Equation (2) applies to angles theta in the range [0°, 90°]. For larger angles, it is replaced by two other equations.
These are both solved explicitly giving, on [90°, 135°] equation (3) or its simplified form

and on [135°, 180°], a modified secant function:

Mandelbrot and Frame's Figure (9) shows the graph of the critical r, the solution of equation (2) and these other two equations, as a function of theta. My version of that graph is shown at the right (theirs shows nicer detail, being printed on nice magazine paper).
Looking at the graph, we might make the following guesses:

1. r is continuous as a function of theta

2. lim(r)=.5 at both ends (theta approaching 0° or 180°)

3. derivative r'=0 at both ends

4. r is increasing on the interval [0°, 90°]

5. r is smooth (differentiable) on each closed interval [90/n, 90/(n-1)]
   and on the intervals [90°, 135°] and [135°, 180°]

6. r' has an abrupt decrease at each end point of these intervals

7. the graph of r is concave-up in each of these intervals

1. We have already observed that the cosine equation makes it easy to prove that r is continuous on [0°, 90°]. A little algebra shows that r is continuous at 90° and 135°, with value 1 / sqrt(2) at both of these angles.

2. & 3. The formula

gives value  r = .5  at  theta = 180, and the derivative there is zero.
At the other end (theta = 0) we show (elsewhere) that the r-values are asymptotic to 1 / 2cos(theta) and so r has the same limit one half) and has a zero derivative at r=0, even though the derivative of r has points of discontinuity approaching 0.

4. & 5. We already observed (without using calculus) that r is increasing on [0, 90], and that it is smooth in the intervals [90/n, 90/(n-1)]. The two formulas which apply for larger angles are certainly differentiable.

6. For angles in the range [0, 90], the derivative of r is given by the implicit function theorem:

The key here is to remember that the sums are taken over exactly those successive k-values for which the cosine values are positive. As theta increases past an angle  90/n  (n>1), the sums in the top and bottom of this fraction both lose their last terms. At that point the cosine in the lost term is zero so the value of the bottom is unchanged. However the sine value is one, so the value of the top (and hence the fraction) decreases, exhibiting a jump discontinuity.

7. It seems that it would be straightforward to show that the second derivative of r is positive, but I have not yet wrestled this problem into submission.