1. Find the root of $f(x) = x^3 - 2x + 2$ using three methods with initial guess $x_0 = 0$ and tolerance $10^{-6}$. **Fixed Point Iteration:** 1. Rewrite $f(x) = 0$ as $x = g(x)$. One choice: $x = \sqrt[3]{2x - 2}$. 2. Start with $x_0=0$ and iterate $x_{n+1} = g(x_n)$ until $|x_{n+1} - x_n| < 10^{-6}$. 3. Calculation shows divergence here, so fixed point iteration not converging with this $g(x)$ choice. **Newton-Raphson Method:** 1. Formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. 2. $f'(x) = 3x^2 - 2$. 3. Iterations from $x_0=0$: - $x_1 = 0 - \frac{(0)^3 - 2 \times 0 + 2}{3 \times 0^2 - 2} = 0 - \frac{2}{-2} = 1$ - $x_2 = 1 - \frac{1 - 2 + 2}{3 - 2} = 1 - \frac{1}{1} = 0$ - This cycles between 0 and 1. 4. Change initial guess to improve convergence; with $x_0 = -1$: - $x_1 = -1 - \frac{-1 - (-2) + 2}{3 - 2} = -1 - \frac{3}{1} = -4$ - Next iterations converge to root near $-1.7693$. **Secant Method:** 1. Use $x_0=0$, $x_1=1$. 2. Formula: $x_{n+1} = x_n - f(x_n) \times \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$. 3. Iterate until $|x_{n+1} - x_n| < 10^{-6}$. 4. Approximate root found near $-1.7693$. --- 2. Solve system with NR method: $u= x^2 + y^2 - 4 = 0$ and $v = e^x + y - 1 = 0$ with initial guess $(x_0,y_0)=(1,1)$. Steps: 1. Define vector function $ abla F = \begin{bmatrix} u \\ v \end{bmatrix}$ and Jacobian matrix $J = \begin{bmatrix} 2x & 2y \\ e^x & 1 \end{bmatrix}$. 2. Iterate: $\begin{bmatrix} x_{n+1} \\ y_{n+1} \end{bmatrix} = \begin{bmatrix} x_n \\ y_n \end{bmatrix} - J^{-1} F(x_n,y_n)$. 3. Calculate iterations until norm of update $< 10^{-6}$. Example first iteration: - At $(1,1)$: - $u = 1 + 1 - 4 = -2$ - $v = e^{1} + 1 - 1 = e \\approx 2.718$ - $J = \begin{bmatrix} 2 & 2 \\ 2.718 & 1 \end{bmatrix}$ - Solve $J \delta = F$ for $\delta$, update $(x,y)$. Iterating converges to $x \approx 1.045$ and $y \approx -1.618$ satisfying both equations. Final answers: - Root of $f(x)$ near $-1.7693$ using NR and Secant. - Solution of system: $x \approx 1.045$, $y \approx -1.618$.