Graphics Reference
In-Depth Information
(g) In order to appeal to qn 39, the number used must be
1.
Because of part (b) the number must be greater than
.Ifwe
want to use
, then
k
11.
41
(i)
a
/
a
(1
1/
n
)
, when
n
10.
(ii)
a
/
a
2/(
n
1)
, when
n
2.
(iii)
a
/
a
1/(1
1/
n
)
, for all
n
1.
(iv)
a
/
a
(0.9)(1
1/
n
)
0.95, when
n
55.
42 When
1
x
1, (
x
) is a null sequence by qn 39(d). Now
x
x
,so(
x
) is a null sequence, and using the absolute
valuerul, (qn 33), (
x
) is also a null sequence.
43
(iii) Take
N
max(
N
,
N
).
(iv)
a
b
a
·
b
a
, from qns 2.58 and 2.11.
(v)
a
b
c
is theproduct of
a
b
and
c
. Product of
k
terms by
induction.
44
(iii) Take
N
max(
N
,
N
).
(iv)
a
b
a
b
, from qns 2.61 and 2.10.
(v)
a
b
c
is thesum of
a
b
and
c
. Sum of
k
terms by
induction.
45 (
c
·
a
) are both null sequences from the scalar rule, qn
32. Their sum is a null sequence from the sum rule, qn 44.
The difference rule follows by putting
c
1 and
d
1.
) and (
d
·
b
46
(i) (a) From the difference rule (qn 45).
(b) 0
b
a
c
a
and the squeeze rule, qn 34.
(c)
b
(
b
a
)
a
. Now usethesum rul, qn 44.
(ii) (
a
) and (
c
) are null sequences by the subsequence rule,
qn 36(a).
So (
b
) is a null sequence from part (i). Now (
b
) is a null
sequence by the shift rule, qn 36(c).
47 For example
a
(
1)
.
48 Terms tend to 1. (
a
1) is null.
(i)
n
/(2
n
1/2(2
n
49
1)
1), so (
a
) is a subsequence of
(1/
n
).
(ii) (2
n
1)/(
n
1)
2
1/(
n
1), so (
a
2) is a scalar
multiple of a subsequence of (1/
n
).
50 The constant sequence (
b
a
) is null. Now
b
a
0, for if not
b
a
gives a contradiction.
51
(i) 6.2, 6.25, 6.25, 6.25.
(ii) 0.3, 0.33, 0.333, 0.3333.
(iii) 1.4, 1.41, 1.414, 1.4142.
