External rays for primary bulbs

In [1] Devaney give some statements about external rays for primary bulbs (tangent to the main cardioid).
Remind that a bulb Bp/q consists of c-values for which the quadratic map has an attracting q-cycle. The root point of this bulb is the landing point of exactly 2 M-rays, and the angles of each of these rays have period q under doubling. The two corresponding parameter rays have the same angles as the two dynamic rays which bound the critical value sector S1. The root point of the p/q bulb of M divides M into two sets. The component containing this bulb is called the p/q limb.

For example, the large bulb directly to the left of the main cardioid (the image to the right) is the 1/2 bulb, so two rays with period 2 under doubling must land there. Now the only angles with period 2 under doubling are 1/3 and 2/3, so these are the angles of the rays that land at the root point 1/2. To the left two dynamical rays with angles 1/3 and 2/3 lands at unstable fixed point z2. The green sector S1 contains the point z = c.
Now consider the 1/3 bulb atop the main cardioid. This bulb lies between the rays 0 and 1/3. There are only two angles between 0 and 1/3 that have period 3 under doubling, namely 1/7 and 2/7, so these are the rays that land at the root point 1/3.
The 2/5 bulb lies between the 1/3 and 1/2 bulbs. Hence two rays must have period 5 under doubling and lie between 2/7 and 1/3. The only angles that have this property are 9/31 and 10/31, so these rays must land at the root of the 2/5 bulb.

Rays landing on the p/q bulb

Let Rp/q denote rotation of the unit circle through p/q turns, i.e.,
    Rp/q(θ) = e2πi(θ+p/q).
We will consider the itineraries of points in the unit circle under R using two different partitions of the circle. The lower partition of the circle is defined as
    I0- = (0, 1-p/q]   and   I1- = (1-p/q, 1].
We then define s-(p/q) to be the lower itinerary of p/q under Rp/q relative to this partition. For example, s-(1/3) = 001 since I0- = (0, 2/3], I1- = (2/3, 1] and the orbit 1/3 → 2/3 → 1 → 1/3... lies in I0-, I0-, I1- respectively. Similarly s-(2/5) = 01001.
The upper partition of the circle is
    I0+ = [0, 1-p/q)   and   I1+ = [1-p/q, 1).
The upper itinerary s+(p/q), is then the itinerary of p/q relative to this partition. Note that I0+, I1+ differ from I0-, I1- only at the endpoints. For example, s+(1/3) = 010 since the orbit is 1/3 → 2/3 → 0... and I0+ = [0, 2/3), I1+ = [2/3, 1). For 2/5, we have s+(2/5) = 01010.

Theorem. The two rays landing at the root point of the p/q bulb are 0.(s-(p/q)) and 0.(s+(p/q)).
Here 0.(s) means the binary expansion with repeated string s. E.g. 0.(001) = 0.001001... = 0012/1112 = 1/7, 0.(010) = 2/7 and 0.(01001) = 10012/111112 = 9/31, 0.(01010) = 10/31.

Note that s-(p/q) and s+(p/q) differ only in their last two digits (provided q ≥ 2). Indeed we may write
    s-(p/q) = s1...sq-2 0 1
    s+(p/q) = s1...sq-2 1 0

The reason for this is that the upper and lower itineraries are the same except at Rp/qq-2(p/q) = -p/q and Rp/qq-1(p/q) = 0, which form the endpoints of the two partitions of the circle.
If we define the size of the p/q limb to be the length of the interval [0.(s-(p/q)), 0.(s+(p/q))] then we may compute it explicitly by using the fact that s±(p/q) differ only in the last two digits.
Theorem. The size of the p/q limb is 1/(2q-1).
Theorem. Suppose α/β < γ/δ are the Farey parents of p/q. Then the size of the p/q limb is larger than the size of any other limb between the α/β and γ/δ limbs.
See literature below for the proof.

[1] Robert L. Devaney The Mandelbrot Set and The Farey Tree

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updated 25 Mar 08