﻿ s18af Method
s18af returns a value for the modified Bessel function ${I}_{1}\left(x\right)$.

# Syntax

C#
```public static double s18af(
double x,
out int ifail
)```
Visual Basic
```Public Shared Function s18af ( _
x As Double, _
<OutAttribute> ByRef ifail As Integer _
) As Double```
Visual C++
```public:
static double s18af(
double x,
[OutAttribute] int% ifail
)```
F#
```static member s18af :
x : float *
ifail : int byref -> float
```

#### Parameters

x
Type: System..::..Double
On entry: the argument $x$ of the function.
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

s18af returns a value for the modified Bessel function ${I}_{1}\left(x\right)$.

# Description

s18af evaluates an approximation to the modified Bessel function of the first kind ${I}_{1}\left(x\right)$.
Note:  ${I}_{1}\left(-x\right)=-{I}_{1}\left(x\right)$, so the approximation need only consider $x\ge 0$.
The method is based on three Chebyshev expansions:
For $0,
 $I1x=x∑′r=0arTrt, where ​t=2x42-1;$
For $4,
 $I1x=ex∑′r=0brTrt, where ​t=x-84;$
For $x>12$,
 $I1x=exx∑′r=0crTrt, where ​t=212x-1.$
For small $x$, ${I}_{1}\left(x\right)\simeq x$. This approximation is used when $x$ is sufficiently small for the result to be correct to machine precision.
For large $x$, the method must fail because ${I}_{1}\left(x\right)$ cannot be represented without overflow.

# 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:
${\mathbf{ifail}}=1$
x is too large. On failure the method returns the approximate value of ${I}_{1}\left(x\right)$ at the nearest valid argument. See also the Users' Note for your implementation.
${\mathbf{ifail}}=-9000$
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:
 $ε≃xI0x-I1xI1xδ.$
Figure 1 shows the behaviour of the error amplification factor
 $xI0x-I1xI1x.$
Figure 1
However, if $\delta$ is of the same order as 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 errors. However the method must fail for quite moderate values of $x$ because ${I}_{1}\left(x\right)$ would overflow; hence in practice the loss of accuracy for large $x$ is not excessive. Note that for large $x$, the errors will be dominated by those of the standard function exp.

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.

Example program (C#): s18afe.cs

Example program data: s18afe.d

Example program results: s18afe.r