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  "body": "\n## Statement\n\nLet $0 \u003c k \u003c 1$, let $k' = \\sqrt{1 - k^2}$, and let\n$$k_1 = \\frac{1 - k'}{1 + k'}.$$\nLet $F(\\varphi, k)$ be the incomplete [[elliptic integral|NXMCYGCB59]] of the first kind and $K(k) = F(\\pi/2, k)$ the [[complete one|RM54CPPE12]].\n\n1. For real $\\varphi$, put $\\varphi_1 = \\varphi + \\arctan(k'\\tan\\varphi)$, where the branch of $\\arctan$ is chosen so that $\\varphi_1$ depends continuously on $\\varphi$ and $\\varphi_1 = 0$ at $\\varphi = 0$. Then\n$$F(\\varphi, k) = \\frac{1 + k_1}{2} \\, F(\\varphi_1, k_1).$$\n2. In particular $K(k) = (1 + k_1) \\, K(k_1)$.\n\n## Remarks\n\nPart 2 is part 1 at $\\varphi = \\pi/2$, where $\\varphi_1 = \\pi$ and $F(\\pi, k_1) = 2K(k_1)$.\nProof idea for part 1: differentiate both sides in $\\varphi$ and check, with $\\tan(\\varphi_1 - \\varphi) = k'\\tan\\varphi$, that the integrands agree; both sides vanish at $\\varphi = 0$.\nCheck at small $k$: then $k' \\approx 1$ and $k_1 \\approx k^2/4 \\approx 0$, so $\\varphi_1 \\approx 2\\varphi$ and the right side is about $\\frac12 \\cdot 2\\varphi = \\varphi = F(\\varphi, 0)$.\nThe modulus shrinks fast: $k_1 \u003c k^2$ for $0 \u003c k \u003c 1$, since $k_1 = k^2/(1 + k')^2$.\nIterating part 2 sends the modulus to $0$ quadratically and gives $K(k) = \\frac{\\pi}{2}\\prod_{n \\ge 1} (1 + k_n)$, where $k_{n+1}$ is obtained from $k_n$ as $k_1$ from $k$.\nIn terms of the integral $I(a, b) = \\int_0^{\\pi/2} d\\theta / \\sqrt{a^2\\cos^2\\theta + b^2\\sin^2\\theta}$, the transformation reads $I(a, b) = I\\big(\\frac{a + b}{2}, \\sqrt{ab}\\big)$, which is the step behind the [[arithmetic–geometric mean formula|E0V3N93DX8]].\n\n## Explanations\n\n### L1 human\n\nLanden's transformation changes the modulus of an elliptic integral into a much smaller one, at the cost of a simple factor and a new angle.\nDoing it again and again makes the modulus tend to $0$, where the integral is just the angle itself.\nThis is an efficient way to compute elliptic integrals.\n\n### L2 human\n\nThe complete form $K(k) = (1 + k_1)K(k_1)$ is equivalent to the identity $K(k) = \\pi / (2M(1, k'))$ with the arithmetic–geometric mean $M$.\nIndeed $k_1' = 2\\sqrt{k'}/(1 + k')$, and the homogeneity $M(\\lambda a, \\lambda b) = \\lambda M(a, b)$ gives $M(1, k_1') = \\frac{2}{1 + k'} M(1, k')$.\nRead backwards, with $k$ in place of $k_1$, it is the ascending Landen transformation, which pushes the modulus toward $1$.\nFor the theta-function side, Landen's transformation corresponds to replacing the nome $q$ by $q^2$, which doubles the ratio $K'/K$.\n",
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