Word Count : 482

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We are given the iteration method: $$ x_{n+1} = \frac{1}{8}\Bigg[6x_n + \frac{3N}{x_n} - \frac{x_n^3}{N}\Bigg], \quad n=0,1,2,\dots $$ We are to find: 1. The quantity to which it converges. 2. The rate of convergence. --- ### Step 1: Determine the quantity to which the iteration converges Let the iteration converge to a fixed point $\alpha$. Then: $$ x_{n+1} = x_n = \alpha $$ So, $$ \alpha = \frac{1}{8} \Bigg[ 6\alpha + \frac{3N}{\alpha} - \frac{\alpha^3}{N} \Bigg] $$ Multiply throughout by 8: $$ 8\alpha = 6\alpha + \frac{3N}{\alpha} - \frac{\alpha^3}{N} $$ Simplify: $$ 2\alpha = \frac{3N}{\alpha} - \frac{\alpha^3}{N} $$ ______ _______ _____ _________ ____ ___ __________.
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