Graphics Programs Reference
In-Depth Information
1.00
l 2
0.50
l 1
0.00
l 3
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
3.00
x
Figure 3.2. Example of quadraticcardinalfunctions.
To provethat the interpolating polynomial passes through the datapoints, we
substitute x
=
x j into Eq. (3.1a) and thenutilize Eq. (3.2). The result is
n
n
P n 1 ( x j )
=
y i i ( x j )
=
y i δ i j =
y j
i
=
1
i
=
1
It can be shown that the errorinpolynomial interpolationis
( x
x 1 )( x
x 2 )
...
( x
x n )
f ( n ) (
f ( x )
P n 1 ( x )
=
ξ
)
(3.3)
n !
where
x n );its value is otherwise unknown. It is
instructivetonote that the fartheradatapoint isfrom x , the more itcontributes to
the errorat x .
ξ
lies somewhere in the interval( x 1 ,
Newton's Method
Evaluation of polynomial
Although Lagrange's methodisconceptually simple, it does not lend itself to an effi-
cient algorithm. Abetter computational procedure isobtainedwith Newton'smethod,
where the interpolating polynomial is writteninthe form
P n 1 ( x )
=
a 1
+
( x
x 1 ) a 2
+
( x
x 1 )( x
x 2 ) a 3
+···+
( x
x 1 )( x
x 2 )
···
( x
x n 1 ) a n
This polynomiallends itself to an efficientevaluationprocedure.Consider,for
example, four datapoints ( n
=
4). Here the interpolating polynomial is
=
a 1 +
x 1 ) a 2 +
x 2 ) a 3 +
P 3 ( x )
( x
( x
x 1 )( x
( x
x 1 )( x
x 2 )( x
x 3 ) a 4
=
a 1
+
( x
x 1 )
{
a 2
+
( x
x 2 ) [ a 3
+
( x
x 3 ) a 4 ] }
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