MÉTODO NEWTON RAPHSON Problemas Minero Sabrina

MÉTODO NEWTON RAPHSON PROBLEMA 1

Encuentre por el método de Newton Raphson la raíz de

− x

e − x , tomando

0

como valor inicial  x = 0 . − x

e − x  tenemos:

Derivando

− x '  f  ( x )=−e −1

Sustituimos en la ecuación de Newton Raphson para obtener las iteraciones:

 x i +1= x i−

 ( x i)  f  ( ' 

f  ( x i) − x

−0

e − x e −0 =0.5  x i +1=0 − − x = −0 −e −1 e + 1 −0

e −0  x i+1=0.5 + −0 = 0.5663 e +1 −0.5663

 x i +1=0.5663 +

e

 x i +1=0.5671 +

e

−0.5663 =0.5671 +1 e −0.5663

−0.5671

− 0.5671 = 0.5671 +1 e −0.5671

PROBLEMA 2.

pro!ime la solución

cos ( x )− x =0  con " decimales # una apro!imación

π  inicial de  x 0= 4 = 0.785398 . Derivando tenemos: '  f  ( x )=−sen ( x )−1

Sustitu#endo en la ecuación de newton raphson para obtener las iteraciones:

 x i +1= x i−

 ( x i)  f  ( ' 

f  ( x i)

 x i +1=0.785398 +

 x i +1=0.739536 +

 x i +1=0.739083 +

 x i +1=0.739085 +

cos ( 0.785398 ) −0.785398

sen ( 0.785398 ) + 1 cos ( 0.739536 )−0.739536

sen ( 0.739536 ) + 1 cos ( 0.739083 ) −0.739083

sen ( 0.739083 ) + 1 cos ( 0.739085 ) −0.739085

sen ( 0.739085 ) + 1

=0.739536

=0.739083

=0.739085

=0.739085

$E%&D& DE NE'%&N R()S&N $*+%-R+E Realice / iteraciones con el método de método multivariable para resolver el sistema de ecuaciones no lineales # utilice como vector inicial

 x =(−1,3 ) . 0

f 1 ( x , y )=7 [ x / ln ( y ) ] − 2 x + 1=0 2

f  2 ( x , y )= 9 x  y −5 x  y + y = 0 2

2

2

0

Evaluamos con  x f 1 y f  2  # tenemos:

f 1 ( x , y )=7 [−1 / ln ( 3) ]− 2 (−1 ) + 1=−7.371 2

2

2

2

f  2 ( x , y )= 9 (−1 ) ( 3 )−5 ( −1 ) ( 3 ) + ( 3 )=−15 Derivadas parciales

∂ f 1

7

∂x

ln ( y )

∂ f 1 ∂y

 =

 =

∂x

 =

∂x

−7 x 2 [ ln ( y )]  y

Evaluar con  x ∂ f 1

∂ f 2

− 4 x

0

 = 18 xy −10 x y 2

∂ f 2

 =9 x 2−10 x2  y + 1

∂y

 en las derivadas parciales.

7 −4 (−1 )=10.37 ln ( 3 )

∂ f 2 ∂x

 = 18 (−1 ) ( 3 )−10 (−1 ) ( 3 )2 =36

∂ f 1 ∂y

 =

−7 (−1 ) =1.933 2 [ ln ( 3 ) ] (3 )

∂ f 2 ∂y

 =9 (−1 )2−10 (−1 )2 ( 3 )+ 1=−20

0alcular el coe1ciente 2acobiano. J =

[

10.37 1.933

0alcular

36 −20

]=−

277

∆ x y ∆ y  :

∆ x=

−15 ( 1.933 ) −(−7.3716 ) (−20 ) =0.6369 −277

∆ y=

−7.3716 ( 36 )−(−15 ) ( 10.37 ) =0.3964 −277

 3 obtenemos  x y y  de la primera iteración: ( 1)

 x =0.6369 + (−1 )=−0.3631 ( 1)

 y =0.3964 + ( 3 ) = 3.3964

Se4unda iteración: f 1 ( x , y )=7 [−0.3631 / ln ( 3.3964 ) ] −2 (−0.3631 ) + 1=−1.342 2

2

2

2

f  2 ( x , y )= 9 (−0.3631 ) ( 3.3964 )− 5 (−0.3631 ) ( 3.3964 ) + ( 3.3964 ) =−0.1778

∂ f 1 ∂x

 =

7 −4 (−0.3631 ) =7.177 ln ( 3.3964 )

∂ f 2 ∂x

 = 18 (−0.3631 ) ( 3.3964 ) −10 ( 0.3631 ) ( 3.3964 )2 =19.68

∂ f 1

 =

∂y

∂ f 2

−7 ( 0.3631) =0.50 2 [ ln ( 3.3964 ) ] (3.3964 ) 2

 =9 ( 0.3631)2−10 ( 0.3631 ) ( 3.3964 ) +1=−2.291

∂y

J =

[

7.17 19.68

0.50 −2.29

]=−

26.25

∆ x=

−0.177 ( 0.50 )−(−1.3424 ) (−2.29 ) =0.1205 −26.25

∆ y=

−1.3424 ( 19.68 )−(−0.177 ) (7.17 ) =0.9581 −26.25

( 1)

 x =0.1205 + (−0.3631 ) =−0.2426 ( 1)

 y =0.9581+ ( 3.3964 )= 4.3545

 %ercera iteración: f 1 ( x , y )=7 [−0.2426 / ln ( 4.3545 ) ] −2 (−0.2426 ) + 1=−0.271 2

2

2

2

f  2 ( x , y )= 9 (−0.2426 ) ( 4.3545 )−5 (−0.2426 ) ( 4.3545 ) + ( 4.3545 )=1.097

∂ f 1 ∂x

 =

7 − 4 (−0.2426 )=5.72 ln ( 4.3545 )

∂ f 2 ∂x

 = 18 (−0.2426 ) ( 4.3545 )−10 (−0.2426 ) ( 4.3445 )2=26.91

∂ f 1

 =

∂y

−7 (−0.2426 ) =0.1797 2 ( ) [ ln 4.3545 ] (4.3545 )

∂ f 2

 =9 (−0.2426 )2−10 (−0.2426 )2 ( 4.3545 ) + 1=−1.03

∂y

J =

[

5.72 26.91

0.1797 −1.03

]=−

10.77

∆ x=

∆ y=

1.097 ( 0.1797 ) − (−0.271 ) (−1.0328 )

−10.77

=0.00763

−0.271 (26.91 ) −( 1.097 ) ( 5.72 ) = 1.259 −10.77

( 1)

 x =0.00763 + (−0.2426 )=−0.2349 ( 1)

 y =1.259+ ( 4.3545 )= 5.613

METODO NEWTON RAPHSON MULTIVARIABLE

Resuelva el si4uiente sistema de ecuaciones no lineal con el método ( 0) newton raphson multivariable con el vector inicial  x =( 0.5, 1.5) ,

realice / iteraciones. 2

2

100 x1− x 1− x 2= 80 2

100 x2− x 1 x2 − x1 =80

2

2

f 1 =100 ( 0.5 )1−( 0.5 )1− (1.5 )2−80=−32.5 2

f  2=100 ( 1.5 )2−( 0.5 )1 ( 1.5 ) 2−( 0.5 )1−80 =68.37

∂ f 1 ∂x

 = 100−2 ( 0.5)= 99

∂ f 2 ∂x

 =−(1.5 )2−1=−5.5

∂ f 1

 =−2 ( 1.5 )=−3

∂y

∂ f 2

 =100−2 ( 1.5 )( 0.5 )= 98.5

∂y

J =

[−

99 5.5

∆ x=

∆ y=

]

−3 =9735

98.5

68.37 (−3 )− (−32.5 ) ( 98.5 ) 9735

−32.5 (−5.5 ) −( 68.37 ) ( 99 ) 9735

=0.307

=−0.6

( 1)

 x =0.307 + ( 0.5 )= 0.807 ( 1)

 y =−0.6 + ( 1.5 )=0.806

Se4unda iteración: 2

2

f 1 =100 ( 0.807 )1−( 0.807 )1− ( 0.806 ) 2− 80=−0.756 2

f  2=100 ( 0.806 )2−( 0.807 )1 ( 0.806 )2 −( 0.807 )1−80 =−0.731

∂ f 1 ∂x

 = 100−2 ( 0.807 )=98.38

∂ f 2 ∂x

2

 =−( 0.806 ) −1=−2.29

∂ f 1

 =−2 ( 0.806 )=−1.61

∂y

∂ f 2

 =100−2 ( 0.806 )( 0.807 )=98.69

∂y

J =

[−

98.38 2.29

∆ x=

]

−1.61 =9706.02 98.69

−0.731 ( 98.38 )−(−0.756 ) (−2.29 ) 9706.02

=7.74 × 10−3

∆ y=

−0.756 (−1.61 ) −(−0.731 )( 98.69 ) 9706.02

( 1)

−3

( 1)

−3

=7.57 × 10−3

 x =7.74 × 10 + ( 0.807 ) =0.8123  y =7.57 × 10 + ( 0.806 ) =0.8131

 %ercera iteración: 2

2

f 1 =100 ( 0.8123 )1−( 0.8123 )1−( 0.8131 )2 − 80=−0.75 2

f  2=100 ( 0.8131 )2−( 0.8123 )1 ( 0.8131 )2 −( 0.8123 )1−80 =−0.73

∂ f 1 ∂x

 = 100−2 ( 0.8123 )= 98.37

∂ f 2 ∂x

 =−( 0.8131 )2−1=−2.32

∂ f 1

 =−2 ( 0.8131 )=−1.62

∂y

∂ f 2

 =100−2 ( 0.8131)( 0.8123 )= 98.67

∂y

J =

[−

98.37 2.32

∆ x=

∆ y=

( 1)

]

−1.62 =9702.22 98.67

−0.73 ( 98.37 )−(−0.75 ) (−2.32 ) 9702.22

=−9.1 × 10−4

−0.75 (−1.62 ) −(−0.73 ) ( 98.37 ) 9702.22

−4

 x =−9.1 × 10 + ( 0.8123 )= 0.811

=2.31 × 10−5

( 1)

−5

 y =2.31 × 10 + ( 0.8131 ) =0.813

Resuelva el si4uente sistema de ecuaciones por el método newton ( 0) raphson multivariable con el valor inicial  x =( 1,0.5 ) , realice dos

iteraciones. 2

 x = x +  y 2

2

 y = x − y

2

2

2

2

2

f 1 =(1 ) +( 0.5 ) −1= 0.25 f  2=(1 ) +( 0.5 ) −0.5 =0.25

∂ f 1 ∂x

 = 2 ( 1 ) −1=1

∂ f 2 ∂x

 = 2 (1 ) =2

∂ f 1

 =−2 ( 0.5 ) =−1

∂y

∂ f 2

 =−2 ( 0.5 ) −1=−2

∂y

J =

[

1 2

∆ x=

∆ y=

]

−1 =−4 −2

0.25 (−1 ) −( 0.25 ) (−2)

−4 0.25 ( 2 )−( 0.25 ) ( 1)

−4

=−0.0625

=−0.0625

( 1)

 x =−0.0625 + ( 1 )= 0.9375 ( 1)

 y =−0.0625 + ( 0.5 ) =0.1875

Se4unda iteración: 2

2

2

2

f 1 =(0.9375 ) +( 0.1875 ) − 0.9375=−0.0234 f  2=(0.9375 ) +( 0.1875) − 0.1875= 0.7265

∂ f 1 ∂x

 = 2 ( 0.9375 )−1= 0.875

∂ f 2 ∂x

 = 2 ( 0.9375 )=1.785

∂ f 1

 =−2 ( 0.1875 ) =−0.375

∂y

∂ f 2

 =−2 ( 0.1875 ) −1=−1.375

∂y

J =

[

0.875 1.785

∆ x=

∆ y=

]

−0.375 =−1.8724 −1.375

−0.0234 (−0.375 )−( 0.7265 ) (−1.375 ) =−0.5381 −1.8724 0.7265 ( 1.785 )−(−0.0234 ) ( 0.875 )

−1.8724

( 1)

 x =−0.5381 + ( 0.9375 ) =0.3994 ( 1)

 y =−0.7035 + ( 0.1875 )=−0.516

=−0.7035