Graphics Programs Reference
In-Depth Information
ϕ
n
(
x
) are polynomials of the degree indicatedbythe
where
Q
n
−
1
(
x
),
R
n
−
1
(
x
) and
subscripts.
11
Therefore,
b
b
b
w
(
x
)
P
2
n
−
1
(
x
)
dx
=
w
(
x
)
Q
n
−
1
(
x
)
dx
+
w
(
x
)
R
n
−
1
(
x
)
ϕ
n
(
x
)
dx
a
a
a
But according to Eq. (6.23) the second integralon the right hand-side vanishes,
so that
b
b
w
(
x
)
P
2
n
−
1
(
x
)
dx
=
w
(
x
)
Q
n
−
1
(
x
)
dx
(c)
a
a
Because apolynomialofdegree
n
−
1is uniquelydefinedby
n
points, it is always
possible to find
A
i
such that
b
n
w
(
x
)
Q
n
−
1
(
x
)
dx
=
A
i
Q
n
−
1
(
x
i
)
(d)
a
i
=
1
In order to arrive at Eq. (a), we must choose for the nodal abscissas
x
i
the roots of
ϕ
n
(
x
)
=
0.According to Eq. (b) we then have
P
2
n
−
1
(
x
i
)
=
Q
n
−
1
(
x
i
),
i
=
1
,
2
,...,
n
(e)
which togetherwith Eqs. (c) and (d)leadsto
b
b
n
w
(
x
)
P
2
n
−
1
(
x
)
dx
=
w
(
x
)
Q
n
−
1
(
x
)
dx
=
A
i
P
2
n
−
1
(
x
i
)
a
a
=
i
1
Thiscompletes the proof.
Theorem
b
A
i
=
w
(
x
)
i
(
x
)
dx
,
i
=
1
,
2
,...,
n
(6.24)
a
where
i
(
x
) are
the Lagrange'scardinalfunctionsspanning the nodes at
x
n
. These functions were definedinEq. (3.2).
Proof
Applying Lagrange'sformula, Eq. (3.1a), to
Q
n
−
1
(
x
) yields
x
1
,
x
2
,...
n
Q
n
−
1
(
x
)
=
Q
n
−
1
(
x
i
)
i
(
x
)
i
=
1
which upon substitutioninEq. (d) gives us
n
Q
n
−
1
(
x
i
)
b
a
i
(
x
)
dx
n
w
(
x
)
=
A
i
Q
n
−
1
(
x
i
)
i
=
1
i
=
1
or
Q
n
−
1
(
x
i
)
A
i
−
b
i
(
x
)
dx
n
w
(
x
)
=
0
a
i
=
1
11
It can be shown that
Q
n
−
1
(
x
) and
R
n
−
1
(
x
) are uniquefor given
P
2
n
−
1
(
x
) and
ϕ
n
(
x
).
Search WWH ::
Custom Search