analisis numerico

República Bolivariana de Venezuela
Ministerio del Poder Popular para la Educación Universitaria
Instituto Universitario Politécnico
“Santiago Mariño”
Cátedra: Análisis Numérico






Ejercicios Propuestos del 10% Resueltos






Realizado por:
Aurimar Casanova
C.I.: 23.758.473



Maracaibo,Diciembre de 2014
1) Usando el método newton con error inferior a 10-2, halla el valor de las raíces de las siguientes ecuaciones:

a)





a) 2X= tgX 2X-tgX = 0
f(X) = 2X – tgX
f’ (X)= 2- sec2 X

X0 =0.5
X1 = X0 – f(X0)
f’(X0)

f(X0) = f(0.5) = 2(0.5) – tg (0.5) = 0.4537
f’(X0) = f’(0.5) = 2- sec2 (0.5) = 0.7016

X1 = 0.5 – 0.4537 = -0.1467
0.7016

f(X1) =f(-0.1467) = 2(-0.1467)- tg(-0.1467) = -0.1456
f’(X1)= f’ (-0.1467) = 2- sec2 (-0.1467) = 0.9782

X2 = -0.1467 – (-0.1456) = 0.0021
(0.9782)
℮ = │X2-X1│= │0.0021 + 0.1467│= 0.1488

f(X2) =f(0.0021) = 2 (0.0021) – tg (0.0021) = 0.0021
f’(X2)= f’ (0.0021) = 2- sec2 (0.0021) = 1

X3 = 0.0021 – 0.0021 = 0
1
℮ = │X3-X2│ = │0-0.0021│= 0.0021

To│ = 10-2 = 0.01
0.0021 < 0.010.0021 < To│

RAIZ X=0

b) senX – ex =0 X0 = 0.5
f(X) = senX – ex
f’ (X)= cosX - ex

X0 = 0.5

f(X0) = f(0.5) = sen(0.5) – e0.5 = - 1.1693
f’(X0) = f’(0.5) = cos (0.5) – e0.5 = -0.7711

X1 = X0 – f(X0)
f’(X0)
X1= 0.5 – (-1.1693) = 1.0164
(-0.7711)
℮ = │X1-X0│ = │-1.0164 – 0.5│ = │- 1.5164│= 1.5164

f(X1) = f( -1.0164) = sen (-1.0164) – e -10164 =-1.2121
f’(X1) = f’( -1.0164) = cos (-1.0164) – e -10164 = 0.1645

X2 = -1.0164 - -1.2121_ = 6.3520
0.1645

f(X2) = f( 6.3520) = sen (6.3520) - ℮-6.3520 = - 573.57
f’(X2) = f’( 6.3520) = cos (6.3520) - ℮-6.3520 = - 572.64

X3 = X2 – f(X2)
f’(X2)

X3= 6.3520 – (-573.57)
(-572.64)

X3= 5.3505

f(X3) = f(5.3504)= sen (5.3504) – e5.3504 = -211.4948
f’(X3) =f(5.3504)= cos (5.3504) – e5.3504 = - 210.0970

X4= X3 – f(X3)
f’(X3)

X4 = 5.35.04 – - 211.4958_
210.0970

X4 = 4.3437

f(X4) = f(4.3437) = sen (4.3437) – e4.3437 = - 77.9247
f’(X4) = f(4.3437) = cos (4.3437) – e4.3437 = -77.3523

X5= X4 – f(X4)
f’(X4)

X5 = 4.3437 - -77.9247_ = 3.3363
-77.3523

f(X5) = f(3.3363) = sen (3.3363) – e3.3363 = -28.3084
f’(X5) =f(3.3363) = cos (3.3363) – e3.3363 = - 29.0960

X6= 3.3363 - -28.3084_ = 2.3634
-29.0960

f(X6)= f(2.3634) = sen ( 2.3634) – e2.3634 = - 9.9250
f’(X6)= f(2.3634) = cos ( 2.3634) – e2.3634 = - 11.3392




X7 = X6 – f(X6)
f’(X6)

X7 = 2.3634 - -9.9250 _ = 1.4881
-11.3392

f(X7) = f (1.4881) =sen (1.4881) – e1.4881= -3.4321
f’(X7) = f (1.4881) =cos (1.4881) – e1.4881= -4.3461

X8 = X7 – f(X7) = 1.4881 - -3.4321 = 0.6984
f’(X7) -4.3461

f(X8) = f(0.6984) = sen ( 0.6984) – e0.6984 = -1.3675
f’(X8) = f(0.6984) = cos ( 0.6984) – e0.6984 = - 1.2447

X9 = 0.6984 - - 1.3675_ = - 0.4003
1.2447

f(X9) = f(- 0.4003) = sen ( -0.4003) – e-0.4003 = -1.0598
f’(X9) = f(- 0.4003) = cos ( -0.4003) – e-0.4003 = 0.2508

X10 = X9 – f(X9) = -0.4003 - -1.0598 = 3.8254
f’(X9) 0.2508

f(X10) = f( 3.8254) = sen ( 3.8254) – e3.8254 = - 46.4829
f’(X10) = f( 3.8254) = cos ( 3.8254) – e3.8254 = - 46.6263

X11 = 3.8254 - - 46.4829 = 2.8285
46.6263

f(X11) = f( 2.8285) = sen (2.8285) – e2.8285 = -16.6121
f’(X11) = f( 2.8285) = cos (2.8285) – e2.8285 = - 17.8714

X12 = 2.8285 - -16.6121 = 1.8990
-17.8714f(X12) = f(1.8990) = sen (1.8990) – e1.8990 = - 5.7326
f’(X12) = f(1.8990) = cos (1.8990) – e1.8990 = - 7.0016

X13 = 1.8990 - -5.7326 = 1.080
-7.0016

f(X13) = f(1.080) = sen (1.080) – e1.080 = - 2.0627
f’(X13) = f(1.080) = cos (1.080) – e1.080 = - 2.4734

X14 = 1.080 - -2.0627 = 0.2460
-2.4734

f(X14) = f(0.2460) = sen (0.2460) – e0.2460 = -1.0354
f’(X14) =...