Trabajo de Aalisis Numerico Problemas y Teoria
Enviado por Ensa05 • 15 de Febrero de 2018 • 2.874 Palabras (12 Páginas) • 355 Visitas
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F(a) = F(0.2) = (0.2)3 – 7(0.2)2 + 14(0.2) – 6 = –3.472
F (b) = F(1) = (1)3 – 7(1)2 + 14(0.1) – 6 = 2
F(a)*F(b)
Ahora comenzaremos con las iteraciones:
- Iteración :
C= (0.2+1)/2 = 0.6
F(c) = F(0.6) = (0.6)3 – 7(0.6)2 + 14(0.6) – 6 = 0.096
X
a
c
b
F(x)
-
+
+
a = 0.2
b = 0.6
C = r = 0.6
- Iteración :
C = (0.2 + 0.6)/2 = 0.4
F(c) = F(0.4) = (0.4)3 – 7(0.4)2 + 14(0.4) – 6 = -1.456
X
a
c
b
F(x)
-
-
+
a = 0.4
b = 0.6
Error=|(0.4-0.6)/0.4|*100=50% de error.
C = r = 0.4
- Iteración :
C = (0.4 + 0.6)/2 = 0.5
F(c) = F(0.5) = (0.5)3 – 7(0.5)2 + 14(0.5) – 6 = -0.625
X
a
c
b
F(x)
-
-
+
a=0.5
b=0.6
Error=|(0.5-0.4)/0.5|*100=20% de error.
C = r = 0.5
- Iteración :
C = (0.5 + 0.6)/2 = 0.55
F(c) = f(0.55) = (0.55)3 – 7(0.55)2 + 14(0.55) – 6 = -0.251125
x
a
c
b
F(x)
-
-
+
a = 0.55
b = 0.6
Error=|(0.55 -0.5)/ 0.55|*100=9,0909% de error.
C = r = 0.55
- Iteración :
C = (0.55 + 0.6)/2 = 0.575
F(c) = F(0.575) = (0.575)3 – 7(0.575)2 + 14(0.575) – 6 = -0.074265625
x
a
c
b
F(x)
-
-
+
a = 0.575
b = 0.6
Error=|(0.575-0.55)/ 0.575|*100=4.347826% de error.
C = r = 0.575
- Iteración :
C = (0.575 + 0.6)/2 = 0.5875
F(c) = F(0.5875) = (0.5875)3 – 7(0.5875)2 + 14(0.5875) – 6
F(c) = 0.0116855468
x
a
c
b
F(x)
-
+
+
a = 0.575
b = 0.5875
Error=|(0.5875-0.575)/ 0.5875|*100=2.1276595% de error.
C = r = 0.5875
- Iteración :
C = (0.575 + 0.5875)/2 = 0.58125
F(c) = F(0.58125) = (0.58125)3 – 7(0.58125)2 + 14(0.58125) – 6
F(c) = -0.0310847168
x
a
c
b
F(x)
-
-
+
a = 0.58125
b = 0.5875
Error=|(0.58125-0.5875)/ 0.58125|*100=1.075268817% de error.
C = r = 0.58125
- Iteración :
C = (0.58125 + 0.5875)/2 = 0.584375
F(c) = F(0.584375) = (0.584375)3 – 7(0.584375)2 + 14(0.584375) – 6
F(c) = -0.009648345947
x
a
c
b
F(x)
...