Graphics Reference
In-Depth Information
m
n
Then
m
(1
x
)
(
n
1)
(
x
)
1
and hence (1
x
)
n
1
m
m
n
1
(
x
)
m
1
thus (1
x
)
(
x
)
.
n
This establishes the result by induction.
44 Radius of convergence for both series
x
1 by qn 5.107.
x
45 By qn 5.98, the series is convergent for
1, so
f
is a pointwiselimit
function. Within the circle of convergence the convergence is uniform (qn
40) so, using qn 42,
f
(
x
)
a
n
1
(
n
1)
x
.
Now
n
a
(
a
1)...(
a
n
1)
1)
a
(
a
1) . . . (
a
n
)
(
n
(
n
n
!
1)!
a
(
a
1) . . . (
a
n
1)
(
n
1)!
a
n
n
1
.
So (
x
1)
f
(
x
)
af
(
x
).
g
(
x
)
a
(1
x
)
f
(
x
)
(1
x
)
f
(
x
)
g
x
)
af
(
x
)
x
)
f
(
x
)
(1
(
(1
(
x
))
g
(
x
)
0.
By the Mean Value Theorem (qn 9.17)
g
is a constant function.
Since
f
(0)
1,
g
(0)
1, so
f
(
x
)
(1
x
)
.
46
(i) [
x
1]
[
x
]
1, so
f
(
x
)
f
(
x
1). [
m
]
m
.
0
f
(
x
)
1.
(ii)
s
are half and quarter-sized versions of
s
.
(iii) When
f
(
x
)
1, 0
1
f
(
x
)
and
s
.
s
(
x
)
s
(
x
)
3/4,
(
x
)
7/8; though these are not least upper bounds.
(iv)
s
is linear on segments and contiguous, so
s
is continuous for all
x
.
Now thecompositeof two continuous functions and a scalar
multiplegivs thecontinuity of
s
s
(
x
)
s
(
x
)
s
.
(v)
s
(
x
)
(1/2
)
s
(2
x
)
1/2
.
b
is uniformly convergent by the
Weierstrass
M
-test, and continuous since the components are
continuous.
s
(2
ยท 2
x
)
(vi)
b
(2
x
)
s
(2
x
)
2
