Graphics Reference
In-Depth Information
a
]
a
a
]
[10
10
[10
1 gives the inequality.
a
]/10
a
by the squeeze rule,
(1/10
) is null by qn 39, so ([10
)
qn 34.
Every term of ([10
a
]/10
) is rational.
52 (
a
)
a
(
a
a
) is null (by definition)
(
a
a
) is null (by
theshift rul, qn 36(c))
(
a
)
a
(by definition).
53
(i) and (ii) giveno basis for a guss.
(iii) With
k
10
,(
a
2) is null, so (
a
2) is null by theshift
rule, and (
a
)
2.
54
(i) (
a
)
a
(
a
a
) null
(
c
· (
a
a
)) null
(
c
·
a
c
·
a
) null
(
c
·
a
)
c
·
a
.
(ii) (
a
a
) is null,
by the subsequence rule, qn 36(a)
every subsequence of
(
a
)
a
(
a
a
) null
every subsequence of (
a
)
a
.
(v) Put
c
1 and
d
1 in the linear rule.
(vi) (
a
(
b
a
)) are both null sequences by the scalar
rule(qn 32), and ((
a
b
)) and (
b
(
a
b
)) is a null sequence by the
product rule(qn 43). Now usethesum rule(qn 44).
(vii) (
a
a
)(
b
a
) null
(
a
a
) null, by theabsolutevaluerule(qn
33).
Now 0
a
a
a
a
from qn 2.63, and so by the
squeeze rule (qn 34), (
a
a
) is null, and by the
absolutevaluerule(qn 33), (
a
a
) is null, so (
a
)
a
.
(viii)
a
c
b
a
a
c
a
b
a
. Now the squeeze rule
with a shift (qn 46(ii)) gives (
c
)
a
.
55
(i)
a
b
c
is thesum of
a
b
and
c
. Thesum of
k
terms
by induction.
(ii)
a
b
c
is theproduct of
a
b
and
c
. Theproduct of
k
terms by
induction.
(iii) By repeated use the product rule, the scalar rule and the sum
rule, the limit is 1
2
a
3
a
4
a
5
a
.
56 Each tends to 1.
57
(c) 0
a
(
b
1)/
n
. Now usethescalar rule(qn 32) and the
squeeze rule (qn 34).
(
a
b
)
b
·
((
a
/
b
)
1), so
b
(
a
b
)
b
·
(
a
/
b
1)
b
·
2. Now useqn 57, the
scalar rule (qn 32) and the squeeze rule (qn 54(viii)).
58
(c)
59
(b) For
n
2, the third term in the expansion of (1
a
)
is
)
.
(c) For
n
2, (1/
(
n
1)) is a subsequence of a null sequence by
qn 30. So (
a
n
(
n
1)(
a
) is null by the scalar rule and the squeeze rule.
