How Far East Can the Mandelbrot Set Go?

At a glance

Let

The disk |𝑧 −𝑎| ≤15 is invariant for the iteration with parameter 𝑐0, and the orbit enters it at 𝑧1 =𝑐0. Hence 𝑐0 is in the Mandelbrot set and 𝑅 ≥Re⁡(𝑐0) =725.

For an arbitrary 𝑐 =𝑥 +𝑦𝑖, the second iterate satisfies

If 𝑥 >1, then |𝑧2| >2, so every such parameter is outside the set. Therefore

Key idea

A lower bound and an upper bound demand opposite certificates. To push the lower bound east, it is enough to exhibit one bounded orbit with a large real coordinate. To push the upper bound west, one must exclude every possible imaginary coordinate beyond a vertical line. An invariant disk supplies the first certificate; a sum of nonnegative terms supplies the second.

Worked solution, part by part

1. Create the candidate point

First square 𝑎:

Therefore

In particular, Re⁡(𝑐0) =725.

2. Trap the orbit in a disk

The magnitude of 𝑎 is

Since 𝑧1 =𝑐0 =𝑎 −𝑎2,

Thus the orbit has entered the disk 𝐷. Now use 𝑐0 =𝑎 −𝑎2:

Assume |𝑧𝑛 −𝑎| ≤15. Write 𝑧𝑛 +𝑎 =(𝑧𝑛 −𝑎) +2𝑎. The triangle inequality gives

Consequently,

Every point of the orbit already in 𝐷 produces another point in 𝐷. Since 𝑧1 ∈𝐷, all later iterates stay there. The disk is bounded, so 𝑐0 ∈𝑀. Therefore

3. Build the eastern wall

For 𝑐 =𝑥 +𝑦𝑖,

Thus

The coefficient 2𝑥2 +2𝑥 +1 =2(𝑥 +12)2 +12 is positive, so the last two terms are nonnegative. Therefore

If 𝑥 >1, then 𝑥2 +𝑥 >2. Hence |𝑧2|2 >4, so |𝑧2| >2 and the orbit has escaped by its second iterate. No point of 𝑀 has real part greater than 1, which proves 𝑅 ≤1.

4. Close the cage

The invariant disk produces one member of 𝑀 with real part 725, so the greatest real part cannot be smaller than 725. The second-iterate estimate excludes every 𝑥 +𝑦𝑖 with 𝑥 >1, regardless of 𝑦, so the greatest real part cannot be larger than 1. Together,

Proof and completeness

Neither certificate can replace the other. A bounded example does not say how far the set might continue beyond that example. An exclusion wall does not prove that the set reaches anywhere near the wall. Their combination is what produces a genuine two-sided bound.

Check

Numerically, 725 =0.28. The candidate really is nonreal, because its imaginary part is 3√325. This is why the earlier endpoint 14 =0.25 on the positive real axis does not control the full set’s easternmost real coordinate.