bases/dimension Q

[sickplaya]

Let f(X) be a polynomial of degree n in Pn(R). Prove that for any g(x) ∈ Pn(R) there exists scalars C0, C1, C2,....., such that g(x) = C0f(x) + C1f'(x)+ .........Cnf(n)(x) where f(n)(x) is the nth derivative of f(x).

So I prove that f(x) and all its derivatives form a basis. Can I say that g(x) contains this basis and that the dimension of f(x) and g(x) are equal since they have the same number of basis elements and I am not really sure how to prove the existence of these scalars.

 

[pka]

What is being said is that given any \(\displaystyle f \in P_{ n} (x)\quad \& \quad \deg (f) = n\) then the set \(\displaystyle \left\{ {f,f',f'', \cdots ,f^{\left( n \right)} } \right\}\) forms a basis for \(\displaystyle P_{ n} (x)\).
It is a matter of showing the set is linearly independent.

 

[daon]

If you show that B = {f(x), f'(x), f''(x), ... , fⁿ(x)} forms a basis for \(\displaystyle P_n\(\mathbb{R}\)\), then you're basically done (as far as I can see).

Since g(x) is an element of \(\displaystyle P_n\(\mathbb{R}\)\), by the definition of a basis, it must be a linear combination of the elements of B (hence there are such scalars). Although maybe you can't make these assumptions?