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Chebyshev, Fibonacci & Generating Functions

Chebyshev polynomials obey a linear recurrence T_{n+1}=2xT_n-T_{n-1}. Fibonacci numbers satisfy F_n=F_{n-1}+F_{n-2}; their generating function packs the whole sequence into a rational function — Extension-flavoured sequences at their richest.

Learning outcomes

  • Generate Chebyshev polynomials from the recurrence
  • Evaluate Chebyshev polynomials at given x
  • Use Fibonacci-type recurrences and identities
  • Interpret a generating function as an encoding of a sequence

Why these topics appear in the Sequences booklet

Beyond routine AP/GP drills, HSC Extension work rewards comfort with linear recurrences. Chebyshev polynomials and Fibonacci numbers are two clean families: each new term is a fixed linear combination of earlier terms. Generating functions then turn the recurrence into algebra.

Chebyshev polynomials of the first kind

Defined by T0(x)=1T_0(x)=1, T1(x)=xT_1(x)=x, and

Tn+1(x)=2xTn(x)Tn1(x).T_{n+1}(x)=2x\,T_n(x)-T_{n-1}(x).

Worked example 1 — Build T2T_2 and T3T_3

T2(x)=2xT1(x)T0(x)=2x21,T_2(x)=2x\cdot T_1(x)-T_0(x)=2x^2-1, T3(x)=2x(2x21)x=4x33x.T_3(x)=2x(2x^2-1)-x=4x^3-3x.

Evaluate T3(12)T_3\bigl(\frac12\bigr):

T3(12)=418312=1232=1.T_3\bigl(\tfrac12\bigr)=4\cdot\tfrac18-3\cdot\tfrac12=\tfrac12-\tfrac32=-1.

Trigonometric view (extension insight): Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) explains roots and composition identities such as Tm(Tn(x))=Tmn(x)T_m(T_n(x))=T_{mn}(x).

Fibonacci and companions

With F1=1F_1=1, F2=1F_2=1, Fn=Fn1+Fn2F_n=F_{n-1}+F_{n-2}, many identities follow from the recurrence alone (e.g. links to Lucas numbers). Treat them as sequence problems: write a few terms, conjecture, prove by induction or by constructing an auxiliary sequence.

Generating functions

A generating function packages a sequence (an)(a_n) into a power series S(x)=anxnS(x)=\sum a_n x^n (indexing conventions vary). For Fibonacci, algebraic manipulation of SxSx2SS-xS-x^2S using the recurrence produces a rational closed form.

Worked example 2 — Fibonacci generating function (sketch)

If S(x)=n=1Fnxn+1S(x)=\sum_{n=1}^{\infty} F_n x^{n+1} (as in the booklet/blog convention), the recurrence cancels interior coefficients and yields

S(x)=x21xx2S(x)=\frac{x^2}{1-x-x^2}

(for x|x| small enough that the series converges). Substituting x=0.1x=0.1 gives S(0.1)=189S(0.1)=\frac{1}{89}, which is why Fibonacci digits appear in the decimal expansion of 189\frac{1}{89} — with carries explaining digits that are not raw Fibonacci numbers.

Worked example 3 — Discrete calculus teaser

The forward difference Δan=an+1an\Delta a_n=a_{n+1}-a_n turns summation into a discrete analogue of integration. Sums of powers and telescoping with factorial powers live in the booklet appendices; they connect AP-space problems in Part 2 to a broader toolkit. Open the PDF appendices when you need the full formal development.

Common mistakes

  • Off-by-one index errors in the Chebyshev recurrence (Tn+1T_{n+1} vs TnT_n).
  • Evaluating the polynomial before simplifying.
  • Generating functions: mismatched powers of xx when subtracting xSxS and x2Sx^2S.
  • Forgetting convergence restrictions when substituting a numeric xx.

Practice (hints only)

  1. Find T4(x)T_4(x) from the recurrence, then T4(1)T_4(1).
  2. Explain one digit carry in 189=0.0112359\frac{1}{89}=0.0112359\ldots using F6=8F_6=8 and F7=13F_7=13.

Where next

Return to the full Sequences study guide for the TOC, or open the PDF booklet for every worked solution in Part 1 and Part 2. Related teaching articles are linked below and on Vu's QED.

Open section in PDF (p. 12)Full study guideMore on the blog

Related worked solutions