V = C^2[0,1] is defined as the space of twice-differentiable real-valued functions on [0,1]. I'm wondering what it means to be on [0,1]. I think it has to do with the inner product. (i.e. the bounds of integration for the inner product are 0,1). What would be the most general function from V?
"\(V = \mathcal{C}^2 \ [0,1]\) is defined as the space of twice-differentiable real-valued functions on \([0,1]\)." means V is the set of all functions which are twice-differentiable on the closed interval \(0\leq x \leq 1\). That means that if you have some function \(f\) on this interval, then \(f' \ \text{ and } \ f''\) exist and are well-defined on this interval as well. Some examples: f(x) = x f(x) = x^2 f(x) = (any polynomial) f(x) = sin(x) f(x) = cos(x) f(x) = e^x
Thanks for the response. That makes sense, but the reason I asked for the most general function is that I am asked to compute two linearly independent eigenvectors for a genral eigenvalue x > 0.
eigenvectors in this case are functions (the FUNCTIONS THEMSELVES are considered vectors in this vector space...) which are not constant multiplies of one another. Usually \(f(x)=c_1\sin(x)+c_2\cos(x)\) (the function f is come linear combination of two linearly independent functions sin and cosine) is a classic example of this. \(f(x) = c_1x+c_2x^2\) is another one.
Also \[\Huge f(x) = c_1e^{mx}+c_2e^{nx}\] where \(m\neq n\)
This is very clear now. Thank you.
Question -- have you studied linear algebra?
I have taken an introductory linear algebra course, and I am now in an applied linear course. I'm used to calculating eigenvalues/eigenvectors for matrices and transformations that have explicit matrix representations, however this question requires a deeper understanding.
Check out Ch 02 (Vector Spaces) from Jim Hefferon's FREE text (with lots of examples, and SOLVED problems) http://joshua.smcvt.edu/linalg.html/ it might be helpful. :)
Wow this looks great. It will make a good complement to the Linear Algebra Schaum's outline that we use for the course.
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