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

CHAPTER 11 PROBLEMS PLUS D 107
l
9. We start with the geometric series f; x" = - 1 - , lxl < 1, and differentiate:
n.=O 1- X
}: nxn- l = -d }: xn = -d - 1
_ = ( 1
- ) 2
for lxl < 1
oo d(
00
) d(1) 1
n=l X n=O X X X
for lxl < 1. Differentiate again:
oo oo X
=> L nxn = X L nxn-l = ..,.---~
n =1 n =l (1 - x)2
00
_ d x (1 - x? - x · 2(1 - x)( - 1) x + 1 ~ 2 ,.. x 2 + x
}: n 2 x" 1 - - - = => L- n x = =>
n= 1 - dx (1 - x)2 - (1 - x) 4 (1 - x)3 n =l (1 - x)S
f; n3xn- 1 = !!:.._ x 2 + x = (1 - x?(2x + 1) - (x 2 + x)3(1 - x) 2 ( - 1) = x 2 + 4x + 1 ==>
n=l dx (1 - x) 3 (1 - x ) 6 (1 - x)~
f; n 3 x" = x3
t
n = l
4 2
x
1- X
)~ x , lx l < 1. The radius of convergence is 1 because that is the radius of convergence for the
geometric series we started with. If x = ± 1, the series is }:n 3 (±1)n, which diverges by the Test For Divergence, so the
interval of convergence is (-1, 1).
11. ln(1- : 2
) = ln( n~~ 1 ) ,;, In (~+ ll~n - 1 ) = ·in[(n +1)(n- 1)]- lnn 2
= ln(n + 1) + ln(n - 1) - 2ln n = ln(n - 1) - Inn- ln n + In(n + 1)
n- 1 n - 1 n
= In - n - - [lnn - ln(n+ 1)] = In - n - -In n+ 1
.
k ( 1) k ( n-1 n )
Let Sk = E In 1 - 2 = E In - - - In - - 1
fork ~ 2. Then
n = 2 n n = 2 n n+
1 2) ( 2 3) ( k - 1 k ) 1 k
. . ( 1 k ) 1
Sk =
(
ln - - In - + In - - In - + · · · + In - - - In - - = In- - In - - so
2 3 3 4 k k+1 2 k+ 1 '
00 (
}: In 1 - - 1) = lim Sk = hm In- -In-- = In - -ln 1 = ln 1 -In 2 - In 1 = - In 2.
n = 2 n 2 k-oo k-oo 2 k + 1 2
·I
13. (a) The x-intercepts of the curve occur where sin x = 0 x = mr,
nan integer. So using the formula for disks (and either .a CAS or
l-+-l-~~+-''r7"""'-'...,.._---t 40 sin 2 x = ~ (1 - cos 2x) and Formula 99 to evaluate the integral),
the volume of the nth bead is
- I v. f"'' ( - :r:/10 . )2 dx rmr -:r:/ 5 . 2 d
n = rr J (,.-l) .. e sm x = rr J (n- 1 ) .. e sm x x
= 2;g,rr (e- (n-l)lf/5- e- mr/5)
(b) The total volume is
00
rr .{ 0
e- :r:/S sin 2 x dx = f V,, = 2 ~it f; [e- (n- 1 )"/ 5 - e-n"/ 5 ] = ;g; 2 [telescoping sum].
n =l n=l
Another method: If the volume in part (a) has been written as Vn = 2 ~g; e-n"/ 5 (e"/ 5 -
as a geometric series with a= 2 ~g; (1 - e-"1 5 ) and r = e-"/ 5 .
1), then we recognize f; Vn
· n=l
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