Chebyshev composition

Chebyshev polynomials are a classic family defined by a simple recurrence that hides a surprising trigonometric identity. In this problem, we use that identity to prove a composition property that turns the composition of two polynomials into a single polynomial with multiplied index—a result that feels almost magical. Working through this proof will sharpen your ability to recognise structural shortcuts and apply them confidently, a valuable skill for advanced HSC sequence problems.

Problem Statement

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

$$T_0(x)=1,\quad T_1(x)=x,\quad T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).$$

It is given that $T_n(\cos\theta)=\cos(n\theta)$ for all $n\ge0$ and $\theta\in\R$. (Do NOT prove this fact.)

For first-kind Chebyshev polynomials, prove that

$$T_m(T_n(x))=T_{mn}(x)$$

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

$$T_2(T_3(x)).$$

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Hints

Use $T_n(\cos\theta)=\cos(n\theta)$. For the explicit polynomial, use $T_6(x)$.

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Solutions

We are given that for $x = \cos\theta$, $T_n(x) = \cos(n\theta)$. So we can substitute the trigonometric form directly into the composition.

Let $x=\cos\theta$. Then

$$T_m(T_n(x))=T_m(\cos(n\theta))=\cos(mn\theta)=T_{mn}(\cos\theta)=T_{mn}(x).$$

Therefore

$$T_2(T_3(x))=T_6(x).$$

We can now find $T_6(x)$ explicitly using the recurrence. With $n=5$, $T_6(x) = 2x T_5(x) - T_4(x)$. Substituting the known expressions for $T_5(x)$ and $T_4(x)$ gives

$$T_6(x)=2xT_5(x)-T_4(x)=32x^6-48x^4+18x^2-1.$$

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Takeaways

• The trigonometric form $T_n(\cos\theta) = \cos(n\theta)$ turns composition into simple angle multiplication.
• Composition of first-kind Chebyshev polynomials multiplies the indices: $T_m \circ T_n = T_{mn}$.
• Recognising structural simplifications can help you avoid messy polynomial algebra.

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Further Readings

HSC Vectors, HSC Mechanics, HSC Last Resorts, HSC Distributions