SPECIAL FUNCTIONS
The normalisation, whenm=n, is easily found by making the substitution
x=cosθand using (18.55). We immediately obtain
∫ 1− 1Tn(x)Tn(x)(1−x^2 )−^1 /^2 dx={
π forn=0,
π/2forn=1, 2 , 3 ,.... (18.62)The orthogonality and normalisation conditions mean that any (reasonable)
functionf(x) can be expanded over the interval|x|<1inaseriesoftheform
f(x)=^12 a 0 +∑∞n=1anTn(x),where the coefficients in the expansion are given by
an=2
π∫ 1− 1f(x)Tn(x)(1−x^2 )−^1 /^2 dx.For the Chebyshev polynomials of the second kind,Un(x), we see from (18.58)that (1−x^2 )^1 /^2 Un(x)=Vn+1(x) satisfies Chebyshev’s equation (18.54) withν=
n+ 1. Thus, the orthogonality relation for theUn(x), obtained by replacingTi(x)
byVi+1(x) in equation (18.61), reads
∫ 1− 1Un(x)Um(x)(1−x^2 )^1 /^2 dx=0 ifn=m.The corresponding normalisation condition, whenn=m, can again be found by
making the substitutionx=cosθ, as illustrated in the following example.
Show thatI≡∫ 1
− 1Un(x)Un(x)(1−x^2 )^1 /^2 dx=π
2.
From (18.58), we see that
I=
∫ 1
− 1Vn+1(x)Vn+1(x)(1−x^2 )−^1 /^2 dx,which, on substitutingx=cosθ,gives
I=
∫ 0
πsin(n+1)θsin(n+1)θ1
sinθ(−sinθ)dθ=π
2.
The above orthogonality and normalisation conditions allow one to expandany (reasonable) function in the interval|x|<1 in a series of the form
f(x)=∑∞n=0anUn(x),