Slide 1

Roots of PolynomialsTo find the roots of polynomial equations of the general form:fn(x) = a0 + a1x + a2x2 + +anxn

Where n = the order of the polynomial and the as = constant coefficients. Although the coefficients can be complex numbers, we will limit our discussion to cases where they are real. For such cases, the roots can be real and/or complex.The roots of such polynomials follow these rules:For an nth-order equation, there are n real or complex roots. It should be noted that these roots will not necessarily be distinct. If n is odd, there is at least one real root.If complex roots exist, they exist in conjugate pairs (that is, + i and i), where i = 1.

Mullers MethodMllers method takes a similar approach, butprojects a parabola through three points.

The method consists of deriving the coefficients of the parabola that goes through the three points. These coefficients can then be substituted into the quadratic formula to obtain the point where the parabola intercepts the x axisthat is, the root estimate. The approach is facilitated by writing the parabolic equation in a convenient form,

f2(x) = a (x x2)2 + b (x x2) + c

The coefficients of equation above can be evaluated by substituting each of the three points to givef(x0) = a(x0 x2)2 + b(x0 x2) + c f(x1) = a(x1 x2)2 + b(x1 x2) + c f(x2) = a(x2 x2)2 + b(x2 x2) + c

Because we have three equations, we can solve for the three unknown coefficients, a, b, and c.

c = f (x2)f(x0) f(x2) = a(x0 x2)2 + b(x0 x2) f(x1) f(x2) = a(x1 x2)2 + b(x1 x2) Algebraic manipulation can then be used to solve for the remaining coefficients, a and b. One way to do this involves defining a number of differences,

h0 = x1 x0 h1 = x2 x1 f(x1)-f(x0) x1-x0

0 = f(x2)-f(x1) x2-x1

1 = Which can be solved for a, b and c. The results can be summarized as :1 0 x1-x0a=ah1 + 1b= c= f(x2)To find the root, we apply the quadratic formula. However, because of potential round-off error, rather than using the conventional form, we use the alternative formulation -2cbb2-4acx3 = x2 +Note that the use of the quadratic formula means that both real and complex roots can be located. This is a major benefit of the method.the error can be calculated as: Ea = X 100 %x3 - x2 x3 Once x3 is determined, the process is repeated. This brings up the issue of which point is discarded. Two general strategies are typically used:

1. If only real roots are being located, we choose the two original points that are nearest the new root estimate, x3.

2. If both real and complex roots are being evaluated, a sequential approach is employed.That is, just like the secant method, x1, x2, and x3 take the place of x0, x1, and x2.Bairstows MethodBairstows method is an iterative approach related loosely to both the Mller and Newton- Raphson methods. Before launching into a mathematical description of the technique, recall the factored form of the polynomial,fn(x) = a0 + a1x + a2x2 + +anxnCan be divided by the factor x t to yield a second polynomial that is one order lower,fn1(x) = b1 + b2x ++b3xn2+bnxn1with a remainder R = b0, where the coefficients can be calculated by the recurrence relationshipbn = anbi = ai + bi+1t for i = n 1 to 0Note that if t were a root of the original polynomial, the remainder b0 would equal zero.To permit the evaluation of complex roots, Bairstows method divides the polynomial by a quadratic factor x2 rx s. The result is a new polynomialfn2(x) = b2 + b3x ++bn-1xn-3 + bnxn2with a remainder:R = b1(x r ) + b0 As with normal synthetic division, a simple recurrence relationship can be used to perform the division by the quadratic factor:

bn = an

bn1 = an1 + rbn

bi = ai + rbi+1 + sbi+2 for i = n 2 to 0

If x2 rx s is an exact divisor of the polynomial, complex roots can be determined by the quadratic formula.

Thus, the method reduces to determining the values of r and s that make the quadratic factor an exact divisor. In other words, we seek the values that make the remainder term equal to zero.Bairstows method uses a strategy similar to the Newton- Raphson approach. Because both b0 and b1 are functions of both r and s, they can be expanded using a Taylor series b1(r + r,s + s) = b1 + b1r

r +b1s

sb0(r + r,s + s) = b0 + b0r

r +b0s

swhere the values on the right-hand side are all evaluated at r and sThe changes, r and s , needed to improve our guesses can be estimated by setting equal to zero to give

b1r

r + b1r

s = - b1b0r

r + b0r

s = - b0If the partial derivatives of the bs can be determined, these are a system of two equations that can be solved simultaneously for the two unknowns, r and s.

Bairstow showed that the partial derivatives can be obtained by a synthetic division of the bs in a fashion similar to the way in which the bs themselves were derived:cn = bn

cn1 = bn1 + rcn

ci = bi + rci+1 + sci+2 for i = n 2 to 1Where:b0 r

b0 s

b1 r

b1 s

= c3= = c2= c1Thus, the partial derivatives are obtained by synthetic division of the bs.22Then the partial derivatives can be substituted and yieldc2 r + c3 s = b1

c1 r + c2 s = b0These equations can be solved for r and s, which can in turn be employed to improve the initial guesses of r and s. At each step, an approximate error in r and s can be estimated, as in Ea,r = X 100 %r r Ea,s = X 100 %s sWhen both of these error estimates fall below a prespecified stopping criterion s, the values of the roots can be determined by r2+4s2 r

x = At this point, three possibilities exist:2. The quotient is a quadratic. 3. The quotient is a first-order polynomial. For this case, the remaining single root can be evaluated simply as x =

Sr1. The quotient is a third-order polynomial or greater. For this case, Bairstows method would be applied to the quotient to evaluate new values for r and s. The previous values of r and s can serve as the starting guesses for this application.