Here is an improved question that was asked before.
Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Let our inner product be defined by: $$ \langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1). $$
Find the angle between the polynomials given by $p(t)=-4+9t-6t^2+t^3$ and $q(t)=3-6t+3t^2$.
Can someone please help me to solve this? I have been trying to solve this for days and can not come up with an answer. I really need someone to help me with this answer please.
2 Answers
The angle $\theta$ is given by $\cos\theta=\frac{\langle p,q\rangle}{\langle p,p\rangle^{1/2}\langle q,q\rangle^{1/2}}$
Hint: The angle between $p$ and $q$ is $$\arccos \dfrac{\langle p, q \rangle}{\sqrt{\langle p, p \rangle}\sqrt{\langle q, q \rangle}}.$$
