A rational recurrence with period four · source reference

Printed question

A sequence $\{u_n\}$ is defined by

$$u_{n+1}=\frac2{2-u_n},\quad n\in\mathbb Z^+,\qquad u_1\ne0,1,2.$$

(a) Prove that the sequence does not converge. [3]

(b) Given $u_1=4$: (i) write down the next four terms [1]; (ii) prove algebraically that $u_{n+4}=u_n$. [3]

Printed working

(a) Suppose the sequence converges to $L$. The printed argument is

$$L=\frac2{2-L}\Rightarrow2L-L^2=2\Rightarrow L^2-2L+2=0\Rightarrow(L-1)^2+1=0.$$

There is no real solution for $L$, so the sequence does not converge.

(b)(i) $u_1=4,u_2=-1,u_3=\frac23,u_4=\frac32,u_5=4$.

(b)(ii)

$$u_{n+2}=\frac2{2-u_{n+1}}=\frac2{2-2/(2-u_n)}=\frac{2(2-u_n)}{2-2u_n}=\frac{2-u_n}{1-u_n}.$$ $$u_{n+4}=\frac{2-u_{n+2}}{1-u_{n+2}}=\frac{2-(2-u_n)/(1-u_n)}{1-(2-u_n)/(1-u_n)}=\frac{-u_n}{-1}=u_n.$$

Teacher notes

Sequence converges to $L$ means its limit is a finite value $L$. As $n\to\infty$, both $u_n\to L$ and $u_{n+1}\to L$. Substitute these in the recurrence formula.

Printed answer key

(b)(i) $u_1=4,u_2=-1,u_3=2/3,u_4=3/2,u_5=4$

Mathematical review and qualifications

(a) A finite limit $L=2$ is impossible because $2/(2-u_n)$ would be unbounded. For $L \ne2$, continuity gives $L^2-2L+2=0$, or $(L-1)^2+1=0$, with no real root. Thus no finite real limit exists.

(b)(i) The next terms are $-1,2/3,3/2,4$. (ii) For $F(x)=2/(2-x)$, direct simplification gives $F^2(x)=(2-x)/(1-x)$, $F^3(x)=2-2/x$, $F^4(x)=x$. The exclusions $0,1,2$ avoid the poles along the cycle. Hence $u_{n+4}=u_n$.

The limit substitution must first exclude the pole L=2; dividing by a denominator tending to zero is invalid.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2015/TJC/Promo/6] Question PDF pages 2; solution PDF pages 4, 5. The original files remain private QA records outside this native content.