Newton-Raphson Method

Description

Algorithm

  1. Define the initial approximations.

  2. Evaluate f1f_1, f2f_2, ∂f1∂x\dfrac{\partial{f_1}}{\partial{x}}, ∂f1∂y\dfrac{\partial{f_1}}{\partial{y}}, ∂f2∂x\dfrac{\partial{f_2}}{\partial{x}}, ∂f2∂y\dfrac{\partial{f_2}}{\partial{y}}.

  3. Find Δx\Delta x and Δy\Delta y:

    Δx=[−f1(xi,yi)∂f1∂y∣xi,yi−f2(xi,yi)∂f2∂y∣xi,yi∂f1∂x∣xi,yi∂f1∂y∣xi,yi∂f2∂x∣xi,yi∂f2∂y∣xi,yi]=−f1∂f2∂y+f2∂f1∂y∂f1∂x∂f2∂y−∂f2∂x∂f1∂yΔy=[∂f1∂x∣xi,yi−f1(xi,yi)∂f2∂x∣xi,yi−f2(xi,yi)∂f1∂x∣xi,yi∂f1∂y∣xi,yi∂f2∂x∣xi,yi∂f2∂y∣xi,yi]=−f2∂f1∂x+f1∂f2∂x∂f1∂x∂f2∂y−∂f2∂x∂f1∂y\newcommand{\diff}[2]{\dfrac{\partial#1}{\partial#2}} \newcommand{\diffto}[4]{\left.\diff{#1}{#2}\right|_{#3, #4}} \begin{aligned} \Delta x & = & \left[ \begin{array}{rr} -f_1(x_i, y_i) & \diffto{f_1}{y}{x_i}{y_i} \\ -f_2(x_i, y_i) & \diffto{f_2}{y}{x_i}{y_i} \\ \hline \diffto{f_1}{x}{x_i}{y_i} & \diffto{f_1}{y}{x_i}{y_i} \\ \diffto{f_2}{x}{x_i}{y_i} & \diffto{f_2}{y}{x_i}{y_i} \end{array} \right] & = & \dfrac{-f_1\diff{f_2}{y} + f_2\diff{f_1}{y}} {\diff{f_1}{x}\diff{f_2}{y} - \diff{f_2}{x}\diff{f_1}{y}} \\ \text{} \\ \Delta y & = & \left[ \begin{array}{rr} \diffto{f_1}{x}{x_i}{y_i} & -f_1(x_i, y_i) \\ \diffto{f_2}{x}{x_i}{y_i} & -f_2(x_i, y_i) \\ \hline \diffto{f_1}{x}{x_i}{y_i} & \diffto{f_1}{y}{x_i}{y_i} \\ \diffto{f_2}{x}{x_i}{y_i} & \diffto{f_2}{y}{x_i}{y_i} \end{array} \right] & = & \dfrac{-f_2\diff{f_1}{x} + f_1\diff{f_2}{x}} {\diff{f_1}{x}\diff{f_2}{y} - \diff{f_2}{x}\diff{f_1}{y}} \end{aligned}

  4. Solve for xi+1x_{i+1} and yi+1y_{i+1} in Δx\Delta x and Δy\Delta y:

    Δx=xi+1−xiΔy=yi+1−yi\begin{aligned} \Delta x & = x_{i+1} - x_i \\ \Delta y & = y_{i+1} - y_i \\ \end{aligned}

  5. Calculate the relative error, if it’s sufficiently small, the calculation stops, otherwise, continue:

    ep=∣xi+1−xixi+1∣×100e_p = |\dfrac{x_{i + 1} - x_i}{x_{i + 1}}| \times 100

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