s17ah returns a value of the Airy function, $\mathrm{Bi}\left(x\right)$.

# Syntax

C# |
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public static double s17ah( double x, out int ifail ) |

Visual Basic |
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Public Shared Function s17ah ( _ x As Double, _ <OutAttribute> ByRef ifail As Integer _ ) As Double |

Visual C++ |
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public: static double s17ah( double x, [OutAttribute] int% ifail ) |

F# |
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static member s17ah : x : float * ifail : int byref -> float |

#### Parameters

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

- ifail
- Type: System..::..Int32%
*On exit*: ${\mathbf{ifail}}={0}$ unless the method detects an error or a warning has been flagged (see [Error Indicators and Warnings]).

#### Return Value

s17ah returns a value of the Airy function, $\mathrm{Bi}\left(x\right)$.

# Description

s17ah evaluates an approximation to the Airy function $\mathrm{Bi}\left(x\right)$. It is based on a number of Chebyshev expansions.

For $x<-5$,

where $z=\frac{\pi}{4}+\frac{2}{3}\sqrt{-{x}^{3}}$ and $a\left(t\right)$ and $b\left(t\right)$ are expansions in the variable $t=-2{\left(\frac{5}{x}\right)}^{3}-1$.

$$\mathrm{Bi}\left(x\right)=\frac{a\left(t\right)\mathrm{cos}\u200az+b\left(t\right)\mathrm{sin}\u200az}{{\left(-x\right)}^{1/4}}\text{,}$$ |

For $-5\le x\le 0$,

where $f$ and $g$ are expansions in $t=-2{\left(\frac{x}{5}\right)}^{3}-1$.

$$\mathrm{Bi}\left(x\right)=\sqrt{3}\left(f\left(t\right)+xg\left(t\right)\right)\text{,}$$ |

For $0<x<4.5$,

where $y$ is an expansion in $t=4x/9-1$.

$$\mathrm{Bi}\left(x\right)={e}^{11x/8}y\left(t\right)\text{,}$$ |

For $4.5\le x<9$,

where $v$ is an expansion in $t=4x/9-3$.

$$\mathrm{Bi}\left(x\right)={e}^{5x/2}v\left(t\right)\text{,}$$ |

For $x\ge 9$,

where $z=\frac{2}{3}\sqrt{{x}^{3}}$ and $u$ is an expansion in $t=2\left(\frac{18}{z}\right)-1$.

$$\mathrm{Bi}\left(x\right)=\frac{{e}^{z}u\left(t\right)}{{x}^{1/4}}\text{,}$$ |

For $\left|x\right|<\mathit{machineprecision}$, the result is set directly to $\mathrm{Bi}\left(0\right)$. This both saves time and avoids possible intermediate underflows.

For large negative arguments, it becomes impossible to calculate the phase of the oscillating function with any accuracy so the method must fail. This occurs if $x<-{\left(\frac{3}{2\epsilon}\right)}^{2/3}$, where $\epsilon $ is the machine precision.

For large positive arguments, there is a danger of causing overflow since Bi grows in an essentially exponential manner, so the method must fail.

# References

Abramowitz M and Stegun I A (1972)

*Handbook of Mathematical Functions*(3rd Edition) Dover Publications# Error Indicators and Warnings

Errors or warnings detected by the method:

- ${\mathbf{ifail}}=1$
- x is too large and positive. On failure, the method returns zero. (see the Users' Note for your implementation for details)

- ${\mathbf{ifail}}=2$
- x is too large and negative. On failure, the method returns zero. See also the Users' Note for your implementation.

# Accuracy

For negative arguments the function is oscillatory and hence absolute error is the appropriate measure. In the positive region the function is essentially exponential-like and here relative error is appropriate. The absolute error, $E$, and the relative error, $\epsilon $, are related in principle to the relative error in the argument, $\delta $, by

In practice, approximate equality is the best that can be expected. When $\delta $, $\epsilon $ or $E$ is of the order of the machine precision, the errors in the result will be somewhat larger.

$$E\simeq \left|x{\mathrm{Bi}}^{\prime}\left(x\right)\right|\delta ,\epsilon \simeq \left|\frac{x{\mathrm{Bi}}^{\prime}\left(x\right)}{\mathrm{Bi}\left(x\right)}\right|\delta \text{.}$$ |

For small $x$, errors are strongly damped and hence will be bounded essentially by the machine precision.

For moderate to large negative $x$, the error behaviour is clearly oscillatory but the amplitude of the error grows like amplitude $\left(\frac{E}{\delta}\right)\sim \frac{{\left|x\right|}^{5/4}}{\sqrt{\pi}}$.

However the phase error will be growing roughly as $\frac{2}{3}\sqrt{{\left|x\right|}^{3}}$ and hence all accuracy will be lost for large negative arguments. This is due to the impossibility of calculating sin and cos to any accuracy if $\frac{2}{3}\sqrt{{\left|x\right|}^{3}}>\frac{1}{\delta}$.

For large positive arguments, the relative error amplification is considerable:

This means a loss of roughly two decimal places accuracy for arguments in the region of $20$. However very large arguments are not possible due to the danger of causing overflow and errors are therefore limited in practice.

$$\frac{\epsilon}{\delta}\sim \sqrt{{x}^{3}}\text{.}$$ |

# Parallelism and Performance

None.

# Further Comments

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#): s17ahe.cs