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Derivative Matrices 133
Taking ν = 0 in (7.2.12), we recover the entries of ˜ Dn in the Legendre case:
⎧
(7.2.13)
˜ d (1)
ij =
⎪⎨
⎪⎩
−1 4n(n + 1) i = j = 0,
Pn(η (n)
i )
Pn(η (n)
j )
η (n)
i
1
− η(n)
j
0 ≤ i ≤ n, 0 ≤ j ≤ n, i = j,
0 1 ≤ i = j ≤ n − 1,
1
4n(n + 1) i = j = n.
Similarly, (1.5.1) and (3.1.15) lead to the Chebyshev case (ν = −1 2 ):
⎧
(7.2.14)
˜ d (1)
ij =
⎪⎨
⎪⎩
− 1
6 (2n2 + 1) i = j = 0,
1
2 (−1)i /(1 + η (n)
i ) 1 ≤ i ≤ n − 1, j = 0,
1
2 (−1)n i = n, j = 0,
−2 (−1) j /(1 + η (n)
j ) i = 0, 1 ≤ j ≤ n − 1,
(−1) i+j
η (n)
i
− η(n)
j
−η (n)
i
2 (1 − (η (n)
i ) 2 )
1 ≤ i ≤ n − 1, 1 ≤ j ≤ n − 1, i = j,
1 ≤ i = j ≤ n − 1,
2 (−1) j+n /(1 − η (n)
j ) i = n, 1 ≤ j ≤ n − 1,
− 1
2 (−1)n i = 0, j = n,
− 1
2 (−1)i+n /(1 − η (n)
i ) 1 ≤ i ≤ n − 1, j = n,
1
6 (2n2 + 1) i = j = n.