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

The contributions of the terms in $\mathop{\mathrm{ker}_{\nu}\/}\nolimits x$, $\mathop{\mathrm{kei}_{\nu}\/}\nolimits x$, ${\mathop{\mathrm{ker}_{\nu}\/}\nolimits^{\prime}}x$, and ${\mathop{\mathrm{kei}_{\nu}\/}\nolimits^{\prime}}x$ 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 $\displaystyle{\mathop{\mathrm{ber}\/}\nolimits^{2}}x+{\mathop{\mathrm{bei}\/}% \nolimits^{2}}x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(1+\frac{1}{4\sqrt{2}}\frac{% 1}{x}+\frac{1}{64}\frac{1}{x^{2}}-\frac{33}{256\sqrt{2}}\frac{1}{x^{3}}-\frac{% 1797}{8192}\frac{1}{x^{4}}+\cdots\right),$ 10.67.10 $\displaystyle\mathop{\mathrm{ber}\/}\nolimits x{\mathop{\mathrm{bei}\/}% \nolimits^{\prime}}x-{\mathop{\mathrm{ber}\/}\nolimits^{\prime}}x\mathop{% \mathrm{bei}\/}\nolimits x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(\frac{1}{\sqrt{2}}+\frac{1}% {8}\frac{1}{x}+\frac{9}{64\sqrt{2}}\frac{1}{x^{2}}+\frac{39}{512}\frac{1}{x^{3% }}+\frac{75}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.11 $\displaystyle\mathop{\mathrm{ber}\/}\nolimits x{\mathop{\mathrm{ber}\/}% \nolimits^{\prime}}x+\mathop{\mathrm{bei}\/}\nolimits x{\mathop{\mathrm{bei}\/% }\nolimits^{\prime}}x$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(\frac{1}{\sqrt{2}}-\frac{3}% {8}\frac{1}{x}-\frac{15}{64\sqrt{2}}\frac{1}{x^{2}}-\frac{45}{512}\frac{1}{x^{% 3}}+\frac{315}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.12 $\displaystyle\left({\mathop{\mathrm{ber}\/}\nolimits^{\prime}}x\right)^{2}+% \left({\mathop{\mathrm{bei}\/}\nolimits^{\prime}}x\right)^{2}$ $\displaystyle\sim\frac{e^{x\sqrt{2}}}{2\pi x}\left(1-\frac{3}{4\sqrt{2}}\frac{% 1}{x}+\frac{9}{64}\frac{1}{x^{2}}+\frac{75}{256\sqrt{2}}\frac{1}{x^{3}}+\frac{% 2475}{8192}\frac{1}{x^{4}}+\cdots\right).$ 10.67.13 $\displaystyle{\mathop{\mathrm{ker}\/}\nolimits^{2}}x+{\mathop{\mathrm{kei}\/}% \nolimits^{2}}x$ $\displaystyle\sim\frac{\pi}{2x}e^{-x\sqrt{2}}\left(1-\frac{1}{4\sqrt{2}}\frac{% 1}{x}+\frac{1}{64}\frac{1}{x^{2}}+\frac{33}{256\sqrt{2}}\frac{1}{x^{3}}-\frac{% 1797}{8192}\frac{1}{x^{4}}+\cdots\right),$ 10.67.14 $\displaystyle\mathop{\mathrm{ker}\/}\nolimits x{\mathop{\mathrm{kei}\/}% \nolimits^{\prime}}x-{\mathop{\mathrm{ker}\/}\nolimits^{\prime}}x\mathop{% \mathrm{kei}\/}\nolimits x$ $\displaystyle\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}-\frac{1% }{8}\frac{1}{x}+\frac{9}{64\sqrt{2}}\frac{1}{x^{2}}-\frac{39}{512}\frac{1}{x^{% 3}}+\frac{75}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.15 $\displaystyle\mathop{\mathrm{ker}\/}\nolimits x{\mathop{\mathrm{ker}\/}% \nolimits^{\prime}}x+\mathop{\mathrm{kei}\/}\nolimits x{\mathop{\mathrm{kei}\/% }\nolimits^{\prime}}x$ $\displaystyle\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}+\frac{3% }{8}\frac{1}{x}-\frac{15}{64\sqrt{2}}\frac{1}{x^{2}}+\frac{45}{512}\frac{1}{x^% {3}}+\frac{315}{8192\sqrt{2}}\frac{1}{x^{4}}+\cdots\right),$ 10.67.16 $\displaystyle\left({\mathop{\mathrm{ker}\/}\nolimits^{\prime}}x\right)^{2}+% \left({\mathop{\mathrm{kei}\/}\nolimits^{\prime}}x\right)^{2}$ $\displaystyle\sim\frac{\pi}{2x}e^{-x\sqrt{2}}\left(1+\frac{3}{4\sqrt{2}}\frac{% 1}{x}+\frac{9}{64}\frac{1}{x^{2}}-\frac{75}{256\sqrt{2}}\frac{1}{x^{3}}+\frac{% 2475}{8192}\frac{1}{x^{4}}+\cdots\right).$