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

CHAPTER 11 PROBLEMS PLUS 0 109
x x 2 · x 3 x 4
21. Let f (x) denote the left-hand side of the equation l + 2T + 4
! + 6f + 8f + .. · = 0. lfx;:::: ~· then f(x) ;:::: 1 and t~ere are
no solutions oftheequation. Note that / (- x 2 ) = 1 - 2! + 4
! -
x2 x4 x s xs
6
! + 8f - · · · = cosx. The solutions of cosx = 0 for
x < 0 are given by x = i - 1rk, where k is a positive integer. Thus, tl1e solutions off ( x) = 0 are x = - (i - 1r k) 2 , where
k is a positive integer.
23. Call the series S. We group the terms according to the number of digits in their denominators:
Now in the group gn , since we have 9 choices for each of then digits in the denominator, there are 9n terms.
Furthermore, each term in 9n is less than
10 ~_ 1 [except for the first term in gl]. So 9n < gn ·
10 ~_ 1 = 9( { 0
)"'- 1 •
00
Now L: 9( 1
9
n = l
0
) n - l is a geometric series witl1 a = 9 and r = fa < 1. Therefore, by the Comparison Test,
oo oo 9 n - 1 9
S = L: 9n < L: 9(w) = 1 _ 9110
= 90.
n = l n=l
Use the Ratio Test to show that the series for u, v, and w have positive radii of convergence (oo in each case), so
Theorem 11'.9.2 applies, and hence, we may differentiate each of these series:
du 3x 2 6x 5 9x 8 x 2 x 5 x 8
- = -+-+-+··· = - +-+ - +· ·· = w
dx 3! 6! 9! 2! 5! 8!
. . dv x 3 x 6 x 9 dw x 4 x 7 x 10
Similarly, dx = 1 + 3! + 6f + 9! + · · · = u, and dx = x + 4
! + 7T + lO! + · · · = v.
So u' = w, v' = u, and w' = v. Now differentiate the left-hand side of the desired equation:
d
dx (u 3 + v 3 + w 3 -
3uvw) = 3u 2 u' + 3v 2 v' + 3w 2 w' - 3(u'vw + uv'w + uvw')
= 3u 2 w + 3v 2 u + 3w 2 v - 3(vw 2 + u 2 w + u~ 2 ) = 0 =>
u 3 + v 3 + wa - 3uvw = C. To find the value of the constant C, we put x = 0 in the last equation and get
1 3 + 0 3 + 0 3 - 3(1 · 0 · 0) = C => C = 1, so u 3 + •v 3 + w 3 - 3uvw = 1.
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