[crntaylor]

The point is that when you're considering the Taylor series for a dual number argument, you don't lose any precision, because higher powers of the "imaginary" part of the dual number vanish. The example he gives is

    f(a + be) = f(a) + f'(a)be + 0.5 * f''(a) b^2 e^2 + O(e^3)
              = f(a) + f'(a)be

because e^n=0 for all n>1. This isn't an approximation - it's an exact relationship for dual numbers!

You will lose some precision by using floating point numbers instead of an arbitrary-precision real number type, but this is a limitation of the machine you're working on. The method is exact.