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