Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 11 PROBLEMS PLUS ¤ 513
19. Let f(x) = ∞ c m x m and g(x) =e f(x)
= ∞ d n x n .Theng 0
(x) = ∞ nd n x n−1 ,sond n occurs as the coefficient
m=0
n=0
of x n−1 .Butalso
∞
g 0 (x) =e f(x) f 0
∞
(x) = d n x n
mc m x m−1
n=0
m=1
= d 0 + d 1x + d 2x 2 + ···+ d n−1x n−1 + ··· c 1 +2c 2x +3c 3x 2 + ···+ nc nx n−1 + ···
so the coefficient of x n−1
is c 1d n−1 +2c 2d n−2 +3c 3d n−3 + ···+ nc nd 0 = n
ic id n−i. Therefore, nd n = n ic id n−i.
21. Call the series S. We group the terms according to the number of digits in their denominators:
S = 1
+ 1 + ···+ 1 +
1 + 1
1 2 8 9
+ ···+
1 + 1
11 99
+ ···+
1 + ···
111 999
g 1 g 2 g 3
Now in the group g n ,sincewehave9 choices for each of the n digits in the denominator, there are 9 n terms.
Furthermore, each term in g n is less than
Now ∞
n=1
S = ∞
g n < ∞
n=1
n=0
1
[except for the first term in g
10 n−1 1 ]. So g n < 9 n 1 · =9
9 n−1
.
10 n−1 10
9
9 n−1
is a geometric series with a =9and r = 9 < 1. Therefore, by the Comparison Test,
10
10
n=1
9 9
10
n−1
=
9
=90.
1 − 9/10
i=1
i=1
23. u =1+ x3
3! + x6
6! + x9
x4
+ ···, v = x +
9! 4! + x7
7! + x10
x2
+ ···, w =
10! 2! + x5
5! + x8
8! + ···.
Use the Ratio Test to show that the series for u, v, andw have positive radii of convergence (∞ in each case), so
Theorem 11.9.2 applies, and hence, we may differentiate each of these series:
du
dx = 3x2 + 6x5 + 9x8 + ···= x2
3! 6! 9! 2! + x5
5! + x8
8! + ···= w
Similarly, dv
dx =1+x3 3! + x6
6! + x9
dw
+ ···= u,and
9! dx = x + x4
4! + x7
7! + x10
+ ···= v.
10!
So u 0 = w, v 0 = u,andw 0 = v. Now differentiate the left hand side of the desired equation:
d
dx (u3 + v 3 + w 3 − 3uvw) =3u 2 u 0 +3v 2 v 0 +3w 2 w 0 − 3(u 0 vw + uv 0 w + uvw 0 )
=3u 2 w +3v 2 u +3w 2 v − 3(vw 2 + u 2 w + uv 2 )=0 ⇒
u 3 + v 3 + w 3 − 3uvw = C. Tofind the value of the constant C, weputx =0in the last equation and get
1 3 +0 3 +0 3 − 3(1 · 0 · 0) = C ⇒ C =1,sou 3 + v 3 + w 3 − 3uvw =1.