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  "body": "\n## Example\n\nLet $d_0, d_1 \\colon \\C \\to \\C$ be two [[contraction mappings|0QN69YA357]] with fixed points $p_0$ and $p_1$, and assume $d_0(p_1) = d_1(p_0)$.\nThe *de Rham curve* of $(d_0, d_1)$ is the [[continuous|0X3WGX95FE]] map $c \\colon [0, 1] \\to \\C$ that solves the system of [[functional equations|SRAN4EKWS0]]\n$$c\\left(\\frac{x}{2}\\right) = d_0(c(x)), \\qquad c\\left(\\frac{1 + x}{2}\\right) = d_1(c(x)) \\qquad \\text{for all } x \\in [0, 1].$$\nSetting $x = 0$ in the first equation and $x = 1$ in the second gives $c(0) = p_0$ and $c(1) = p_1$.\nBoth equations give a value at $1/2$: the first with $x = 1$ gives $d_0(p_1)$, the second with $x = 0$ gives $d_1(p_0)$.\nThis is why the condition $d_0(p_1) = d_1(p_0)$ is needed.\n\nA concrete case: let $a = \\frac12 + \\frac{\\sqrt3}{6} i$, $d_0(z) = a \\bar z$ and $d_1(z) = a + (1 - a) \\bar z$.\nBoth maps are contractions with ratio $\\abs{a} = \\abs{1 - a} = 1/\\sqrt3 \u003c 1$.\nThe fixed point of $d_0$ is $p_0 = 0$, and the fixed point of $d_1$ is $p_1 = 1$, since $d_1(1) = a + 1 - a = 1$.\nThe condition holds: $d_0(1) = a = d_1(0)$.\nThe resulting curve from $0$ to $1$ is the Koch curve.\n\nThe simplest case is $d_0(z) = z/2$ and $d_1(z) = (z + 1)/2$, for which $c(x) = x$: indeed $c(x/2) = x/2 = d_0(x)$ and $c((1+x)/2) = (1 + x)/2 = d_1(x)$.\n\n## Remarks\n\nThe solution exists and is unique.\nThe system says that $c$ is a fixed point of an operator on the complete metric space of continuous maps $[0, 1] \\to \\C$ with $c(0) = p_0$ and $c(1) = p_1$.\nThis operator is a contraction for the uniform metric, so the [[Banach fixed point theorem|0VK6VEH345]] gives exactly one solution.\nOn dyadic rationals the curve can be computed directly: $c(0) = p_0$, $c(1) = p_1$, $c(1/2) = d_0(p_1)$, and so on.\n\n## Explanations\n\n### L1 human\n\nA de Rham curve is built by a simple rule.\nThe first half of the curve is a shrunken copy of the whole curve, and so is the second half.\nThe two shrinking maps say how each copy sits in the plane.\nIf they shrink without twisting, the curve is a straight segment.\nWith a reflection and a tilt, the same rule produces the Koch snowflake curve, which has a corner at every scale.\n",
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