27.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS
root byξand the values of successive approximations byx 1 ,x 2 ,...,xn,.... Then,
for any particular method to be successful,
lim
n→∞xn=ξ, wheref(ξ)=0. (27.2)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 values
ofxnbe close to that ofξfor alln>N,whereNis a relatively low number;
exactly how low it is naturally depends on the computing resources available and
the 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 the
value
ξ=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 form
x=φ(x), (27.4)whereφ(x)isaslowlyvarying function ofx, then an iteration scheme
xn+1=φ(xn) (27.5)will often produce a fair approximation to the rootξafter a few iterations,
as follows. Clearly,ξ=φ(ξ), sincef(ξ) = 0; thus, whenxnis close toξ,
the next approximation,xn+1, will differ little fromxn, the actual size of the
difference giving an order-of-magnitude indication of the inaccuracy inxn+1
(when compared withξ).
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 insensitive
to the value ofxused to compute it, and so the form (27.6) fits the general
requirements for the method to work satisfactorily. It remains only to choose a
starting approximation. It is easy to see from figure 27.1 that the valuex=1. 5
would be a good starting point, but, so that the behaviour of the procedure at
values some way from the actual root can be studied, we will make a poorer
choice,x 1 =1.7.
With this starting value and the general recurrence relationshipxn+1=(2x^2 n+3)^1 /^5 , (27.7)