Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 11.10 TAYLOR AND MACLAURIN SERIES D 87
1.5
= 1- ~x '' + fixs- 7~oxl2 + ...
The series for cos x converges for all x , so the same is true of the series for
f(x), that is, R = c;o· Notice that, as n increases, Tn(x) becomes a better
approximation to f(x).
- 1.5 t--r.'-----+---~ ..:--11.5
- 1.5
\ . T 8 = T 9 = T 10
= T 11
•
\ 't
\
.
oo n oo ( )" oo n
:r;
41 . e
(11)
=
""
L- I,
X
so e
- x
=
""
L-
....=::___1 ·= ""{-1)"
L-
.:._,SO
n =O n. n =O n. n =O n 1
f (x) = xe- "' = f: (-1)" ~ x"+l ·
n=O n.
The series for e"' converges for all x, so the same is true of the series
for f(x); that is, R = oo. From the graphs off and the first few Taylor
polynomials, we see that T,. ( x) provides a closer fit to f ( x) near 0 a!; n increases.
43. 5° = 5°
(
7T ) 7r . oo x2" x2 x4 x6
1800
= 36
radians and cos x = .. ~o ( -1t ( 2
n)! = 1 - 2f + 41
- 6f + ·· ·,so
7T (7r/ 36) 2 (7r/ 36) 4 (7r/ 36) 6 (7r/ 36) 2 . (7r/ 36)~
cos - = 1 - --- + - + ... Now 1 - ~ 0 99619 and addmg 2 4 X w-G
36 2! 4! 6! . 21 . 4! ~ .
does not affect the fifth decimal place, so cos 5° ~ 0.99619 by the Alternating Series Estimation Theorem.
(b)sin- lx= l 1 dx=C + x + f: 1·3 · 5 ·· .. · (2n-1)x 2 n+l
V1- x 2 n = l (2n + 1)2n · n !
= x + f; 1 · 3 · 5 · · · · · { 2 n - 1 ) x 2 n+l since 0 = s in- 1 0 = C.
n=l (2n + 1)2". nl
(16) 00 x 2n
47. COSX = "~ 0 (- 1 )" { 2 n)! I
oo
X Gn+2
x cos(x 3 ) dx = C + .~o ( - 1)" ( 6
n + 2
)( 2
n)! , with R = oo.
(16) oo n x2" oo x2"
49. cosx = .~o ( -1) ( 2 n)! => cosx - 1 = .~
1 ( - 1)" { 2 n)! . 1 oo 2n- 1
cos x - = 2: {-1)" .::_____ =>
X n=l (2n)!
I
COS X - 1 oo n X 2n •
dx = C + 2:::{-1) 2
· ( 2
) 1 , wtth R = oo.
X n =l n n .
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