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

96 D CHAPTER 11 INFINITE SEQUENCES AND SERIES
(b) From the table, the first term in the Maclaurin series of
tanhx is x, so ifthe water is shallow, we can approximate
n
0
1
j (n) (x)
tanhx
sech 2 x
f (nl(o)
0
1
2
-2sech 2 xtanhx
0
3
2sech 2 x (3tanh 2 x- 1)
- 2
(c) Since tanh x is an odd function, its Maclaurin series is alternating, so the error in the approximation
2trd 2trd. . . IJ"'(O)I (2trd) 3 1 (2trd) 3
tanh L ~ L IS less than the first neglected term, wh1ch IS --
3
,- L =
3 L .
If L > lOd, then ~ ( 2 ~d) 3 < ~ ( 2tr · 1
1
0 ) 3
= ;
3
5 , so the error in the approximation v 2 = gd is less
. .
gL tr3
than 7r • ~ 0.0132gL.
2 375
37. (a) Lis the length of the arc subtended by the angle B, soL = RB =>
B=L/R.NowsecB=(R+C)/R. => RsecB=R+C =>
C = RsecB- R = Rsec(L/R)- R.
(b) First we'll find a Taylor polynomial n(x) for f(x) = secx at X = 0.
n
/(nl(x)
0 secx
1 secx tanx
2 secx(2tan 2 x + 1)
3 secxtanx(6tan 2 x + 5)
4 secx(24tan 4 x + 28tan 2 x + 5)
/(n) (0)
1
0
1
0
5
Thus, f(x) = secx ~ T4(x) = 1 + ~(x- o? + -fi(x-W = 1 + ~x 2
+ f4x 4 • By part (a),
(c) Taking L = 100 km and R = 6370 km, the formula in part (a) says that
C = Rsec(L/ R)- R = 6370 sec(100/6370)- 6370 ~ 0.785 009 96544 km.
. . £ 2 5£ 4 100 2 5. 100 4 .
The formula m part (b) says that C ~ 2
R + 24
R 3
= 2
. 6370
+ 24
. 63703
~ 0.785 009 957 36 km.
The difference between these two results is only 0.000 000 008 08 km, or 0.000 008 08 m!
39. Using f(x) = Tn(x) + Rn(x) with n = 1 and x = r, we have J(r) = T1(r) + R 1(r), where T1 is the first-degree Taylor ,
polynomial off at a. Because a= Xn, J(r) = f(xn) + f'(xn)(r- x,.. ) + R1(r). But r is a root off, so f(r) = 0
and we have 0 = f ( Xn) + J' ( Xn) ( r -
Xn) + R1 ( r). Taking the first two terms to the left side gives us
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