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Derivative Matrices 127
(7.1.7) c (1)
i
= (2i + 1)
n
j=i+1
i+j odd
cj, 0 ≤ i ≤ n − 1.
For the Chebyshev basis, we still get c (1)
n = 0. Besides, one has
(7.1.8) c (1)
i =
⎧
n
jcj if i = 0,
j=1 ⎪⎨ j odd
n
2 jcj if 1 ≤ i ≤ n − 1.
⎪⎩
j=i+1
i+j odd
The linear mapping which transforms the vector {cj}0≤j≤n into the vector {c (1)
i }0≤i≤n
is clearly expressed by a (n + 1) × (n + 1) matrix which is upper triangular and has a
vanishing diagonal . Therefore, all its eigenvalues are zero. By applying k times (k ≥ 1)
the derivative matrix to the vector {cj}0≤j≤n we obtain the vector {c (k)
i }0≤i≤n, which
represents the Fourier coefficients of the k-th derivative of the polynomial p. This implies
that, since d n+1
dx n+1 p ≡ 0 ∀p ∈ Pn, the n + 1 power of the matrix has all its entries equal
to zero.
For 2 ≤ k ≤ n, an analytic formula relating the coefficients c (k)
i , 0 ≤ i ≤ n, to the
coefficients cj, 0 ≤ j ≤ n, is given in karageorghis and phillips(1989). Here
we just present such a formula for k = 2, respectively for Legendre and Chebyshev
expansions. In these two cases, c (2)
n = c (2)
n−1
(7.1.9) c (2)
i
=
(7.1.10) c (2)
i
i + 1
2
⎧
=
1
2
⎪⎨
⎪⎩
n
j=i+2
i+j even
n
j=2
j even
n
j=i+2
i+j even
j 3 cj
= 0, and
(j(j + 1) − i(i + 1)) cj, 0 ≤ i ≤ n − 2,
j(j 2 − i 2 )cj
if i = 0,
if 1 ≤ i ≤ n − 2.