s15adf calculates an approximate value for the complement of the error function
Let
$\hat{x}$ be the root of the equation
$\mathrm{erfc}\left(x\right)-\mathrm{erf}\left(x\right)=0$ (then
$\hat{x}\approx 0.46875$). For
$\left|x\right|\le \hat{x}$ the value of
$\mathrm{erfc}\left(x\right)$ is based on the following rational Chebyshev expansion for
$\mathrm{erf}\left(x\right)$:
where
${R}_{\ell ,m}$ denotes a rational function of degree
$\ell $ in the numerator and
$m$ in the denominator.
For
$\left|x\right|>\hat{x}$ the value of
$\mathrm{erfc}\left(x\right)$ is based on a rational Chebyshev expansion for
$\mathrm{erfc}\left(x\right)$: for
$\hat{x}<\left|x\right|\le 4$ the value is based on the expansion
and for
$\left|x\right|>4$ it is based on the expansion
For each expansion, the specific values of
$\ell $ and
$m$ are selected to be minimal such that the maximum relative error in the expansion is of the order
${10}^{-d}$, where
$d$ is the maximum number of decimal digits that can be accurately represented for the particular implementation (see
x02bef).
For
$\left|x\right|\ge {x}_{\mathrm{hi}}$ there is a danger of setting underflow in
$\mathrm{erfc}\left(x\right)$ (the value of
${x}_{\mathrm{hi}}$ is given in the
Users' Note for your implementation).. For
$x\ge {x}_{\mathrm{hi}}$,
s15adf returns
$\mathrm{erfc}\left(x\right)=0$; for
$x\le -{x}_{\mathrm{hi}}$ it returns
$\mathrm{erfc}\left(x\right)=2$.
There are no failure exits from
s15adf. The argument
ifail has been included for consistency with other routines in this chapter.
If
$\delta $ and
$\epsilon $ are relative errors in the argument and result, respectively, then in principle
That is, the relative error in the argument,
$x$, is amplified by a factor
$\frac{2x{e}^{-{x}^{2}}}{\sqrt{\pi}\mathrm{erfc}\left(x\right)}$ in the result.
The behaviour of this factor is shown in
Figure 1.
It should be noted that near
$x=0$ this factor behaves as
$\frac{2x}{\sqrt{\pi}}$ and hence the accuracy is largely determined by the
machine precision. Also, for large negative
$x$, where the factor is
$\text{}\sim \frac{x{e}^{-{x}^{2}}}{\sqrt{\pi}}$, accuracy is mainly limited by
machine precision. However, for large positive
$x$, the factor becomes
$\text{}\sim 2{x}^{2}$ and to an extent relative accuracy is necessarily lost. The absolute accuracy
$E$ is given by
so absolute accuracy is guaranteed for all
$x$.
Internal changes have been made to this routine as follows:
For details of all known issues which have been reported for the NAG Library please refer to the
Known Issues.