Respuesta :
Answer:
The basis is <1/√3, -2/5 x + 1 /15, 0.4825x^2 - 0.6466 x -0.3748>
Step-by-step 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 [tex] \frac{-7}{\sqrt{147}} = \frac{1}{\sqrt{3}} .[/tex]
we define
[tex] \tilde{g}(x) = g(x) - <g,v1> * v1[/tex]
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,
[tex] \tilde{g}(x) = -4x+5-35/9 = -4x + 2/3[/tex]
Now, lets calculate the norm, for that
[tex] \tilde{g}(-1) = 14/3[/tex]
[tex] \tilde{g}(0) = 2/3[/tex]
[tex] \tilde{g}(1) = -10/3[/tex]
As a result, [tex] < \tilde{g}, \tilde{g} > = (14/3)^2+(2/3)^2+(-10/3)^2 = 100, hence [tex] || \tilde{g} || = 10 [/tex]
We take [tex] v2 = \tilde{g} / 10 = -2/5 x + 1 /15 [/tex]
Finally, we take
[tex] \tilde{h}(x) = h(x) - v1 <h,v1> - v2 <h,v2> [/tex]
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
[tex] \tilde{h}(x) = 4x^2-8x-5 +7/3 + 66/25x -11/25 = 4x^2-5.36x-233/75 [/tex]
since
[tex]\tilde{h} (-1) = 469/75; \tilde{h} (0) = 233/75: tilde{h}(1) = -67/15 [/tex]
we obtain that [tex] ||\tilde{h}|| = \sqrt{<\tilde{h},\tilde{h}>} = \sqrt{86.706} = 8.288 [/tex]
Therefore, [tex] v3 = \tilde{h}/8.288 = (4x^2-5.36x-233/75)/8.288 = 0.4825x^2 - 0.6466 x -0.3748 [/tex]
The basis is <1/√3, -2/5 x + 1 /15, 0.4825x^2 - 0.6466 x -0.3748>
