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I have got a question concerning the Airy functions in relation to the Bessel function.

From Wiki, it is possible to see how

$$ Ai(x)=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{1/3}\left(\frac{2}{3}x^{\frac{3}{2}}\right) $$

The question is: how can the Airy function retrieve a 0.3550 value when evaluated in $ x = 0 $, if

$$ K_{1/3}\left(\frac{2}{3}0^{\frac{3}{2}}\right) = \infty $$

It's probably a naive question but I would say that

$$ Ai(0) = 0*\infty = NaN$$

looking at the above equivalence.

I thank you in advance for supporting.

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3 Answers 3

up vote 3 down vote accepted

For fixed $\nu>0\;$ and from the reference DLMF you have the equivalence for $z$ near $0$ : $$\operatorname{K}_{\nu}(z)\sim \frac{\Gamma(\nu)}2\left(\frac z2\right)^{-\nu}$$ From this you may deduce that near $0$ the Airy function $\operatorname{Ai}(x)$ is equivalent to : \begin{align} \frac{1}{\pi}\sqrt{\frac{x}{3}}\operatorname{K}_{1/3}\left(\frac{2}{3}x^{\frac{3}{2}}\right)&\sim \frac{\Gamma(1/3)}{2\pi}\sqrt{\frac{x}{3}}\left(\frac{1}{3}x^{\frac{3}{2}}\right)^{-1/3}\\ &\\ &\sim \frac{\sqrt[3]{3}\ \Gamma(1/3)}{\sqrt{3}\;2\,\pi}\\ &\\ &\sim 0.355028053887817239260063186\cdots \end{align} Getting even a closed form for the limit.

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You can formulate your problem in the following way: compute the limit of of $\frac{f(x)}{g(x)}$ as $x\to0$ where $$f(x)=\sqrt x\qquad \text{and}\qquad g(x)=\frac{1}{K_{\frac13}(x^{\frac32})}.$$

To study this, you can for example use a Taylor expansion at $x=0$ for $f$ and $g$.

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that means using standard analysis techniques such as De l'Hopital theorem or infinitesimal orders? –  fpe Jun 21 '13 at 19:53

Computationally, $0 * \infty$ is $NaN$. But we're talking math here, not computers and floating point arithmetic. (Or, at least, I assume that's what we're talking.) You need to take the limit,

$\displaystyle \lim_{x\rightarrow 0} \text{Ai}(x)$

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