Increase speed of Python STA/LTA by about 8 times

This patch increases the speed of the Python implementation of STA/LTA from about **20** times the duration of the C implementation to about **2.5** times the duration of the C implementation:

```python
In [68]: tr = read()[0]

In [75]: %timeit classicSTALTA(tr.data, 5, 60)
10000 loops, best of 3: 44 µs per loop

In [76]: %timeit classicSTALTAPy_old(tr.data, 5, 60)
1000 loops, best of 3: 842 µs per loop

In [77]: %timeit classicSTALTAPy_new(tr.data, 5, 60)
10000 loops, best of 3: 110 µs per loop
```

The result stays the same:

```python
In [69]: stalta1 = classicSTALTA(tr.data, 5, 60)

In [70]: stalta2 = classicSTALTAPy_old(tr.data, 5, 60)

In [71]: stalta3 = classicSTALTAPy_new(tr.data, 5, 60)

In [72]: np.allclose(stalta1, stalta2)
Out[72]: True

In [73]: np.allclose(stalta1, stalta3)
Out[73]: True

In [74]: np.allclose(stalta2, stalta3)
Out[74]: True
```

Since the old function contiained a comment about improving the speed of the function, I figured I might just as well submit this. :-)

Yes, but there are subtractions between there and here... That being said, now that I think about it, a cumulative sum of the squares would guarantee that the end is larger than the beginning, so the difference there is positive too.