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=
which upon substitutioninto Eq. (A1)togetherwith
a
1yields
1)
2
1)
3
1)
4
(
x
−
2!
(
x
−
3!
(
x
−
ln(
x
)
=
0
+
(
x
−
1)
−
+
−
+···
2!
3!
4!
1
2
(
x
1
3
(
x
1
4
(
x
1)
2
1)
3
1)
4
=
(
x
−
1)
−
−
+
−
−
−
+···
EXAMPLE A2
Use the first fiveterms of the Taylor series expansion of
e
x
about
x
=
0:
x
4
4!
+···
togetherwith the error estimate to find the boundsof
e
.
x
2
2!
+
x
3
3!
+
e
x
=
+
+
1
x
Solution
1
2
+
1
6
+
1
24
+
65
24
+
e
=
1
+
1
+
E
4
=
E
4
e
ξ
5!
,0
)
h
5
5!
f
(4)
(
E
4
=
ξ
=
≤
ξ
≤
1
The boundson the truncation errorare
e
0
5!
e
1
5!
1
120
e
120
(
E
4
)
min
=
=
(
E
4
)
max
=
=
Thus the lowerbound on
e
is
65
24
+
1
120
163
60
e
min
=
=
and the upperbound is givenby
65
24
+
e
max
120
e
max
=
which yields
119
120
e
max
=
65
24
325
119
e
max
=
Therefore,
163
60
325
119
≤
e
≤
EXAMPLE A3
Compute the gradient and the Hessian matrix of
ln
x
2
f
(
x
,
y
)
=
+
y
2
at the point
x
=−
2,
y
=
1.
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