The iterative method to solve f(x)=0, given by the fixed-point method g(x)=x, where pn=g(pn-1)=pn-1-(f(pn-1))/(f^'(pn-1))-(f^''(pn-1))/(2 f^'(pn-1))[(f(pn-1))/(f^'(pn-1))]^2, for n=1,2,3, …has g^'(p)=g^''(p)=0. This will generally yield cubic ( α=3 ) convergence. Expand the analysis of Example 1 to compare quadratic and cubic convergence. | Numerade

High-Low Method Accounting | Definition, Formula & Examples

SOLVED: Consider the following fixed-point iteration algorithm: Xi 1 = g(xi) = Xi 1. From a graphical solution of x = g(x), you know that X = 2.24 is approximately the fixed point of x = g(x).If you start the fixed-point algorithm with Xo = 2.4, how many iterations do you need to perform at least in order to reach an error less than 10^-7?Use your knowledge about the order of convergence of the fixed-point method to conduct your analysis.Answer: