ABSTRACT
Um+1(x) = 2xUm(x) − Um−1(x) (D.1) with initial conditions U0(x) = 1 and U1(x) = 2x.
Proof By Definition D.1 we have that U0(cos θ) = 1 and U1(cos θ) = sin(2θ)/ sin(θ) = 2 cos θ, which implies that U0(x) = 1 and U1(x) = 2x. By using the trigonometric identity
sin((m+ 2)θ) = sin((m+ 1)θ) cos θ + cos((m+ 1)θ) sin θ
we obtain that Um+1(x)− xUm(x) = cos((m+ 1)θ). (D.2)
On the other hand, by the trigonometric identity
cos((m+ 1)θ) = cos(mθ) cos θ − sin(mθ) sin θ together with (D.2), we have that
Um+1(x)− xUm(x) = x(Um(x) − xUm−1(x)) − Um−1(x)(1 − x2), which implies that Um(x) satisfies (D.1). By induction on m we can show that Um(x) is a polynomial of degree m with integer coefficients, as required.
