Untitled - Cdm.unimo.it

26 Polynomial Approximation of Differential Equations
(2.2.13) (Laguerre)
(2.2.14) (Hermite)
+∞
0
L (α)
2 n (x) x α e −x dx =
+∞
−∞
H 2 n(x) e −x2
Proof - Taking u = v = un in (2.2.2), if b ≡ 0 we have
(2.2.15) un 2 w = λ −1
n u ′ n 2 a.
Γ(n + α + 1)
,
n!
dx = 2 n n! √ π.
Let us consider first the Jacobi case. We get λn = n(n + α + β + 1) by theorem
1.3.1. Moreover, recalling (1.3.6) and substituting in (2.2.15), by iterating n times this
procedure one obtains
(2.2.16)
= (n + α + β + 1) · · · (n + α + β + n)
= n + α + β + 1
1
−1
4n
P (α,β)
n
1
−1
2 (x)
(1 − x) α (1 + x) β dx
P (α+1,β+1)
n−1
4 n n!
=
1
−1
2 (x)
Γ(2n + α + β + 1)
2 2n n! Γ(n + α + β + 1)
(1 − x) α+1 (1 + x) β+1 dx = · · · =
P (α+n,β+n)
0
2 (x) (1 − x) α+n (1 + x) β+n dx
1
(1 − x)
−1
α+n (1 + x) β+n dx.
We arrive at (2.2.10) by using (1.2.3) with y = α + n + 1 and x = β + n + 1. Then,
setting α = β = 0 we have (2.2.11). By choosing α = β = −1/2 and considering
(1.5.1) we have (2.2.12) (otherwise recall (1.5.6) and take x = cos θ). To prove (2.2.13)
and (2.2.14) we argue exactly as we did for the Jacobi case. This time we use the
relations (1.6.6) and (1.7.8) respectively.