The equation x3−x2+3=0x^3 - x^2 + 3 = 0x3−x2+3=0 has exactly one real root.
Show that, for this equation, the Newton-Raphson formula can be written
xn+1=2xn3−xn2−33xn2−2xn x_{n+1} = \frac{2x_n^3 - x_n^2 - 3}{3x_n^2 - 2x_n} xn+1=3xn2−2xn2xn3−xn2−3Using the formula given in part (a) with x1=−1x_1 = -1x1=−1, find the values of x2 x_2\,x2 and x3x_3x3
Explain why, for this question, the Newton-Raphson method cannot be used with x1=0x_1 = 0x1=0
329 exam-style questions on Edexcel A Level Maths Numerical Methods, covering 10.1 Locating Roots, 10.2 Iteration, 10.3 The Newton-Raphson Method, and 10.4 Applications to Modelling. Each one has a worked solution and a mark scheme showing where the marks go.