Thomas Calculus 13th [Solutions]

782 Chapter 10 Infinite Sequences and Series
56.
tan
n 1 2n
1
1
3 5 7 9 ( 1) x
x x x x x x
and
3 5 7 9 2n
1
2 n 1
2 2 1
lim x 2n 1 lim 2n
1 tan
2n
1 x 2n
1
n
x n
2n
1
x x
converges for | x | 1; when x 1 we have
n 1
( 1)
n 1
2n
1
which is a convergent series
n
( 1)
2n
1
n 1
which is a convergent series; when x 1 we have
the series representing
1
tan x diverges for | x| 1
57.
n 1 2n
1
1
3 5 7 9 ( 1)
tan
x x x x
x
1
x x and when the series representing 48 tan 1 has an
3 5 7 9 2n
1
18
6
error less than 1
3 10 , then the series representing the sum 1 1 1 1 1
48 tan 32 tan 20 tan 1 also
18 57 239
has an error of magnitude less than
terms
6
10 ; thus
1
2n
1
18
1
2n
1 310
error 48 n 4 using a calculator 4
6
58. (a)
m k m k 1 m k 1
f ( x) 1 k x f ( x) k k x (1 x) f ( x) (1 x)
k k x
k 1 k 1 k 1
m k 1 m k 1 m k 1 m k
k k x x k k x k k x k k x
k 1 k 1 k 1 k 1
0 1 1
1
m (1) x m k k x k m k k x k m m k k x k m k k x
k
k 2 k 1 k 2 k 1
Note that:
k
m k 1 m k
k k x k 1 k x
2 k 1
( 1) .
Thus,
1
x f x m m k k x k m k k x k m m k 1 k x k m k k x
k
k 2 k 1 k 1 k 1
(1 ) ( ) ( 1)
(b) Let
m k m k m m k
m k 1 k x k k x m k 1 k k k x
k 1 k 1
Note that:
( 1) ( 1) .
m
m m ( m 1) m ( k 1) 1 m ( m 1) ( m k 1)
k 1 ( k 1) k k ( k 1)
k
( k 1)! k!
m ( m 1) ( m k ) m ( m 1) ( m k 1) ( 1) ( 1) ( 1) ( 1)
k m m m k m k k m m m m k m
m k! k! k! k!
k
( ) .
m m k m k m k
Thus, (1 x ) f ( x ) m k 1
( k 1) k k x m m k x m m k x
k 1
m
k
k
k 1 k 1 k 1
m f ( x)
(1 x)
m 1 x m f ( x ) f ( x ) if 1 x 1.
m m 1
m
g x x f x g x m x f x x
( ) (1 ) ( ) ( ) (1 ) ( ) (1 )
m 1 m m f ( x)
m 1 m 1
f x m x f x x m x f x x m f x
(1 x)
( ) (1 ) ( ) (1 ) (1 ) ( ) (1 ) ( ) 0.
(c)
m
m
g ( x) 0 g( x) c (1 x) f ( x) c f ( x) c c(1 x ) . Since
m
(1 x)
f ( x) 1
k
1
m
k
k
x
m k m m
f k
c c f x x
k 1
(0) 1 (0) 1 0 1 (1 0) 1 1 ( ) (1 ) .
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