Asignacion Analisis Num II Corte 10
Páginas: 7 (1533 palabras)
Publicado: 22 de abril de 2015
INSTITUTO UNIVERSITARIO POLITÉCNICO
“SANTIAGO MARIÑO “
AMPLIACIÓN MARACAIBO
Asignatura: Análisis numérico
Estudiante: Br. Danny Mavarez
C.I. 16.917.891
Actividad del 10%
II Corte
ASIGNACIÓN: ANÁLISIS NUMÉRICO (10%)
2
1. Usando el Método de Newton, con error inferior a 10
, halla el valor de las raíces de las siguientes
ecuaciones:
a)
Solución:
2X= tgX
2XtgX =0
f(X) = 2X – tgX
2
f’ (X)= 2 sec
X
X
=0.5
0
X
= X
–
f(X
)
1
0
0
f’(X
)
0
f(X
) = f(0.5) = 2(0.5) – tg (0.5) = 0.4537
0
2
f’(X
) = f’(0.5) = 2 sec
(0.5) = 0.7016
0
X
= 0.5 –
0.4537
= 0.1467
1
0.7016
f(X) = f(0.1467) = 2(0.1467) tg(0.1467) = 0.1456
1
2
f’(X
)= f’ (0.1467) = 2 sec
(0.1467) = 0.9782
1
X
= 0.1467 –
(0.1456)
= 0.0021
2
(0.9782)
℮ = │X
X
│= │0.0021 + 0.1467│ = 0.1488
2
1
Página 1
de 15
f(X
) =f(0.0021) = 2 (0.0021) – tg (0.0021) = 0.0021
2
2
f’(X
)= f’ (0.0021) = 2 sec (0.0021) = 1
2
X
= 0.0021 –
0.0021
= 0
3
1
℮
= │X
X
│ = │00.0021│ = 0.0021
3
2
2
To│ = 10
= 0.01
0.0021 < 0.01
0.0021 < To│
Página 2
de 15
b)
= 0
Solución:
x
senxe
=0
x
f(x)=senxe
x
f‘(x)=cosxe
x
=0.5
0
0.5
f(x
) = f(0.5 ) = sen(0.5 ) e= 1.1693
0
0.5
f‘(x
) = f’(0.5 )= cos(0.5 ) e
= 0.7711
0
x
=x
– f(x
)
1
0
0
f’(x
)
0
x
= 0.5 – (1.1693) =1.0164
1
(0.7711)
e = Ix
x
I = I1.01640.5I=I1.5164I=1.5164
1
0
1.0164
f(x
) = f(1.0164 ) = sen(1.0164 ) – e
= 1.2121
1
1.0164
f’(x
) = f’(1.0164 ) = cos(1.0164 ) – e
= 0.1645
1
x= 1.0164 – (1.2121)
2
6.3520
( 0.1645)
6.3520
f(x
) = f(6.3520 ) = sen(6.3520 ) – e
= 573.57
2
6.3520
f’(x
) = f’(6.3520) = cos(6.3520 ) – e
= 572.64
2
Página 3
de 15
x
=0.5
0
x
= x
f(x
)
3
2
2
f‘(x
)
2
x
= 6.3520 (573.57)
3
(572.64)
x
= 5.3504
3
5.3504
f(x
) = f(5.3504) = sen(5.3504) – e= 211.4958
3
5.3504
f’(x
) = f’(5.3504) = cos(5.3504) – e
= 210.0970
3
x
= x
f(x
)
4
3
3
f‘(x
)
3
x
= 5.3504 ( 211.4958)
4
( 210.0970)
x
= 4.3437
4
4.3437
f(x
) = f(4.3437) = sen(4.3437) – e
= 77.9247
4
4.3437
f’(x
) = f’(4.3437) = cos(4.3437) – e
= 77.3523
4
x
= x
f(x
)
5
4
4
f‘(x
)
4
x
= 4.3437 – ( 77.9247) = 3.3363
5
(77.3523)
3.3363
f(x
) = f(3.3363) = sen(3.3363) – e
= 28.3084
5
3.3363
f’(x
) = f’(3.3363) = cos(3.3363) – e
= 29.0960
5
Página 4
de 15
x
= 3.3363 (28.3084)= 2.3634
6
(29.0960)
2.3634
f(x
) = f(2.3634) = sen(2.3634) – e
= 9.9250
6
2.3634
f’(x) = f’(2.3634) = cos(2.3634) – e
= 11.3392
6
x
= x
f(x
)
7
6
6
f’(x
)
6
x
= 2.3634 (9.9250) = 1.4881
7
(11.3392)
1.4881
f(x
) = f(1.4881) = sen(1.4881) – e
= 3.4321
7
1.4881
f’(x
) = f’(1.4881) = cos(1.4881) – e
= 4.3461
7
x
= x
f(x
) =1.4881(3.4321) =0.6984
8
7
7
f’(x
) (4.3461)
7
0.6984f(x
) = f(0.6984) = sen(0.6984) – e
= 1.3675
8
0.6984
f’(x
) = f’(0.6984) = cos(0.6984) – e
= 1.2447
8
x
= 0.6984 (1.3675) =0.4003
9
(1.2447)
0.4003
f(x
) = f(0.4003) = sen(0.4003) – e
= 1.0598
9
0.4003
f’(x
) = f’(0.4003) = cos(0.4003) – e
= 0.2508
9
x
= x
– f(x...
“SANTIAGO MARIÑO “
AMPLIACIÓN MARACAIBO
Asignatura: Análisis numérico
Estudiante: Br. Danny Mavarez
C.I. 16.917.891
Actividad del 10%
II Corte
ASIGNACIÓN: ANÁLISIS NUMÉRICO (10%)
2
1. Usando el Método de Newton, con error inferior a 10
, halla el valor de las raíces de las siguientes
ecuaciones:
a)
Solución:
2X= tgX
2XtgX =0
f(X) = 2X – tgX
2
f’ (X)= 2 sec
X
X
=0.5
0
X
= X
–
f(X
)
1
0
0
f’(X
)
0
f(X
) = f(0.5) = 2(0.5) – tg (0.5) = 0.4537
0
2
f’(X
) = f’(0.5) = 2 sec
(0.5) = 0.7016
0
X
= 0.5 –
0.4537
= 0.1467
1
0.7016
f(X) = f(0.1467) = 2(0.1467) tg(0.1467) = 0.1456
1
2
f’(X
)= f’ (0.1467) = 2 sec
(0.1467) = 0.9782
1
X
= 0.1467 –
(0.1456)
= 0.0021
2
(0.9782)
℮ = │X
X
│= │0.0021 + 0.1467│ = 0.1488
2
1
Página 1
de 15
f(X
) =f(0.0021) = 2 (0.0021) – tg (0.0021) = 0.0021
2
2
f’(X
)= f’ (0.0021) = 2 sec (0.0021) = 1
2
X
= 0.0021 –
0.0021
= 0
3
1
℮
= │X
X
│ = │00.0021│ = 0.0021
3
2
2
To│ = 10
= 0.01
0.0021 < 0.01
0.0021 < To│
Página 2
de 15
b)
= 0
Solución:
x
senxe
=0
x
f(x)=senxe
x
f‘(x)=cosxe
x
=0.5
0
0.5
f(x
) = f(0.5 ) = sen(0.5 ) e= 1.1693
0
0.5
f‘(x
) = f’(0.5 )= cos(0.5 ) e
= 0.7711
0
x
=x
– f(x
)
1
0
0
f’(x
)
0
x
= 0.5 – (1.1693) =1.0164
1
(0.7711)
e = Ix
x
I = I1.01640.5I=I1.5164I=1.5164
1
0
1.0164
f(x
) = f(1.0164 ) = sen(1.0164 ) – e
= 1.2121
1
1.0164
f’(x
) = f’(1.0164 ) = cos(1.0164 ) – e
= 0.1645
1
x= 1.0164 – (1.2121)
2
6.3520
( 0.1645)
6.3520
f(x
) = f(6.3520 ) = sen(6.3520 ) – e
= 573.57
2
6.3520
f’(x
) = f’(6.3520) = cos(6.3520 ) – e
= 572.64
2
Página 3
de 15
x
=0.5
0
x
= x
f(x
)
3
2
2
f‘(x
)
2
x
= 6.3520 (573.57)
3
(572.64)
x
= 5.3504
3
5.3504
f(x
) = f(5.3504) = sen(5.3504) – e= 211.4958
3
5.3504
f’(x
) = f’(5.3504) = cos(5.3504) – e
= 210.0970
3
x
= x
f(x
)
4
3
3
f‘(x
)
3
x
= 5.3504 ( 211.4958)
4
( 210.0970)
x
= 4.3437
4
4.3437
f(x
) = f(4.3437) = sen(4.3437) – e
= 77.9247
4
4.3437
f’(x
) = f’(4.3437) = cos(4.3437) – e
= 77.3523
4
x
= x
f(x
)
5
4
4
f‘(x
)
4
x
= 4.3437 – ( 77.9247) = 3.3363
5
(77.3523)
3.3363
f(x
) = f(3.3363) = sen(3.3363) – e
= 28.3084
5
3.3363
f’(x
) = f’(3.3363) = cos(3.3363) – e
= 29.0960
5
Página 4
de 15
x
= 3.3363 (28.3084)= 2.3634
6
(29.0960)
2.3634
f(x
) = f(2.3634) = sen(2.3634) – e
= 9.9250
6
2.3634
f’(x) = f’(2.3634) = cos(2.3634) – e
= 11.3392
6
x
= x
f(x
)
7
6
6
f’(x
)
6
x
= 2.3634 (9.9250) = 1.4881
7
(11.3392)
1.4881
f(x
) = f(1.4881) = sen(1.4881) – e
= 3.4321
7
1.4881
f’(x
) = f’(1.4881) = cos(1.4881) – e
= 4.3461
7
x
= x
f(x
) =1.4881(3.4321) =0.6984
8
7
7
f’(x
) (4.3461)
7
0.6984f(x
) = f(0.6984) = sen(0.6984) – e
= 1.3675
8
0.6984
f’(x
) = f’(0.6984) = cos(0.6984) – e
= 1.2447
8
x
= 0.6984 (1.3675) =0.4003
9
(1.2447)
0.4003
f(x
) = f(0.4003) = sen(0.4003) – e
= 1.0598
9
0.4003
f’(x
) = f’(0.4003) = cos(0.4003) – e
= 0.2508
9
x
= x
– f(x...
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