Recent content by Padfoot89

I would argue that taking the first two conditions along with not R is perfectly valid in this problem, as the question is ultimately "given these two conditions, prove this implication" and how you prove an implication is to assume the premise. However I do admit it's less formal and more a...

Anytime you are asked to prove an implicated statement, in this case, "(Not R) imples P", you may assume the first part to prove the second part. In this case, your professor has added additional stipulations that you must assume as well, so I'd argue that 5 is not only valid, but entirely...

You need to provide a little more background into your question, I think.

Yes. As you can see, there's some built in ambiguity in satisfying equation 3, but by choosing equation 4, you've chosen the simplest way of satisfying 3. So it works out really nicely. This ratio (equation 3) being equal to w^2 is what ensures that you have normality of the vector, as well...

Yes, I just dropped the _t. Since I basically explained the answer to this in my second reponse, I think that rather than give you the solution, it will be universally more beneficial to you for you to work it out yourself. However, here's an outline of the solution. Start with K R = w^2 M...

We get (1) from assuming that R_t is an eigenvector of of the matrix (M^-1 * K) with eigenvalue w^2. 1 is the definition of this. The matrix acting on R_t returns R_t multiplied by w^2. As you get further into Linear Algebra, and understanding notation, (2) will eventually make your stomach...

I'm not sure if this is the answer you're looking for, but it might provide some insight into what's going on. So, with this normalization condition, we require that the eigenvectors R_s,R_t are normalized. Hence we want the inner product of the vectors with themselves to be 1, or<R_s|R_s> = 1...

Mathematically, assuming you're looking for vectors R that solve this equation, as every positive definite matrix is invertible, any vector that solves that equation will also solve M^{-1} K R = w^2 R. Hence R must an eigenvector of M^{-1}K with eigenvalue w. So then solving it is merely a task...

You're making use of the fact that the set of { \cos \frac{n \pi x}{L} } for all integers n serves as a complete basis for even functions on the interval L. Without the n = 0 term, you don't have a complete basis and thus can't span the space of even continuous on this interval (in reality...