* and the arithmetic mean is always higher than the geometric mean.
     *
     * b) this iteration converges to floor(sqrt(x)). In fact, the number of correct digits doubles
     * with each iteration, so this algorithm takes O(log(digits)) iterations.
     *
     * We start out with a double-precision approximation, which may be higher or lower than the
     * true value. Therefore, we perform at least one Newton iteration to get a guess that's