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Let $$ x_{n+1} = \left\{ \begin{array}{c} x_{n}^2 & \mbox{if $b_{n+1}=1$} \\ \alpha x_n & \mbox{if $b_{n+1}=0$}\end{array} \right. , $$ for $n\ge 0$ and $\alpha > 1$. Can we write $x_k$ in terms of $x_0$ and $P$ where $$ P = \sum_{i=1}^{k} b_i. $$ For $\alpha=1$, $$ x_k = x_0^{\left( 2^P \right)}. $$ I am unable to figure out the expression for any $\alpha>1$.

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I'm sure $x_k$ depends on the order of the $b_i$s if $\alpha^2 \ne 1$ – Cocopuffs Jun 23 '12 at 17:26
exactly! and that's why I thought that $P$ alone may not be sufficient to describe it. – ubaabd Jun 23 '12 at 17:28
What's the connection between $x_n$ and $b_n$? – Mercy King Jun 23 '12 at 17:42
they are independent – ubaabd Jun 23 '12 at 17:46
what is $b_n$ then? – Mercy King Jun 23 '12 at 18:10
up vote 1 down vote accepted

No. Let $\alpha > 1$ and $x_0 \ne 0$. For $b_1 = 0, b_2 = 1$ you have $x_2 = \alpha^2 x_0^2$ and for $b_1 = 1, b_2 = 0$ you have $x_2 = \alpha x_0^2$. These aren't equal, so $x_2$ doesn't depend on $x_0$ and $P = b_1 + b_2$ alone.

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