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In the most general setting, <math>\operatorname{am}(u,m)</math> is a [[multivalued function]] (in <math>u</math>) with infinitely many [[Branch point|logarithmic branch points]] (the branches differ by integer multiples of <math>2\pi</math>), namely the points <math>2sK(m)+(4t+1)K(1-m)i</math> and <math>2sK(m)+(4t+3)K(1-m)i</math> where <math>s,t\in\mathbb{Z}</math>.<ref name="sala">{{cite journal |last=Sala |first=Kenneth L. |date=November 1989 |title=Transformations of the Jacobian Amplitude Function and Its Calculation via the Arithmetic-Geometric Mean|url=https://epubs.siam.org/doi/abs/10.1137/0520100 |journal=SIAM Journal on Mathematical Analysis|volume=20|issue=6|pages=1514–1528|doi=10.1137/0520100 }}</ref> This multivalued function can be made single-valued by cutting the complex plane along the line segments joining these branch points (the cutting can be done in non-equivalent ways, giving non-equivalent single-valued functions), thus making <math>\operatorname{am}(u,m)</math> [[Analytic function|analytic]] everywhere except on the [[Branch point#Branch cuts|branch cuts]]. In contrast, <math>\sin\operatorname{am}(u,m)</math> and other elliptic functions have no branch points, give consistent values for every branch of <math>\operatorname{am}</math>, and are [[meromorphic function|meromorphic]] in the whole complex plane. Since every elliptic function is meromorphic in the whole complex plane (by definition), <math>\operatorname{am}(u,m)</math> (when considered as a single-valued function) is not an elliptic function.
However, a particular cutting for <math>\operatorname{am}(u,m)</math> can be made in the <math>u</math>-plane by line segments from <math>2sK(m)+(4t+1)K (1-m)i</math> to <math>2sK(m)+(4t+3)K(1-m)i</math> with <math>s,t\in\mathbb{Z}</math>; then it only remains to define <math>\operatorname{am}(u,m)</math> at the branch cuts by continuity from some direction. Then <math>\operatorname{am}(u,m)</math> becomes single-valued and singly-periodic in <math>u</math> with the minimal period <math>4iK(1-m)</math> and it has
Let
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A function which solves the above pendulum differential equation with initial angle <math>\theta_0</math> is
:<math>\theta=2\arcsin (k\operatorname{cd}(\sqrt{c}t,k^2)),\quad k=\sin\frac{\theta_0}{2},\quad c>0,\,t\in\mathbb{R}.</math>
===Derivatives with respect to the second variable===
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