There is plenty of heuristic evidence that the (r, theta) graph has zero slope at theta = 0. Our best proof of that fact is still a bit long, we would be interested in seeing any nicer proof.

Letting

we observed elsewhere that S is a continuous, increasing function of r with

Hence there is, for each theta, a unique value of r in the interval [0,1] such that S(r) = 1/2

We want to show that the r-value is at least 1/2 (actually it lies in the interval from 1/2 to the square root of 1/2). We will use the inequalities:

which prevail in the range of angles in which we are interested.

The only part of this multiple inequality which is less than obvious is

We have closed form expressions for both sums, the inequality is equivalent to

We want to know that the (common) denominator of these two fractions is positive.

We are interested in r-values near zero, and for small enough values of r, the subtracted term 2rcos(theta) < 1 so we are safe.

The right-hand expression is just the left expression with two extra terms. It turns out that both extra terms are positive, because we choose N so that

This finishes the proof of the inequality, now we use it to study the left end of the (r, theta) curve.

We solve and get r = -1 and r = 1/2.

Then we solve and get r = 1 , r = -1 and r = 1/(2cos(theta)).

Hence, for any theta, the solution r of S(r) = 1/2 is trapped between 1/2 and 1/(2cos(theta)). This is enough to imply that (in the limit) r(0)=1/2 and that r has a derivative (with respect to theta) which is zero, at theta = 0.

It appears that the r-graph has very high order contact (at theta=0) with the graph of r = 1/(2cos(theta)), even to the extent the two share derivatives of all orders.