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.