Notebooks · posts/pi-in-the-mandelbrot-set

Pi from escape counts at the cusp of the Mandelbrot set

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

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 $π$ .

c = \frac{1}{4} + ε ⟹ N (ε) ε \to π

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 $\int_{- \infty}^{\infty} d w / (w^{2} + ε) = π / ε$ .

Look inside this program

Atlas
The program’s map: every function and which calls which.
There’s no recorded run of this program yet.

Shift+Enter in a cell runs it and moves to the next. Running a cell runs every cell above it too, as one program.

[1]lines 19–64

pi = 3.141592653589793

#| Steps of z -> z^2 + c, from z = 0, until z > 2. For real c > 1/4 it always gets there.
def escape_steps(c) {
    z = 0.0
    n = 0.0
    while z <= 2.0 {
        z = z * z + c
        n = n + 1.0
    }
    return n
}

K = 5
eps_all = zeros(K)
estimate = zeros(K)
for k to K {
    eps = pow(10.0, -2.0 * (k + 1.0))
    eps_all[k] = eps
    estimate[k] = escape_steps(0.25 + eps) * sqrt(eps)
}
show estimate
error = abs(estimate - pi)

figure("Steps to escape, times $\\sqrt{\\epsilon}$", "$\\epsilon$", "$N\\sqrt{\\epsilon}$")
xscale("log")
plot(eps_all, estimate, "$N\\sqrt{\\epsilon}$")
plot(eps_all, zeros(K) + pi, "$\\pi$")

test "at c = 1/4 the map has a fixed point at z = 1/2, where its slope 2z is exactly 1" {
    z_fixed = 0.5
    assert_near(z_fixed * z_fixed + 0.25, z_fixed, 0.0)
    assert_near(2.0 * z_fixed, 1.0, 0.0)
}
test "at eps = 1e-10 the count times sqrt(eps) is pi to four decimals" {
    assert_near(eps_all[4], 1.0e-10, 1.0e-24)
    assert_near(estimate[4], pi, 1.0e-4)
}
test "and every hundredfold cut in eps cuts the error about tenfold: it goes like sqrt(eps)" {
    for k to K - 1.0 {
        assert_lt(error[k + 1.0], 0.2 * error[k])
    }
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors

pi = 3.141592653589793

#| Steps of z -> z^2 + c, from z = 0, until z > 2. For real c > 1/4 it always gets there.
def escape_steps(c) {
    z = 0.0
    n = 0.0
    while z <= 2.0 {
        z = z * z + c
        n = n + 1.0
    }
    return n
}

K = 5
eps_all = zeros(K)
estimate = zeros(K)
for k to K {
    eps = pow(10.0, -2.0 * (k + 1.0))
    eps_all[k] = eps
    estimate[k] = escape_steps(0.25 + eps) * sqrt(eps)
}
show estimate
error = abs(estimate - pi)

figure("Steps to escape, times $\\sqrt{\\epsilon}$", "$\\epsilon$", "$N\\sqrt{\\epsilon}$")
xscale("log")
plot(eps_all, estimate, "$N\\sqrt{\\epsilon}$")
plot(eps_all, zeros(K) + pi, "$\\pi$")

test "at c = 1/4 the map has a fixed point at z = 1/2, where its slope 2z is exactly 1" {
    z_fixed = 0.5
    assert_near(z_fixed * z_fixed + 0.25, z_fixed, 0.0)
    assert_near(2.0 * z_fixed, 1.0, 0.0)
}
test "at eps = 1e-10 the count times sqrt(eps) is pi to four decimals" {
    assert_near(eps_all[4], 1.0e-10, 1.0e-24)
    assert_near(estimate[4], pi, 1.0e-4)
}
test "and every hundredfold cut in eps cuts the error about tenfold: it goes like sqrt(eps)" {
    for k to K - 1.0 {
        assert_lt(error[k + 1.0], 0.2 * error[k])
    }
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors