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Numerical Integration 49
(3.4.2) w (n)
j
(2n + α + β) Γ(n + α) Γ(n + β)
= 2α+β
n! Γ(n + α + β + 1)
×
P (α,β)
n−1 (ξ(n) j ) d
dx
(α,β)
P n (ξ (n)
−1 j ) , 1 ≤ j ≤ n.
For the proof we argue as follows. By equating the leading coefficients, (1.3.5) yields
(3.4.3)
un(x)
x − ξ (n)
j
= (2n + α + β)(2n + α + β − 1)
2n (n + α + β)
un−1(x) + qj(x), 1 ≤ j ≤ n,
where qj ∈ Pn−2, 1 ≤ j ≤ n, and un = P (α,β)
n . Therefore, recalling (3.4.1) and (3.2.4),
(3.4.4) w (n)
j =
=
=
1
un−1(ξ (n)
j )
1
1
−1
(un−1u ′ n)(ξ (n)
j )
1
un−1(ξ (n)
j )
n
i=1
l (n)
j un−1w dx =
(l (n) (n)
j un−1)(ξ i ) w (n)
i
1
(un−1u ′ n)(ξ (n)
j )
(2n + α + β)(2n + α + β − 1)
2n (n + α + β)
Finally, (3.4.2) is obtained with the help of (2.2.10).
1
−1
un
x − ξ (n)
j
un−1w dx
un−1 2 w, 1 ≤ j ≤ n.
Legendre case - To obtain an expression for the weights, we simply set α = β = 0 in
(3.4.2). This gives
(3.4.5) w (n)
j
2
=
n
Pn−1(ξ (n)
j )P ′ n(ξ (n)
−1 j ) , 1 ≤ j ≤ n.
Chebyshev case - This is the most remarkable case. We recall (1.5.1), (3.1.16) and the
relation Tn−1(ξ (n)
j ) = (−1) j+n
(3.4.2), one obtains
(3.4.6) w (n)
j
1 − [ξ (n)
j ] 2 , 1 ≤ j ≤ n. By taking α = β = − 1
2 in
π
= , 1 ≤ j ≤ n.
n