An exponential denominator creates a stable limit · source reference

Printed question

The diagram shows $y=\frac{x}{2^x}-\frac34x$. The two roots of $\frac{x}{2^x}-\frac34x=0$ are $0,\alpha$.

(i) Find $\alpha$ exactly. [2]

A sequence of positive real numbers satisfies $x_{n+1}=\frac{x_n}{2^{x_n}}+\frac14x_n$, $n\ge1$.

(ii) Considering $x_{n+1}-x_n$, show $x_{n+1}>x_n$ if $0<x_n<\alpha$. [2]

(iii) For $0<x_1<\alpha$, use (i),(ii) to find the exact limit. [3]

Printed working

(i)

$$\frac{x}{2^x}-\frac34x=0\Rightarrow x(2^{-x}-3/4)=0\Rightarrow x=0\text{ or }2^x=\frac43.$$ $$x=\frac{\lg(4/3)}{\lg2}.$$

Since $\alpha>0$, $\alpha=\frac{\lg(4/3)}{\lg2}$.

(ii)

$$x_{n+1}-x_n=\frac{x_n}{2^{x_n}}+\frac14x_n-x_n=\frac{x_n}{2^{x_n}}-\frac34x_n.$$

From the graph, if $0<x_n<\alpha$, $\frac{x_n}{2^{x_n}}-\frac34x_n>0$, hence $x_{n+1}>x_n$.

(iii) The printed solution starts “Since the sequence converges”. As $n\to\infty$, $x_n\to L,x_{n+1}\to L$.

$$L=\frac L{2^L}+\frac14L.$$

From (i), $L=0,\alpha$. From (ii), $x_{n+1}>x_n$ if $0<x_1<x_2<\cdots\le\alpha$, so it is increasing. Hence $L=\alpha=\frac{\lg(4/3)}{\lg2}$.

Teacher notes

If $0<x<\alpha$, the graph of $y=\frac{x}{2^x}-\frac34x$ is above the $x$-axis, so $y>0$. Replace $x$ by $x_n$ in this inequality.

Source diagrams redrawn

Printed answer key

(i) $\alpha=\frac{\lg(4/3)}{\lg2}$; (iii) $L=\frac{\lg(4/3)}{\lg2}$

Mathematical review and qualifications

(i) $h(x)=x(2^{-x}-3/4)$, so $\alpha=\ln(4/3)/\ln2$.

(ii) For $0<x_n<\alpha$, $2^{-x_n}>3/4$, hence $x_{n+1}-x_n=h(x_n)>0$.

(iii) Set $F(x)=x2^{-x}+x/4$. On $[0,\alpha]$, $\frac{dF}{dx}=2^{-x}(1-x\ln2)+1/4>0$, since $\alpha\ln2=\ln(4/3)<1$. With $F(0)=0,F(\alpha)=\alpha$, this maps $(0,\alpha)$ into itself. Thus the sequence increases and is bounded above by $\alpha$. Its limit solves $L(2^{-L}-3/4)=0$; as $L\ge x_1>0$, $L=\alpha$.

The denominator is 2 raised to x_n. Establishing an upper bound is necessary; increasing alone does not imply convergence.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/YJC/Prelim/I/1] Question PDF pages 4; solution PDF pages 11, 12. The original files remain private QA records outside this native content.