Untitled - Cdm.unimo.it

Orthogonality 25
As we promised, we give the proof of theorem 1.1.1.
Proof - For n = 2 we have to determine three degrees of freedom ρ2, σ2, τ2, in order
to evaluate a second degree polynomial. Therefore, (1.1.2) is trivially satisfied. If n > 2,
we write p := un − ρnxun−1, where ρn is such that p is a polynomial whose degree is
at most n − 1. Thus, due to the orthogonality condition
I
pumw dx =
I
unumw dx − ρn
I
un−1(xum)w dx = 0, ∀m < n − 2
(note that the degree of xum is less than n − 1). This implies that p is only a linear
combination of the terms un−1 and un−2, i.e., p = σnun−1 + τnun−2. The proof is
concluded.
Finally, norms can be also evaluated.
Theorem 2.2.2 - For any n ∈ N, we have
(2.2.10) (Jacobi)
(2.2.11) (Legendre)
(2.2.12) (Chebyshev)
=
⎧
⎨
⎩
1
−1
P (α,β)
n
2 (x)
2 α+β+1 Γ(α+1) Γ(β+1)
Γ(α+β+2)
2 α+β+1
(2n+α+β+1) n!
1
−1
1
−1
T 2 n(x)
(1 − x) α (1 + x) β dx
Γ(n+α+1) Γ(n+β+1)
Γ(n+α+β+1)
P 2 n(x) dx =
dx
√ 1 − x 2 =
2
2n + 1 ,
if n = 0,
if n > 0,
⎧
⎨π
if n = 0,
⎩
π
2 if n > 0,