Metdodo De Biseccion
Enviado por TOLUCO1312 • 3 de Marzo de 2015 • 1.417 Palabras (6 Páginas) • 239 Visitas
METODO DE BISECCION
Aproximar por el método de Bisección la función f(x)= x3+2x2+10x-20 en el intervalo (1,2) con una tolerancia de de .001% y con un truncamiento de 4 cifras significativas.
f (x)= x^3 + 2x^2 +10x -20 siendo (1,2) (a,b)
Paso 1
f (a) = 〖(1)〗^3 + 2〖(1)〗^2 +10(1) -20
f (a) = -7
f (b) = 〖(2)〗^3 + 2〖(2)〗^2 +10(2) -20
f(b) = 8+8+20-20
f (b) = 16
f(a)*f(b) = -7*16 = -112
f(a)*f(b) < 0
Paso2
f (p) = con p = (a+b)/2
p = (1+2)/2 = 1.5
f (p) = (〖1.5)〗^3 + 2〖(1.5)〗^2 +10(1.5) -20
f (p) = 3.37+4.5+15-20
f(p) = 2.87
f(a)*f(p)=-7(2.87) = - 20.09
Paso3
/b-a/ = /2-1 / =1
Iteración a b f(a) p f(p) F(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
Paso4
p= (1 +1.5)/2 = 1.25
f (p) = (〖1.25)〗^3 + 2〖(1.25)〗^2 +10(1.25) -20
f (p) = 1.9531 + 3.125 + 12.5 -20
f (p) = -2.4219
f(a)*f(p)=-7(-2.4219) = 16.9533
/b-a/ = /1.5-1 / =0.5
Iteración a b f(a) p f(p) F(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.4219 16.9533 0.5
Dado que f(a)*f(p)= >0 (p,b) contiene la raíz entonces a=p y f(a) = f(p)
f (a) = 〖(1.25)〗^3 + 2〖(1.25)〗^2 +10(1.25) -20 f (a) = -2.4219
p= (1.25 +1.5)/2 = 1.375
f (p) = (〖1.375)〗^3 + 2〖(1.375)〗^2 +10(1.375) -20
f (p) = 2.5996 + 3.7812 + 13.75 -20
f (p) = 0.1308
f(a)*f(p)=-2.4219(0.1308) = -0.3169
/b-a/ = /1.5-1.25 / =0.25
Iteración a b f(a) p f(p) f(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.43 17.01 0.5
3 1.25 1.5 -2.4219 1.375 0.1308 -0.3169 0.25
Dado que f(a)*f(p)= <0 (a,b) contiene la raíz entonces b=p
p= (1.25 +1.375)/2 = 1.3125
f (p) = (〖1.3125)〗^3 + 2〖(1.3125)〗^2 +10(1.3125) -20
f (p) = 2.2609 + 3.4453 + 13.125 -20
f (p) = -1.1688
f(a)*f(p)=-2.4219(-1.1688) = 2.8607
/b-a/ = /1.375-1 .25/ =0.125
Iteración a b f(a) p f(p) f(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.43 17.01 0.5
3 1.25 1.5 -2.4219 1.375 0.1308 -0.04 0.25
4 1.25 1.375 -2.4219 1.3125 -1.1688 2.8607 0.125
Dado que f(a)*f(p)= >0 (p,b) contiene la raíz entonces a=p y f(a) = f(p)
f (a) = 〖(1.3125)〗^3 + 2〖(1.3125)〗^2 +10(1.3125) -20 f (a) = -1.1688
p= (1.3125 +1.375)/2 = 1.3437
f (p) = (〖1.3437)〗^3 + 2〖(1.3437)〗^2 +10(1.3437) -20
f (p) = 2.4260 + 3.6110 + 13.437 -20
f (p) = -0.526
f(a)*f(p)=-1.1688(-0.526) = 0.6147
/b-a/ = /1.375-1.3125 / =0.0625
Iteración a b f(a) p f(p) f(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.43 17.01 0.5
3 1.25 1.5 -2.4219 1.375 0.1308 -0.04 0.25
4 1.25 1.375 -2.4219 1.3125 -1.1688 2.8607 0.125
5 1.3125 1.375 -1.1688 1.3437 -0.526 0.6147 0.0625
Dado que f(a)*f(p)= >0 (p,b) contiene la raíz entonces a=p y f(a) = f(p)
f (a) = (〖1.3437)〗^3 + 2〖(1.3437)〗^2 +10(1.3437) -20 f (a) = -0.526
p= (1.3437 +1.375)/2 = 1.3593
f (p) = (〖1.3593)〗^3 + 2〖(1.3593)〗^2 +10(1.3593) -20
f (p) = 2.5115 + 3.6953 + 13.593 -20
f (p) = -0.2002
f(a)*f(p)=-0.526(-0.2002) = 0.1053
/b-a/ = /1.375-1.3437 / =0.0313
Iteración a b f(a) p f(p) f(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.43 17.01 0.5
3 1.25 1.5 -2.4219 1.375 0.1308 -0.04 0.25
4 1.25 1.375 -2.4219 1.3125 -1.1688 2.8607 0.125
5 1.3125 1.375 -1.1688 1.3437 -0.526 0.6147 0.0625
6 1.3437 1.375 -0.526 1.3593 -0.2002 0.1053 0.0313
Dado que f(a)*f(p)= >0 (p,b) contiene la raíz entonces a=p y f(a) = f(p)
f (a) = (〖1.3593)〗^3 + 2〖(1.3593)〗^2 +10(1.3593) -20 f (a) = -0.2002
p= (1.3593 +1.375)/2 = 1.3671
f (p) = (〖1.3671)〗^3 + 2〖(1.3671)〗^2 +10(1.3671) -20
f (p) = 2.5550 + 3.7379 + 13.671 -20
f (p) = -0.0361
f(a)*f(p)=-0.2002(-0.0361) = 0.0072
/b-a/ = /1.375-1.3593/ =0.0157
Iteración a b f(a) p f(p) f(a)*f(p) /b-a/
1 1 2 -7 1.5 2.87 -20.09 1
2 1 1.5 -7 1.25 -2.43 17.01 0.5
3 1.25 1.5 -2.4219 1.375 0.1308 -0.04 0.25
4 1.25 1.375 -2.4219 1.3125 -1.1618 2.8607 0.125
5 1.3125 1.375 -1.1688 1.3437 -0.526 0.6147 0.0625
6 1.3437 1.375 -0.526 1.3593 -0.2002 0.1053 0.0313
7 1.3593 1.375 -0.2002 1.3671 -0.0361 0.0072 0.0157
Dado que f(a)*f(p)=
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