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

SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS 0 93
15.
n f (n) (x) J (n){1)
0 x2/3 1
1 1x- l /3 2
3 3
2 _1x- 4/3 2
9
- g-
3 .§..x- 7/3 8
27 27
4
- ~ x- 10 / 3
81
(c) 0.00006
. 2/9 8/27
(a) f(x) = x 2 1 3 ~ Ta(x) = 1 + ~(x- 1) - 2!(x - 1) 2 + 3!(x - 1) 3
= 1 + ~(x- 1) - · ~(x-I?+ 8
4 1 (x - 1) 3
(b) IR3(x)l5 ~ lx - 11 4 , where J j < 4 >(~)~5 M. Now 0.8 5 x 5 1.2 =?
lx- l l 5 0.2 ::}
lx - 11 4 5 0.0016. Since jJ < 4 >(x) l is decreasing
on [0.8, 1.2], we can take M =I j < 4 >(o.s)l = ~(0 . 8) - 10 1 3 , so
IR3(x)l 5
~(0 8)-10/3
81
' (0.0016) ~ 0.000096 97.
24
From the graph of IRa(x)l = Jx 2 / 3 -
Ta(x) I• it seems that the
error is less than 0.000 053 3 on [0.8, 1.2).
17.
n
f(n) (x)
J (n){O)
(a) f(x) = secx ~ T2(x) = 1 + ~x 2
0 secx
1
1 secx tanx
2 secx (2sec 2 x- 1)
0
1
3 sec x tan x ( 6 sec 2 x - 1)
(b) IR2(x)l 5 ~ lx l 3 , where I t < 3 > (x) l 5 M . Now - 0.2 5 x 5 ~ . 2 =:> lx l 5 0.2 =? lxl 3 5 (0.2) 3 .
f(3) (x) is an odd function and it is increasing on (0, 0.2) since secx and tanx are i"ncreasing on [0, 0.2) ,
so I J C3>(x) l 5 J < 3 >(o.2) ~ 1.085 158892. Thus, IR2(x)l 5 J < 3 >i~· 2 ) (0.2) 3 ~ 0.001447.
(c)
0.0004
· From the graph ofiR2(x)l = isecx - T2(x)l, it seems that the
error is less than 0.000 339 on [-0.2, 0.2).
19.
n·
0
J(x)J 5 M. Now 0 5 x $ 0.1 =?
x 4 5 (0.1) 4 , and letting x = 0.1 gives
3
ex 2 (12x + 8x 3 )
0
IR3(x)l $ eo.Dl (12 + ~!8 + 0.0016) (0.1)4 ~ 0.00006.
4
e"' 2 (12 + 48x 2 + 16x 4 )
® 2012 Ccngngc Lc:uning. AU Ri~tsRc.scn·cd. M.D.)' not be .scrutncd. coried, orduplicattd. or posted to a publicly acc