Ejercicios de métodos numéricos
Enviado por ALADIN0 • 21 de Mayo de 2023 • Prácticas o problemas • 1.964 Palabras (8 Páginas) • 36 Visitas
7. Aplique el método de bisección para encontrar soluciones exactas dentro de para el siguientes problema. [pic 2][pic 1]
para 0.2 ≤ x ≤ 0.3 y para 1.2 ≤ x ≤ 1.3[pic 3]
*Para 0.2 ≤ x ≤ 0.3 [pic 4]
[pic 5]
[pic 6]
[pic 7]
0.2 0.3
a1 b1[pic 8]
Iteración 1: a1 = 0.2 f(a1) [pic 10][pic 9]
b1 = 0.3 f(b1) = [pic 12][pic 11]
x1 = = 0.25 f(x1) [pic 13][pic 14]
Iteración 2: a2 = x1 = 0.25 f(a2) [pic 16][pic 15]
b2 = b1 = 0.3 f(b2) .0199959 > 0[pic 18][pic 17]
x2 = = 0.275 f(x2) [pic 21][pic 19][pic 20]
Iteración 3: a3 = x2 = 0.275 f(a3) [pic 23][pic 22]
b3 = b2 = 0.3 f(b3) .0199959 > 0[pic 25][pic 24]
x3 = = 0.2875 f(x3) [pic 28][pic 26][pic 27]
Iteración 4: a4 = x3 = 0.2875 f(a4) [pic 30][pic 29]
b4 = b3 = 0.3 f(b4) .0199959 > 0[pic 32][pic 31]
x4 = = 0.29375 f(x4) 0[pic 35][pic 33][pic 34]
Iteración 5: a5 = a4 = 0.2875 f(a5) [pic 37][pic 36]
b5 = x4 = 0.29375 f(b5) [pic 39][pic 38]
x5 = = 0.290625 f(x5) [pic 42][pic 40][pic 41]
Iteración 6: a6 = x5 = 0.290625 f(a6) [pic 44][pic 43]
b6 = b5 = 0.29375 f(b6) [pic 46][pic 45]
x6 = = 0.2921875 f(x6) [pic 49][pic 47][pic 48]
Iteración 7: a7 = x6 = 0.2921875 f(a7) [pic 51][pic 50]
b7 = b6 = 0.29375 f(b7) [pic 53][pic 52]
x7 = = 0.29296875 f(x7) [pic 56][pic 54][pic 55]
Iteración 8: a8 = a7 = 0.2921875 f(a8) [pic 58][pic 57]
b8 = x7 = 0.29296875 f(b8) [pic 60][pic 59]
x8 = = 0.292578125 f(x8) [pic 63][pic 61][pic 62]
Iteración 9: a9 = x8 = 0.292578125 f(a9) [pic 65][pic 64]
b9 = b8 = 0.29296875 f(b9) [pic 67][pic 66]
x9 = = 0.2927734375 f(x9) [pic 70][pic 68][pic 69]
Iteración 10: a10 = x9 = 0.2927734375 f(a10) [pic 72][pic 71]
b10 = b9 = 0.29296875 f(b10) [pic 74][pic 73]
x10 = = 0.2928710938 f(x10) [pic 77][pic 75][pic 76]
Iteración 11: a11 = x10 = 0.2928710938 f(a11) [pic 79][pic 78]
b11= b10 = 0.29296875 f(b11) [pic 81][pic 80]
x11 = = 0.2929199219 f(x11) [pic 84][pic 82][pic 83]
...