ama1007 (calculus and linear algebra) …...the hong kong polytechnic university department of...
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The Hong Kong Polytechnic University
Department of Applied Mathematics
AMA1007 (Calculus and Linear Algebra)
Assignment 04 (Solution)
Question 1
( a ) 1 1 1
1 1 1
2 2 2n n nn n n
n
n n
1
1
2nn n
converges by the comparison test: 1 1
2 2n nn and
1
1
2nn
is a
convergent geometric series. 1
1
2nn
n
n
converges.
( b ) Diverges by the comparison test: 1 1 1
ln ln(ln )n n n and
3
1
n n
diverges.
( c ) Converges by the ratio test:
1
1 1
1 3 5 (2 1) (2 1) 4 2 !lim lim
4 2 ( 1)! 1 3 5 (2 1)
n n
n
n nn nn
a n n n
a n n
2 1 1lim 1
8( 1) 4n
n
n
( d ) Diverges by the ratio test:
1 3 3
1
3 1 3
3 2 3 3lim lim lim 1
( 1) 2 3 2( 1) 2
n n
n
n nn n nn
a n n
a n n
( e ) Converges by the integral test:
1 1
ln32 23 ln3
1/limsec sec (3)
2(ln ) ln 1 1
t
t
n dudx u
n n u u
( f ) Converges by the alternating series test:
Since ( 1 ) 2
1 1 2( ) '( ) 0
1 2 ( 1)
x x xf x f x
x x x
( )f x is decreasing.
and ( 2 ) 1
lim lim 01
nn n
na
n
.
( g ) Converges absolutely by the ratio test:
1
1 (100) ! 100lim lim lim 0 1
( 1)! (100) 1
n
n
nn n nn
a n
a n n
( h ) Converges by the integral test:
1
2 2 2 01 1 0
1 1/lim tan
(1 ln ) 1 ln 1 2
t
t
n dudx dx u
n n n u
( i ) Converges by the alternating series test:
Since ( 1 ) 1 1
( ) ln 1 '( ) 0( 1)
f x f xx x x
for 0x .
( )f x is decreasing for 0x
and ( 2 ) 1 1
lim limln 1 ln lim 1 ln1 0nn n n
an n
.
Question 2
( a ) ( ) 2xf x (1) 2f
'( ) ln 2 2xf x '(1) ln 2 2f 2''( ) (ln 2) 2xf x 2''(1) (ln 2) 2f 3'''( ) (ln 2) 2xf x 3'''(1) (ln 2) 2f
( ) ( ) (ln 2) 2n n xf x ( ) (1) (ln 2) 2n nf
Taylor series at 1x : 2 3
2 3
0
2(ln 2) 2(ln 2) 2(ln 2) ( 1)2 (2ln 2)( 1) ( 1) ( 1)
2! 3! !
n n
n
xx x x
n
( b ) 3 2( ) 2 3 8f x x x x (1) 2f 2'( ) 6 2 3f x x x '(1) 11f
''( ) 12 2f x x ''(1) 14f
'''( ) 12f x '''(1) 12f
( ) ( ) 0nf x for 4n ( ) (1) 0nf for 4n
Taylor series at 1x :
2 3 2 314 122 11( 1) ( 1) ( 1) 2 11( 1) 7( 1) 2( 1)
2! 3!x x x x x x
Question 3
( a ) 2( ) (1 ) xf x x e (0) 1f 2'( ) (3 2 ) xf x x e '(0) 3f 2''( ) (8 4 ) xf x x e ''(0) 8f
2'''( ) (20 8 ) xf x x e '''(0) 20f
Maclaurin’s polynomial (up to degree 3):
2 3 2 3''(0) '''(0) 10(0) '(0) 1 3 4
2 6 3
f ff f x x x x x x
( b ) ( ) ln(3 )xf x e (0) ln 4f
'( )3
x
x
ef x
e
1'(0)
4f
2
3''( )
(3 )
x
x
ef x
e
3''(0)
16f
2
3
9 3'''( )
(3 )
x x
x
e ef x
e
3'''(0)
32f
Maclaurin’s polynomial (up to degree 3):
2 3 2 3''(0) '''(0) 1 3 1(0) '(0) ln 4
2 6 4 32 64
f ff f x x x x x x
Question 4
( a ) By Sarrus’ Rule:
2 1 4
3 5 7 ( 20 7 72) (20 84 6) 65
1 6 2
( b ) Expand along the second row:
1 3 2 51 3 5 1 3 2
0 0 3 2( 3) 1 5 0 (2) 1 5 4
1 5 4 01 2 1 1 2 1
1 2 1 1
By Sarrus’ Rule:
( 3) (5 0 10) (25 0 3) (2) (10 8 3) (5 12 4) 39
( c ) Expand along the third row:
0 1 2 11 2 1 0 1 2
4 3 3 5(1) 3 3 5 ( 1) 4 3 3
1 0 0 11 0 1 1 1 0
1 1 0 1
By Sarrus’ Rule:
(1) (3 10 0) ( 3 0 6) (1) (0 3 8) ( 6 0 0) 5
Question 5
4 1 01 0 4 0 4 1
det( 4 ) 0 4 1 (0) ( 4) (1) 017 4 4 4 4 17
4 17 4
A I .
4 is a root of det( ) 0x A I .
1 01 0 0 1
0 1 (0) ( ) (1)17 8 4 8 4 17
4 17 8
xx x
x xx x
x
3 2 2( )( )(8 ) (1)(17 4) 8 17 4 ( 4)( 4 1)x x x x x x x x x x
2det( ) ( 4)( 4 1) 0 4, 2 3x x x x x A I .
Question 6
( a )
1
2
3
2 4 6 1
4 6 2 3
6 2 4 5
x
x
x
.
2 4 6
4 6 2 144
6 2 4
( 0) .
1
1 4 6
3 6 2 132
5 2 4
, 2
2 1 6
4 3 2 12
6 5 4
, and 3
2 4 1
4 6 3 12
6 2 5
.
By Cramer’s rule, we have 11
132 11
144 12x
, 22
12 1
144 12x
, and
33
12 1
144 12x
.
( b )
1 2 3
1 2 3
1 2 3
2 2 3
3 4 2 1
8 3 4
x x x
x x x
x x x
11
491
49x
, 22
491
49x
, and 33
491
49x
.
Question 7
( a ) 3 3 3 2 2 2( )( ) ( )( ) I I I A I A I A A I A I A A 1 2( ) I A I A A
( b ) 3 3 2 1 2( ) A A I 0 I A A A I A A I A
Question 8
1 2
1 2
( ) ( )
( ) ( )
f x f xdW d
g x g xdx dx
1 2 2 1( ) ( ) ( ) ( )d
f x g x f x g xdx
1 2 1 2 2 1 2 1'( ) ( ) ( ) '( ) '( ) ( ) ( ) '( )f x g x f x g x f x g x f x g x
1 2 2 1 1 2 2 1'( ) ( ) '( ) ( ) ( ) '( ) ( ) '( )f x g x f x g x f x g x f x g x
1 2 1 2
1 2 1 2
'( ) '( ) ( ) ( )
( ) ( ) '( ) '( )
f x f x f x f x
g x g x g x g x
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