Question

Let f(x)=−7, g(x)=−4x+5 and h(x)=4x2−8x−5. Consider the inner product 〈p,q〉=p(−1)q(−1)+p(0)q(0)+p(1)q(1) in the vector space P2 of polynomials of degree at most 2. Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of P2 spanned by the polynomials f(x), g(x) and h(x).

Answer 1

The basis is <1/√3, -2/5 x + 1 /15, 0.4825x^2 - 0.6466 x -0.3748>

Explanation:

First, we calculate the norm of f

||f||² = <f,f> = f(-1)²+f(0)²+f(1)² ) = 3*(-7)² = 147

Therefore, ||f|| = √147

We take as the first element of the basis
`(-7)/(√(147)) = (1)/(√(3)) .`

we define

`\tilde{g}(x) = g(x) - <g,v1> * v1`

lets calculate <g,v1>

g(-1) = 9

g(0) = 5

g(1) = 1

v1(-1) = v1(0) = v1(1) = 1/√3

Then <g,v1> = 9*7/√(147)+5*7/√(147)+1*7/√(147) = 15*7/√(147) = 105/√147

and <g,v1>v1 = 105/√(147) * 7/√(147) = 735/147 = 35/9

Therefore,

`\tilde{g}(x) = -4x+5-35/9 = -4x + 2/3`

Now, lets calculate the norm, for that

`\tilde{g}(-1) = 14/3`

`\tilde{g}(0) = 2/3`

`\tilde{g}(1) = -10/3`

As a result,
`< \tilde{g}, \tilde{g} > = (14/3)^2+(2/3)^2+(-10/3)^2 = 100, hence [tex] || \tilde{g} || = 10`

We take
`v2 = \tilde{g} / 10 = -2/5 x + 1 /15`

Finally, we take

`\tilde{h}(x) = h(x) - v1 <h,v1> - v2 <h,v2>`

Note that

h(-1) = 7

h(0) = -5

h(1) = -9

v1(-1) = v1(0) = v1(1) = 7/√(147) = 1/√3

v2(-1) = 7/15

v2(0) = 1/15

v3(1) = -1/3

Thus,

<h,v1>v1 = (7-5-9)*(7/√(147))² = -7/3

<h,v2>v2 = ((7*7/15) + (-5*1/15) + (-9*-1/3)) * (-2/5 x + 1 /15) = -66/25 x + 11/25

As a consecuence, we have that

`\tilde{h}(x) = 4x^2-8x-5 +7/3 + 66/25x -11/25 = 4x^2-5.36x-233/75`

since

`\tilde{h} (-1) = 469/75; \tilde{h} (0) = 233/75: tilde{h}(1) = -67/15`

we obtain that
`||\tilde{h}|| = \sqrt{<\tilde{h},\tilde{h}>} = √(86.706) = 8.288`

Therefore,
`v3 = \tilde{h}/8.288 =  (4x^2-5.36x-233/75)/8.288 = 0.4825x^2 - 0.6466 x -0.3748`

The basis is <1/√3, -2/5 x + 1 /15, 0.4825x^2 - 0.6466 x -0.3748>