Analisis Numerico

EJERCICIOS PROPUESTOS
Nombre: Evelyn Ivette Ascencio Acharte Codigo: 097241I

1.- Resolver numéricamente las ecuaciones utilizando el método y las condiciones indicadas.
C) x3-x-4=0 , reglafalsa, error <2% en
Solución:
f1=13-1-4=-4
f2=23-2-4=2
r=1-((-4) (2-1))/ (2-(-4))=1.666666667
fr=f1.66=(1.66)3-(1.66)-4=-1.037037037035
a=1 r=1.66 b=2
a=r=1.66
2ºiteración
r=1.66--1.037037037035(2-1.66)2-1.037037037035=-1.780487803
fr=f1.8059=(1.8059)3-(1.8059)-4
a=1.66 r=1.8059 b=2
b=r=1.8059
error = (1.66-1.859)1.8059*100%= 8.07%

3º iteraciónr=1.66--1.504(1.8059-1.66)0.083636-(-1.504)=1.7982
fr=f1.7982=(1.7982)3-(1.7982)-4=0.01632
a=1.66 r=1.7982 b=1.8059 b=r= 1.7982
error = (1.8059 - 1.7982) 1.7982*100%= 0.42%Resumen
Iter. | ri | error |
1 | 1.666666667 | - |
2 | 1.780487805 | 6.3926940525% |
3 | 1.794473652 | 0.7793843607% |

d) ln(x)+ x2-4=0, regla falsa, en[1,2], error<1%f1=-3
f2=0.69314
r=1--3(2-1)0.69314-(-3)=1.8123

a=1 r=1.8123 b=2 a=r=1.8123
2º iteración
r=1.8123--0.1209(0.1877)0.69314-(-0.1209)=-1.8401fr2=ln1.8401+1.84012-4=-0.00421
a=1.8123 r=1.8401 b=2
error = (1.8123-1.8401)1.8401*100%= 1.51%

3º iteración
r=1.8401--0.00421(0.1599)0.69314-(-0.00421)=1.84106f1.84106=ln1.84106+1.84106-4=-0.0001565
a=1.8401 r=1.84106 b=2
error = (1.8401 - 1.84106) 1.84106*100%= 0.05%

Resumen
Iter. | ri | error |
1 | 1.8123 | - |
2 | 1.8401 | 1.51% |
3| 1.84106 | 0.05% |

e) (x-1)(ex-1)=0 , newton raphson, comenzando con x=2.
Solución:
(x-1)( ex-1)= f (x)
f' (x)=(x-1)'(ex-1) + (x-1)( ex-1)'
f' (x)=(ex-1)+ (x-1) (ex-1)
f' (x)=(ex-1)(1+x-1)
f' (x)=( ex-1)x

f' (2)=(2-1)'(e2-1) = 2.718281828
f' (2)=(e2-1) 2 = 5.436563657
r2=r1- fr1f' r1
r2= 2-207182818285.436563657 = 1.5

error = (2 - 1.5) 1.5*100%= 83.333%

II. Aproximar...