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Recurrence, Limits & Iteration

Recursive sequences model loans and fixed-point iteration. Find a candidate limit L from the steady-state equation, then prove convergence with a contraction or monotone-bounded argument — standard Ext 1 / Ext 2 technique.

Learning outcomes

  • Model reducing-balance loans with geometric sums
  • Find candidate limits from x = f(x)
  • Prove convergence via contraction or squeeze ideas
  • Interpret iteration geometrically (cobweb intuition)

Recursive sequences in the HSC

A recursive sequence defines each term from previous ones: xn+1=f(xn)x_{n+1}=f(x_n). Two big exam themes:

  1. Financial maths — interest then repayment produces a linear recurrence solved by a finite GP.
  2. Limits of recursion — assume xnLx_n\to L, solve L=f(L)L=f(L), then prove the limit rigorously.

Plain convergence theorems (monotone + bounded, algebra of limits, squeeze theorem) still apply when the sequence is given explicitly; recursion adds the fixed-point step.

Worked example 1 — Reducing-balance loan

A loan of LL grows by factor (1+r)(1+r) each period, then a repayment PP is subtracted. If MnM_n is the balance after nn periods:

M1=L(1+r)P,M_1=L(1+r)-P,

and unfolding yields

Mn=L(1+r)nPk=0n1(1+r)k.M_n=L(1+r)^n-P\sum_{k=0}^{n-1}(1+r)^k.

The sum is a finite GP with first term 11 and ratio 1+r1+r:

Mn=L(1+r)nP((1+r)n1)r(r0).M_n=L(1+r)^n-\frac{P\bigl((1+r)^n-1\bigr)}{r}\quad (r\neq 0).

Car loan: L=20000L=20000, r=0.01r=0.01, n=24n=24, set M24=0M_{24}=0 and solve for PP:

P=Lr(1+r)n(1+r)n1.P=\frac{Lr(1+r)^n}{(1+r)^n-1}.

Substitute the numbers carefully (calculator ok in exams when allowed).

Worked example 2 — Fixed point and contraction

Let x1=16x_1=16 and xn+1=2+1xnx_{n+1}=2+\dfrac{1}{x_n}.

If xnL>0x_n\to L>0, then L=2+1LL=2+\frac{1}{L}, so L22L1=0L^2-2L-1=0 and L=1+2L=1+\sqrt{2} (positive root).

Error identity:

xn+1L=xnLLxn.|x_{n+1}-L|=\frac{|x_n-L|}{L x_n}.

For n2n\ge 2 one can show xn+1L<14xnL|x_{n+1}-L|<\frac14|x_n-L|, hence the error shrinks geometrically and xn1+2x_n\to 1+\sqrt{2}.

Method: candidate limit → error relation → geometric bound → convergence.

Worked example 3 — Iteration intuition

Plotting y=f(x)y=f(x) and y=xy=x, the recursion xn+1=f(xn)x_{n+1}=f(x_n) is a cobweb: vertical to the curve, horizontal to the line y=xy=x. Attracting fixed points pull the cobweb in; repelling points push it away. You rarely need a full diagram in Ext 1, but the picture explains why f(L)<1|f'(L)|<1 suggests local attraction.

For explicit sequences such as averages of squares over n3+kn^3+k, prefer inequalities and the squeeze theorem rather than recursion.

Common mistakes

  • Solving L=f(L)L=f(L) but never proving xnx_n actually converges.
  • In loans: subtracting PP before multiplying by (1+r)(1+r) when the question says interest is added first.
  • Using SS_\infty formulae on a finite annuity sum.
  • Sign errors when rr is a percentage vs a decimal (1%1\% is r=0.01r=0.01).

Practice (hints only)

  1. Write M3M_3 from the loan recurrence and match it to the closed form.
  2. For xn+1=2+xnx_{n+1}=\sqrt{2+x_n} with a suitable x1x_1, find LL and outline a monotone-bounded proof.

Where next

Advanced booklet topics connect recursion to special polynomials and generating functions: Chebyshev, Fibonacci, and generating functions.

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