A Better High-Order Low-Pass Filter?
You can implement very efficient high order low pass filter using cumulative summing. The theory is as follows. A perfect integrator has the recursive definition 

$$y[n] = y[n-1]+x[n]$$

If you take the delayed difference of the output as in
$$y'[n] = y[n] - y[n-d]$$

you are realizing a boxcar filter of length $d-1$, as can be seen from explicitly resolving the recursion for $y[n]$ in $y'[n]$:

$$ y'[n] = (y[n-1]+x[n]) - y[n-d]$$

$$ y'[n] = ((y[n-2] + x[n-1]) + x[n]) - y[n-d]$$

$$ \dots$$

$$ y'[n] = y[n-d] + \sum_{k=1}^{d-1}x[n-k] - y[n-d] = \sum_{k=1}^{d-1}x[n-k]$$

That means you have a low pass filter of effective order $d-1$ from just one addition and one subtraction per sample.

Of course, boxcars are only useful up to a certain point. But the same idea can be generalized  to a larger class of filters with a trade off between filter shape and computational effort. And that is done by concatenating several integrator-difference stages.

Two in series will generate a triangular impulse response, three a parabolic one, and so on. Each stage convolves the output with yet another boxcar. In the frequency domain the original sinc() shaped response is taken to the n-th power (or if you have different delay lengths for the different stages, multiplied by other differently scaled sincs). 

But both response sequences in time and frequency domain behave very predictably, because the central limit theorem of statistics predicts that the time domain response quickly approaches a Gaussian distribution. And because the fourier transform of a Gaussian is also Gaussian, you'll get a very good approximation of a Gaussian frequency response low pass filter after only a few stages.

One note about numerical precision though. Because you are integrating, the content of the delay buffer can be become very large and the difference precision suffers. This can be resolved by making the integrator leaky, so that it doesn't accumulate a constant function up to infinity. The leaky integrator difference equation is

$$ y[n] = (1-\epsilon) y[n-1] + x[n] $$

for some sufficiently small positive $\epsilon$.