Graphics Programs Reference
In-Depth Information
1930
T
. Thus the interpolant and its deri-
which yields
a
=
−
0
.
7714 1
.
5075
−
0
.
vatives are
1903
x
2
P
2
(
x
)
=−
0
.
+
1
.
5075
x
−
0
.
7714
P
2
(
x
)
=−
0
.
3860
x
+
1
.
5075
P
2
(
x
)
=−
0
.
3860
which gives us
f
(2)
≈
P
2
(2)
=−
0
.
3860(2)
+
1
.
5075
=
0
.
7355
f
(2)
P
2
(2)
=−
0
.
3860
≈
Solution of Part (2)
We must first determine the second derivatives
k
i
of the spline
at its knots, afterwhich the derivatives of
f
(
x
)can becomputed fromEqs. (5.10) and
(5.11). The first partcan becarried out by the following small program:
% Example 5.4 (Curvatures of cubic spline at the knots)
xData = [1.5; 1.9; 2.1; 2.4; 2.6; 3.1];
yData = [1.0628; 1.3961; 1.5432; 1.7349; 1.8423; 2.0397];
k = splineCurv(xData,yData)
The outputoftheprogram,consisting of
k
1
to
k
6
, is
>>k=
0
-0.4258
-0.3774
-0.3880
-0.5540
0
Since
x
=
2 lies between knots 2 and 3, we must use Eqs. (5.10) and (5.11) with
i
=
2. This yields
3(
x
x
3
)
x
3
)
2
k
2
6
−
f
(2)
≈
f
2
,
3
(2)
=
−
(
x
1
−
x
2
−
x
3
3(
x
x
3
)
x
2
)
2
k
3
6
−
y
2
−
y
3
−
−
(
x
2
−
+
x
2
−
x
3
x
2
−
x
3
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