Approximation Error

For $I_0$, the error term in equation 14 is $E = \frac{15}{8} I_6$. The absolute value of this term is bounded above by:

$$\vert E \vert < \frac{15}{8} L^{-6} \vert I_0 \vert$$ (17)

for $x > L$. Therefore, the relative error in $I_0$ is given by:

$$\vert \epsilon \vert = \vert \frac{E}{I_0} \vert < \frac{15}{8} L^{-6}.$$ (18)

So, since $p$ is proportional to $I_0$, the relative error is the same. Furthermore, if $\epsilon$ is small, then it is also equal to the absolute error in $\log(p)$, since:

$$\log ( p + E_p ) = \log ( p ( 1 + \epsilon ) ) \approx \log(p) + \epsilon.$$ (19)

To obtain a relative error of $\vert \epsilon \vert < 10^{-3}$ therefore requires $\frac{15}{8} L^{-6} < 10^{-3}$ or $L > 3.51$, which corresponds to $Z > 4.966$. However, this is an upper bound on the error, and in practice a relative error of $10^{-3}$ is achieved for $Z>4.8$ (as measured in MATLAB).