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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 recent_update_frequencies(self):
 """ Returns the 10 most recent update frequencies. The given frequencies are computed as short-term frequencies! The 0th element of the list corresponds to the most recent frequency. """

        return list(reversed([(1.0 / p) for p in numpy.diff(self._recent_updates)]))