Mathematical Methods for Physics and Engineering - Matematica.NET

27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONSroot by ξ and the values of successive approximations by x 1 , x 2 , ..., x n , ... . Then,for any particular method to be successful,lim x n = ξ, where f(ξ) =0. (27.2)n→∞However, success as defined here is not the only criterion. Since, in practice,only a finite number of iterations will be possible, it is important that the valuesof x n be close to that of ξ for all n>N,whereN is a relatively low number;exactly how low it is naturally depends on the computing resources available andthe accuracy required in the final answer.So that the reader may assess the progress of the calculations that follow, werecord that to nine significant figures the real root of equation (27.1) has thevalueξ =1.495 106 40. (27.3)We now consider in turn four methods for determining the value of this root.27.1.1 Rearrangement of the equationIf equation (27.1), f(x) = 0, can be recast into the formx = φ(x), (27.4)where φ(x) isaslowly varying function of x, then an iteration schemex n+1 = φ(x n ) (27.5)will often produce a fair approximation to the root ξ after a few iterations,as follows. Clearly, ξ = φ(ξ), since f(ξ) = 0; thus, when x n is close to ξ,the next approximation, x n+1 , will differ little from x n , the actual size of thedifference giving an order-of-magnitude indication of the inaccuracy in x n+1(when compared with ξ).In the present case, the equation can be writtenx =(2x 2 +3) 1/5 . (27.6)Because of the presence of the one-fifth power, the RHS is rather insensitiveto the value of x used to compute it, and so the form (27.6) fits the generalrequirements for the method to work satisfactorily. It remains only to choose astarting approximation. It is easy to see from figure 27.1 that the value x =1.5would be a good starting point, but, so that the behaviour of the procedure atvalues some way from the actual root can be studied, we will make a poorerchoice, x 1 =1.7.With this starting value and the general recurrence relationshipx n+1 =(2x 2 n +3) 1/5 , (27.7)987