A textbook of Computer Based Numerical and Statiscal Techniques part 6. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 36 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Then we bisect the interval and continue the process till the root is found to be desired accuracy. In the above figure f x1 is positive therefore the root lies in between a and x1. The second approximation to the root now is x2 a x1 . If f x2 is negative as shown in the figure then the root lies in between x2 and xv and the third approximation to the root is x3 x2 x1 2 and so on. This method is simple but slowly convergent. It is also called as Bolzano method or Interval halving method. Procedure for the Bisection Method to Find the Root of the Equation f x 0 Step 1 Choose two initial guess values approximation a and b where a b such that f a . f b 0. Step 2 Evaluate the mid point x1 of a and b given by x1 a b and also evaluate f x1 . Step 3 If f a . f x1 0 then set b x1 else set a x1. Then apply the formula of step 2. Step 4 Stop evaluation when the difference of two successive values of x1 obtained from step 2 is numerically less than the prescribed accuracy. Order of Convergence of Bisection Method In Bisection Method the original interval is divided into half interval in each iteration. If we take mid points of successive intervals to be the approximations of the root one half of the current interval is the upper bound to the error. In Bisection Method e 1 or 1 ei ALGEBRAIC AND TRANSCENDENTAL EQUATION 37 Here e and e 1 are the errors in ith and i 1 th iterations respectively. Comparing the above equation with lim i - ei 1 e i A We get k 1 and A . Thus the Bisection Method is first order convergent or linearly convergent. Example 1. Find the root of the equation x3 - x - 1 0 lying between 1 and 2 by bisection method. Sol. Let f x x3 - x - 1 0 Since f 1 13 - 1 - 1 - 1 which is negative and f 2 23 - 2 - 1 5 which is positive Therefore f 1 is negative and f 2 is positive so at least one real root will lie between 1 and 2. First iteration Now using Bisection Method we can take first .
