# §10.69 Uniform Asymptotic Expansions for Large Order

Let $U_{k}(p)$ and $V_{k}(p)$ be the polynomials defined in §10.41(ii), and

 10.69.1 $\xi=(1+ix^{2})^{\ifrac{1}{2}}.$ Defines: $\xi$ (locally) Symbols: $x$: real variable A&S Ref: 9.10.41 Permalink: http://dlmf.nist.gov/10.69.E1 Encodings: TeX, pMML, png See also: Annotations for 10.69

Then as $\nu\to+\infty$,

 10.69.2 $\displaystyle\mathop{\mathrm{ber}_{\nu}\/}\nolimits\!\left(\nu x\right)+i% \mathop{\mathrm{bei}_{\nu}\/}\nolimits\!\left(\nu x\right)$ $\displaystyle\sim\frac{e^{\nu\xi}}{(2\pi\nu\xi)^{\ifrac{1}{2}}}\left(\frac{xe^% {3\pi i/4}}{1+\xi}\right)^{\nu}\sum_{k=0}^{\infty}\frac{U_{k}(\xi^{-1})}{\nu^{% k}},$ 10.69.3 $\displaystyle\mathop{\mathrm{ker}_{\nu}\/}\nolimits\!\left(\nu x\right)+i% \mathop{\mathrm{kei}_{\nu}\/}\nolimits\!\left(\nu x\right)$ $\displaystyle\sim e^{-\nu\xi}\left(\frac{\pi}{2\nu\xi}\right)^{\ifrac{1}{2}}% \left(\frac{xe^{3\pi i/4}}{1+\xi}\right)^{-\nu}\sum_{k=0}^{\infty}(-1)^{k}% \frac{U_{k}(\xi^{-1})}{\nu^{k}},$ 10.69.4 $\displaystyle\mathop{\mathrm{ber}_{\nu}\/}\nolimits'\!\left(\nu x\right)+i% \mathop{\mathrm{bei}_{\nu}\/}\nolimits'\!\left(\nu x\right)$ $\displaystyle\sim\frac{e^{\nu\xi}}{x}\left(\frac{\xi}{2\pi\nu}\right)^{\ifrac{% 1}{2}}\left(\frac{xe^{3\pi i/4}}{1+\xi}\right)^{\nu}\sum_{k=0}^{\infty}\frac{V% _{k}(\xi^{-1})}{\nu^{k}},$ 10.69.5 $\displaystyle\mathop{\mathrm{ker}_{\nu}\/}\nolimits'\!\left(\nu x\right)+i% \mathop{\mathrm{kei}_{\nu}\/}\nolimits'\!\left(\nu x\right)$ $\displaystyle\sim-\frac{e^{-\nu\xi}}{x}\left(\frac{\pi\xi}{2\nu}\right)^{% \ifrac{1}{2}}\left(\frac{xe^{3\pi i/4}}{1+\xi}\right)^{-\nu}\*\sum_{k=0}^{% \infty}(-1)^{k}\frac{V_{k}(\xi^{-1})}{\nu^{k}},$

uniformly for $x$ $\in(0,\infty)$. All fractional powers take their principal values.

All four expansions also enjoy the same kind of double asymptotic property described in §10.41(iv).

Accuracy in (10.69.2) and (10.69.4) can be increased by including exponentially-small contributions as in (10.67.3), (10.67.4), (10.67.7), and (10.67.8) with $x$ replaced by $\nu x$.