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Tpy6a [65]
3 years ago
15

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)defin

es 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= ?
Mathematics
1 answer:
Natasha2012 [34]3 years ago
4 0

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

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