2
$\begingroup$

What is the behavior of, $$\lim_{x \rightarrow \infty} K_n(x)$$ at least for $n \in [0, 1, 2]$?

Based on manually plotting these functions, it looks like in each case, $$K_n(x) \rightarrow a \frac{e^{-x}}{x^{1/2}}, \,\,\, a \sim 5/4$$ But I can't find any references to confirm this, or give an exact value of $a$.


I have a numerical calculation which involves the ratio of modified Bessel functions of the second kind: $K_n(x)$.

Something like,

$$f = \frac{K_0(x_0) + K_1(x_1)}{K_2(x_2)}$$

Where all of the $x$ variables are actually each dependent on the same independent variable $r$, i.e. $x_i = x_i(r)$, and all $x_i$ have similar (but not identical) behavior as $r\rightarrow \infty$.

The issue is that the Bessel functions very rapidly approach zero with increasing $x$, causing numerical issues... but for physical reasons I think the ratio is well-behaved and non-zero finite.

$\endgroup$

1 Answer 1

2
$\begingroup$

From https://dlmf.nist.gov/10.40#iii $$K_{\nu}\left(z\right)=\left(\frac{\pi}{2z}\right)^{\frac{1}{2}}e^{-z}\left(% \sum_{k=0}^{\ell-1}\frac{a_{k}(\nu)}{z^{k}}+R_{\ell}(\nu,z)\right)$$ or from Abramowitz/Stegun 9.7.2 (http://people.math.sfu.ca/~cbm/aands/page_378.htm) $$K_{\nu}\left(z\right)=\left(\frac{\pi}{2z}\right)^{\frac{1}{2}}e^{-z}\left( 1+\frac{\mu-1}{8z}+\frac{(\mu-1)(\mu-9)}{2!(8z)^2} + \cdots \right),\quad \mu = 4 \nu^2$$ So the value of your $a$ is $$a = \left(\frac{\pi}{2}\right)^{\frac{1}{2}}\approx 1.2533$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .