Recent content by Quantumpencil

That's the impression I get... there was substantial grade inflation, and my GPA was far lower than anyone else I knew in college. The grad school standard seemed to be around a 3.8, and most of those students took a lot of graduate math classes which I could not handle.

I went to Harvard as an undergrad, but I was exceedingly poorly prepared in more or less every respect. I ended up finishing with a 3.3 GPA and a 3.1 concentration GPA (My pure math classes are higher (3.45), the lower grade is because I counted four physics classes towards my major that I...

There are two groups of order 21, even though it's isomorphic to the direct product of the Cyclic Group of Order 3 and the Cyclic Group of order 7. 3 and 7 are co-prime. EDIT: Nevermind, didn't read the "abelian" in the problem. Your proof is good.

You have an integrand that is the product of two functions. There is a way to rewrite that integral, by splitting up the parts.

inspect 0 more. what does an open neighborhood (0, x) map into?

I think your proof is fine. I remember doing this problem last year and I think that's basically the solution I used; there may have been some subtlety I overlooked, but I don' think so.

The statement is pretty simple. It says that if you have a Group of order n, where n has factoriaztion p^k * m, where p is a prime number, then there exists a subgroup of order p^k. The proof is a bit tricky, since it's in direct. What about it is confusing you?

You need to show that b* A(v_1) + ... b_n A(v_n) = 0 implies b_1 ... b_n equals zero, right? Well, you know since A is invertible, what is it's kernal?

Yeah, I like that one.

Follow the hint. Think about for example, the The circle minis one point being homeomorphic to the line . You should be able to be explicit about this homeomorphism. It makes sense right? take away one point, and then "fold"" the surface out. Show that R is homeomorphic to any open interval...

Are you saying that your null space is spanned by {-3, 0, 1} and {5, 1, 0}? Your notation is a bit confusion to me; but I believe that they are asking you to describe what kind of "Space" this is. What do two vectors span?

So, |f(x)| ≤ S'(x-a) ≤ S'(x_0 - a) ≤ SM (x_0 - a). Hence, S = 0 if M(x_0 - a) < L. Therefore, f(x) = 0 on 0 ≤ X ≤ L. I don't think this is obvious enough to just state. It works if (M*x_o-a) is less than one, but this requires a specific choice of x_o, and perhaps not the one which equals b.

Yeah. So here is Rudin's Hint for doing this problem: If you fix x_o < L, then using mean value you know that f'(c)(x - a)=f'(x)- f(a), for some c < x_o. and x in 0<x<L If S is the supremum of f(x), and S' is the supremum of f'(x), you've shown f(x)<S'(x - a) < SM(x - a), x<x_o From...

right, blah, that's what I meant. Mean value theorem would be my guess. I might try to work though it.

You'd be done if you could show that f'(x) = C, for any constant C right? for then f(0) would imply that f(x)=0.