Integer type:  int32  int64  nag_int  show int32  show int32  show int64  show int64  show nag_int  show nag_int

Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_specfun_bessel_k1_real_vector (s18ar)

## Purpose

nag_specfun_bessel_k1_real_vector (s18ar) returns an array of values of the modified Bessel function ${K}_{1}\left(x\right)$.

## Syntax

[f, ivalid, ifail] = s18ar(x, 'n', n)
[f, ivalid, ifail] = nag_specfun_bessel_k1_real_vector(x, 'n', n)

## Description

nag_specfun_bessel_k1_real_vector (s18ar) evaluates an approximation to the modified Bessel function of the second kind ${K}_{1}\left({x}_{i}\right)$ for an array of arguments ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
Note:  ${K}_{1}\left(x\right)$ is undefined for $x\le 0$ and the function will fail for such arguments.
The function 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 $x>4$,
 $K1x=e-xx ∑′r=0erTrt, where ​t=9-x 1+x .$
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 $x$ it is impossible to calculate $\frac{1}{x}$ without overflow and the function 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

## Parameters

### Compulsory Input Parameters

1:     $\mathrm{x}\left({\mathbf{n}}\right)$ – double array
The argument ${x}_{\mathit{i}}$ of the function, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
Constraint: ${\mathbf{x}}\left(\mathit{i}\right)>0.0$, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.

### Optional Input Parameters

1:     $\mathrm{n}$int64int32nag_int scalar
Default: the dimension of the array x.
$n$, the number of points.
Constraint: ${\mathbf{n}}\ge 0$.

### Output Parameters

1:     $\mathrm{f}\left({\mathbf{n}}\right)$ – double array
${K}_{1}\left({x}_{i}\right)$, the function values.
2:     $\mathrm{ivalid}\left({\mathbf{n}}\right)$int64int32nag_int array
${\mathbf{ivalid}}\left(\mathit{i}\right)$ contains the error code for ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
${\mathbf{ivalid}}\left(i\right)=0$
No error.
${\mathbf{ivalid}}\left(i\right)=1$
${x}_{i}\le 0.0$, ${K}_{1}\left({x}_{i}\right)$ is undefined. ${\mathbf{f}}\left(\mathit{i}\right)$ contains $0.0$.
${\mathbf{ivalid}}\left(i\right)=2$
${x}_{i}$ is too small, there is a danger of overflow. ${\mathbf{f}}\left(\mathit{i}\right)$ contains zero. The threshold value is the same as for ${\mathbf{ifail}}={\mathbf{2}}$ in nag_specfun_bessel_k1_real (s18ad), as defined in the Users' Note for your implementation.
3:     $\mathrm{ifail}$int64int32nag_int scalar
${\mathbf{ifail}}={\mathbf{0}}$ unless the function detects an error (see Error Indicators and Warnings).

## Error Indicators and Warnings

Errors or warnings detected by the function:

Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.

W  ${\mathbf{ifail}}=1$
On entry, at least one value of x was invalid.
${\mathbf{ifail}}=2$
Constraint: ${\mathbf{n}}\ge 0$.
${\mathbf{ifail}}=-99$
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.

## 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:
 $ε≃ x K0x- K1x K1x δ.$
Figure 1 shows the behaviour of the error amplification factor
 $xK0x - K1 x K1x .$
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 function will return zero. Note that for large $x$ the errors will be dominated by those of the standard function exp.
Figure 1

None.

## Example

This example reads values of x from a file, evaluates the function at each value of ${x}_{i}$ and prints the results.
```function s18ar_example

fprintf('s18ar example results\n\n');

x = [0.4; 0.6; 1.4; 1.6; 2.5; 3.5; 6; 8; 10; 1000];

[f, ivalid, ifail] = s18ar(x);

fprintf('     x           K_1(x)   ivalid\n');
for i=1:numel(x)
fprintf('%12.3e%12.3e%5d\n', x(i), f(i), ivalid(i));
end

```
```s18ar example results

x           K_1(x)   ivalid
4.000e-01   2.184e+00    0
6.000e-01   1.303e+00    0
1.400e+00   3.208e-01    0
1.600e+00   2.406e-01    0
2.500e+00   7.389e-02    0
3.500e+00   2.224e-02    0
6.000e+00   1.344e-03    0
8.000e+00   1.554e-04    0
1.000e+01   1.865e-05    0
1.000e+03   0.000e+00    0
```