public static double s09aa(
	double x,
	out int ifail
)

Public Shared Function s09aa ( _
	x As Double, _
	<OutAttribute> ByRef ifail As Integer _
) As Double

public:
static double s09aa(
	double x, 
	[OutAttribute] int% ifail
)

static member s09aa : 
        x : float * 
        ifail : int byref -> float 

s09aa returns the value of the inverse circular sine, $\arcsin x$. The value is in the principal range $\left(-\pi /2,\pi /2\right)$.

s09aa calculates an approximate value for the inverse circular sine, $\arcsin x$. It is based on the Chebyshev expansion where $-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$ and $t=4x^2-1$.

For $x^2\le \frac{1}{2}\text{, \hspace{1em}}\arcsin x=x\times y\left(x\right)$.

For $\frac{1}{2}<x^2\le 1\text{, \hspace{1em}}\arcsin x=\mathrm{sign} x\left\{\frac{\pi }{2}-\arcsin \sqrt{1-x^2}\right\}$.

For $x^2>1\text{, \hspace{1em}}\arcsin x$ is undefined and the method fails.

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

Errors or warnings detected by the method:

If $\delta$ and $\epsilon$ are the relative errors in the argument and result, respectively, then in principle That is, a relative error in the argument $x$ is amplified by at least a factor $\frac{x}{\arcsin x\sqrt{1-x^2}}$ in the result.

The equality should hold if $\delta$ is greater than the machine precision ($\delta$ is a result of data errors etc.) but if $\delta$ is produced simply by round-off error in the machine it is possible that rounding in internal calculations may lose an extra figure in the result.

This factor stays close to one except near $\left|x\right|=1$ where its behaviour is shown in the following graph.

For $\left|x\right|$ close to unity, $1-\left|x\right|\sim \delta$, the above analysis is no longer applicable owing to the fact that both argument and result are subject to finite bounds, ($\left|x\right|\le 1$ and $\left|\arcsin x\right|\le \frac{1}{2}\pi$). In this region $\epsilon \sim \sqrt{\delta }$; that is the result will have approximately half as many correct significant figures as the argument.

For $\left|x\right|=1$ the result will be correct to full machine precision.

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.