The diagram above shows the curve of ${y=\frac{2x}{{{\operatorname{e}}^{x}}}}$. Two points $A$ a

Question

The diagram above shows the curve of ${y=\frac{2x}{{{\operatorname{e}}^{x}}}}$. Two points $A$ and $B$ on the curve have coordinates $\left( \alpha ,\frac{1}{2} \right)$ and $\left( \beta ,\frac{1}{2} \right)$ respectively. A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$… satisfies the recurrence relation ${{{x}_{n+1}}=\frac{1}{4}{{\mathrm{e}}^{{{x}_{n}}}}}$, for ${n\ge 1}$. Show algebraically that if the sequence converges, then it converges to either $\alpha $ or $\beta $.[3] Show that ${{x}_{n+1}}<\alpha $ if ${{x}_{n}}<\alpha $.[2] Show that ${{x}_{n+1}}>{{x}_{n}}$ if ${{x}_{n}}<\alpha $.[2] Explain briefly how the results in (ii) and (iii) may be used to deduce that the sequence converges to $\alpha $ when ${{x}_{1}}=0$.[2]

Accepted Answer

(i) If the sequence converges to $L$, then $L\in\{\alpha,\beta\}$. (ii) $x_nx_n$. (iv) When $x_1=0$, the sequence is strictly increasing, bounded above by $\alpha$, and converges to $\alpha$.

2012 TJC P2 Q2

2012 TJC P2 Q2

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