The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 7 Roots of Polynomials

The roots of polynomials such as Follow these rules: 1. For an n th order equation, there are n real or complex roots. 2. If n is odd, there is at least one real root. 3. If complex root exist in conjugate pairs (that is, ( +  i and -  i ), where.

The efficiency of bracketing and open methods depends on whether the problem being solved involves complex roots. If only real roots exist, these methods could be used. Finding good initial guesses complicates both the open and bracketing methods, also the open methods could be susceptible to divergence. Conventional Methods

Special methods have been developed to find the real and complex roots of polynomials: - Müller method - Bairstow methods Conventional Methods

Müller’s method obtains a root estimate by projecting a parabola to the x axis through three function values. Roots of Polynomials: Müller’s Method Secant Method Muller’s Method

The method consists of deriving the coefficients of parabola that goes through the three points: 1. Write the equation in a convenient form:

Muller’s Method 2. The parabola should intersect the three points [x o, f(x o )], [x 1, f(x 1 )], [x 2, f(x 2 )]. The coefficients of the polynomial can be estimated by substituting three points to give

Muller’s Method 3. Three equations can be solved for three unknowns, a, b, c. Since two of the terms in the 3 rd equation are zero, it can be immediately solved for c = f(x 2 ).

Muller’s Method Solving the above equations

Muller’s Method Roots can be found by applying an alternative form of quadratic formula: The error can be calculated as ± term yields two roots. This will result in a largest denominator, and will give root estimate that is closest to x 2.

Muller’s Method: Example Use Muller’s method to find roots of f(x)= x x - 12 Initial guesses of x 0, x 1, and x 2 of 4.5, 5.5 and 5.0 respectively. (Roots are -3, -1 and 4) Solution - f(x o )= f(4.5)=20.626, - f(x 1 )= f(5.5)= and, - f(x 2 )= f(5)= h o = = 1, h 1 = = -0.5  o = ( ) /( ) =  1 = ( )/ (5-5.5) = 69.75

Muller’s Method: Example - a = ( )/(-0.5+1) = 15 - b =15(-0.5) = c = 48 ±(b 2 -4ac) 0.5 = ± Choose sign similar to the sign of b (+ve) x 3 = 5 + (-2)(48)/( ) = The error estimate is  t =|( )/( )|*100 = 25.7% The second iteration will have x 0 =5.5 x 1 =5 and x 2 =

Müller’s Method: Example Iteration x r Error %