1/x(n+ 1)= f( 1/xn)

Only n iterations of f(x) are needed.

To do this, find a fixed point.

X = ax 2+2ax, and X = 0 is the only fixed point.

a= 1/2，f(x)=2x/(x+2)

N iterations of f are one.

1/an = 1/f(an- 1)= 1/a(n- 1)+ 1/2

So {1/an} is arithmetic progression.

a 1=2x/(x+2)， 1/an = 1/a 1+ 1/2 *(n- 1)

an=2x/(nx+2)

Substitution xn

1/xn = 2 * 1/ 1000/[(n- 1)* 1/ 1000+2]= 2/(n+ 1999)

xn=(n+ 1999)/2

x2007=2003