Graphics Reference
In-Depth Information
by qn 3.71, (
h
(
n
))
0 and since
g
is strictly decreasing,
g
(
x
)
0as
x
.
36
f
(
x
)
E
(
x
log
x
).
f
is continuous sincelog and
E
arecontinuous.
f
(
x
)
E
(
x
log
x
) · (log
x
x
· 1/
x
)
0 when log
x
1or
x
1/e. On
(0, 1/e),
f
1/
n
(
x
)
0.
f
(1/
n
)
(1/
n
)
1 by qn 3.59. Since
f
is
strictly decreasing on this domain,
f
(
x
)
1as
x
0
. Minimum is
f
(1/e).
37 See qn 9.38.
38 Useqn 10.60 (integration by substitution) to show that
dt
t
dt
t
.
1
dt
(
t
)
dt
1
(
t
)
...
(
t
)
dt
t
)
t
)
.
1
(
1
(
K
/(
K
1)
t
1/(
K
1)
t
1
1/(
t
1)
K
1
t
/(
t
1)
t
(
K
1).
Also
(
K
x
K
1)
1
1
x
1
if
1.
n
n
But
K
is constant, so ((
K
1)/(
n
1))
0as
n
.
Expansion is valid when
1
x
1.
39
f
(
x
)
(1
x
).
40 Squaring both sides of the inequality makes the comparison trivial.
((
a
b
)
(
f
(
a
)
f
(
b
))
).
41
f
is continuous by qns 6.29, thecontinuity of polynomials, 7.27, the
continuity of root functions, and 6.40, thecontinuity of composite
functions.
f
(
x
)
x
/
(1
x
)on(
1, 1).
42
f
(
b
)
f
(
a
)
f
(
c
) · (
b
a
), so
((
b
a
)
(
f
(
b
)
f
(
a
))
)
(
b
a
)
(1
(
f
(
c
))
)
b
a
((
b
a
)
(
f
(
b
)
f
(
a
))
)
)
.
(1
c
43 Adding a point to the subdivision replaces one part of the polygonal
arc length by two new ones using the original end points, this increases
the polygonal arc length by the triangle inequality.
44 On [
1, 0]
f
is strictly increasing, so the polynomial arc length
x
x
f
(
x
)
f
(
x
)
1
1.
