Let's solve each part step by step with explanations. --- ## a) What will the density be after four years? ### **Step 1: Identify the recursive relation and initial condition** The system is: \[ x_{t+1} = \frac{7x_t}{1 + 3x_t^2} \] with \( x_ = .9 \). ### **Step 2: Calculate \( x_1, x_2, x_3, x_4 \)** - \( x_1 = \frac{7 \times .9}{1 + 3 \times (.9)^2} \) - \( x_2 = \frac{7 \times x_1}{1 + 3 x_1^2} \) - \( x_3 = \frac{7 \times x_2}{1 + 3 x_2^2} \) - \( x_4 = \frac{7 \times x_3}{1 + 3 x_3^2} \) Let's do the calculations one by one: #### \( x_1 \): \[ x_1 = \frac{7 \times .9}{1 + 3 \times (.9)^2} = \frac{6.3}{1 + 3 \times .81} = \frac{6.3}{1 + 2.43} = \frac{6.3}{3.43} \approx 1.837 \] #### \( x_2 \): \[ x_2 = \frac{7 \times 1.837}{1 + 3 \times (1.837)^2} = \frac{12.859}{1 + 3 \times 3.374} = \frac{12.859}{1 + 10.122} = \frac{12.859}{11.122} \approx 1.156 \] #### \( x_3 \): \[ x_3 = \frac{7 \times 1.156}{1 + 3 \times (1.156)^2} = \frac{8.092}{1 + 3 \times 1.336} = \frac{8.092}{1 + 4.008} = \frac{8.092}{5.008} \approx 1.617 \] #### \( x_4 \): \[ x_4 = \frac{7 \times 1.617}{1 + 3 \times (1.617)^2} = \frac{11.319}{1 + 3 \times 2.616} = \frac{11.319}{1 + 7.848} = \frac{11.319}{8.848} \approx 1.279 \] **Explanation:** Each value is calculated by plugging the previous value into the recurrence relation. This tracks the density year by year. **Final answer:** \[ x_4 \approx 1.279 \] --- ## b) What is the updating function \( f(x) \)? ### **Step 1: Identify the function from the recurrence** From the recurrence, \[ x_{t+1} = \frac{7x_t}{1 + 3x_t^2} \] So, \[ f(x) = \frac{7x}{1 + 3x^2} \] **Explanation:** The updating function takes the current value and gives the next value. **Final answer:** \[ f(x) = \frac{7x}{1 + 3x^2} \] --- If you'd like to see the rest of the answers (c, d, e), just let me know!

## c) What are the biologically relevant equilibria? ### **Step 1: Find equilibria by solving \( f(x) = x \)** To find the equilibria, set: \[ f(x) = x \implies \frac{7x}{1 + 3x^2} = x \] ### **Step 2: Rearrange and solve** Rearranging gives: \[ 7x = x(1 + 3x^2) \] \[ 7x = x + 3x^3 \] \[ 3x^3 - 6x = 0 \] Factoring out \( x \): \[ x(3x^2 - 6) = 0 \] Thus, we have: \[ x = 0 \quad \text{or} \quad 3x^2 - 6 = 0 \implies x^2 = 2 \implies x = \sqrt{2} \text{ or } x = -\sqrt{2} \] **Biologically relevant equilibria are**: - \( x = 0 \) - \( x = \sqrt{2} \) (since density cannot be negative) **Explanation:** Equilibria represent states where the density does not change. We solved for \( x \) where the updating function equals the current density. **Final answer:** \[ 0; \sqrt{2} \] --- ## d) Compute \( f(z) \). ### **Step 1: Use the previously determined function** We already found: \[ f(x) = \frac{7x}{1 + 3x^2} \] ### **Step 2: Confirm the calculation** To compute \( f(z) \), substitute \( z \) into the function: \[ f(z) = \frac{7z}{1 + 3z^2} \] **Explanation:** The function \( f(z) \) remains the same as previously derived. It maps the current density \( z \) to the next density. **Final answer:** \[ f(z) = \frac{7z}{1 + 3z^2} \] --- ## e) Use the derivative test to determine the stability properties for each of the two equilibria. ### **Step 1: Compute the derivative \( f'(x) \)** Using the quotient rule: \[ f'(x) = \frac{(1 + 3x^2)(7) - 7x(6x)}{(1 + 3x^2)^2} = \frac{7 + 21x^2 - 42x^2}{(1 + 3x^2)^2} = \frac{7 - 21x^2}{(1 + 3x^2)^2} \] ### **Step 2: Evaluate \( f'(x) \) at equilibria** 1. At \( x = 0 \): \[ f'(0) = \frac{7 - 21(0)^2}{(1 + 3(0)^2)^2} = 7 \] 2. At \( x = \sqrt{2} \): \[ f'(\sqrt{2}) = \frac{7 - 21(2)}{(1 + 3(2))^2} = \frac{7 - 42}{(1 + 6)^2} = \frac{-35}{49} = -\frac{5}{7} \] **Stability analysis:** - If \( |f'(p)| < 1 \): stable - If \( |f'(p)| > 1 \): unstable #### Stability for \( x = 0 \): - \( f'(0) = 7 \) (unstable) #### Stability for \( x = \sqrt{2} \): - \( f'(\sqrt{2}) = -\frac{5}{7} \) (stable) **Explanation:** The derivative indicates how the density responds to small changes. If the absolute value is greater than 1, small perturbations grow, indicating instability. **Final answers:** - For \( x = 0 \): \( f'(0) = 7 \) (unstable) - For \( x = \sqrt{2} \): \( f'(\sqrt{2}) = -\frac{5}{7} \) (stable)