Graphics Programs Reference
In-Depth Information
3(2
2)
1)
2
−
0
.
4258
)
6
−
2
.
(
=
−
(
−
0
.
−
.
(
0
2)
3(2
2)
9)
2
(
−
0
.
3774)
6
−
1
.
1
.
3961
−
1
.
5432
−
−
−
.
+
(
0
(
−
0
.
2)
(
−
0
.
2)
=
0
.
7351
k
2
x
−
x
3
k
3
x
−
x
2
f
(2)
≈
f
2
,
3
(2)
=
x
3
−
x
2
−
x
2
−
x
3
4258
)
2
−
2
.
1
3774)
2
−
1
.
9
=
−
0
.
−
(
−
0
.
=−
0
.
4016
(
(
−
0
.
2)
(
−
0
.
2)
Note that the solutionsfor
f
(2) in parts (1) and (2) differ onlyinthe fourth significant
figure, but the values of
f
(2) are much farther apart. This is not unexpected,consid-
ering the general rule: the higher the order of the derivative, the lower the precision
with which itcan becomputed. It is impossible to tell which of the two results is
betterwithoutknowing the expression for
f
(
x
). In this particular problem, the data
points fall on the curve
f
(
x
)
x
2
e
−
x
/
2
,sothat the “correct”values of the derivatives
=
are
f
(2)
7358 and
f
(2)
=
0
.
=−
0
.
3679.
EXAMPLE 5.5
Determine
f
(0) and
f
(1)from the following noisydata
.
.
.
x
0
0
2
0
4
0
6
f
(
x
)
1
.
9934
2
.
1465
2
.
2129
2
.
1790
x
0
.
8
1
.
0
1
.
2
1
.
4
f
(
x
)
2
.
0683
1
.
9448
1
.
7655
1
.
5891
Solution
We used the program listed in Example 3.10tofind the best polynomial fit
(in the least-squares sense) to the data. The results were:
degree of polynomial = 2
coeff =
-7.0240e-001
6.4704e-001
2.0262e+000
sigma =
3.6097e-002
degree of polynomial = 3
coeff =
4.0521e-001
Search WWH ::
Custom Search