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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_specfun_kelvin_bei_vector (s19ap)

## Purpose

nag_specfun_kelvin_bei_vector (s19ap) returns an array of values for the Kelvin function $\mathrm{bei}x$.

## Syntax

[f, ivalid, ifail] = s19ap(x, 'n', n)
[f, ivalid, ifail] = nag_specfun_kelvin_bei_vector(x, 'n', n)

## Description

nag_specfun_kelvin_bei_vector (s19ap) evaluates an approximation to the Kelvin function $\mathrm{bei}{x}_{i}$ for an array of arguments ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
Note:  $\mathrm{bei}\left(-x\right)=\mathrm{bei}x$, so the approximation need only consider $x\ge 0.0$.
The function is based on several Chebyshev expansions:
For $0\le x\le 5$,
 $bei⁡x = x24 ∑′r=0 ar Tr t , with ​ t=2 x5 4 - 1 ;$
For $x>5$,
 $bei⁡x = e x/2 2πx 1 + 1x a t sin⁡α - 1x b t cos⁡α$
 $+ e x/2 2π x 1 + 1x c t cos⁡β - 1x d t sin⁡β$
where $\alpha =\frac{x}{\sqrt{2}}-\frac{\pi }{8}$, $\beta =\frac{x}{\sqrt{2}}+\frac{\pi }{8}$,
and $a\left(t\right)$, $b\left(t\right)$, $c\left(t\right)$, and $d\left(t\right)$ are expansions in the variable $t=\frac{10}{x}-1$.
When $x$ is sufficiently close to zero, the result is computed as $\mathrm{bei}x=\frac{{x}^{2}}{4}$. If this result would underflow, the result returned is $\mathrm{bei}x=0.0$.
For large $x$, there is a danger of the result being totally inaccurate, as the error amplification factor grows in an essentially exponential manner; therefore the function must fail.

## 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 $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.

### 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
$\mathrm{bei}{x}_{i}$, 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.
${\mathbf{ivalid}}\left(i\right)=1$
$\mathrm{abs}\left({x}_{i}\right)$ is too large for an accurate result to be returned. ${\mathbf{f}}\left(\mathit{i}\right)$ contains zero. The threshold value is the same as for ${\mathbf{ifail}}={\mathbf{1}}$ in nag_specfun_kelvin_bei (s19ab), 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

Since the function is oscillatory, the absolute error rather than the relative error is important. Let $E$ be the absolute error in the function, and $\delta$ be the relative error in the argument. If $\delta$ is somewhat larger than the machine precision, then we have:
 $E≃ x2 - ber1⁡x+ bei1⁡x δ$
(provided $E$ is within machine bounds).
For small $x$ the error amplification is insignificant and thus the absolute error is effectively bounded by the machine precision.
For medium and large $x$, the error behaviour is oscillatory and its amplitude grows like $\sqrt{\frac{x}{2\pi }}{e}^{x/\sqrt{2}}$. Therefore it is impossible to calculate the functions with any accuracy when $\sqrt{x}{e}^{x/\sqrt{2}}>\frac{\sqrt{2\pi }}{\delta }$. Note that this value of $x$ is much smaller than the minimum value of $x$ for which the function overflows.

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 s19ap_example

fprintf('s19ap example results\n\n');

x = [0.1; 1; 2.5; 5; 10; 15; -1];

[f, ivalid, ifail] = s19ap(x);

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

```
```s19ap example results

x           bei(x)   ivalid
1.000e-01   2.500e-03    0
1.000e+00   2.496e-01    0
2.500e+00   1.457e+00    0
5.000e+00   1.160e-01    0
1.000e+01   5.637e+01    0
1.500e+01  -2.953e+03    0
-1.000e+00   2.496e-01    0
```