Let’s address each part of the problem systematically:

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(a) What is $Q(3)$?

The given dynamical system is: $Q_{t+1} = 9Q_t + 8$ with the initial condition $Q_0 = 5$.

We compute $Q_1$, $Q_2$, and $Q_3$ step by step:

1. Step 1: $Q_1 = 9Q_0 + 8 = 9(5) + 8 = 45 + 8 = 53$.
2. Step 2: $Q_2 = 9Q_1 + 8 = 9(53) + 8 = 477 + 8 = 485$.
3. Step 3: $Q_3 = 9Q_2 + 8 = 9(485) + 8 = 4365 + 8 = 4373$.

Thus, $Q(3) = 4373$.

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(b) Backwards Discrete-Time Dynamical System, $Q_t = f(Q_{t+1})$

From the forward equation: $Q_{t+1} = 9Q_t + 8,$ we solve for $Q_t$ in terms of $Q_{t+1}$: $Q_t = \frac{Q_{t+1} - 8}{9}.$

This is the backward discrete-time dynamical system: $Q_t = \frac{Q_{t+1} - 8}{9}.$

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(c) Equilibrium Solution $Q^*$

At equilibrium, the value of $Q_t$ remains constant, meaning $Q_{t+1} = Q_t = Q^*$. Substitute $Q^*$ into the equation: $Q^* = 9Q^* + 8.$ Rearrange: $0 = 8 + 8Q^* \implies Q^* = -1.$

Thus, the equilibrium solution is: $Q^* = -1.$

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(d) Two-Step Dynamical System, $Q_{t+2} = f(Q_t)$

Start from the forward equation: $Q_{t+1} = 9Q_t + 8.$ Substitute $Q_{t+1}$ into another instance of the same equation: $Q_{t+2} = 9Q_{t+1} + 8 = 9(9Q_t + 8) + 8.$ Simplify: $Q_{t+2} = 81Q_t + 72 + 8 = 81Q_t + 80.$

Thus, the two-step dynamical system is: $Q_{t+2} = 81Q_t + 80.$

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Summary of Results

1. $Q(3) = 4373$.
2. Backward system: $Q_t = \frac{Q_{t+1} - 8}{9}$.
3. Equilibrium solution: $Q^* = -1$.
4. Two-step system: $Q_{t+2} = 81Q_t + 80$.

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Do you have any questions or want further details? Here are 5 follow-up questions:

1. How does the equilibrium solution relate to the stability of the system?
2. Can we derive a general formula for $Q_t$ in terms of $t$?
3. What happens if the initial condition $Q_0$ changes?
4. Is the system stable, and how can we analyze its long-term behavior?
5. How does the two-step system's growth rate compare to the one-step system?

Tip: For dynamical systems, always verify stability by checking how solutions behave as $t \to \infty$.