impl. of sqrtf_neon_hpf()

[GoogleCodeExporter]

sqrtf_neon_hpf() first computes the inverse of the square root, and then the 
reciprocal, i.e.

t = 1/sqrt(x)
r = 1/t

it might be easier/faster to compute the inverse of the square root, and then 
multiply by the original value, i.e.

t = 1/sqrt(x)
r = x * t

Original issue reported on code.google.com by [email protected] on 15 Sep 2011 at 11:53