s18ac Method
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s18ac returns the value of the modified Bessel Function K0x.

Syntax

C#
public static double s18ac(
	double x,
	out int ifail
)
Visual Basic (Declaration)
Public Shared Function s18ac ( _
	x As Double, _
	<OutAttribute> ByRef ifail As Integer _
) As Double
Visual C++
public:
static double s18ac(
	double x, 
	[OutAttribute] int% ifail
)
F#
static member s18ac : 
        x:float * 
        ifail:int byref -> float 

Parameters

x
Type: System..::.Double
On entry: the argument x of the function.
Constraint: x>0.0.
ifail
Type: System..::.Int32 %
On exit: ifail=0 unless the method detects an error (see [Error Indicators and Warnings]).

Return Value

s18ac returns the value of the modified Bessel Function K0x.

Description

s18ac evaluates an approximation to the modified Bessel Function of the second kind K0x.
Note:  K0x is undefined for x0 and the method will fail for such arguments.
The method is based on five Chebyshev expansions:
For 0<x1,
K0x=-lnx'r=0arTrt+'r=0brTrt,   where ​t=2x2-1.
For 1<x2,
K0x=e-x'r=0crTrt,   where ​t=2x-3.
For 2<x4,
K0x=e-x'r=0drTrt,   where ​t=x-3.
For x>4,
K0x=e-xx 'r=0erTrt,where ​ t=9-x 1+x .
For x near zero, K0x-γ-ln x2 , where γ denotes Euler's constant. This approximation is used when x is sufficiently small for the result to be correct to machine precision.
For large x, where there is a danger of underflow due to the smallness of K0, the result is set exactly to zero.

References

Error Indicators and Warnings

Accuracy

Let δ and ε be the relative errors in the argument and result respectively.
However, if δ is of the same order as machine precision, then rounding errors could make ε slightly larger than the above relation predicts.
For small x, the amplification factor is approximately 1lnx , which implies strong attenuation of the error, but in general ε can never be less than the machine precision.
For large x, εxδ and we have strong amplification of the relative error. Eventually K0, which is asymptotically given by e-xx , becomes so small that it cannot be calculated without underflow and hence the method will return zero. Note that for large x the errors will be dominated by those of the exponential function.

Further Comments

Example

See Also