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?

  • $\begingroup$ $\partial$ is not a standard notation for the degree of a polynomial and you certainly mean $\partial(p) \leq 2$ when you wrote $\partial(p) < 2$. Also, $0$ is already a polynomial of degree at most $2$. $\endgroup$ Apr 7, 2019 at 23:41

1 Answer 1


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, $$\langle p, q \rangle = p(0)q(0) + p'(0)q'(0) + p''(0)q''(0).$$

  • $\begingroup$ thank you, like taylor polynomial $\endgroup$
    – Joãonani
    Apr 7, 2019 at 23:48
  • $\begingroup$ Exactly! Note that they wrote $\frac{x^2}{2!}$, instead of the (simpler) $\frac{x^2}{2}$. $\endgroup$ Apr 7, 2019 at 23:49

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