s11ab returns the value of the inverse hyperbolic sine, $\mathrm{arcsinh} x$.

# Syntax

C#
```public static double s11ab(
double x
)```
Visual Basic
```Public Shared Function s11ab ( _
x As Double _
) As Double```
Visual C++
```public:
static double s11ab(
double x
)```
F#
```static member s11ab :
x : float -> float
```

#### Parameters

x
Type: System..::..Double
On entry: the argument $x$ of the function.

#### Return Value

s11ab returns the value of the inverse hyperbolic sine, $\mathrm{arcsinh} x$.

# Description

s11ab calculates an approximate value for the inverse hyperbolic sine of its argument, $\mathrm{arcsinh} x$.
For $\left|x\right|\le 1$ it is based on the Chebyshev expansion
 $arcsinh x=x×yt=x∑′r=0crTrt, where ​t=2x2-1.$
For $\left|x\right|>1$ it uses the fact that
 $arcsinh x=sign x×lnx+x2+1.$
This form is used directly for $1<\left|x\right|<{10}^{k}$, where $k=n/2+1$, and the machine uses approximately $n$ decimal place arithmetic.
For $\left|x\right|\ge {10}^{k}$, $\sqrt{{x}^{2}+1}$ is equal to $\left|x\right|$ to within the accuracy of the machine and hence we can guard against premature overflow and, without loss of accuracy, calculate
 $arcsinh x=sign x×ln 2+lnx.$

# References

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

None.

# Accuracy

If $\delta$ and $\epsilon$ are the relative errors in the argument and the result, respectively, then in principle
 $ε≃x1+x2arcsinh xδ.$
That is, the relative error in the argument, $x$, is amplified by a factor at least $\frac{x}{\sqrt{1+{x}^{2}}\mathrm{arcsinh} x}$, in the result.
The equality should hold if $\delta$ is greater than the machine precision ($\delta$ due to data errors etc.) but if $\delta$ is simply due to round-off in the machine representation it is possible that an extra figure may be lost in internal calculation round-off.
The behaviour of the amplification factor is shown in the following graph:
Figure 1
It should be noted that this factor is always less than or equal to one. For large $x$ we have the absolute error in the result, $E$, in principle, given by
 $E∼δ.$
This means that eventually accuracy is limited by machine precision.

None.

None.

# Example

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

Example program (C#): s11abe.cs

Example program data: s11abe.d

Example program results: s11abe.r