Pi from escape counts at the cusp of the Mandelbrot set

Adam Zettel · September 21, 2026 · written with an agent

c = \frac{1}{4} + ε ⟹ N (ε) ε \to π

Start just right of the cusp, at $c = \frac{1}{4} + ε$ , and count the steps of $z \to z^{2} + c$ until $z > 2$ . Multiply the count by $ε$ . It tends to $π$ .

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Atlas
The program’s map: every function and which calls which.
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At $c = \frac{1}{4}$ the map has a fixed point at $z = \frac{1}{2}$ where its slope is exactly 1, and the orbit from $z = 0$ climbs toward it in ever smaller steps without ever passing it. Raise $c$ by $ε$ and the fixed point disappears, but near $z = \frac{1}{2}$ each step still moves $z$ only a little. Write $w = z - \frac{1}{2}$ : each step is $w \to w + w^{2} + ε$ , which for small steps is the differential equation $d w / d n = w^{2} + ε$ . The number of steps to get through is $\int_{- \infty}^{\infty} d w / (w^{2} + ε) = π / ε$ .

Printed

estimate: [3, 3.12, 3.14, 3.1414, 3.14157]

Tests 3 of 3 pass

at c = 1/4 the map has a fixed point at z = 1/2, where its slope 2z is exactly 1 (passes)
at eps = 1e-10 the count times sqrt(eps) is pi to four decimals (passes)
and every hundredfold cut in eps cuts the error about tenfold: it goes like sqrt(eps) (passes)