DOC: Update the pandas.DataFrame.abs docstring
Improve description and returns, and added notes and examples.

Checklist for the pandas documentation sprint (ignore this if you are doing
an unrelated PR):

- [x] PR title is "DOC: update the <your-function-or-method> docstring"
- [x] The validation script passes: `scripts/validate_docstrings.py <your-function-or-method>`
- [x] The PEP8 style check passes: `git diff upstream/master -u -- "*.py" | flake8 --diff`
- [x] The html version looks good: `python doc/make.py --single <your-function-or-method>`
- [x] It has been proofread on language by another sprint participant

Please include the output of the validation script below between the "```" ticks:

```
################################################################################
####################### Docstring (pandas.DataFrame.abs) #######################
################################################################################

Return a Series/DataFrame with absolute numeric value of each element.

This function only applies to elements that are all numeric.

Returns
-------
abs
    Series/DataFrame containing the absolute value of each element.

Notes
-----
For ``complex`` inputs, ``1.2 + 1j``, the absolute value is
:math:`\sqrt{ a^2 + b^2 }`. See the Python

Examples
--------
Absolute numeric values in a ``Series``.

>>> s = pd.Series([-1.10, 2, -3.33, 4])
>>> s.abs()
0    1.10
1    2.00
2    3.33
3    4.00
dtype: float64

Absolute numeric values in a ``Series`` with ``complex`` numbers.

>>> s = pd.Series([1.2 + 1j])
>>> s.abs()
0    1.56205
dtype: float64

Select rows with data closest to certian value using argsort (from
`StackOverflow <https://stackoverflow.com/a/17758115>`__).

>>> df = pd.DataFrame({
...     'a': [4, 5, 6, 7],
...     'b': [10, 20, 30, 40],
...     'c': [100, 50, -30, -50]
... })
>>> df
     a    b    c
0    4   10  100
1    5   20   50
2    6   30  -30
3    7   40  -50
>>> df.loc[(df.c - 43).abs().argsort()]
     a    b    c
1    5   20   50
0    4   10  100
2    6   30  -30
3    7   40  -50

See Also
--------
numpy.absolute : calculate the absolute value element-wise.

################################################################################
################################## Validation ##################################
################################################################################

Docstring for "pandas.DataFrame.abs" correct. :)
```

I feel there aren't any similar methods for the See Also section.