Post on 28-Jan-2016
description
Deber 3
Ejercicio 1. Sean los vectores: 1, x, x2⊂C(−1, 1), ortogonalizar este sistema, o sea encontrar, e1, e2, e3→S.O,
encontrar e1̃, e2̃, e3̃→S.O.N .
Para el vector P =3−5x+x2. Comprobar que para P =α1e1̃+α2e2̃+α3e3̃, se cumple que ‖P ‖2=α12+α2
2+α32.
e1=1
e2= x− (x, 1)
‖1‖2 · (1)= x−α · (1)
α=(x, 1)
‖1‖2
(x, 1)=
∫
−1
1
(x) dx=0
‖1‖2=∫
−1
1
(1) dx=2
entonces:α=(x, 1)
‖1‖2 =0
2e2= x
e3= x2− (x2, 1)
‖1‖2 · (1)− (x2, x)
‖x‖2 · (1)=x2−α · (1)− β(x)
α=(x2, 1)
‖1‖2
β=(x2, x)
‖x‖2
(x2, 1)=
∫
−1
1
(x2) dx=2/3
‖1‖2=∫
−1
1
(1) dx=2
α=1/3
(x2, x) =
∫
−1
1
(x3) dx=0
‖x‖2=∫
−1
1
(x2) dx=2/3
β=0
entonces: e3= x2− (1/3) · (1)− 0 · (x) =x2− 1
3
SistemaOrtogonal: e1=1, e2= x, e3= x2− 1
3.
(e1, e2)= (1, x) =
∫
−1
1
(x) dx=0
(e1, e3) =
(
1, x2− 1
3
)
=
∫
−1
1(
x2− 1
3
)
dx=0
(e2, e3) =
(
x, x2− 1
3
)
=
∫
−1
1 (
x3− x
3
)
dx=0
1
Para ortonormalizar el sistema encontramos: ‖1‖=∫
−1
1
(1) dx
√
= 2√
‖x‖=∫
−1
1
(x2) dx
√
= 6√
/3
∥
∥
∥
∥
x2− 1
3
∥
∥
∥
∥
=
∫
−1
1(
x2− 1
3
)
2
dx
√
=2 10√
/15, entonces el sistema ortonormal es:
e1̃=1
2√ , e2̃=
3x
6√ , e3̃=
15(
x2− 1
3
)
2 10√ ,
(e1̃, e2̃)=
(
1
2√ ,
3x
6√
)
=
∫
−1
1(
3x
2 3√
)
dx=0
(e1̃, e3̃)=
(
1
2√ ,
15(
x2− 1
3
)
2 10√
)
=
∫
−1
1
(
15(
x2− 1
3
)
4 5√
)
dx=0
(e2̃, e3̃) =
(
3x
6√ ,
15(
x2− 1
3
)
2 10√
)
=
∫
−1
1
(
45x(
x2− 1
3
)
4 15√
)
dx=0
****************************************************************************************
3− 5x+x2=α1
(
1
2√
)
+α2
(
3x
6√
)
+α3
(
15(
x2− 1
3
)
2 10√
)
, encontramosα1, α2, α3
α1
(
1
2√)
=3⇒α1=3 2√
α2
(
3x
6√
)
=−5x⇒α2=−5 6√
3
α3
(
15(
x2− 1
3
)
2 10√
)
=x2⇒α3=2 10√
x2
15(
x2− 1
3
),
‖P ‖2=α12+α2
2+α32
‖P ‖2=(
3 2√ )
2+
(
−5 6√
3
)
2
+
(
2 10√
x2
15(
x2− 1
3
)
)
2
‖P ‖2= 18+50
3+
8x4
45(
x2− 1
3
)
2
2