Graphics Reference
In-Depth Information
33 Let
f
(
b
)
f
(
a
)
(
b
a
)
f
(
a
)
(
b
a
)
f
(
a
)
K
,
(
b
a
)
and let
F
(
x
)
f
(
b
)
f
(
x
)
(
b
x
)
f
(
x
)
(
b
x
)
f
(
x
)
K
(
b
x
)
. The
application of Rolle's Theorem to
F
gives
K
f
-
(
c
)/6 for some
c
, with
a
c
b
.
f
(
a
h
)
f
(
a
)
hf
(
a
)
(
h
/2!)
f
(
a
)
(
h
/3!)
f
-
(
a
h
).
36 For some
,0
1,
h
2!
f
h
(
n
h
n
!
f
f
(
a
h
)
f
(
a
)
hf
(
a
)
(
a
)
...
1)!
f
(
a
)
(
a
h
).
x
n
!
exp(
x
), for some
,0
1.
38
R
(
x
)
The function exp is monotonic (see chapter 11) so if
k
max(1, exp(
x
)),
R
(
x
)
x
/
n
!
·
k
. Now (
x
/
n
!
) is a null sequence by
qn 3.74(ii). So (
R
(
x
))
0as
n
by qn 3.34 (squeeze rule). The
n
th
partial sum of the power series
exp(
x
)
R
(
x
)
exp(
x
)as
n
.
39 If
f
(
x
)
sin
x
,
f
(
x
)
cos
x
sin(
x
), so
f
(
x
)
sin(
x
n
).
R
(
x
)
(
x
/2
n
!)· sin(
x
n
). So
R
(
x
)
x
/2
n
!
. This shows that
(
R
(
x
))
0as
n
. As in qn 38, this enables us to prove that the series
converges to the function for each value of
x
.
40 If
f
(
x
)
cos
x
,
f
(
x
)
sin
x
cos(
x
), so
f
cos(
x
n
). The
argument proceeds as in qn 38.
x
·(
1)
(
n
1)!
x
41
R
(
x
)
(
1)
.
n
!(1
x
)
n
(1
x
)
x
R
x
/
n
(i) For 0
(
x
)) is a null sequence, and
the series converges to the function, as in qn 38.
1,
(
x
)
and so (
R
(ii)
R
(1)
1/
n
(1
)
1/
n
, and again (
R
(
x
)) is a null sequence.
(iii) For
x
0,
x
1
x
,so
x
/(1
x
)
1 and
therefore
R
(
x
)
1/
n
, so that (
R
(
x
)) is a null sequence.
42 From qn 5.95 (or 5.69) the radius of convergence is 1. Now consider a
particular
x
1.
The
n
th partial sum of series
x
)
R
(
x
)) were a null
sequence, the sequence of
n
th partial sums would converge.
log(1
(
x
). If (
R
43 The two series coincide. Radius of convergence
1.
For
n
2,
the remainder after the
n
th term
a
/(1
)
a
,
a
0.06 . . .,
a
0.039 . . .. So four terms needed.
