Recent content by factor

I'm trying to show that the Frenet frame structural equation for a curve in R_3 can be written in the following form for a vector W(s): T'(s) = W(s) x T(s) N'(s) = W(s) x N(s) B'(s) = W(s) x B(s) The problem I'm having here is that I define first that T(s) should be the unit tangent at...

I'm wondering exactly why |G|/2 of them form a subgroup of G. By Lagrange's theorem the order of any subgroup must divide the order of the group, but suppose your group has order p where p is a prime, then it clearly can't have a subgroup of order 2 unless p is in fact equal to 2 in which case...

Well if they all contain rational points, then by the fact that they're all disjoint we can choose one from each (by the axiom of choice) and this rational uniquely identifies the set because it's only in that single set. So this list of points in S is at worst countably infinite.

This has to do with defining an equivalence relation on the set right? Defining two elements to be equivalent if and only if there is some open interval in your set that they share? The relation partitions the points of the set into open intervals which must contain rational points by their...

The proof that they're disjoint seems simple enough. Pick any x and y, where x =/= y, in S and then look at the intervals (x,s(x)) and (y,s(y)), and suppose the contrary. We can assume without loss of generality that x < y. Then there exists some a which belongs to both intervals. But since...

Those intervals should have empty intersections with S by the definition of the function s(.). But it seems like at least one of them should have a non-empty intersection by the uncountability of the set S.

Dick, all I can get from that right off the bat is perhaps a contradiction with the fact that S is uncountable. Because if S is well ordered, it certainly has a smallest element a. So now I pick the immediate successor to a, s(a). As a result any subset of S, which does not include a, has s(a)...

Homework Statement Prove that any well-ordered subset (under the natural order) of the real numbers is countable. Homework Equations None. The Attempt at a Solution My attempt thus far has been to prove by contradiction. I didn't see a very clear way to get from well-ordered subset...

Thanks a lot Iblis, your proof made a lot of sense. We never really discussed much of the theory of Fourier series, it was mostly a couple of lectures intended to demonstrate an application of the study of Hilbert spaces and orthogonal fuctions.

I see, and then taking these series at x = pi will give me the sum after some simple rearrangement and reduction. Thanks a lot. I can't believe I didn't even mess around with that series when I found it originally, I just kind of dismissed it even though it was simple enough to use. Much...

That was actually my first instinct when my first attempts failed, I thought perhaps the original function was a typo. The integral to find the Fourier coefficients of x^4 on this interval (which I obtained from Wolfram as I didn't want to do all the integration by parts on paper) is: [...

Find the Fourier series for f(x) = x^2 on [-pi,pi] and use it to find sum(n=1..infinity) 1/n^4

I honestly don't know, we proved that identity in class, and it's a relatively trivial problem once you have the Fourier series so I don't think he intended for us to find it again.

That's the trouble I'm having, I can't really figure out how to go about it. And all the attempts I've made so far leave me with series and sums that are interesting but of no real use in solving the problem.

Homework Statement I'm trying to find the sum of the infinite series: 1/n^4 using the Fourier series of x^2 on [-pi, pi] which I have as PI^2/3 + 4*sum(n=1..infinity) (-1)^n/n^2*cos(nx) Homework Equations The Attempt at a Solution So far all my attempts have been focused...