(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) \( P_{n+1}=\sqrt{\sqrt{\sqrt{256 P_{n}}}} \)
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Let's find the equilibrium points of the given systems. (i) To find the equilibrium point for the equation \( P_{n+1} = \sqrt{\frac{2 P_n + \frac{1}{P_n}}{3}} \), we set \( P_{n+1} = P_n = P \). This gives us the equation: \[ P = \sqrt{\frac{2P + \frac{1}{P}}{3}} \] Squaring both sides, we have: \[ P^2 = \frac{2P + \frac{1}{P}}{3} \] Multiplying through by 3 to clear the fraction: \[ 3P^2 = 2P + \frac{1}{P} \] Rearranging gives: \[ 3P^3 - 2P^2 - 1 = 0 \] This is a cubic equation. To solve it, you can use numerical methods or the Rational Root Theorem to find possible rational roots. However, we can also apply a numerical method or software to find roots. It appears there is one real solution, which can be approximated. (ii) For the second equation \( P_{n+1} = \sqrt{\sqrt{\sqrt{256 P_n}}} \), we set \( P_{n+1} = P_n = P \): \[ P = \sqrt{\sqrt{\sqrt{256 P}}} \] This simplifies to: \[ P = (256P)^{1/8} \] Raising both sides to the power of 8 yields: \[ P^8 = 256P \] Since \( 256 = 2^8 \), we can change the equation to: \[ P^8 - 256P = 0 \] Factoring out P gives: \[ P(P^7 - 256) = 0 \] This gives two solutions: 1. \( P = 0 \) 2. \( P^7 = 256 \) ⇒ \( P = 2 \) Thus, the equilibrium points are \( P = 0 \) and \( P = 2 \).
