Let $V=\{p(x); \partial(p(x)<2\} \cup\{0\}$ a vector space, determine a inner product such that the basis $\{1,x,\frac{x^2}{2!}\}$ is orthonormal.
My solution:
I found the inner product: $<a_0 + a_1x+a_2x^2, b_0+b_1x+b_2x^2> = a_0b_0 + a_1b_1 + 4a_2b_2.$
this way we have $\{1,x,\frac{x^2}{2!}\}$ orthonormal basis...
But I would to solve this exercise using integrals or differential equations, how I do it?
Given $p(x) = a_0 + a_1 x + a_2 x^2$, we have $p(0) = a_0$, $p'(0) = a_1$, and $p''(0) = 2a_2$. Hence, $$\langle p, q \rangle = p(0)q(0) + p'(0)q'(0) + p''(0)q''(0).$$