Question

f p(x) and q(x) are arbitrary polynomials of degreeat most 2, then the mapping< p,q >= p(-2)q(-2)+ p(0)q(0)+ p(2)q(2)defines an inner product in P3.Use this inner product to find < p,q >, llpll, llqll, and the angletetha, between p(x) and q(x) forp(x) = 2x^2+6x+1 and q(x) = 3x^2-5x-6.< p;q >= ?llpll = ?llqll= ?

Answer 1

We're given an inner product defined by

$\langle p,q\rangle=p(-2)q(-2)+p(0)q(0)+p(2)q(2)$

That is, we multiply the values of $p(x)$ and $q(x)$ at $x=-2,0,2$ and add those products together.

$p(x)=2x^2+6x+1$

$q(x)=3x^2-5x-6$

The inner product is

$\langle p,q\rangle=-3\cdot16+1\cdot(-6)+21\cdot(-4)=-138$

To find the norms $\|p\|$ and $\|q\|$, recall that the dot product of a vector with itself is equal to the square of that vector's norm:

$\langle p,p\rangle=\|p\|^2$

So we have

$\|p\|=\sqrt{\langle p,p\rangle}=\sqrt{(-3)^2+1^2+21^2}=\sqrt{451}$

$\|q\|=\sqrt{\langle q,q\rangle}=\sqrt{16^2+(-6)^2+(-4)^2}=2\sqrt{77}$

Finally, the angle $\theta$ between $p$ and $q$ can be found using the relation

$\langle p,q\rangle=\|p\|\|q\|\cos\theta$

$\implies\cos\theta=\dfrac{-138}{22\sqrt{287}}\implies\theta\approx1.95\,\mathrm{rad}\approx111.73^\circ$