(b) For the following systems, find the equilibrium point(s), if they exist: (i) \( P_{n+1}=\sqrt{\frac{2 P_{n}+\frac{1}{P_{n}}}{3}} \), where \( P_{n} \neq 0 \) (ii) \( P_{n+1}=\sqrt{\sqrt{\sqrt{256 P_{n}}}} \).

To determine the equilibrium points for each system, we set \( P_{n+1} = P_n = P \) and solve for \( P \). ### (i) \( P_{n+1} = \sqrt{\frac{2P_n + \frac{1}{P_n}}{3}} \), where \( P_n \neq 0 \) Set \( P_{n+1} = P \): \[ P = \sqrt{\frac{2P + \frac{1}{P}}{3}} \] Square both sides to eliminate the square root: \[ P^2 = \frac{2P + \frac{1}{P}}{3} \] Multiply both sides by 3P to clear the denominator: \[ 3P^3 = 2P^2 + 1 \] Rearrange the equation: \[ 3P^3 - 2P^2 - 1 = 0 \] To find the real solutions, we can test possible rational roots (factors of the constant term divided by factors of the leading coefficient). Testing \( P = 1 \): \[ 3(1)^3 - 2(1)^2 - 1 = 3 - 2 - 1 = 0 \] Thus, \( P = 1 \) is a real root. Factoring out \( (P - 1) \): \[ 3P^3 - 2P^2 - 1 = (P - 1)(3P^2 + P + 1) \] The quadratic equation \( 3P^2 + P + 1 = 0 \) has no real solutions (discriminant \( 1 - 12 = -11 < 0 \)). Therefore, the only real equilibrium point is: \[ P = 1 \] ### (ii) \( P_{n+1} = \sqrt{\sqrt{\sqrt{256P_n}}} \) Interpreting the nested square roots, we have: \[ P_{n+1} = (256P_n)^{1/8} \] Set \( P_{n+1} = P \): \[ P = (256P)^{1/8} \] Raise both sides to the 8th power: \[ P^8 = 256P \] Assuming \( P \neq 0 \), divide both sides by \( P \): \[ P^7 = 256 \] Thus, \[ P = 256^{1/7} \quad \text{or} \quad P = 2^{8/7} \] Additionally, \( P = 0 \) is also a valid equilibrium point since: \[ P_{n+1} = \sqrt{\sqrt{\sqrt{256 \cdot 0}}} = 0 \] **Final Answer:** (i) The only real equilibrium is P = 1; (ii) All equilibria are P = 0 and P equal to the seventh root of 256. Thus, • P = 0 • P = 256¹⁄⁷