Graphics Reference
In-Depth Information
f
(
a
).
For non-constant
f
, the domain must be disconnected.
So
f
(
x
)
18
f
(
y
)
f
(
x
)
x
f
(
c
).
y
f
(
c
)
L
gives the result.
a
a
f
(
a
)
f
(
a
)
L
·
a
a
L
·
b
a
.
a
a
f
(
a
)
f
(
a
)
L
·
a
a
L
·
a
a
L
·
b
a
.
Further steps by induction. Now 0
L
1, so (
L
) is a null sequence.
So for suMciently large
n
,
a
a
, and (
a
) is a Cauchy sequence
and so, by qn 4.57, is convergent.
A
and
B
liein [
a
,
b
] by qn 3.78, since the interval is closed.
(
a
)
A
implies (
a
)
A
,so(
f
(
a
))
A
. But
f
is continuous at
A
,so
f
(
A
)
A
and similarly
f
(
B
)
B
.
b
a
f
(
b
)
f
(
a
)
L
·
b
a
...
L
·
b
a
.
So (
b
a
) is a null sequence and by qn 3.54(v) (
a
) and (
b
) havethesame
limit.
19 Let
f
(
x
)
1.
f
(
x
)
sin
x
.So
f
is a contraction mapping on [0, 1] and, by qn 18,
f
(
x
)
x
has a uniquesolution. Obviously no solution on [1,
].
20 If 1
x
2, then 3
x
2
4, so
3
(
x
2)
2 which implies
1
(
x
2)
2. Thus
f
has domain and co-domain [1, 2].
f
cos
x
. Then if 0
x
,0
cos
x
(
x
)
1/2
(
x
2). 1
x
2
f
(
x
)
1/2
3
f
(
x
)
1. So
f
is a
contraction mapping on [1, 2]. Therefore if
a
[1, 2], the sequence defined
f
(
a
by
a
) is convergent.
21 See figure 9.3.
22 Let
f
(
c
)
l
0. Then, for some
,
f
(
x
)
f
(
c
)
x
c
l
f
(
x
)
x
c
0
x
c
l
l
l
,
since
f
0.
Thus
c
x
c
f
(
c
)
0.
Now, by the Mean Value Theorem, for any
x
satisfying
c
x
c
,
(
x
)
0, and
c
x
c
f
(
x
)
f
(
x
)
f
(
c
)
x
c
f
(
d
)
for some
d
(
x
,
c
), so
f
(
d
)
0 and
f
(
x
)
f
(
c
).
Likewise for
c
x
c
,
f
(
x
)
f
(
c
), so
f
has local minimum at
c
.
