Topic 10.1: Bisection Method (Examples)
ece.uwaterloo.ca › bisection › examplesExample 1. Consider finding the root of f ( x) = x2 - 3. Let ε step = 0.01, ε abs = 0.01 and start with the interval [1, 2]. Table 1. Bisection method applied to f ( x ) = x2 - 3. Thus, with the seventh iteration, we note that the final interval, [1.7266, 1.7344], has a width less than 0.01 and |f (1.7344)| < 0.01, and therefore we chose b ...
Solution of Nonlinear Equations - uap-bd.edu
www.uap-bd.edu › ce › anamBisection Method The basic principle of this method is the theorem that ‘If f(x) is continuous in an interval x 1 ≤ x 3 ≤ x 2 and if f(x 1) and f(x 2) are of opposite signs, then f(x 3) = 0 for at least one number x 3 such that x 1 x 3 x 2’. This method is implemented using the following steps 1. Assume two values of x (i.e., x 1 and x ...