External rays for primary bulbs
In  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.
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.
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.
Rays landing on the p/q bulb
Let Rp/q denote rotation of the unit circle through 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
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
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.
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.
 Robert L. Devaney
Mandelbrot Set and The Farey Tree
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updated 25 Mar 08