Second-kind Chebyshev recurrence

The Chebyshev polynomials are a famous family of orthogonal polynomials that appear in approximation theory, numerical analysis, and even in the design of filters. For HSC Mathematics Extension 1 students, this problem offers valuable practice with recurrence relations – a core topic in sequences and series. By carefully applying the given recurrence, you’ll see how to build higher-order terms from the initial values, while sharpening your algebraic manipulation.

Problem Statement

The second-kind Chebyshev polynomials satisfy

$$U_0(x)=1,\qquad U_1(x)=2x,\qquad U_{n+1}(x)=2xU_n(x)-U_{n-1}(x).$$

Use this recurrence to find $U_3(x)$ and $U_4(x)$.

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Hints

First find $U_2(x)$, then continue to $U_3(x)$ and $U_4(x)$.

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Solutions

We begin by finding $U_2(x)$, which is the first new term generated by the recurrence. Using $n=1$ with $U_1(x)=2x$ and $U_0(x)=1$:

$$U_2(x)=2x(2x)-1=4x^2-1.$$

Now that we have $U_2(x)$, we can step forward to $U_3(x)$ by setting $n=2$ and feeding in $U_2(x)$ and $U_1(x)$:

$$U_3(x)=2x(4x^2-1)-2x=8x^3-4x.$$

Finally, to obtain $U_4(x)$, we let $n=3$ and substitute the expressions for $U_3(x)$ and $U_2(x)$:

$$U_4(x)=2x(8x^3-4x)-(4x^2-1)=16x^4-12x^2+1.$$

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Takeaways

• The first-kind and second-kind Chebyshev polynomials share the same recurrence relation; it's the initial conditions that set them apart.
• The starting values determine which family is produced.
• Keep your recurrence calculations tidy: substituting correctly and simplifying line by line prevents small mistakes.

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Further Readings

HSC Vectors, HSC Proofs, HSC Last Resorts, HSC Mechanics