Question

(1 point) If p(x) and (x) are arbitrary polynomials of degree at most 2, then the mapping =p(-1)q(-1) + p(070) +p(3)q(3) defines an inner product in P3- Use this inner product to find ||P||||9||, and the angle between p(x) and (x) for p(x) = 2x2 + 6 and 7(x) = 4x2 - 4x. < p, >= ||pl= ||ql| = A = radians.

Answer 1

If p(x) and q(x) are arbitrary polynomials of degree at most 2 then

||p||||q|| = 26($\sqrt{640}$) and angle between p(x) and q(x) is 0.233.

Given that

<p,q> = p(-1)q(-1) + p(0) q(0) + p(3)q(3)

and p(x) = 2x²+ 6 , q(x)= 4x²-4x

then the values of p and q at x = -1,0,3 are given as;

x = -1,

p(-1) = 2(-1)² + 6 = 8   ,    q(-1) = 4(-1)² - 4(-1) = 8

x = 0,

p(0) = 2(0)² + 6 = 6   ,    q(0) = 4(0)² - 4(0) = 0

x = 3,

p(3) = 2(3)² + 6 = 24  ,  q(3) = 4(3)²- 4(3) = 24.

<p,q> = p(-1)q(-1) + p(0)q(0) + p(3)q(3)

         = (8)(8) + (6)(0) + 24(24)

         = 64 + 0 + 576

<p,q> = 640

Now we have to find ||p|| ||q||, for this we'll find ||p|| and ||q||

||p|| = $\sqrt{ < p,p > }$

     = $\sqrt{8(8) + 6(6) + 24(24)}$

     = $\sqrt{676}$

||p|| = 26

and

||q|| = $\sqrt{ < q,q > }$

      =$\sqrt{8(8) + 0(0) + 24(24)}$

||q||  =$\sqrt{640}$

∴||p||||q|| = 26($\sqrt{640$)

Now we have to find angle between p(x) and q(x),

∴ α = cos⁻¹$\frac{ < p,q > }{||p||||q||}$

      = cos ⁻¹ $\frac{640}{26(\sqrt{640}) }$

      = cos ⁻¹ $\frac{4\sqrt{10} }{13}$

  α  = 13.34°

In radian

α = 0.233.

To know more about Inner product here

brainly.com/question/14185022

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