Specified error (epsilon) not met because of np.isclose()

[OliRom]

The np.isclose() condition is sometimes used to return specific results, either to avoid division by zero errors or to skip calculations in trivial cases. By default, np.isclose() uses a tolerance of 1e-8. However, if the user specifies a decomposition tolerance of 1e-10, the resulting error may exceed the requested precision.

These occurrences of np.isclose() should either be replaced with a more robust check (e.g., verifying that the returned value lies within the allowed error margin using the operator norm before returning it) or should adapt its tolerance dynamically based on the maximum error specified by the user.

Moreover, this special case should be covered in the tests.

For example, in rz_approx.z_rotational_approximation() we currently have:

    # Checks if the angle is trivial
    exponent = round(2 * theta / math.pi)
    if np.isclose(0, theta):
        return np.array(
            [
                [Domega.from_ring(1), Domega.from_ring(0)],
                [Domega.from_ring(0), Domega.from_ring(1)],
            ],
            dtype=object,
        )
    elif np.isclose(2 * theta / math.pi, exponent):
        T = np.array(
            [
                [Domega(-D(1, 0), D(0, 0), D(0, 0), D(0, 0)), Domega.from_ring(0)],
                [Domega.from_ring(0), Domega(D(0, 0), D(0, 0), D(1, 0), D(0, 0))],
            ],
            dtype=object,
        )
        M = T**exponent
        return M