Let S be the set of points (x, y) in the plane such that the sequence an defined by a0 = x, an+1 = (an2 + y2)/2 converges. What is the area of S?
Answer: 4 + π.
We claim that:
(1) the sequence diverges for |y| > 1;
(2) the sequence diverges for |y| < 1 and |x| > 1 + √(1-y2);
(3) the sequence converges for |y| < 1 and |x| < 1 + √(1-y2).
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 an+1 - an = ( (an - 1)2 - (1 - y2 ) )/2 (*). So if |y| > 1, then an+1 - an > (y2 - 1)/2 > 0, which establishes (1).
In case (2), take ε > 0, such that |x| > 1 + √(1 - y2) + ε. Then (x - 1)2 > 1 - y2 + ε2, so a1 - x > ε2/2 (using (*) ). Hence a1 > x > 1 + √(1 - y2) + ε. So by a simple induction an+1 - an > ε2/2 for all n, which establishes (2).
In case (3), assume first that x is positive. x < 1 + √(1-y2) implies that (x - 1)2 < 1 - y2, so by (*) a1 < x. So by a simple induction an+1 < an for all x. But all an 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.
53rd Putnam 1992
© John Scholes
7 Jan 2001