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∞

Example 10. Find the sum of the series ∑ nxn−1 ( | x |< 1).

Solution. It is needed to find a sum

n=1

S(x) = 1+ 2x + 3x2 + 4x3 + …+ nxn−1 + … .

We’ll consider a geometrical series

1+ x + x2 + x3 + …+ xn + ….

convergent

for

| x |< 1,

thus its

sum

S (x)

1− x

We use the property of term-by-term differentiation of power series and get

S(x) =

(S (x))′

′ =

(1− x)

1− x

Example 11. Expand the functions into Maclaurin’s series:

а) f (x) = ex ; b) g(x) = x2e2 x .

Solution. а) Function

f (x) = ex

infinitely

differential,

thus

ex = (ex )′ =

= (e

)

′′

= …= (e

)

(n)

for

arbitrary n N.

Therefore

′

= f (0)

′′

= …

f (0) = f (0)

... = f (0)(n) = 1 and

= 1

+ …+

+ … .

(2.12)

We’ll find the radius of convergence of this series:

R = lim	an		(n + 1)!	= lim(n + 1) = ∞.
	an	= lim	(n + 1)!
	a	= lim	n!
n→∞	a	n→∞	n!	n→∞
	n+1
	n+1

Thus a series converges in an interval (−∞; ∞).					x (−r; r) , where r
It remains to prove that lim Rn = 0.				For all	x (−r; r) , where r	is any
		n→∞
positive number, inequality		f (n) (x)	= ex	< er = M	is fulfilled.
Then by the theorem 2 lim Rn = 0.
	n→∞
Consequently, at any	interval (−r; r) (−∞; ∞), that is at the					interval

(−∞; ∞), the function ex is expanded into Maclaurin’s series by the formula (2.12).

b) To expand the function

g(x)

into Maclaurin’s series, we replace x by

2x in a formula (2.12) and multiply a result by x2, then we’ll get

g(x) = x

= x

(2x)2

(2x)3

(2x)n

+ …+

+ …

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= x	2	+	2x3	+	22 x4	+	23 x5	+ …+	2n xn+1	+ … .
			1!		2!		3!		n!

This series converges to the given function on the whole numerical line. Example 12. Expand the functions into Taylor’s series in powers of x −1 :

а)

f (x) =

; b)

g(x) =

x + 2

(x + 2)2

Solution. а) Let

x − 1 = t,

then

x +

+ 1+ 2

+ t

Using formula 7 (Table 2.2) we get

∞

n t

∞

∑

(−1)

= ∑ (−1)

3n+1

3 n=0

n=0

After replacement

x −1 , we’ll obtain a final answer

−

x + 2

x − 1

(x − 1)

n (x − 1)

∞

n (x −

−

− …+ (−1)

+ …=

∑

(−1)

n+1

n=0

This equality is true if

x − 1

< 1( −2 < x < 4 ).

′

= −

. We use the property of term-by-term differentiation

(x + 2)

of a power series and get

− 1

(x −1)2

n (x −1)n

′

= −

−

−…+ (−1)

+ … =

(x +

n+1

2(x − 1)

n n(x − 1)n−1

= − −

−…+ (−1)

+ … =

n+1

∞

(−1)n+1

n−1

= ∑

n(x − 1)

, x (−2; 4).

n+1

n=1

Example 13. Expand the function f (x) = arctg x

into Maclaurin’s series.

Solution. Replacing in a formula 7 (Table 2.2) x by x2 , we’ll write down an equality

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	1	= 1− x2 + x4 − x6 + …+ (−1)n x2n + …, x (−1; 1) .
	1+ x2	= 1− x2 + x4 − x6 + …+ (−1)n x2n + …, x (−1; 1) .
	1+ x2
We use the property of term-by-term integration of power series and get
x	dt	x	x	x	x	x
∫	dt	= ∫ dt − ∫t2 dt + ∫t4 dt − ∫t6dt + …+ (−1)n ∫ t2n dt + … .
∫	2	= ∫ dt − ∫t2 dt + ∫t4 dt − ∫t6dt + …+ (−1)n ∫ t2n dt + … .
0	1+ t	0	0	0	0	0

x2n+1

arctg x = x −

−

+…+ (−1)n

+… .

2n +1

This equality is true for

x = ±1 too. So the series converges for

x [−1; 1] .

Example 14. Expand the function

f (x) = ln(x2 + 3x + 2) into Maclaurin’s

series.

Solution. We shall transform logarithmic function

ln(x2 + 3x + 2) = ln(x + 2)(x + 1) = ln(x + 2) + ln(x + 1) =

+ ln(1+ x) = ln 2 + ln

(1+ x).

= ln 2 1

+ ln

We use the expansion 4 (Table 2.2) for logarithmic functions ln

and

ln(1+ x). Then

n−1 xn

ln(x

+ 3x + 2) = ln 2 +

−

− …

(−1)

+ …

∞

−1

n+1

n−1 x

+ ∑

(

)

+ x −

− …+ (−1)

+ …

= ln 2

)

(

)

n=1

(

n+1

∞

−1

∞

−1

+ ∑

= ln 2 +

∑

+ 1 x .

n=1

Let’s find the domain of convergence of this series. Maclaurin’s series for

		x			x			−2 < x ≤ 2,
function	ln 1+		converges at −1 <			≤ 1,	i. e			and a series for
function	ln 1+		converges at −1 <		2	≤ 1,	i. e			and a series for
		2			2
ln(1+ x)	converges on an interval			−1 < x ≤ 1.			Then the received expansion is
convergent for		−1 < x ≤ 1, i.e. for		those x,			at	which both		series converge
simultaneously.
Example 15. Expand the function f (x) =							x2		into Maclaurin’s series.
Example 15. Expand the function f (x) =									into Maclaurin’s series.
							9 + x2
										53

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Solution. We shall transform function

f (x)

so:

9 + x

Using the formula 6 (Table 2.2) for

m = −0,5,

we’ll expand the function

= (1+ t )

−

into a series

1+ t

−

− 1

−

− 2

= 1

+ −

t +

1+ t

− n

−

− 1

−

− 2

...

−

+ 1

+…+

+ ... =

1 3

1 3 5

1 3 ... (2n − 1)

= 1

−

t3 +

... + (−1)n

tn + ... .

22 2!

23 3!

2n n!

We replace t

in this equality and get

n 1 3 ... (2n −1)

= 1

−

1 3

−

1 3 5

+ ...+

(

−1

+ ....

18 2!

18 3!

Finally the required expansion is

1 3

1 3 5

−

+ ...

x2 + 9

n 1 3 ... (2n −1)

2n+ 2

...+ (−1)

+ ...

Domain

convergence

the

series

the

interval −1 <

≤ 1 , or

−3 < x ≤ 3.

Example 16. Calculate

with accuracy ε = 0,001.

Solution. Using expansion of the function receive

1					1		1			1
e = e	2	= 1	+			+			+
				2	1!
						22		2! 23 3!

ex into Maclaurin’s series we

+ …+ 2n1n! + … .

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Let’s define such n at which an error of the approximate equality

e ≈ 1+	1	+	1	+	1		+ …+	1
e ≈ 1+	2 1!	+	22 2!	+	23 3!		+ …+	2n n!

doesn’t exceed the set accuracy. For this purpose we’ll estimate the remainder

+ …=

2n+1 (n + 1)!

2n+ 2 (n + 2)!

2n+3 (n + 3)!

+ …

n+1

2(n

+ 2)

+ 2)(n

+ 3)

(n + 1)!

…

n+1

2(n + 2)

(n + 2)

(n +

(n + 1)!

2(n + 2)

n + 2

2n+1 (n + 1)!

+1 (n + 1)!

2n (2n + 3)(n + 1)!

1−

2n + 3

2(n + 2)

The

inequality

rn <

n + 2

< 0,001

carried

out

starting with

2n (2n + 3)(n + 1)!

n = 4. So,

e ≈ 1+

≈

2 1!

22 2! 23

3! 24

≈ 1+ 0,5 + 0,125 + 0,0208 + 0,0026 = 1,6484.

For calculation of logarithms we use the series

ln(1+ x) = x −

− …+ (−1)

n−1

+ … ( −1 < x ≤ 1).

But it is slowly convergent. In practice the following formula is used:

1+ x

x2n−1

x (−1; 1).

= 2 x +

+ …

+ … ,

(2.13)

− x

− 1

The error of this formula is estimated by inequality

| rn (x) |= 2

x2n+1

x2n+3

|x|2n+1

(1+ x2 + x4 + …) = 2

|x|2n+1

+ …

< 2

2n + 1

2n + 3

2n + 1

(2n + 1)(1− x2 )

Let’s

x =

t N

the

formula (2.13). We

get

the

formula for

2t + 1

calculation of logarithms of natural numbers:

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ln(t + 1) = ln t + 2	1		1			1
		+			+ …+			+ …	(2.14)
	2t + 1	+	3(2t + 1)	3	+ …+	(2n − 1)(2t + 1)	2n−1	+ …	(2.14)
	2t + 1		3(2t + 1)			(2n − 1)(2t + 1)

and

				\| rn (t) \|<											1						.
				\| rn (t) \|<							2(2n + 1)t(t + 1)(2t + 1)2n−1										.
Example 17. Calculate ln 2 with accuracy ε = 0,001.
Solution. In the formula (2.14) we put t = 1. We get
						1				1						1
		ln 2 = 2						+						+ …+						+		… .
		ln 2 = 2				3		+		3				+ …+	(2n − 1) 3			2n−1		+		… .
						3				3 3					(2n − 1) 3
By selection we define n so that the following inequality is true:
								rn (1) <							1		.
								rn (1) <						4(2n + 1) 32n−1			.
We have: r (1) <				1			=				1		, r (1) <		1		=		1			< 0,001.
We have: r (1) <				4 5 33			=		180				, r (1) <		4 7	35	=	6804				< 0,001.
2				4 5 33					180					3	4 7	35		6804
So, n = 3 and for calculation ln 2 the approximate equality is received
	1		1					1
ln 2 ≈ 2		+			+							≈ 2(0,3333 + 0,0123 + 0,0008) ≈ 0,693.
ln 2 ≈ 2	3	+	3		+				5			≈ 2(0,3333 + 0,0123 + 0,0008) ≈ 0,693.
	3		3 3			5 3
							1
							2
Example 18. Calculate							∫ e− x2 dx with accuracy ε = 0,001.
							0

Solution. The indefinite integral ∫ e− x2 dx is not expressed through elemen-

tary functions. For calculation of integral we expand the integrand in power series and use the property of term-by-term integration of power series.

We receive

x2n

− x2

∫ e

dx =

∫

− x

−

+ …

+ (−1)

+ … dx

n x2n+1

x −

−

+ …+ (−1)

+ …

5 2!

7 3!

(2n + 1) n!

−

+ …+ (−1)n

+ ….

23 3 25

2! 27 7 3!

22n+1 (2n + 1) n!

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This series is alternating and satisfies conditions of Leibniz’ theorem. We use a corollary from this theorem and determine the least number n for which the following inequality is true:

	1		< 0,001, or 22n+1 (2n + 1) n! > 1000.
	22n+1 (2n + 1) n!		< 0,001, or 22n+1 (2n + 1) n! > 1000.
	22n+1 (2n + 1) n!

This inequality is fulfilled starting with n = 3. Therefore, we take the first three members of the series and obtain

1		1			1				1
2	− x2	1			1				1
∫ e dx ≈			−				+			≈ 0,5 − 0,0417 + 0,0031 ≈ 0, 461.
∫ e dx ≈		2	−	2	3	3	+	5	5 2!	≈ 0,5 − 0,0417 + 0,0031 ≈ 0, 461.
0		2		2		3	2		5 2!

Example 19. Find the approximate solution of the Cauchy’s problem y′ = x2 + y3 , y(0) = 1 using the first four nonzero members of expansion of this solution in power series.

Solution. We’ll find the solution in the form of Maclaurin’s series

′

′′

′′′

(n)

(0)

y = y(0) +

y (0)

x +

y (0)

(0)

x3... +

xn + ... .

From the condition of the task we can find the first two coefficients

y(0) = 1,

y′(0) = 02 + 13 = 1.

We differentiate the initial equation

y′′ = 2x + 3y2 y′ .

Then we substitute x = 0, y(0) = 1

and

y′(0) = 1 into this equation and get

the factor y′′(0) = 0 + 3 = 3. Now we pass to the equation

y′′′ = 2+ 3(2y(y′)2 + y2 y′′).

Then y′′′(0) = 2 + 3(2 + 3) = 17.

So, the approximate

solution of

Cauchy’s

problem is defined by the formula

y ≈ 1+ x + 32 x2 + 176 x3. This formula is more precise if x tends to zero.

Micromodule 2

CLASS AND HOME ASSIGNMENTS

Find the domain of convergence of functional series.

	∞			1					∞		1				∞			x	n
1.	∑			1			.	2.	∑		1	.		3.	∑			x				.
1.	∑		2			4	.	2.	∑		1	.		3.	∑					2n		.
	n=1 n			+ x					n=1 n						n=1	1	+ x
	n=1 n			+ x					n=1 n		sin x				n=1	1	+ x
		1								x	n(n−1)									1
	∞	1							∞	x	2				∞					1
4.	∑				.			5.	∑				.	6.	∑								.
4.	∑		nx		.			5.	∑				.	6.	∑								.
	n=1	2							n=1		n!				n=1 n			(x			+ 2)

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∞	(1− n) x		∞	n		∞	1	x + 1	n+1
7. ∑ e		.	8. ∑ n ln		x.	9. ∑			.

n=1			n=1			n=1 n		x −1

Investigate the functional series for uniform convergence within the given interval.

	∞			1
10.	∑								,		x (−∞; ∞).
10.	∑	5	n	+ x			2		,		x (−∞; ∞).
	n=1	5		+ x
	∞			x	2
12.	∑			x						,	x (−∞; ∞).
12.	∑				2	x		2		,	x (−∞; ∞).
	n=1	1+ n				x
14.	∞			+ 1)xn ,							x (−1; 1).
14.	∑ (n			+ 1)xn ,							x (−1; 1).

n=1

11.	∑ cos nx6 ,	x (−∞; ∞).
	∞
	n=1 n!+ x
13.	∞
	∑ e− nx , x (0; ∞).
	n=1
15.	∞	x [0; ∞).
	∑ nxe− nx ,

n=1

Find the domain of convergence of power series.

∞

16.

∑

17. ∑

18. ∑

19. ∑

n=1 n

n=1 4 n + 1

n=1

(2n + 1)!

n=1 n ln(n + 1)

∞

(x − 3)n

∞

n+1

∞

20.

∑

21.

∑

+ 2)

22.

∑

(2n + 3) 5

−1)

n+1

n=1

∞

x3n

∞

2n2 −1

23.

∑

24.

∑

25.

∑

n=0 2n

n=0

+ 1

n=1

Find the interval of convergence of power series.

∞

n − 1

∞

x2n

∞

(n!)2

26. ∑

(x − e)

27. ∑

28. ∑

(2n)!

n=1

n + 1

n=0

Find the sum of the power series.

∞

29. ∑ n2 xn .

n=1

∞	x	n		∞	x	n+1
30. ∑			.	31. ∑			.
	n				n!
n=1				n=1

Expand the given functions into Taylor’s series in powers of x − x0 :

32.	1		, x = 2.		33.	1	,	x	= −1.

		x	0			3x + 1		0

34.	ln(x2 + 2x + 2), x = −1.				35.	1		,	x = −2.
						(x + 3)2
				0					0
36.			1	, x = −2.	37.	xe2x−1,		x	= 1.
		(x + 3)3
				0				0

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Expand the given functions into Maclaurin’s series:

38.	2x cos2 x.									39.	x	.				40.					1		.
				2x + 3						1	− x3									2x − 3
41.				2x + 3					.	42. ln(3 + 6x).						43. (1+ x2 )e− x .
41.		x2 + 3x + 2							.	42. ln(3 + 6x).						43. (1+ x2 )e− x .
		x2 + 3x + 2
Evaluate the values of functions with accuracy ε.
44.	3 130,							ε = 0,0001.					45.	1			,	ε = 0,001.
44.	3 130,							ε = 0,0001.					45.				,	ε = 0,001.
															3 e			1
46.	cos10°,							ε = 0,0001.					47.	arctg				1		,	ε = 0,001.
46.	cos10°,							ε = 0,0001.					47.	arctg				3		,	ε = 0,001.
48.	ln 3, ε = 0,0001.												49.					3			ε = 0,0001.
48.	ln 3, ε = 0,0001.												49.	ln 0,98,							ε = 0,0001.
Evaluate the definite integrals with accuracy ε.
		1		sin x										1
	2			sin x										1					2
50.		∫				dx,			ε = 0,0001.				51.		∫ cos x					dx,		ε = 0,0001.
50.		∫		x		dx,			ε = 0,0001.				51.		∫ cos x					dx,		ε = 0,0001.
	0			x										0
		1		dx											1		dx
	2			dx										3			dx
52.		∫					, ε = 0,001.						53.		∫						,	ε = 0,001.
52.		∫	1+ x		4		, ε = 0,001.						53.		∫	1+ x4					,	ε = 0,001.
	0		1+ x											0		1+ x4

Find the approximate solution of the Cauchy’s problem using the first four nonzero members of expansion of this solution in power series.

54.	y′ = xy + ey ,	y(0) = 0 .
55.	y′ = y3 + x	y, y(0) = 1.
56.	xy′ = x2 y2 − y + 1, y(0) = 2.
57.	y′′ = yy′ − x2 , y(0) = 1, y′(0) = 1.

Answers

1. (−∞; ∞). 2. (2πk; π + 2πk). 3. x ≠ ±1. 4. (0;∞). 5. [−1; 1]. 6. (−∞; ∞). 7. (0;∞). 8.

(e−1; e). 9. (−∞; 0]. 10. Uniformly convergent. 11. Uniformly convergent. 12. Uni-

formly convergent. 13. Is not uniformly convergent. 14. Is not uniformly convergent. 15. Is not uniformly convergent. 16. [−1; 1]. 17. [−1; 1). 18. [−1; 1]. 19. [−1; 1). 20. [1; 5).

21. x = −2. 22. (−0,2; 0,2) . 23. (−5; 5). 24. (−2; 2). 25. [−0,5; 0,5). 26. (0; 2e).

27. (− 2e;	2e). 28.	(−4; 4). 29. x(x + 1) /(1− x)3,		\| x \|< 1. Hint. Write this series as
∞	∞		∞	∞
∑n2xn =	= x2 ∑n(n − 1)xn−2		+x∑nxn−1 . Consider the series ∑ xn and use term-by-
n=1	n=1		n=1	n=0
				59

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∞

(x − 2)n

term

differentiation.

30.

− ln |1 − x | .

31.

xex ,

x R.

32. ∑ (−1)n

, x (0; 4).

n+1

n=0

∞ 3n (x + 1)n

∞

n+1 (x + 1)2n

∞

n−1

n+1

33. − ∑

34. ∑(−1)

, x [−2; 0]. 35. ∑(−1)

x −

;

−

n(x + 2)

n+1

n=0

3 3

n=1

∞

(−1)

n(n − 1)

∞

n−1

(n +

x (

−

1).

36.

∑

n−2

− −

1).

37.

∑

−

, x (

e 1

n=1

22 x3

24 x5

22n x2n+1

∞

x R.

38.

2x −

− …+ (−1)n

+ … x R.

39. ∑ x3n+1,

−1 < x < 1.

(2n)!

n=0

∞

(2x)n

∞

n−1 2n xn

40.

− ∑

| x|<

41.

∑

(−1)

1 +

−1 < x < 1.

42. ln3 + ∑(−1)

n+1

n=0

n=1

∞

x (−0,5; 0,5].

43.

1 − x + ∑(−1)n

xn,

x R.

44.

5,0658. 45.

0,716.

46.

0,9948.

47.

0,321.

48.

n=2

− 2)!

1,0986. 49. – 0,0202. 50. 0,4931. 51. 0,9045. 52. 0,494.

53. 0,333.

54.

y ≈1+

. 55. y ≈1+ x +2x2

+19 x3.

56.

y ≈1++

2x4

17x6

. 57.

y ≈ 1 + x +

315

Micromodule 2

SELF-TEST ASSIGNMENTS

2.1. Find the domain of convergence of functional series.

∞

∑∞ (2x − 5)

n+1

2.1.1. а)

∑

;

∑

xn sin

;

4 + x

n n

n 1

n=1

2.1.2. а) ∑

;

∑ xn tg

x ;

∑

1− x

∞

n 1

1− 2x

2.1.3.а)

2.1.4.а)

2.1.5.а)

2.1.6.а)

	(		)			n+1
∞			−1
∑				n	x			;
n=1				n
∞									x
∑	2n+1 sin								x	;
∑	2n+1 sin								n	;
n 1								3
=	1
∞	1				n
∑				tg			(3 x) ;
∑		3		tg			(3 x) ;
n 1 n
=
∞			1
∑								;
∑		n! x				n		;
n=1		n! x

∞	n		3n		x		∞	arctg(nx)
b) ∑ 2		x		arcsin		;	c) ∑				.
b) ∑ 2		x		arcsin	n	;	c) ∑	n ln	4	n	.
n=1					n		n=2	n ln		n

	∞		x	4
b)	∑		x		;
b)	∑	1		2n	;
	n=1	1	+ x
	∞		n + 1
b)	∑					;
b)	∑					;
	n=1 x nx+2
b)	∞
b)	∑ lnn (2x −1) ;

n=1

c) ∑∞ (5x + 2)n .
	n 1		7n+2
	=		1
	∞		1
c)	∑				.
c)	∑	(3n + 1)		x	.
	n 1	(3n + 1)
	=
c)	∑		x + 1	n+1
c)	∑		x + 1	.
	∞
	n 1		x − 2
	=

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