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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _approximate_common_period(periods: List[float], approx_denom: int = 60, reject_atol: float = 1e-8) -> Optional[float]: """Finds a value that is nearly an integer multiple of multiple periods. The returned value should be the smallest non-negative number with this property. If `approx_denom` is too small the computation can fail to satisfy the `reject_atol` criteria and return `None`. This is actually desirable behavior, since otherwise the code would e.g. return a nonsense value when asked to compute the common period of `np.e` and `np.pi`. Args: periods: The result must be an approximate integer multiple of each of these. approx_denom: Determines how the floating point values are rounded into rational values (so that integer methods such as lcm can be used). Each floating point value f_k will be rounded to a rational number of the form n_k / approx_denom. If you want to recognize rational periods of the form i/d then d should divide `approx_denom`. reject_atol: If the computed approximate common period is at least this far from an integer multiple of any of the given periods, then it is discarded and `None` is returned instead. Returns: The approximate common period, or else `None` if the given `approx_denom` wasn't sufficient to approximate the common period to within the given `reject_atol`. """

    if not periods:
        return None
    if any(e == 0 for e in periods):
        return None
    if len(periods) == 1:
        return abs(periods[0])
    approx_rational_periods = [
        fractions.Fraction(int(np.round(abs(p) * approx_denom)), approx_denom)
        for p in periods
    ]
    common = float(_common_rational_period(approx_rational_periods))

    for p in periods:
        if p != 0 and abs(p * np.round(common / p) - common) > reject_atol:
            return None

    return common