Unformatted text preview: Section 9.2
55. Given an
0, by definition of convergence there
and
corresponds an N such that for all n N, L1 an
L 2 an
. (There is one such number for each series,
and we may let N be the larger of the two numbers.) Now
L2 L1
L2 an an L1
an L1
L 2 an 1 L1, lim ai(n) L2, and L1 n→ ak(n) L1 , and an N2 such that for i(n) ai(n) L2 0 N1, max {N1, N2}. Then for n an L1 and an implies that lim an L2 N2, N, we have that for infinitely many n. This L1 and lim an n→ n→ L2 where L1 (b) The line y L2. 1. f (x)
f (x)
f (x)
f (x)
f (n)(x) 1 (x f (x) 3x
x 1
, which means lim f (x)
1
x→ 3. an for all positive integers n,
n→ s Section 9.2 Taylor Series (pp. 469–479)
Exploration 1 Designing a Polynomial to
Specifications
1, we know that the constant coefficient is 1. Since P (0) 1) 4 ( 1)nn!(x 1) 1)! x f (x) nx n f (x) n(n 1)x n (The 2 in the denominator is needed to cancel the factor of f (x) n(n 1)(n 2 that results from differentiating x 2.) Similarly, we find the f (k)(x) 4
5
coefficients of x 3 and x 4 to be and .
6
24
32
43
54
Thus, P(x) 1 2x
x
x
x.
2
6
24 n for n 1 xn 5. f (x) 3
3, we know that the coefficient of x is .
2 (n 1) 3x
3x ln 3
3x(ln 3)2
3x(ln 3)3
3x(ln 3)n 4. f (x) ln x
f (x) x 1
f (x)
x2
f (x) 2x 3
f (4)(x)
6x 4
f (n)(x) ( 1)n 1(n 2, we know that the coefficient of x is 2. Since P (0) 3. f (x)
f (x)
f (x)
f (x)
f (n)(x) 3 1) 6(x f (n)(x) 2 1) 2(x f (x) it follows that lim an must also be 3. 1. Since P(0) x f (x) 3 is a horizontal asymptote of the graph of Because f (n) e2x
2e2x
4e2x
8e2x
2ne2x
1 2. f (x) 3 the function f (x) Approximating sin 13 4. 20 terms. an does not converge and hence diverges.
1
1 x6
6! x6
, and the Taylor series is
6!
2n
… ( 1)n x
….
(2n)! 1. 0.4201670368… Since the limit of a sequence is unique (by Exercise 55), 3n
57. (a) lim
n→ n x4
4! x4
4! Quick Review 9.2 . Assume an converges. Let N x2
2 x2
2 Exploration 3 L2. Given an there corresponds an N1 such that for k(n) 1 2. A clever shortcut is simply to differentiate the previouslydiscovered series for sin x termbyterm! 56. Consider the two subsequences ak(n) and ai(n), where
lim ak(n) A Power Series for the Cosine 1. cos (0) 1
cos (0)
sin (0) 0
cos (0)
cos (0)
1
cos(3) (0) sin (0) 0
etc.
The pattern 1, 0, 1, 0 will repeat forever. Therefore,
P6(x) 2.
L2 L1
2 says that the difference between two fixed
values is smaller than any positive number 2 . The only
nonnegative number smaller than every positive number
is 0, so L2 L1 0 or L1 L2. n→ Exploration 2 371 2 f (n)(x) 1
2 2)x n 3 n!
xn k
(n k)!
n! 0
x
n!
0! 6. dy
dx d xn
dx n! nx n
n! 7. dy
dx d 2n(x a)n
n!
dx 1 xn 1
(n 1)!
2nn(x a)n
n! 1 2n(x
(n a)n
1) ! 1 372 8.
9.
10. Section 9.2
d
x 2n 1
( 1)n
dx
(2n 1)! dy
dx 2n dy
dx d (x a)
dx (2n)! dy
dx 2n(x d (1 x)n
n!
dx n(1 ( 1)n (2n 1)x 2n
(2n 1)! 2n 1 ( 1)nx 2n
(2n)! 5. cos (x (x a)
(2n 1)! x)n 1( 1)
n! (1 x)n 1
( n 1) ! (cos 2) at the end of Section 9.2.
(2x)
3!
4x 3
3 2x
2x 5 2n 1 (2x)
… ( 1)n (2x)
5!
(2n 1)!
4x 5
( 1)n(2x)2n 1
…
…
15
(2n 1)! … x for x in the Maclaurin series for ln (1 x) x) x5
…
5!
(cos 2)x 2
2! (sin 2)x ( x)2
( x)3
…
2
3
n
( x)
…
( 1)n 1
n
2
3
n
x
x
…x
…
x
2
3
n 1 x 1, so the interval … ( 1)n(cos 2)
where 2n
(2n)! 2n 1 k. Thus the coefficient is B sin 2 if n is even and cos 2 if n is odd. 1 cos 2 sin 2
x , 2 cos 2 cos (2 and so on, so the general term is
2 2n 1
(x 2)3
(x 2)5
… ( 1)n (x )
3
5
2n 1
n 4n 2
x6
x 10
… ( 1) x
…
3
5
2n 1 x2
x2 This series converges when x 2 7x(1
7x x
7x 2 … n
x2
…x
…)
2!
n!
3
n1
7x
… 7x
2!
n! This series converges for all real x. ),
n
2 1
cos 2
n! x n. The series converges for all real x.
6. x 2 cos x x2
2! x2 1
x4
2 x2 1, so the interval of convergence is [ 1, 1].
4. 7xe x k. Thus the coefficient is shown at the end of Section 9.2.
x2 x 2n 1
…
(2n 1)!
(sin 2)x 3
(cos 2)x 4
3!
4! ( 1)n Another way to handle the general term is to observe that
3. Substitute x 2 for x in the Maclaurin series for tan 1 … We need to write an expression for the coefficient of x k. and B of convergence is [ 1, 1). tan x 2n
(2n)! ( 1)int[(k 1)/2](cos 2)
( 1)(k 1)/2(sin 2)
, which is the same as
.
k!
(2n 1)!
An
( 1) Bx
n1
, where A int
,
Hence the general term is
n!
2 ( x) This series converges when ( 1)n ( 1)int[(k 1)/2](cos 2)
( 1)k/2(cos 2)
, which is the same as
.
k!
k!
n1
( 1) (sin 2)
where
If k is odd, the coefficient is
(2n 1)! shown at the end of Section 9.2.
ln (1 (sin 2)(sin x) … If k is even, the coefficient is This series converges for all real x.
2. Substitute x4
4!
x3
3! (sin 2)x 5
5! 1. Substitute 2x for x in the Maclaurin series for sin x shown sin 2x x2
2! (sin 2) x Section 9.2 Exercises 3 (cos 2)(cos x) (cos 2) 1 2n a)
(2n)! 2) x6
24 … x4
…
4!
( 1)nx 2n
(2n)! ( 1)n
1 x 2n
(2n)! … … The series converges for all real x.
7. Factor out x and substitute x 3 for x in the Maclaurin series
for 1
1 x shown at the end of Section 9.2.
x 1
1 x3 x[1 1 x3 x x3 x x4 (x 3)2
x7 … The series converges for x 3
convergence is ( 1, 1). …
x 3n (x 3)n
1 …] … 1, so the interval of Section 9.2
2x for x in the Maclaurin series for e x shown at 8. Substitute x1/2 12. f(4) the end of Section 9.2.
2x 1
2x ( 2x)2
…
2!
nnn
… ( 1) 2 x
n! ( 2x) 1 e 2x 2 ( 2x)n
n! … f (4) 9. f (2) 1
xx 2 f (2)
f (2) 2x f (2)
P0(x)
P1(x)
P2(x)
P3(x)
10. f
f
f 4 1 3 x2
4
4
6x
x2 1
2
1
2
1
2
1
2 x f (2)
1
2!
8
3
f (2)
, so
8
3! 4
2 8 x 2 (x 4 2)3
16 (x (b) f (1) 2 /4 2
x cos x 2 /4 f 2
x 2 /4 4 , so 2 2!
f 4 , so 2
12 3! 2 P1(x) 2 P2(x)
P3(x) 2 2 x 2
2 2 2
2 2 4 f
f f cos x 4 4 4 P0(x) 4 x3 2x 6x f (1) 6 P3(x) 3 4 4 2 1 x1 f (1)
2! 3 6, so f (1)
3! 1 1) 3(x 1)2 6, so x1 x1 4 (x 4 2x 3 x2 3x 6x 2 2x 12x 2 f (1) 12 P3(x) 3 2 8 14, so f (1)
2! f (1)
12, so
3! 2 x1 x1 11(x 1) 2 /4 2
x 2 /4 2 /4 2 P2(x) 2 P3(x) 2 7 1)2 7(x 2(x f , so 4 2 2! 4 (b) f (1) x4 1 x1 f , so f (1) 3! 12x 2
24x
1 4(x 1) 16. (a) P3(x) 2
12 4x 3 P3(x) 4 f (1) 4 5x 82
x
2! 4 5x x1 x1 4
12, so f (1)
2! 2
2 2 2
2 2 2
2
2
x
12 x 4
2 x 4 4 4 4 2 x 3 4 x1 6 24, so f (1)
3! 4 2 x 4
2 x 6(x 4x 2 4 f (0.2) P3(0.2) 4.848 1)2 63
x
3! x3 1)3 0, the Taylor 2
x 2
2 11 x1 2 P1(x) 2 x1 15. (a) Since f (0) f (0) f (0) f (0)
polynomial of order 3 is P3(0) 0. 2 1)3 (x 14. (a) Since f is a cubic polynomial, it is its own Taylor
polynomial of order 3.
P3(x) 2x 3 x 2 3x 8 or 8 3x x 2 2x 3 f (1) 2 x 4 2
x 4)3
512 3 f (1) sin x (x x1 4 3 /4 cos x 4
4 f (1) 2
x sin x 4 x 2 x 2 2 2 2
x
12 11. f 4
x 3x 2 (b) f (1) 4 x 2 2 (x 4 2 P0(x) 4)2
64
(x 4)2
64 4 f (1) 2 sin x x 2 /4 x 4
4 f (1) 2 cos x 4 2)2 8 x 2 x 1
512 13. (a) Since f is a cubic polynomial, it is its own Taylor
polynomial of order 3.
P3(x) x 3 2x 4 or 4 2x x 3 2)2 (x 4 4 1
16 2 P3(x) , so 2 x sin x 4 f x2 2 P2(x) 1
4 2 x f (4)
3
, so
3!
256 1
64 2 P1(x) 1
2 f (4)
1
, so
2!
32 3 5/2
x
8
x4 P0(x) The series converges for all real x. 1
4 1 3/2
x
4
x4 f (4) … 2 x4 1 1/2
x
2
x4 f (4) 373 4(x 1)3 374 Section 9.2 16. continued (x)(e x)
d ex 1
x
dx
xe x e x 1
x2
ee1
1
1 (c) g (x) (b) Since the Taylor series of f (x) can be obtained by
differentiating the terms of the Taylor series of f (x), the
g (1)
second order Taylor polynomial of f (x) is given by
5 3x 2. Evaluating at x 8x f (0.2) g (x)
3
(x
2! 4 ( 1)(x 1) 4 (x 3
(x
2 f (1.2) 1) P3(1.2) 2
(x
3! 1)2
1
(x
3 1)2 1)3 3.863 ∑ second order Taylor polynomial of f (x) is given by
1) f (1.2) 1)2. Evaluating at x (x 1.2, n1 n1 n
(n x
ex g(x)
(c) g(x) x2
2! x3
3! x4
4! f (t) t2 … … x
for x in the Maclaurin series for e x shown
2 x
2
x
2 1 ex (b) g(x) x2
8 3 n! … 2 1 xn
2 … 2 G(x) … x
2! … (t 2)3 2t 4 2x 3
3
x)1/2 (1 f (0) f (0) xn
2n n! 1
(1
2 … 2t 6 P4(x) 2x 5
5 (t 2)n 2t 2n …] … x) 5/2
x0 x2
8 x
2 2x 2n 1
2n 1 1
f (0)
, so
4
2! 3/2
x0 x) … … 1
2 1/2
x0 3
(1
8 1 2x 7
7 1 x0 x) 1
(1
4 3
f (0)
, so
8
3! 1
8
1
16 x3
16 x
x2
2! x
2!
x3
3!
x2
3! x2
3! f (x 2), the first four terms are (b) Since g(x)
n x
3! ... x
n! … xn
n! … xn 1
n! … … … xn
(n 1)! … 1 1 x2
2 x4
8 (c) Since h (0) x6
.
16 5, the constant term is 5. The next three terms are obtained by integrating the first three terms of This can also be written as
1 , which means 1 x
2! 1 1)! 0, the constant term is zero and we may 2x f (0) x 1
1
x
1
x
x … find G(x) by integrating the terms of the series for f (x).
xn 1
(n 1)! at the end of Section 9.2
e x/2 n
(n (t 2)2 2t 2 (b) Since G(0) 21. (a) f (0) x2
2 nx n 1
(n 1)! … … 1)! 1
at the end of Section 9.2.
1x
2
1 t2
1
2
1 t2 1 19. (a) Substitute (n series for 2 (b) Multiply each term of f (x) by x. xn … 1. 1)! 2[1
1
.
11 x3
4! 20. (a) Factor out 2 and substitute t 2 for x in the Maclaurin 0.36 1
x
1
18. (a) Since f (0)x
, f (0)
.
2
2!
2!
f (10)(0) 10
10!
x 10 (10)
Since
x
, f (0)
11!
10 !
11! ∑ Therefore, g (1) differentiating the terms of the Taylor series of f (x), the 3(x x2
d
x
1
3!
dx
2!
2x
3x 2
1
3!
4!
2!
nxn 1
1)!
n 1 (n ∑ 1)3 (b) Since the Taylor series of f (x) can be obtained by 1 1)(1) From the series, 0.2, 3.52 17. (a) P3(x) (e x
x2 the answer to part (b). The first four terms of the series
…. for h(x) are 5 x x3
6 x5
.
40 375 Section 9.2
22. (a) a0 1
3
a
10
3
a
21
3
a
32 a1
a2
a3 31 3 3
2 9
2 3 (x 2 a2 we have f (0) 3
n 3
.
n! 1 92
x
2 3x … 93
x
2 (b) Since the series can be written as ∑ n0 3 and 3e 3x 3n
x
n! …
L2(x) 1
(x
2 3 n (3x)
, it represents
n! 1
x
2 3) 3
2 4x and 3, respectively. e.
3e 3 x1 23. First, note that cos 18 ∑( n 3x the function f (x) Using cos x 3) 4, f ( 1
, so the linearizations are L1(x)
2 f ( 3) n0 (c) f (1) 0, f (0) n term by , an
n 1)(4) (4x)(2x)
(x 2 1)2 4 4x 2
,
(x 2 1)2 9
2 Since each term is obtained by multiplying the previous ∑ anx 4x
d
dx x 2 1 29. (a) Since f (x) 0.6603. 1)n n0 x 2n
, enter the following twostep
(2n)! commands on your home screen and continue to hit [ 2, 4] by [ 3, 3] (b) f (a) must be 0 because of the inflection point, so the
second degree term in the Taylor series of f at x a is
zero. ENTER.
30. The series represents tan
tan 1 1 4 . When x tan 1( 1)
The sum corresponding to N 25 is about 0.6582 (not
withing 0.001 of exact value), and the sum corresponding
to N 26 is about 0.6606, which is within 0.001 of the
exact value. Since we began with N 0, it takes a total of
27 terms (or, up to and including the 52nd degree term).
24. One possible answer: Because the end behavior of a
polynomial must be unbounded and sin x is not unbounded.
Another: Because sin x has an infinite number of local
extrema, but a polynomial can only have a finite number.
25. (1) sin x is odd and cos x is even
(2) sin 0 0 and cos 0 1 (3x)5
35 81
so = .
5!
5! 40 x. When x 1, it converges to 1, it converges to . 1
(sin x)
x
2n 1
1
x3
x5
… ( 1)n x
…)
(x
x
3!
5!
(2n 1)!
2
4
2n
x
x
… ( 1)n x
…
1
3!
5!
(2n 1)! 31. (a) f (x) (b) Because f is undefined at x
(c) k 0. 1 32. Note that the Maclaurin series for
1 26. Replace x by 3x in series for sin x. Therefore, we have 4 1 … x2 x 1
1 x is …. If we differentiate this series xn and multiply by x, we obtain the desired Maclaurin series
x 2x 2 … 3x 3 …. Therefore, the desired nx n function is
3 27. Since
is 1
4 3! d
ln x
dx 3 2x 3 , which is 1
at x
4 2, the coefficient 1
.
24 28. The linearization of f at a is the first order Taylor
polynomial generated by f at x a. f (x) x d
1
dx 1 x 33. (a) f (x) (1 x 1
(x 1)2 . x)m f (x) m(1 x)m f (x) m( m 1)(1 f (x) x
x)2 (1 m(m 1 1)(m x)m 2 2)(1 x)m (b) Differentiating f (x) k times gives
f (k)(x) m(m 1)(m 2) … (m
Substituting 0 for x, we have
f (k)(0) m(m 1)(m 2) … (m 3 k 1)(1 k 1). x)m k. ...
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 Spring '08
 ALL
 Taylor Series, Cos

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