guia series funcionales_2014
TRANSCRIPT
I. Escribir los Desarrollos en serie de Potencias de “x” e indicar los intervalos deconvergencia de las siguientes funciones:
1. f(x) =)x21)(x1(
3
;
0n
n1nn ]x.21)([1 ; x < ½
2. f(x) =2)1x()3x2(
; -
0n
n3)x(n ; x < 1
3. f(x) = x. x2e ; x +
2n
n1n-1n
1)!(nx2(-1)
; x <
4. f(x) =2xe ; /n!x1
1n
2n
; x <
5. f(x) = Senh(x) ;
0n
12n
1)!(2nx
; x <
6. f(x) = Cos(2x) ;
1n
2n2nn
(2n)!.x2
1)(1 ; x <
7. f(x) = Cos2(x) ;
1n
2nn
(2n)!(2x)1)(
1/21 ; x <
8. f(x) =2x9
x
;
0n
1n
12nn
9x
1)( ; x < 3
9. f(x) =2x4
1
;
12n
2n
0n 2x
2.4.6...2n1)2n1.3.5....(
; x < 2
10. f(x) = Ln
x1x1
;
0n
12n
12nx
2 ; x < 1
11. f(x) = ax (a > 0) ;
1n
nn
n!.a.Lnx
1 ; x <
12. f(x) = Ln(2 + x) ; Ln(2) + x/2 + x2/2.22 + x3/3.23 + … +n
n1n
2nx
.)1( ; 2 < x 2
13. f(x) = Sen2(x) ;
1n
2n12n1n
(2n)!.x2
1)( ; x <
14. f(x) = Cos(x + a) ; Cos(a) – xSen(a) – x2.Cos(a) + x3 Sen(a) + x4Cos(a) + .... x < 12! 3! 4!
15. f(x) = Ln(x + 2x1 ) ;12n
xn)2.4.6...(2
1)n1.3.5...(21)(
12n
n
n
; x 1
16. f(x) = Arctg(x) ; 12n
1n
1n
x12n
1)(
; [-1, 1]
17. f(x) = Sen(x).Cos(x)
18. f(x) = 3 x8
19. f(x) =x
)x(Cos1
20. f(x) =2x1
1
21. f(x) = x2.ex
22. f(x) = x2Sen(x)
23. f(x) = Cos(x2)
24. f(x) = x(1 + 2x)-2
II. Calcular el Límite, Usando Series de Potencias:
1.)1x(Lnx
)x(Arctg)x(SenLim
20x
=
61
2.)x(Arctgxx2ee
Limxx
0x
= 1
3.)1x(Lnxx)x(Arctg
Lim20x
=31
4.
2
2
20x x1x1
Ln.x1
Lim = 2
5.
1x1x
SenLim2
1x= 2
6.)x(Senx
x)x(TgLim
0x
= 2
7.)x(Arctg24x2
Lim0x
2
1
8.20x x
))x(Cos(LnLim
=
21
9.30x x
)x(Senh)x(TgLim
=61
10.)1e(x
)x(Arcsen)x1(LnLim
x0x
=
21
11. 32
xxxxLim 3 233 23
x
12.)1x(Lnx
)x(Arctg)x(SenLim
20x
=
61
13.30x x
)x(Sec)x(TgLim
= -
14.20x ))x(Cosh1(
)x(SenhxLim
=
15.2023
xexx3))x(Cos2)(x(Sen
Lim5
x3
0x
16.)2/x(Sen)x(Cose
Limx
0x
=
17.)x(Sen
)x(ArcsenxLim
30x
=61
18.416x
39xLim
2
2
0x
=34
19.)x(Sen)x1(Ln
1)x1(Lim
2/1
0x
=
20.28x2
x)x(ArcsenLim
3 20x
=
21. )x(Cotg)x(CoscLim0x
= 0
22.3
2
0x x)xx1(Lnx
Lim
=
61
23.
)x(Cos1
1x1
Lim0x
=
24.x1
x1Lim
1x
=
25. 1)1e(x
)xx1(Ln)xx1(LnLim
x
22
0x
26. 4)x(Cos)x(Cos
)x(Sen2)x(Sen2)x2(SenLim
2
2
0x
27.)x(xSen)x(Cos1
Lim0x
=21
28.x/1
)x/1(Arctg2x/1Limx
= -1
29.
x)x(Cotg
x1
Lim20x
=31
30.xx
)x(Tg3)x(SenLim
3
2
0x
=
31.3x ))x(Tg(
)x(ArctgxLim
=
32.20x x1
))x(Sen1(LnLim
=
x + _1_.x5 + _1.3__.x9 + … + 1. 3.5…(2n – 1)x4n +1 x < 12.5 22.9.2! 2n(4n + 1)n!
III. Calcular las siguientes integrales utilizando desarrollos en series de potencias eindicar los intervalos de Convergencia:
1. t
0 xsen(x)dx
;1)!1)(2n(2n
x1)(
12n
0n
n
; x <
2. dxex
0
x2
;
1n
12nn
1)n!(2nx
1)(x ; x <
3. x
0
x)/x)dx(Ln(1 ;
1n
2
n1n
nx
1)( ; x 1
4.
x
04x1
dx;
5. x
0
Arctg(x)
6. dx)xLn(1x
0
2
IV. Resolver las siguientes Integrales:
1. f(x) = t)dt/(1ex
0
t ; x + x3/6 – x4/12 + 3x5/40 - ….
2. f(x) = x
0
/tArctg(t)dt
3. f(x) = 2x
0
t 1dt/te2
4. f(x) = )dtt.Cos(ex
0
t
5. f(x) = dt))tLn(Cos(x
0
V. Usar una serie de Potencia, encontrar el valor aproximado con una exactitud de 4cifras decimales:
1. dxe1
0
2x ; R = 0.7468
2. 1
02
dxxCos(x)1
R = 0.4864
3. dxx
x)Ln(11/2
0
R = 0.4484
4. 1/2
0
dxx
Arctg(x)R =
5. dxxe11/2
0
x
R =
6. dx)Sen(x1/2
0
2 R = 0.0415
7. 1/2
0
3 )xdx/(1 R = 0.04854
8. dxx
Sen(x)1
0 R = 0.621
9. 1
0
3 Cos(x)dxx R = 0.608
10. dxx11/2
0
3 R = 0.508
11. dxex2x
1
0
2 R = 0.189
12. dxxArctg(x)0.3
0 R = 0.0088
13. 1/2
0
2 )dxArctg(x R =
14. 1
0
4 )dxx.Sen(x R =
VI. Calcular el valor aproximado usando series de Potencias:
1. 24 ; R = 4.899
2. 5 30 ; R =1.974
3. 102 ; R = 10.09995
VII.