The curve of q^2 = p – 1/p, with q as the dependent variable consists of a parabolalike figure with it's axis being the p axis (independent variable axis) and vertex at (1,0) in union with a curve going through (1,0) asymptotically approaching the q axis towards +infinity and also asymptotically approaching the q axis towards infinity.
The curve q = 4/(1/p + p) is valid for all real p (if we define (0,0) as part of the curve, but this doesn't play into the solution). It has negative q for negative p and positive q for positive p and is in fact what's called an odd function; q approaches zero as p becomes larger or as p becomes smaller; q is a minimum at (1,2) and maximum at (1,2). It intersects the other graph in each of the two parts described above. One of the intersections has 2<p<3, 1<q<2, and the other has –1<p<0, 2<q<1. From a graph we can see there are no other solutions, as the first equation has no values of p < 1 and values of q get farther from zero for higher values of p than 1, while values of q approach zero for larger values of p in the other equation.
The following program finds the positive solution:
DEFDBL AZ
p = 2.5: q = 1.4
CLS
DO
prevP = p: prevQ = q
p = ((4 + SQR(16  4 * q * q)) / (2 * q) + p) / 2
q = (SQR(p  1 / p) + q) / 2
PRINT p; q, p * p; q * q
LOOP UNTIL prevP = p AND prevQ = q
Which displays as its final iteration:
2.414213562373095 1.414213562373095 5.82842712474619 2
for p, q, p^2 and q^2. So this solution is q=sqrt(2), p=sqrt(2) + 1, which checks out by plugging in to the equations.
By inspection, guided by the above solution and the graph, the other solution is q=sqrt(2), p=1sqrt(2).
Edited on July 10, 2007, 11:16 am

Posted by Charlie
on 20070710 11:16:04 