public static double s19ab(
	double x,
	out int ifail
)

Public Shared Function s19ab ( _
	x As Double, _
	<OutAttribute> ByRef ifail As Integer _
) As Double

public:
static double s19ab(
	double x, 
	[OutAttribute] int% ifail
)

static member s19ab : 
        x : float * 
        ifail : int byref -> float 

s19ab returns a value for the Kelvin function $\mathrm{bei} x$.

s19ab evaluates an approximation to the Kelvin function $\mathrm{bei} x$.

Note:  $\mathrm{bei}\left(-x\right)=\mathrm{bei} x$, so the approximation need only consider $x\ge 0.0$.

The method is based on several Chebyshev expansions:

For $0\le x\le 5$, For $x>5$, 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 method must fail.

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

Errors or warnings detected by the method:

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: (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.

None.

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