Understanding Engineering Mathematics

wherexn is thenth approximation. Starting with a first approximation x 1 we find
another by:

x 2 =F(x 1 )

and another

x 3 =F(x 2 )

and so on.
This generates a sequence of approximate solutions:

x 1 ,x 2 ,x 3 ,...,xn,...

and under favourable circumstances the limit of this sequence is the solution of the equa-
tion. To get a good enough approximation we then takenlarge enough. Clearly, whether
this is possible will depend on the form of the functionF.
Perhaps the most popular iteration method isNewton’s method. In this method the
(n+ 1 )th approximation,xn+ 1 , to the solution of an equationf(x)=0 is calculated from
thenth approximationxn, subject to certain conditions, using the relation

xn+ 1 =xn−

f(xn)

f′(xn)

Again we generate an infinite sequence of approximate solutions that (we hope) will
eventually converge on the actual solution.

Exercise on 14.8

Rewriting the equationx^3 − 5 x− 3 =0intheform

x=^15 (x^3 − 3 )

and starting with the valuex 0 =−1, use your calculator to generate a sequence of approx-
imate solutions to the equation. Do you think this sequence converges to a solution? What
happens if you tryx 0 =−2?
Now rewrite the equation in the form

x=

5 x+ 3

x^2

and again start withx 0 =−2. What happens?

Answer

Starting withx 0 =−1 the first rearrangement leads quickly to the approximationx=
− 0 .657. Starting withx 0 =−2 leads to a divergent sequence and no solution. Starting with
x 0 =−2 in the second rearrangement however leads quickly to an approximate solution
x=− 1 .83. Note that some other starting valuesx 0 might not lead to a root at all.