Let S be the set of points (x, y) in the plane such that the sequence a_{n} defined by a_{0} = x, a_{n+1} = (a_{n}^{2} + y^{2})/2 converges. What is the area of S?

**Solution**

Answer: 4 + π.

We claim that:

(1) the sequence diverges for |y| > 1;

(2) the sequence diverges for |y| < 1 and |x| > 1 + √(1-y^{2});

(3) the sequence converges for |y| < 1 and |x| < 1 + √(1-y^{2}).

It follows that the area S is the square with sides x = ±1, y = ±1 with a hemispherical cap on each of the vertical sides x = ±1, and hence that the area is 4 + π.

Notice that a_{n+1} - a_{n} = ( (a_{n} - 1)^{2} - (1 - y^{2} ) )/2 (*). So if |y| > 1, then a_{n+1} - a_{n} > (y^{2} - 1)/2 > 0, which establishes (1).

In case (2), take ε > 0, such that |x| > 1 + √(1 - y^{2}) + ε. Then (x - 1)^{2} > 1 - y^{2} + ε^{2}, so a_{1} - x > ε^{2}/2 (using (*) ). Hence a_{1} > x > 1 + √(1 - y^{2}) + ε. So by a simple induction a_{n+1} - a_{n} > ε^{2}/2 for all n, which establishes (2).

In case (3), assume first that x is positive. x < 1 + √(1-y^{2}) implies that (x - 1)^{2} < 1 - y^{2}, so by (*) a_{1} < x. So by a simple induction a_{n+1} < a_{n} for all x. But all a_{n} are positive, so convergence follows. In the case x negative the sequence is the same for all later terms as that started by -x, so we have convergence in this case also.

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001