(a) Compute the values of \( u_{1} \) and \( u_{2} \) when the difference equations and initial values are as given below. (i) \( u_{n+1}=4 u_{n}\left(u_{n}-5\right), u_{0}=7 \) (ii) \( u_{n+1}=\frac{u_{n}}{3}-\frac{1}{2 u_{n}}, u_{0}=1 \).
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To find the values of \( u_{1} \) and \( u_{2} \) for the given difference equations, we will substitute the initial values \( u_{0} \) into the recurrence relations to compute each subsequent term. (i) For the equation \( u_{n+1}=4 u_{n}\left(u_{n}-5\right) \) with \( u_{0}=7 \): 1. **Calculate \( u_{1} \)**: \[ u_{1} = 4 u_{0} (u_{0} - 5) = 4 \cdot 7 (7 - 5) = 4 \cdot 7 \cdot 2 = 56 \] 2. **Calculate \( u_{2} \)**: \[ u_{2} = 4 u_{1} (u_{1} - 5) = 4 \cdot 56 (56 - 5) = 4 \cdot 56 \cdot 51 = 11424 \] Thus, for \( u_{0}=7 \), we have \( u_{1} = 56 \) and \( u_{2} = 11424 \). (ii) For the equation \( u_{n+1}=\frac{u_{n}}{3}-\frac{1}{2 u_{n}} \) with \( u_{0}=1 \): 1. **Calculate \( u_{1} \)**: \[ u_{1} = \frac{u_{0}}{3} - \frac{1}{2 u_{0}} = \frac{1}{3} - \frac{1}{2 \cdot 1} = \frac{1}{3} - \frac{1}{2} \] To get a common denominator (which is 6), we have: \[ u_{1} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} \] 2. **Calculate \( u_{2} \)**: \[ u_{2} = \frac{u_{1}}{3} - \frac{1}{2 u_{1}} = \frac{-\frac{1}{6}}{3} - \frac{1}{2 \left(-\frac{1}{6}\right)} = -\frac{1}{18} + 3 = \frac{-1 + 54}{18} = \frac{53}{18} \] Thus, for \( u_{0}=1 \), we have \( u_{1} = -\frac{1}{6} \) and \( u_{2} = \frac{53}{18} \).
