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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_specfun_fresnel_s_vector (s20aq)

## Purpose

nag_specfun_fresnel_s_vector (s20aq) returns an array of values for the Fresnel integral $S\left(x\right)$.

## Syntax

[f, ifail] = s20aq(x, 'n', n)
[f, ifail] = nag_specfun_fresnel_s_vector(x, 'n', n)

## Description

nag_specfun_fresnel_s_vector (s20aq) evaluates an approximation to the Fresnel integral
 $Sxi=∫0xisinπ2t2dt$
for an array of arguments ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
Note:  $S\left(x\right)=-S\left(-x\right)$, so the approximation need only consider $x\ge 0.0$.
The function is based on three Chebyshev expansions:
For $0,
 $Sx=x3∑′r=0arTrt, with ​ t=2 x3 4-1.$
For $x>3$,
 $Sx=12-fxxcosπ2x2-gxx3sinπ2x2 ,$
where $f\left(x\right)=\underset{r=0}{{\sum }^{\prime }}\phantom{\rule{0.25em}{0ex}}{b}_{r}{T}_{r}\left(t\right)$,
and $g\left(x\right)=\underset{r=0}{{\sum }^{\prime }}\phantom{\rule{0.25em}{0ex}}{c}_{r}{T}_{r}\left(t\right)$,
with $t=2{\left(\frac{3}{x}\right)}^{4}-1$.
For small $x$, $S\left(x\right)\simeq \frac{\pi }{6}{x}^{3}$. This approximation is used when $x$ is sufficiently small for the result to be correct to machine precision. For very small $x$, this approximation would underflow; the result is then set exactly to zero.
For large $x$, $f\left(x\right)\simeq \frac{1}{\pi }$ and $g\left(x\right)\simeq \frac{1}{{\pi }^{2}}$. Therefore for moderately large $x$, when $\frac{1}{{\pi }^{2}{x}^{3}}$ is negligible compared with $\frac{1}{2}$, the second term in the approximation for $x>3$ may be dropped. For very large $x$, when $\frac{1}{\pi x}$ becomes negligible, $S\left(x\right)\simeq \frac{1}{2}$. However there will be considerable difficulties in calculating $\mathrm{cos}\left(\frac{\pi }{2}{x}^{2}\right)$ accurately before this final limiting value can be used. Since $\mathrm{cos}\left(\frac{\pi }{2}{x}^{2}\right)$ is periodic, its value is essentially determined by the fractional part of ${x}^{2}$. If ${x}^{2}=N+\theta$ where $N$ is an integer and $0\le \theta <1$, then $\mathrm{cos}\left(\frac{\pi }{2}{x}^{2}\right)$ depends on $\theta$ and on $N$ modulo $4$. By exploiting this fact, it is possible to retain significance in the calculation of $\mathrm{cos}\left(\frac{\pi }{2}{x}^{2}\right)$ either all the way to the very large $x$ limit, or at least until the integer part of $\frac{x}{2}$ is equal to the maximum integer allowed on the machine.

## 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
$S\left({x}_{i}\right)$, the function values.
2:     $\mathrm{ifail}$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  ${\mathbf{ifail}}=1$
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:
 $ε≃ x sin π2 x2 Sx δ.$
Figure 1 shows the behaviour of the error amplification factor $\left|\frac{x\mathrm{sin}\left(\frac{\pi }{2}{x}^{2}\right)}{S\left(x\right)}\right|$.
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 $x$, $\epsilon \simeq 3\delta$ and hence there is only moderate amplification of relative error. Of course for very small $x$ where the correct result would underflow and exact zero is returned, relative error-control is lost.
For moderately large values of $x$,
 $ε ≃ 2x sin π2 x2 δ$
and the result will be subject to increasingly large amplification of errors. However the above relation breaks down for large values of $x$ (i.e., when $\frac{1}{{x}^{2}}$ is of the order of the machine precision); in this region the relative error in the result is essentially bounded by $\frac{2}{\pi x}$.
Hence the effects of error amplification are limited and at worst the relative error loss should not exceed half the possible number of significant figures.
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 s20aq_example

fprintf('s20aq example results\n\n');

x = [0; 0.5; 1; 2; 4; 5; 6; 8; 10; -1; 1000];

[f, ifail] = s20aq(x);

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

```
```s20aq example results

x           S(x)
0.000e+00   0.000e+00
5.000e-01   6.473e-02
1.000e+00   4.383e-01
2.000e+00   3.434e-01
4.000e+00   4.205e-01
5.000e+00   4.992e-01
6.000e+00   4.470e-01
8.000e+00   4.602e-01
1.000e+01   4.682e-01
-1.000e+00  -4.383e-01
1.000e+03   4.997e-01
```