# §10.67 Asymptotic Expansions for Large Argument

## §10.67(i) $\operatorname{ber}_{\nu}x,\operatorname{bei}_{\nu}x,\operatorname{ker}_{\nu}x,% \operatorname{kei}_{\nu}x$, and Derivatives

Define $a_{k}(\nu)$ and $b_{k}(\nu)$ as in §§10.17(i) and 10.17(ii). Then as $x\to\infty$ with $\nu$ fixed,

 10.67.1 $\displaystyle\operatorname{ker}_{\nu}x$ $\displaystyle\sim e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\*% \sum_{k=0}^{\infty}\frac{a_{k}(\nu)}{x^{k}}\cos\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{k}{4}+\frac{1}{8}\right)\pi\right),$ 10.67.2 $\displaystyle\operatorname{kei}_{\nu}x$ $\displaystyle\sim-e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\*% \sum_{k=0}^{\infty}\frac{a_{k}(\nu)}{x^{k}}\*\sin\left(\frac{x}{\sqrt{2}}+% \left(\frac{\nu}{2}+\frac{k}{4}+\frac{1}{8}\right)\pi\right).$
 10.67.3 $\operatorname{ber}_{\nu}x\sim\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\*% \sum_{k=0}^{\infty}\frac{a_{k}(\nu)}{x^{k}}\cos\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{3k}{4}-\frac{1}{8}\right)\pi\right)-\frac{1}{\pi}(\sin% \left(2\nu\pi\right)\operatorname{ker}_{\nu}x+\cos\left(2\nu\pi\right)% \operatorname{kei}_{\nu}x),$
 10.67.4 $\operatorname{bei}_{\nu}x\sim\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\*% \sum_{k=0}^{\infty}\frac{a_{k}(\nu)}{x^{k}}\sin\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{3k}{4}-\frac{1}{8}\right)\pi\right)+\frac{1}{\pi}(\cos% \left(2\nu\pi\right)\operatorname{ker}_{\nu}x-\sin\left(2\nu\pi\right)% \operatorname{kei}_{\nu}x).$
 10.67.5 $\displaystyle\operatorname{ker}_{\nu}'x$ $\displaystyle\sim-e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\sum% _{k=0}^{\infty}\frac{b_{k}(\nu)}{x^{k}}\*\cos\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{k}{4}-\frac{1}{8}\right)\pi\right),$ 10.67.6 $\displaystyle\operatorname{kei}_{\nu}'x$ $\displaystyle\sim e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\sum% _{k=0}^{\infty}\frac{b_{k}(\nu)}{x^{k}}\*\sin\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{k}{4}-\frac{1}{8}\right)\pi\right).$
 10.67.7 $\operatorname{ber}_{\nu}'x\sim\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\*% \sum_{k=0}^{\infty}\frac{b_{k}(\nu)}{x^{k}}\cos\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{3k}{4}+\frac{1}{8}\right)\pi\right)-\frac{1}{\pi}(\sin% \left(2\nu\pi\right)\operatorname{ker}_{\nu}'x+\cos\left(2\nu\pi\right)% \operatorname{kei}_{\nu}'x),$
 10.67.8 $\operatorname{bei}_{\nu}'x\sim\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\*% \sum_{k=0}^{\infty}\frac{b_{k}(\nu)}{x^{k}}\sin\left(\frac{x}{\sqrt{2}}+\left(% \frac{\nu}{2}+\frac{3k}{4}+\frac{1}{8}\right)\pi\right)+\frac{1}{\pi}(\cos% \left(2\nu\pi\right)\operatorname{ker}_{\nu}'x-\sin\left(2\nu\pi\right)% \operatorname{kei}_{\nu}'x).$

The contributions of the terms in $\operatorname{ker}_{\nu}x$, $\operatorname{kei}_{\nu}x$, $\operatorname{ker}_{\nu}'x$, and $\operatorname{kei}_{\nu}'x$ on the right-hand sides of (10.67.3), (10.67.4), (10.67.7), and (10.67.8) are exponentially small compared with the other terms, and hence can be neglected in the sense of Poincaré asymptotic expansions (§2.1(iii)). However, their inclusion improves numerical accuracy.

## §10.67(ii) Cross-Products and Sums of Squares in the Case $\nu=0$

As $x\to\infty$

 10.67.9 $\displaystyle{\operatorname{ber}^{2}}x+{\operatorname{bei}^{2}}x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(1+\frac{1}{4\sqrt{2}}\frac{% 1}{x}+\frac{1}{64}\frac{1}{x^{2}}-\frac{33}{256\sqrt{2}}\frac{1}{x^{3}}-\frac{% 1797}{8192}\frac{1}{x^{4}}+\cdots\right),$ 10.67.10 $\displaystyle\operatorname{ber}x\operatorname{bei}'x-\operatorname{ber}'x% \operatorname{bei}x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(\frac{1}{\sqrt{2}}+\frac{1}% {8}\frac{1}{x}+\frac{9}{64\sqrt{2}}\frac{1}{x^{2}}+\frac{39}{512}\frac{1}{x^{3% }}+\frac{75}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.11 $\displaystyle\operatorname{ber}x\operatorname{ber}'x+\operatorname{bei}x% \operatorname{bei}'x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(\frac{1}{\sqrt{2}}-\frac{3}% {8}\frac{1}{x}-\frac{15}{64\sqrt{2}}\frac{1}{x^{2}}-\frac{45}{512}\frac{1}{x^{% 3}}+\frac{315}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.12 $\displaystyle\left(\operatorname{ber}'x\right)^{2}+\left(\operatorname{bei}'x% \right)^{2}$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(1-\frac{3}{4\sqrt{2}}\frac{% 1}{x}+\frac{9}{64}\frac{1}{x^{2}}+\frac{75}{256\sqrt{2}}\frac{1}{x^{3}}+\frac{% 2475}{8192}\frac{1}{x^{4}}+\cdots\right).$ 10.67.13 $\displaystyle{\operatorname{ker}^{2}}x+{\operatorname{kei}^{2}}x$ $\displaystyle\sim\frac{\pi}{2x}e^{-x\sqrt{2}}\left(1-\frac{1}{4\sqrt{2}}\frac{% 1}{x}+\frac{1}{64}\frac{1}{x^{2}}+\frac{33}{256\sqrt{2}}\frac{1}{x^{3}}-\frac{% 1797}{8192}\frac{1}{x^{4}}+\cdots\right),$ 10.67.14 $\displaystyle\operatorname{ker}x\operatorname{kei}'x-\operatorname{ker}'x% \operatorname{kei}x$ $\displaystyle\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}-\frac{1% }{8}\frac{1}{x}+\frac{9}{64\sqrt{2}}\frac{1}{x^{2}}-\frac{39}{512}\frac{1}{x^{% 3}}+\frac{75}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.15 $\displaystyle\operatorname{ker}x\operatorname{ker}'x+\operatorname{kei}x% \operatorname{kei}'x$ $\displaystyle\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}+\frac{3% }{8}\frac{1}{x}-\frac{15}{64\sqrt{2}}\frac{1}{x^{2}}+\frac{45}{512}\frac{1}{x^% {3}}+\frac{315}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.16 $\displaystyle\left(\operatorname{ker}'x\right)^{2}+\left(\operatorname{kei}'x% \right)^{2}$ $\displaystyle\sim\frac{\pi}{2x}e^{-x\sqrt{2}}\left(1+\frac{3}{4\sqrt{2}}\frac{% 1}{x}+\frac{9}{64}\frac{1}{x^{2}}-\frac{75}{256\sqrt{2}}\frac{1}{x^{3}}+\frac{% 2475}{8192}\frac{1}{x^{4}}+\cdots\right).$