If y is the order of b, and a^j=1, then y...

Exactly! We know that x divides k in that case. So now look at this: (ab)^k=a^k b^k=1 (since \mathbb{Z}/n\mathbb{Z} is abelian) and use the facts that: 1) gcd(x,y)=1 and 2) the order of ab is the least positive integer such that (ab)^k=1 to finish the problem.

Perhaps the fact that I used "n" in my post was confusing. The "n" that I was using was not the same "n" that you are modding out by. I'll retype my idea. I'm just going to talk about the order of an element, not "ord_n(a)" since this is cumbersome, and we both know that we're working mod n...

Let's just look at one element, a. Suppose a has order k. Now, a^n=1 if and only if...what? What must be true about k and n for a^n=1? Now use that to look at when (ab)^n=1.

Thanks for that information. I can't call and get my scores until May 14th, and I've been wondering what kind of scores should be my "cut off", such that I will decide not to take the GRE again in October. Many of my professors think that I should take it twice no matter what (unless I get an...