Fix some minor problems in boost calculation and improve docstring

[combet]

• prevent rscale from being an optional keyword with default value (1000) as it must have the same units as rvals. I think it's safer for the user to have to provide a value.
• make corresponding changes
  • in utils.py
  • in boost notebooks
  • in test_utils
• Add equations to the docstrings to have a nice rendering in the documentation
• Rename functions correct_sigma_with_boost* as the boost will generally be applied to DeltaSigma or shear and not Sigma.

[coveralls]

coverage: 100.0%. remained the same
when pulling a3d4ae0 on issue/645/improve_boost
into 157bdb0 on main.

[hsinfan1996]

Thank you.
In the notebook demo_boost_factors.ipynb, is it expected to get nan from print(u.compute_nfw_boost(1000, 1000, boost0=0.1)) (cell 10)?

[combet]

Thank you. In the notebook demo_boost_factors.ipynb, is it expected to get nan from print(u.compute_nfw_boost(1000, 1000, boost0=0.1)) (cell 10)?

Ah right, I forgot to remove that line in the notebook... I was checking things out. I think that's not really a problem as the denominator will go to zero when x=1, so that's expected. Question is whether we want to add a warning or something to explicitely deal with the case R_proj = R_s?

[hsinfan1996]

For completeness, yes, we should probably add something to deal with the case R_proj = R_s. My guess is that the value F(x) at x = R_proj / R_s = 1 is just a constant, so we can fill the value in the output array and don't need to show warnings.

[hsinfan1996]

We can do the following to deal with the case R_proj = R_s:

    r_norm = np.array(rvals) / rscale

    def _calc_finternal(r_norm):
        radicand = r_norm**2 - 1

        finternal = (
            -1j
            * np.log(
                (1 + np.lib.scimath.sqrt(radicand) * 1j) / (1 - np.lib.scimath.sqrt(radicand) * 1j)
            )
            / (2 * np.lib.scimath.sqrt(radicand))
        )

        return np.nan_to_num((1 - finternal)/radicand, copy=False, nan=1.0/3.0).real

    return 1.0 + boost0 * _calc_finternal(r_norm)

[combet]

We can do the following to deal with the case R_proj = R_s:

    r_norm = np.array(rvals) / rscale

    def _calc_finternal(r_norm):
        radicand = r_norm**2 - 1

        finternal = (
            -1j
            * np.log(
                (1 + np.lib.scimath.sqrt(radicand) * 1j) / (1 - np.lib.scimath.sqrt(radicand) * 1j)
            )
            / (2 * np.lib.scimath.sqrt(radicand))
        )

        return np.nan_to_num((1 - finternal)/radicand, copy=False, nan=1.0/3.0).real

    return 1.0 + boost0 * _calc_finternal(r_norm)

Thanks, I implemented that and that took care of the issue!

[combet]

For completeness, yes, we should probably add something to deal with the case R_proj = R_s. My guess is that the value F(x) at x = R_proj / R_s = 1 is just a constant, so we can fill the value in the output array and don't need to show warnings.

Done. And docstrings updated accordingly.

[hsinfan1996]

@combet Thank you. I will approve the PR.

[m-aguena]

Looks good! Thanks @hsinfan1996 for doing the hard part of the review!