Método de Newton
Método de Newton
Método de Newton
f ( xn)
x n+1=x n −
f ' ( xn )
Ejemplos:
2
f (x)=x −2
x o=1
Con tol=0.001
Sol:
f ´ ( x )=2 x
2
f ( x o ) =1 −2=−1
f ´ ( x o ) =2∗1=2
f ( xo) −1 3
x 1=x o− '
=1− =
f ( xo ) 2 2
()
2
3 1
f ( x1)= −2=
2 4
2∗3
f ´ ( x1 )= =3
2
1
f ( x1) 3 4
x 2=x 1− = − =1.4167
f ´ ( x1 ) 2 3
Xi F(xi) Error
x1 1.5 0.25 0.5
x2 1.4167 0.00694 0.08333
x3 1.4142 0.00001 0.00245
x4 1.4142 0.00000 0.00000
Sol: x=1.4142
x 0=0.5
Con tol=0.001
Sol:
g ´ ( x )=−sin ( x )−1
g ( xo) 0.3779
x 1=x 1=x o− =0.5− =0.7554
'
g ( xo ) −1.4794
Tabla:
Iteraciones
Paso x F (X) |x(i) - x(i-1)|
x1 0.7552 -0.02710 0.25522
x2 0.7391 -0.00009 0.01608
x3 0.7391 -0.00000 0.00006
Sol: x=0.7391
Método Quasi-Newton:
Método de la Secante:
x n− xn −1
x n+1=x n −f (x n )
f (x n)−f (x n−1 )
2
f (x)=x −2
x o=1 , x 1=2
2
f ( x o ) =1 −2=−1
2
f ( x 1 ) =2 −2=2
x 1−x o 2∗2−1
x 2=x 1−f ( x 1 ) =2− =1.33333333
f ( x 1) −f ( x o ) 2−(−1 )
f ( x 2 ) =( 1.33333333 )2−2=−0.222222222
x 2−x 1 0.22222222∗1.33333333−2
x 3=x 2−f ( x 2 ) =1.33333333+ =1.4
f ( x 2 )−f ( x1 ) −0.2222222−2
Siguientes Iteraciones:
Sol: x=1.4142
Ejercicio 2: Encuentra una aproximación de la raíz de la función usando el
método de la secante entre 1 y 2
g ( x )=cos ( x )−x
x o=0 , x 1=1
g ( x o )=cos ( 0 )−0=1
g ( x 1 )=cos (1 )−1=−0.459697
x 1−x o 1−0
x 2=x 1−g ( x 1) =1−(−0.459697) =0.685073
g ( x1 ) −g ( x o ) −0.459697−1
Continuación:
Método de la Bisección:
Se muestra la tabla:
Paso x F(x) |x(i) - x(i-1)|
x2 1.5 -0.125 0.5
x3 1.75 1.60938 0.25
x4 1.625 0.66602 0.125
x5 1.5625 0.25220 0.0625
x6 1.5313 0.05911 0.03125
x7 1.5156 -0.03405 0.01563
x8 1.5234 0.01225 0.00781
x9 1.5195 -0.01097 0.00391
x10 1.5215 0.00062 0.00195
x11 1.5205 -0.00518 0.00098
Ejemplo 2: Para la función g(x )=cos (x)−x en el intervalo ([0, 1]), seguimos
el mismo proceso. Calculamos c=0.5, evaluamos g(0.5), y determinamos el
nuevo intervalo para la siguiente iteración.
Se muestra la tabla:
Paso x F(x) |x(i) - x(i-1)|
x2 0.5 0.37758 0.5
x3 0.75 -0.01831 0.25
x4 0.625 0.18596 0.125
x5 0.6875 0.08533 0.0625
x6 0.7188 0.03388 0.03125
x7 0.7344 0.00787 0.01563
x8 0.7422 -0.00520 0.00781
x9 0.7383 0.00135 0.00391
x10 0.7402 -0.00192 0.00195
x11 0.7393 -0.00029 0.00098
Método de Fibonacci
F n−k F 6 13
ratio= = =
F n−k−1 F 7 21
I =−5
D=15
F n−1 13
x 2=I + ∗( D−I )=−5+ ∗( 15+5 )=7.3809
Fn 21
x 1=D+ I −x 2=15−5−7.3809=2.6191
f ( x 1 ) =6.8596
f ( x 2 ) =54.4776
I =−5
D=7.3809
Se repite:
k ratio I D x1 x2 f(x1) f(x2) I/D
b−a
c=b−
ϕ
b−a
d=a+
ϕ
2−0
c=2− =0.76393202
1.618033988749895
2−0
d=0+ =1.23606798
1.618033988749895
f ( c ) =0.583
f ( d )=1.5278
Se continua en la tabla:
Se
phi a b c d f© f(d) mantiene
1.61803399 0 2 0.76393202 1.23606798 0.58359214 1.52786405 a
1.61803399 0 1.23606798 0.47213595 0.76393202 0.22291236 0.58359214 a
1.61803399 0 0.76393202 0.29179607 0.47213595 0.08514495 0.22291236 a
1.61803399 0 0.47213595 0.18033989 0.29179607 0.03252248 0.08514495 a
1.61803399 0 0.29179607 0.11145618 0.18033989 0.01242248 0.03252248 a
1.61803399 0 0.18033989 0.06888371 0.11145618 0.00474497 0.01242248 a
1.61803399 0 0.11145618 0.04257247 0.06888371 0.00181242 0.00474497 a
1.61803399 0 0.06888371 0.02631123 0.04257247 0.00069228 0.00181242 a
1.61803399 0 0.04257247 0.01626124 0.02631123 0.00026443 0.00069228 a
1.61803399 0 0.02631123 0.01005 0.01626124 0.000101 0.00026443 a
1.61803399 0 0.01626124 0.00621124 0.01005 3.858E-05 0.000101 a
1.61803399 0 0.01005 0.00383876 0.00621124 1.4736E-05 3.858E-05 a
La solución es x=0.
Se
phi a b c d f© f(d) mantiene
1.61803399 2 3 2.38196601 2.61803399 15.2016261 17.3262379 a
1.61803399 2 2.61803399 2.23606798 2.38196601 5 5.67376208 a
1.61803399 2 2.38196601 2.14589803 2.23606798 4.60487837 5 a
1.61803399 2 2.23606798 2.09016994 2.14589803 4.36881039 4.60487837 a
1.61803399 2 2.14589803 2.05572809 2.09016994 4.22601798 4.36881039 a
1.61803399 2 2.09016994 2.03444185 2.05572809 4.13895366 4.22601798 a
1.61803399 2 2.05572809 2.02128624 2.03444185 4.08559805 4.13895366 a
1.61803399 2 2.03444185 2.01315562 2.02128624 4.05279554 4.08559805 a
1.61803399 2 2.02128624 2.00813062 2.01315562 4.03258858 4.05279554 a
1.61803399 2 2.01315562 2.005025 2.00813062 4.02012525 4.03258858 a
1.61803399 2 2.00813062 2.00310562 2.005025 4.01243212 4.02012525 a
1.61803399 2 2.005025 2.00191938 2.00310562 4.0076812 4.01243212 a
BIBLIOGRAFÍA
López, M. G., & Hernández, P. R. (2021). Análisis Numérico Avanzado: Teoría y Práctica.
McGraw-Hill.