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Кузбасский государственный технический университет

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354 CHAPTER 8. INFINITE SERIES

Exercises 8.4 In problem 1–12, determine the Taylor series expansion for each function f about the given value of a.

1.	f(x) = e−2x, a = 0								2.	f(x) = cos(3x), a = 0
3.	f(x) = ln(x), a = 1								4.	f(x) = (1 + x)−2, a = 0
5.	f(x) = (1 + x)−3/2, a = 0								6.	f(x) = ex, a = 2
7.	f(x) = sin x, a =				π				8.	f(x) = cos x, a =			π
7.	f(x) = sin x, a =					6			8.	f(x) = cos x, a =
						6					4
9.	f(x) = sin x, a =				π				10.	f(x) = x1/3, a = 8
9.	f(x) = sin x, a =					3			10.	f(x) = x1/3, a = 8
						3				f(x) = cos x − 2				, a = 0
11.	f(x) = sin x − 2						, a = 0		12.	f(x) = cos x − 2				, a = 0
				1							1
								n
								Xk		(x − a)k
In problems 13-20, determine f(k)(a)												.
							=0				k!
							=0
13.	f(x) = ex2 , a = 0, n = 3								14.		f(x) = x2e−x, a = 0, n = 3
	1
15.	f(x) =		, a = 0, n = 2						16.		f(x) = arctan x, a = 0, n = 3
15.	f(x) =	1 − x2	, a = 0, n = 2						16.		f(x) = arctan x, a = 0, n = 3
17.	f(x) = e2x cos 3x, a = 0, n = 4								18.		f(x) = arcsin x, a = 0, n = 3
19.	f(x) = tan x, a = 0, n = 3								20.		f(x) = (1 + x)1/2, a = 0, n = 5

8.7Taylor Polynomials and Series

Theorem 8.7.1 (Taylor’s Theorem) Suppose that f, f0, · · · , f(n+1) are all continuous for all x such that |x − a| < R. Then there exists some c between a and x such that

f(x) = Pn(x) + Rn(x)

where

	n	(x − a)k		(x − a)n+1
Pn(x) =	f(k)(a)	(x − a)k	, Rn(x) = f(n+1)(c)	(x − a)n+1	.
Xk		k!		(n + 1)!
	=0

= 1 + mx +

x3 + · · · .

x2 + m(m − 1)(m − 2) 3!

k=1

8.7. TAYLOR POLYNOMIALS AND SERIES

355

The polynomial Pn(x) is called the nth degree Taylor polynomial approximation of f. The term Rn(x) is called the Lagrange form of the remainder.

Proof. We deﬁne a function g of a variable z such that

g(z) = [f(x)

−

f(z)]

−

(z)(x − z)

−

f00(z)(x − z)2

− · · ·

−

f(n)(z)(x − z)n

−

(x)

(x − z)n+1

(x − a)n+1

Then

( n f(k)(a) ) X

g(a) = f(x) − (x − a)k + Rn(x) = 0, k!

k=0

and

g(x) = f(x) − f(x) = 0.

By the Mean Value Theorem for derivatives there exists some c between a and x such that g0(c) = 0. But

g0(z) = −f0(z) − [−f0(z) + f00(z)(x − z)] − −f00(z)(x − z) +

000(

z)(x

z)2

− · · ·

−

− −

fn(z)(x − z)n−1

f(n+1)(z)(x − z)n

+ R

(x)

(n + 1)(x − z)n

(x − a)n+1

−

f(n+1)(z)

(x − z)n

+ R

(x)

(n + 1)(x − z)n

(x − a)n+1

g0(c) = 0 =

−

f(n+1)(c)

(x − c)n

+ R

(x)

(n + 1)(x − c)n

Therefore,

(x − a)n+1

(x) =

(x − a)n+1

f(n+1)(c)

= f(n+1)

(c)

(x − a)n+1

n + 1

(n + 1)!

as required. This completes the proof of this theorem.

Theorem 8.7.2 (Binomial Series) If m is a real number and |x| < 1, then

∞

(1 + x)m = 1 + X m(m − 1) · · · (m − k + 1) xk k!

m(m − 1) 2!

f(x) =

356		CHAPTER 8. INFINITE SERIES
This series is called the binomial series. If we use the notation
	k =	( − 1) · ·k·!( −			k	+ 1)
	m	m m	m		k
m	is called the binomial coe cient and
then k	is called the binomial coe cient and
		∞	k	xk.
	(1 + x)m = 1 + k=1		k	xk.
		X	m
		X
If m is a natural number, then we get the binomial expansion
		m	k	xk.
	(1 + x)m = 1 + k=1		k	xk.
		X	m
		X

Proof. Let f(x) = (1 + x)m. Then for all natural numbers n,

f0(x) = m(1 + x)m−1, f00(x) = m(m − 1)(1 + x)m−2, · · · , f(n)(x) = m(m − 1) · · · (m − n + 1)(1 + x)m−n.

Thus, f(n)(0) = m(m − 1) · · · (m − n + 1), and

∞

X m(m − 1)(m − 2) · · · (m − n + 1) xn

n=0

=X mn xn

n=0

where 0 = 1 and

= m(m − 1) · · · (m − n + 1) is called the nth

binomial coe cient. By the ratio test we get

lim

m(m 1)

(m n)xn+1

−

(n· ·+·

1)!−

· m(m

−

· · ·

−

n + 1)xn

n→∞

= x

lim

−

n + 1

n→∞

= x

lim

−

n→∞

1 + n

= x ,

8.7. TAYLOR POLYNOMIALS AND SERIES

357

and, hence, the series converges for |x| < 1. This completes the proof of the theorem.

Theorem 8.7.3 The following power series expansions of functions are valid.

∞

(1 − x)−1 = 1 +

xk and (1 + x)−1 = 1 + (−1)kxk, |x| < 1.

k=1

∞

k! , e−x = 1 +

k! , |x| < ∞.

ex = 1 +

(−1)k

k=1

∞

x2k+1

sin x =

(−1)k

, |x| < ∞.

(2k + 1)!

∞

x2k

, |x| < ∞.

cos x =

(−1)k

(2k)!

∞

x2k−1

sinh x =

, |x| < ∞.

(2k + 1)!

∞

x2k

cosh x =

, |x| < ∞.

(2k)!

∞

xk+1

, −1 < x ≤ 1.

ln(1 + x) =

(−1)k

k + 1

1 + x

∞

x2k+1

= k=0

, −1 < x < 1.

−

2k + 1

∞

x2k+1

arctan x =

(−1)k

, −1 ≤ x ≤ 1.

2k + 1

∞

1/2

(−1)k

x2k+1

10. arcsin x = k=0

−k

2k + 1 , |x| ≤ 1.

358 CHAPTER 8. INFINITE SERIES

Proof.

Part 1. By the geometric series expansion, for all |x| < 1, we have

∞

−

x = 1 +

1 + x = 1

− −

= 1 +

xk and

( x)

(−1)kxk.

k=1

Part 2.

If f(x) = ex, then f(n)(x) = ex

and f(n)(0) = 1 for each n =

0, 1, 2, · · · . Thus

∞

ex =

n=0

By the ratio test the series converges for all x.

n→∞

xn+1

| n→∞

(n + 1)! ·

n + 1

lim

= x

lim

= 0.

Part 3. Let f(x) = sin

x. Then f0(x) = cos x,

f00(x) =

−

sin x, f

(x) =

− cos x and f

(4)

(x) = sin x. It follows that, for each n = 0, 1, 2, 3, · · · , we

have

f(4n)(0) = 0, f(4n+1)(0) = 1,

f(4n+2)(0) = 0

and f(4n+3)(0) = −1.

Hence,

sin x = x −

− · · ·

∞

=(−1)n

n=0

x2n+1

(2n + 1)!.

By the ratio test, the series converges for all |x| < ∞:

n→∞

−

n+1

x2n+3

(2n + 1)!

(2n + 3)!

x2n+1

lim

(

= x

lim

(2n + 3)(2n + 2)

n→∞

= 0.

Part 4. By term-by-term di erentiation we get

∞	x2n
cos x = (sin x)0 = X(−1)n		, \|x\| < ∞.
	(2n)!

n=0

8.7. TAYLOR POLYNOMIALS AND SERIES						359
Part 5. For all \|x\| < ∞, we get
1		(ex − e−x)
sinh x =
sinh x =	2					!
= 2		n=0	n!	− n=0(−1)n n!		!
1		∞	xn	∞	xn
		X		X

∞ x2n+1

=(2n + 1)!.

n=0

Part 6. By di erentiating term-by-term, we get

∞	x2n
X
cosh x = (sinh x)0 =	(2n)!	, l \|x\| < ∞.
n=0

Part 7. For each |x| < 1, by performing term by integration, we get

ln(1 + x) = Z0	x	1 + x dx
		1
=	∞	∞	(−1)nxn! dx
Z		X

0n=0

∞	xn+1
X

=(−1)n n + 1.

n=0

Part 8. By Part 7, for all |x| < 1, we get

1 + x

[ln(1 + x) − ln(1 − x)]

−

"n=0

−

n + 1

− n=0

−

n + 1

= 1

∞

( 1)n

xn+1

∞

( 1)n

(−x)n+1

= 2

"n=0

(n−+ 1

(1 − (−1)n+1)xn+1#

∞

1)n

∞ x2k+1

=2k + 1.

k=0

	1		1 + x
Recall that arctanh x =		ln		.
	2		1 − x

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