Taylor Series Model Estimator

We may cast the result in equation 15 as a simple quadratic in $\alpha$, ( $0 = a\alpha^2 + b\alpha + c$), with $a = (W''_0)^2 p$, $b = (W_0 W''''_0 - (W''_0)^2)$, and $c = W_0^2 p$.

We may then solve for alpha by using the familiar quadratic formula,

$$\alpha = \ensuremath{\frac{-b + \sqrt{b^2 - 4 a c}}{2 a}}$$ (19)

where we note that only the positive sign is used in the numerator. This is because we require that as $\alpha$ goes to zero (as required by our Taylor series model), our estimate does so also; it may be shown that only by choosing the positive sign will this occur.