Compute the arctangent of x1/x2, choosing the correct quadrant.
JAX implementation of numpy.arctan2()
x1 (ArrayLike) – numerator array.
x2 (ArrayLike) – denomniator array; should be broadcast-compatible with x1.
The elementwise arctangent of x1 / x2, tracking the correct quadrant.
Examples
Consider a sequence of angles in radians between 0 and \(2\pi\):
>>> theta = jnp.linspace(-jnp.pi, jnp.pi, 9) >>> with jnp.printoptions(precision=2, suppress=True): ... print(theta) [-3.14 -2.36 -1.57 -0.79 0. 0.79 1.57 2.36 3.14]
These angles can equivalently be represented by (x, y) coordinates on a unit circle:
>>> x, y = jnp.cos(theta), jnp.sin(theta)
To reconstruct the input angle, we might be tempted to use the identity \(\tan(\theta) = y / x\), and compute \(\theta = \tan^{-1}(y/x)\). Unfortunately, this does not recover the input angle:
>>> with jnp.printoptions(precision=2, suppress=True): ... print(jnp.arctan(y / x)) [-0. 0.79 1.57 -0.79 0. 0.79 1.57 -0.79 0. ]
The problem is that \(y/x\) contains some ambiguity: although \((y, x) = (-1, -1)\) and \((y, x) = (1, 1)\) represent different points in Cartesian space, in both cases \(y / x = 1\), and so the simple arctan approach loses information about which quadrant the angle lies in. arctan2() is built to address this:
>>> with jnp.printoptions(precision=2, suppress=True): ... print(jnp.arctan2(y, x)) [ 3.14 -2.36 -1.57 -0.79 0. 0.79 1.57 2.36 -3.14]
The results match the input theta, except at the endpoints where \(+\pi\) and \(-\pi\) represent indistinguishable points on the unit circle. By convention, arctan2() alwasy returns values between \(-\pi\) and \(+\pi\) inclusive.
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