The Recursive Geometry of Chebyshev Sequences

Polynomial sequences generated by two‑term recurrences appear throughout the HSC Extension 1 and 2 courses, and the Chebyshev family is one of the richest examples. By working through this problem you will see how a simple definition blossoms into a trigonometric identity, pinpoints the exact roots of a family of polynomials, and lets you analyse a numeric sequence using characteristic equations. The techniques here—induction with sum‑to‑product identities, bounding roots with hyperbolic functions, and solving a constant‑coefficient recurrence—are all examinable and deeply connected.

Problem Statement

Let a sequence of functions $(T_n(x))$ be defined for $n \ge 0$ by the recurrence relation

$$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x),$$

with initial terms $T_0(x)=1$ and $T_1(x)=x$.

• Calculate the explicit polynomial expressions for $T_2(x)$ and $T_3(x)$.
• Using the identity

$$\cos(A+B)+\cos(A-B)=2\cos A\cos B,$$

prove by mathematical induction that for all $n \ge 0$,

$$T_n(\cos\theta)=\cos(n\theta).$$

• Let $x_0$ be a root of $T_n(x)$. Show that $|x_0|\le 1$. Hence, determine the exact values of the $n$ distinct roots of $T_n(x)$ in terms of $n$.
• Consider the sequence of values $v_n=T_n(1.5)$. Show that $(v_n)$ satisfies a linear recurrence relation with constant coefficients. Find a closed-form expression for $v_n$ and determine the value of the limit

$$\lim_{n\to\infty}\frac{v_{n+1}}{v_n}.$$

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Hints

• For (i): Substitute the previous two polynomials into the recurrence formula.
• For (ii): Use the sum-to-product identity with $A=k\theta$ and $B=\theta$.
• For (iii): Solve $\cos(n\theta)=0$ for $\theta$, then use $x=\cos\theta$.
• For (iv): Solve the characteristic equation for the recurrence $v_{n+1}=3v_n-v_{n-1}$.

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Solutions

(i) Directly from the recurrence, we feed $T_0$ and $T_1$ in to build the next polynomial:

$$T_2(x)=2x^2-1,\qquad T_3(x)=2x(2x^2-1)-x=4x^3-3x.$$

(ii) We prove the identity by induction on $n$. The base cases $n=0,1$ give $1=\cos0$ and $T_1(\cos\theta)=\cos\theta$, so the statement is true for $n=0,1$. Assume it holds for $k$ and $k-1$:

$$T_k(\cos\theta)=\cos(k\theta),\qquad T_{k-1}(\cos\theta)=\cos((k-1)\theta).$$

Then

$$\begin{aligned} T_{k+1}(\cos\theta) &=2\cos\theta\,T_k(\cos\theta)-T_{k-1}(\cos\theta)\\ &=2\cos\theta\cos(k\theta)-\cos((k-1)\theta)\\ &=\cos((k+1)\theta), \end{aligned}$$

using $\cos((k+1)\theta)+\cos((k-1)\theta)=2\cos(k\theta)\cos\theta$. Hence $T_n(\cos\theta)=\cos(n\theta)$ for all $n\ge0$.

(iii) First we show that any real root must lie in $[-1,1]$. If $x>1$, we write $x=\cosh u$; then $T_n(x)=\cosh(nu)>0$, so $x$ cannot be a root. If $x<-1$, write $x=-\cosh u$; then $T_n(x)=(-1)^n\cosh(nu)\ne0$, again no root. Thus every real root satisfies $|x_0|\le1$.

With this bound we can safely set $x=\cos\theta$. The roots satisfy

$$T_n(\cos\theta)=0\iff \cos(n\theta)=0 \iff n\theta=\frac{(2k-1)\pi}{2}.$$

So the $n$ distinct roots are

$$x_k=\cos\left(\frac{(2k-1)\pi}{2n}\right),\qquad k=1,2,\ldots,n.$$

(iv) To analyse the numeric sequence we substitute $x=1.5$ into the definition of the recurrence, giving a constant‑coefficient recurrence for $v_n=T_n(1.5)$:

$$v_{n+1}=3v_n-v_{n-1},\qquad v_0=1,\quad v_1=\frac32.$$

We solve this with the characteristic equation $\lambda^2-3\lambda+1=0$, which has roots

$$\alpha=\frac{3+\sqrt5}{2},\qquad \beta=\frac{3-\sqrt5}{2}.$$

Hence $v_n=A\alpha^n+B\beta^n$. Using $v_0=1$ and $v_1=\frac32$, we find $A=B=\frac12$, so

$$v_n=\frac12\alpha^n+\frac12\beta^n.$$

Since $|\beta|<\alpha$, the dominant term is $\alpha^n$, and therefore

$$\lim_{n\to\infty}\frac{v_{n+1}}{v_n}=\alpha=\frac{3+\sqrt5}{2}.$$

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Takeaways

• The recurrence produces the Chebyshev polynomials one term at a time.
• The identity $T_n(\cos\theta)=\cos(n\theta)$ turns polynomial roots into trigonometric roots.
• Evaluating a polynomial recurrence at a fixed value gives an ordinary numerical recurrence.

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Further Readings

HSC Functions, HSC Vectors, HSC Polynomials, HSC Probability