Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. Itwouldbefartoomuchworktocompute15 derivatives of f. The key idea is to remember that f (n) (0) occurs in the
coefficient of x n in the Maclaurin series of f. We start with the Maclaurin series for sin: sin x = x − x3
3! + x5
5! − ···.
Then sin(x 3 )=x 3 − x9
3! + x15
− ···, and so the coefficient of x 15 is f (15) (0)
= 1 5!
15! 5! . Therefore,
f (15) (0) = 15!
5!
=6· 7 · 8 · 9 · 10 · 11 · 12 · 13 · 14 · 15 = 10,897,286,400.
3. (a) From Formula 14a in Appendix D, with x = y = θ, wegettan 2θ = 2tanθ
1 − tan 2 θ ,socot 2θ = 1 − tan2 θ
2tanθ
2cot2θ = 1 − tan2 θ
tan θ
tan 1 2 x =cot1 2 x − 2cotx.
=cotθ − tan θ. Replacing θ by 1 2 x,weget2cotx =cot1 2 x − tan 1 2 x,or
⇒
(b) From part (a) with
x
2 n−1 in place of x, tan x 2 n =cot x 2 n − 2cot x
2 n−1 ,sothenth partial sum of ∞
s n = tan(x/2) + tan(x/4) + tan(x/8) + ···+ tan(x/2n )
2 4 8
2
n
cot(x/2)
cot(x/4)
=
− cot x + − cot(x/2) cot(x/8)
+
2
4 2
8
+
− cot(x/4)
+ ···
4
n=1
1
2 n tan x 2 n is
cot(x/2 n )
− cot(x/2n−1 )
= − cot x + cot(x/2n )
[telescoping sum]
2 n 2 n−1 2 n
Now cot(x/2n )
= cos(x/2n )
2 n 2 n sin(x/2 n ) = cos(x/2n ) x/2 n
·
x sin(x/2 n ) → 1 x · 1= 1 x as n →∞since x/2n → 0
for x 6= 0. Therefore, if x 6= 0and x 6=kπ where k is any integer, then
∞ 1
2 tan x
n 2 = lim s n n = lim − cot x + 1
n→∞ n→∞
2 cot x
= − cot x + 1 n 2 n x
n=1
If x =0, then all terms in the series are 0,sothesumis0.
5. (a) At each stage, each side is replaced by four shorter sides, each of length
1
of the side length at the preceding stage. Writing s 3 0 and 0 for the
number of sides and the length of the side of the initial triangle, we
generate the table at right. In general, we have s n =3· 4 n and
n =
1 n
3
, so the length of the perimeter at the nth stage of construction
is p n = s n n =3· 4 n ·
1 n
=3·
4 n
.
3
3
s 0 =3 0 =1
s 1 =3· 4 1 =1/3
s 2 =3· 4 2 2 =1/3 2
s 3 =3· 4 3 3 =1/3 3
.
.
(b) p n =
4n
3 n−1 =4 4
3
n−1
. Since 4 3 > 1, pn →∞as n →∞. 509