s18ad returns the value of the modified Bessel function ${K}_{1}\left(x\right)$.

# Syntax

C#
```public static double s18ad(
double x,
out int ifail
)```
Visual Basic
```Public Shared Function s18ad ( _
x As Double, _
<OutAttribute> ByRef ifail As Integer _
) As Double```
Visual C++
```public:
double x,
[OutAttribute] int% ifail
)```
F#
```static member s18ad :
x : float *
ifail : int byref -> float
```

#### Parameters

x
Type: System..::..Double
On entry: the argument $x$ of the function.
Constraint: ${\mathbf{x}}>0.0$.
ifail
Type: System..::..Int32%
On exit: ${\mathbf{ifail}}={0}$ unless the method detects an error or a warning has been flagged (see [Error Indicators and Warnings]).

#### Return Value

s18ad returns the value of the modified Bessel function ${K}_{1}\left(x\right)$.

# Description

s18ad evaluates an approximation to the modified Bessel function of the second kind ${K}_{1}\left(x\right)$.
Note:  ${K}_{1}\left(x\right)$ is undefined for $x\le 0$ and the method will fail for such arguments.
The method is based on five Chebyshev expansions:
For $0,
 $K1x=1x+xln x∑′r=0arTrt-x∑′r=0brTrt, where ​t=2x2-1.$
For $1,
 $K1x=e-x∑′r=0crTrt, where ​t=2x-3.$
For $2,
 $K1x=e-x∑′r=0drTrt, where ​t=x-3.$
For $x>4$,
 $K1x=e-xx∑′r=0erTrt, where ​t=9-x1+x.$
For $x$ near zero, ${K}_{1}\left(x\right)\simeq \frac{1}{x}$. This approximation is used when $x$ is sufficiently small for the result to be correct to machine precision. For very small $x$ on some machines, it is impossible to calculate $\frac{1}{x}$ without overflow and the method must fail.
For large $x$, where there is a danger of underflow due to the smallness of ${K}_{1}$, the result is set exactly to zero.

# References

Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications

# Error Indicators and Warnings

Errors or warnings detected by the method:
${\mathbf{ifail}}=1$
${\mathbf{x}}\le 0.0$, ${K}_{1}$ is undefined. On failure the method returns zero.
${\mathbf{ifail}}=2$
x is too small, there is a danger of overflow. On failure the method returns approximately the largest representable value. (see the Users' Note for your implementation for details)
${\mathbf{ifail}}=-9000$
An error occured, see message report.

# Accuracy

Let $\delta$ and $\epsilon$ be the relative errors in the argument and result respectively.
If $\delta$ is somewhat larger than the machine precision (i.e., if $\delta$ is due to data errors etc.), then $\epsilon$ and $\delta$ are approximately related by:
 $ε≃xK0x-K1xK1xδ.$
Figure 1 shows the behaviour of the error amplification factor
 $xK0x-K1xK1x.$
However if $\delta$ is of the same order as the machine precision, then rounding errors could make $\epsilon$ slightly larger than the above relation predicts.
For small $x$, $\epsilon \simeq \delta$ and there is no amplification of errors.
For large $x$, $\epsilon \simeq x\delta$ and we have strong amplification of the relative error. Eventually ${K}_{1}$, which is asymptotically given by $\frac{{e}^{-x}}{\sqrt{x}}$, becomes so small that it cannot be calculated without underflow and hence the method will return zero. Note that for large $x$ the errors will be dominated by those of the standard function exp.
Figure 1

None.

None.

# Example

This example reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.