The diagram shows the graph of $y={{2}^{x}}-3x$. The two roots of the equation ${{2}^{x}}-3x=0$

Question

The diagram shows the graph of $y={{2}^{x}}-3x$. The two roots of the equation ${{2}^{x}}-3x=0$ are denoted by $\alpha $ and $\beta $, where $\alpha <\beta $. A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, … satisfies the recurrence relation ${{x}_{n\,+\,1}}=\frac{1}{3}\left( {{2}^{x{{\,}_{n}}}} \right)$ for $n\ge 1$. Prove algebraically that, if the sequence converges, then it converges to either $\alpha $ or $\beta $. For the case ${{x}_{1}}=4$, describe the behaviour of the sequence using a calculator. By considering ${{x}_{n+1}}-{{x}_{n}}$, prove that ${{x}_{n+1}}>{{x}_{n}}$ if ${{x}_{n}}>\beta $. Hence, justify your answer in part (ii).

Accepted Answer

(i) $L=\alpha$ or $L=\beta$. (ii) For $x_1=4$, the sequence is increasing and diverges to $+\infty$. (iii) If $x_n>\beta$, then $x_{n+1}>x_n$, confirming the divergence.

Limit of a Sequence and Behaviour of a Sequence Q2

Limit of a Sequence and Behaviour of a Sequence Q2

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