Proof

Choose such that on , is continuous and strictly bounded by and that with for all values in the interval .

This then means that the successive values in the sequence must lie in the interval as well. Moreover, the sequence then converges to the unique fixed point.

When we expand in a Taylor series about ,

> taylor(g(x),x=p,3);

the remainder term can be written as

> (1/2)*D(D(g))(eta)*(x-p)^2;

where lies between and . Using and , we find

> eq:=g(x)=p+(1/2)*D(D(g))(eta)*(x-p)^2;

or, when evaluated in , bearing in mind that ,

> p[n+1]=p+(1/2)*D(D(g))(eta[n])*(p[n]-p)^2;

with between and . Thus

> expr:=p[n+1]-p;

> expr=(1/2)*D(D(g))(eta[n])*(p[n]-p)^2;

and hence

> limit((abs(p[n+1]-p))/abs(p[n]-p)^2,n=infinity)=abs(D(D(g))(p))/2;

So the sequence is quadratically convergent. Since is strictly bounded by M, we also have that, for sufficiently large values of ,

> expr<(M/2)*(p[n]-p)^2;