i need to find an upper bound (as tightest as possible) of the following recurrence for $\mu^{(j)}$
$$ \left\{ \begin{array}{l} t^{(j-1)} = 2 \mu^{(j-1)} - \nu & j \geq 1\\ \mu^{(j)} = \begin{cases} t^{(j-1)} & \mbox{if } t^{(j-1)} \geq 0 \\ 2 \mu^{(j-1)} & \mbox{otherwise} \end{cases} & j\geq 1 \\ \nu \in \left( 2^{mw} , 2^{mw + 1} \right) \cap \mathbb{Z} \\ \mu^{(0)} = 2^{mw + 1} - \nu \end{array} \right.$$
You can assume $mw$ is integer and $\geq 4$.
Here's my attempt, assuming that $t^{(k)} \geq 0$ for each $k$ then
$$\mu^{(j)} = 2 \mu^{(j-1)} - \nu \Rightarrow \Delta \mu^{(j-1)} = \mu^{(j-1)} - \nu$$
Assuming i solved correctly the difference equation i derived
$$\mu^{(j)} = C \times 2^{j} + \nu$$
Where $C$ is a constant that depends from the initial condition $\mu^{(0)} = 2^{mw + 1} - \nu$.
However the bound is too large, so basically my second attempt was to find a kind of "invariant" between a step that goes between the step $t^{(j-1)} < 0$ and $t^{(j)} \geq 0$ and then use it to prove a bound.
I specifically need a bound for $j = mw + 1$
Any clue?