Linear algebra - best approximation

1. Aug 3, 2009

twotaileddemon

Hi... I have a quick question. I'm given V = ([0,1], <f,g> = interval from 0 to 1 of f(x)g(x)dx, S = {1, 2x-1}, W = lin(s), and P exists in W.

I was determining the best approximation (P) to a function (F). F was some polynomial (3x + 5) and when I did the work I got P to be the same polynomial. (I used projections of both parts of S with F and added them)

The distance between them, d(F,P) was obviously 0, but I was wondering why this was true? I mean.. why it HAD to be true even before I computed P.

Is it because W and F were polynomials of the same degree and p exists in W? Thanks for your help.
Any elaboration is great :) I just want to understand why certain functions already have the best approximation to what is given.

2. Aug 4, 2009

Staff: Mentor

What you've written here is not at all clear. It would help me understand what you're trying to do if you can answer the following questions.

1. What do you mean that V = [0, 1]? Is this just the interval 0 <= x <= 1? If so, why is it named?
2. <f, g> is not an interval; it is an integral, namely $$\int_0^1 f(x)g(x)dx$$
3. What is S? It appears to be a basis with two functions in it.
4. What is W? You show it as equal to lin(s). Do you mean that W is the set of all linear combinations of the vectors in S (not s)? It would be helpful if you said that.
5. How did you calculate the distance from one function to another? I'm guessing that you should be using the inner product you are given.

If an arbitrary function f is in W (that is, f is a linear combination of the functions in S), then I think that an approximation using the functions in W will be exact. If, on the other hand, f is not in W (for example, f(x) = x2), then an approximation will be just that, an approximation.