A logarithmic fixed-point iteration · source reference

Printed question

The diagram shows $y=\ln(x-1)-x+4$. Its roots are $\alpha,\beta$, with $\alpha<\beta$.

(i) Find $\alpha,\beta$ correct to 3 decimal places. [2]

A sequence of real numbers satisfies $x_{n+1}=\ln(x_n-1)+4$, $n\ge1$.

(ii) Prove algebraically that if the sequence converges, it converges to either $\alpha$ or $\beta$. [3]

(iii) By considering the maximum point of $y=\ln(x-1)-x+4$, show $x_{n+1}\le x_n+2$. [3]

Printed working

(i) From GC: $\alpha=1.052,\beta=5.505$.

(ii) If a sequence converges, as $n\to\infty$, $x_n\to L,x_{n+1}\to L$.

$$L=\ln(L-1)+4\Rightarrow\ln(L-1)-L+4=0.$$

Hence $L=\alpha$ or $\beta$.

(iii) Maximum point: $(2,2)$.

$$x_{n+1}-x_n=\ln(x_n-1)-x_n+4.$$

Since $\ln(x-1)-x+4\le2$, $x_{n+1}-x_n\le2$, so $x_{n+1}\le x_n+2$.

Source diagrams redrawn

Printed answer key

(i) $\alpha=1.052,\beta=5.505$

Mathematical review and qualifications

(i) Numerical solution on $(1,2)$ and $(2,\infty)$ gives $\alpha=1.052$, $\beta=5.505$ to 3 d.p.

(ii) For a convergent real sequence defined at every step, $L=1$ is impossible, since $\ln(x_n-1)+4$ would tend to $-\infty$. Hence $L>1$, and continuity gives $L=\ln(L-1)+4$, so $h(L)=0$. Therefore $L=\alpha$ or $\beta$.

(iii) $\frac{dh}{dx}=1/(x-1)-1$, zero only at 2; $\frac{d^2h}{dx^2}=-1/(x-1)^2<0$. The global maximum is $h(2)=2$. Thus $x_{n+1}-x_n=h(x_n)\le2$.

Finding possible limits does not prove convergence; every logarithm requires x_n>1.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2012/MI/Prelim/10] Question PDF pages 2; solution PDF pages 6. The original files remain private QA records outside this native content.