Graphics Programs Reference
In-Depth Information
ξ
,
+
where
h
). Note that the expression for
E
n
is
identical to the first discarded term of the series, but with
x
replacedby
issome point in the interval(
x
x
ξ
.Since the
valueof
is undetermined (onlyits limits areknown), the most wecan get outof
Eq. (A4) are the upper and lowerboundson the truncation error.
If the expression for
f
(
n
+
1)
(
ξ
ξ
) is not available, the information conveyedby
Eq. (A4) is reduced to
(
h
n
+
1
)
E
n
=
O
(A5)
which is a concise way of saying that the truncation erroris
of the order of h
n
+
1
,or
behaves as
h
n
+
1
. If
h
is within the radiusofconvergence, then
(
h
n
)
(
h
n
+
1
)
O
> O
i.e., the erroris always reducedif a term is added to the truncated series (this may not
betruefor the first few terms).
In the specialcase
n
=
1,Taylor's theoremisknown as the
mean value theorem
:
f
(
f
(
x
+
h
)
=
f
(
x
)
+
ξ
)
h
,
x
≤
ξ
≤
x
+
h
(A6)
Function of Several Variables
If
f
is a function of the
m
variables
x
1
,
x
2
,...,
x
m
, thenits Taylor series expansion
x
m
]
T
is
about the point
x
=
[
x
1
,
x
2
,...,
x
x
m
m
m
2
f
∂
f
1
2!
∂
+
=
+
h
i
+
h
i
h
j
+···
f
(
x
h
)
f
(
x
)
(A7)
∂
x
i
∂
x
i
∂
x
j
i
=
1
i
=
1
j
=
1
This issometimes writtenas
1
2
h
T
H
(
x
)
h
f
(
x
+
h
)
=
f
(
x
)
+
∇
f
(
x
)
·
h
+
+···
(A8)
The vector
∇
f
isknown as the
gradient
of
f
and the matrix
H
iscalled the
Hessian
matrix
of
f
.
EXAMPLE A1
Derive the Taylor series expansion of
f
(
x
)
=
ln(
x
) about
x
=
1.
Solution
The derivatives of
f
are
1
x
1
x
2
2!
x
3
3!
x
4
f
(
x
)
f
(
x
)
f
(
x
)
f
(4)
=
=−
=
=−
etc.
Evaluating the derivatives at
x
=
1, we get
f
(1)
f
(1)
f
(1)
f
(4)
(1)
=
1
=−
1
=
2!
=−
3! etc.
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