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Chebyshev composition

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    Name
    Vu Hung
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Problem Statement

The first-kind Chebyshev polynomials Tn(x)T_n(x) are defined by the recurrence

T0(x)=1,T1(x)=x,Tn+1(x)=2xTn(x)−Tn−1(x).T_0(x)=1,\quad T_1(x)=x,\quad T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).

It is given that Tn(cos⁡θ)=cos⁡(nθ)T_n(\cos\theta)=\cos(n\theta) for all n≥0n\ge0 and θ∈R\theta\in\R. (Do NOT prove this fact.)

For first-kind Chebyshev polynomials, prove that

Tm(Tn(x))=Tmn(x)T_m(T_n(x))=T_{mn}(x)

for all xx of the form x=cos⁡θx=\cos\theta. Then use this result to find

T2(T3(x)).T_2(T_3(x)).

Hints

Use Tn(cos⁡θ)=cos⁡(nθ)T_n(\cos\theta)=\cos(n\theta). For the explicit polynomial, use T6(x)T_6(x).

Solutions

Let x=cos⁡θx=\cos\theta. Then

Tm(Tn(x))=Tm(cos⁡(nθ))=cos⁡(mnθ)=Tmn(cos⁡θ)=Tmn(x).T_m(T_n(x))=T_m(\cos(n\theta))=\cos(mn\theta)=T_{mn}(\cos\theta)=T_{mn}(x).

Therefore

T2(T3(x))=T6(x).T_2(T_3(x))=T_6(x).

Using the recurrence after T5(x)=16x5−20x3+5xT_5(x)=16x^5-20x^3+5x,

T6(x)=2xT5(x)−T4(x)=32x6−48x4+18x2−1.T_6(x)=2xT_5(x)-T_4(x)=32x^6-48x^4+18x^2-1.

Takeaways

  • The trigonometric definition makes Chebyshev composition simple.
  • Composition of first-kind Chebyshev polynomials multiplies the indices.
  • Hard-looking polynomial composition can often be avoided by using structure.

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