§10.68 Modulus and Phase Functions

§10.68(i) Definitions

 10.68.1 $\mathop{M_{\nu}\/}\nolimits\!\left(x\right)e^{i\!\mathop{\theta_{\nu}\/}% \nolimits\!\left(x\right)}=\mathop{\mathrm{ber}_{\nu}\/}\nolimits x+i\mathop{% \mathrm{bei}_{\nu}\/}\nolimits x,$
 10.68.2 $\mathop{N_{\nu}\/}\nolimits\!\left(x\right)e^{i\!\mathop{\phi_{\nu}\/}% \nolimits\!\left(x\right)}=\mathop{\mathrm{ker}_{\nu}\/}\nolimits x+i\mathop{% \mathrm{kei}_{\nu}\/}\nolimits x,$

where $\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\,(>0)$, $\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\,(>0)$, $\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$, and $\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ are continuous real functions of $x$ and $\nu$, with the branches of $\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ and $\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ chosen to satisfy (10.68.18) and (10.68.21) as $x\to\infty$. (See also §10.68(iv).)

§10.68(ii) Basic Properties

 10.68.3 $\displaystyle\mathop{\mathrm{ber}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\mathop{\cos\/}% \nolimits\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\mathrm{bei}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\mathop{\sin\/}% \nolimits\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right),$
 10.68.4 $\displaystyle\mathop{\mathrm{ker}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\mathop{\cos\/}% \nolimits\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\mathrm{kei}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\mathop{\sin\/}% \nolimits\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right).$
 10.68.5 $\displaystyle\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=({\mathop{\mathrm{ber}_{\nu}\/}\nolimits^{2}}x+{\mathop{\mathrm{% bei}_{\nu}\/}\nolimits^{2}}x)^{\ifrac{1}{2}},$ $\displaystyle\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=({\mathop{\mathrm{ker}_{\nu}\/}\nolimits^{2}}x+{\mathop{\mathrm{% kei}_{\nu}\/}\nolimits^{2}}x)^{\ifrac{1}{2}},$
 10.68.6 $\displaystyle\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\mathrm{Arctan}\/}\nolimits\!\left(\mathop{\mathrm{bei}_% {\nu}\/}\nolimits x/\mathop{\mathrm{ber}_{\nu}\/}\nolimits x\right),$ $\displaystyle\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\mathrm{Arctan}\/}\nolimits\!\left(\mathop{\mathrm{kei}_% {\nu}\/}\nolimits x/\mathop{\mathrm{ker}_{\nu}\/}\nolimits x\right).$
 10.68.7 $\displaystyle\mathop{M_{-n}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{M_{n}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\theta_{-n}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\theta_{n}\/}\nolimits\!\left(x\right)-n\pi.$

With arguments $(x)$ suppressed,

 10.68.8 $\mathop{\mathrm{ber}_{\nu}\/}\nolimits'x=\tfrac{1}{2}\mathop{M_{\nu+1}\/}% \nolimits\mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}\nolimits-% \tfrac{1}{4}\pi\right)-\tfrac{1}{2}\mathop{M_{\nu-1}\/}\nolimits\mathop{\cos\/% }\nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right)=(% \nu/x)\mathop{M_{\nu}\/}\nolimits\mathop{\cos\/}\nolimits\mathop{\theta_{\nu}% \/}\nolimits+\mathop{M_{\nu+1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(% \mathop{\theta_{\nu+1}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\nu/x)\mathop{M_{% \nu}\/}\nolimits\mathop{\cos\/}\nolimits\mathop{\theta_{\nu}\/}\nolimits-% \mathop{M_{\nu-1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{% \nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right),$
 10.68.9 $\mathop{\mathrm{bei}_{\nu}\/}\nolimits'x=\tfrac{1}{2}\mathop{M_{\nu+1}\/}% \nolimits\mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}\nolimits-% \tfrac{1}{4}\pi\right)-\tfrac{1}{2}\mathop{M_{\nu-1}\/}\nolimits\mathop{\sin\/% }\nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right)=(% \nu/x)\mathop{M_{\nu}\/}\nolimits\mathop{\sin\/}\nolimits\mathop{\theta_{\nu}% \/}\nolimits+\mathop{M_{\nu+1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(% \mathop{\theta_{\nu+1}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\nu/x)\mathop{M_{% \nu}\/}\nolimits\mathop{\sin\/}\nolimits\mathop{\theta_{\nu}\/}\nolimits-% \mathop{M_{\nu-1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{% \nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right).$
 10.68.10 $\displaystyle\mathop{\mathrm{ber}\/}\nolimits'x$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\tfrac{1}{4}\pi\right),$ $\displaystyle\mathop{\mathrm{bei}\/}\nolimits'x$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\tfrac{1}{4}\pi\right).$
 10.68.11 $\mathop{M_{\nu}\/}\nolimits'=(\nu/x)\mathop{M_{\nu}\/}\nolimits+\mathop{M_{\nu% +1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}% \nolimits-\mathop{\theta_{\nu}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\nu/x)% \mathop{M_{\nu}\/}\nolimits-\mathop{M_{\nu-1}\/}\nolimits\mathop{\cos\/}% \nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\mathop{\theta_{\nu}\/}% \nolimits-\tfrac{1}{4}\pi\right),$
 10.68.12 $\mathop{\theta_{\nu}\/}\nolimits'=(\mathop{M_{\nu+1}\/}\nolimits/\mathop{M_{% \nu}\/}\nolimits)\mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}% \nolimits-\mathop{\theta_{\nu}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\mathop{M_% {\nu-1}\/}\nolimits/\mathop{M_{\nu}\/}\nolimits)\mathop{\sin\/}\nolimits\!% \left(\mathop{\theta_{\nu-1}\/}\nolimits-\mathop{\theta_{\nu}\/}\nolimits-% \tfrac{1}{4}\pi\right).$
 10.68.13 $\displaystyle\mathop{M_{0}\/}\nolimits'$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\mathop{\theta_{0}\/}\nolimits-\tfrac{1}{4}\pi\right),$ $\displaystyle\mathop{\theta_{0}\/}\nolimits'$ $\displaystyle=(\mathop{M_{1}\/}\nolimits/\mathop{M_{0}\/}\nolimits)\mathop{% \sin\/}\nolimits\!\left(\mathop{\theta_{1}\/}\nolimits-\mathop{\theta_{0}\/}% \nolimits-\tfrac{1}{4}\pi\right).$
 10.68.14 $\displaystyle\ifrac{\mathrm{d}(x{\mathop{M_{\nu}\/}\nolimits^{2}}\mathop{% \theta_{\nu}\/}\nolimits')}{\mathrm{d}x}$ $\displaystyle=x{\mathop{M_{\nu}\/}\nolimits^{2}},$ $\displaystyle x^{2}\mathop{M_{\nu}\/}\nolimits''+x\mathop{M_{\nu}\/}\nolimits'% -\nu^{2}\mathop{M_{\nu}\/}\nolimits$ $\displaystyle=x^{2}\mathop{M_{\nu}\/}\nolimits{\mathop{\theta_{\nu}\/}% \nolimits'^{2}}.$

Equations (10.68.8)–(10.68.14) also hold with the symbols $\mathop{\mathrm{ber}\/}\nolimits$, $\mathop{\mathrm{bei}\/}\nolimits$, $\mathop{M\/}\nolimits$, and $\mathop{\theta\/}\nolimits$ replaced throughout by $\mathop{\mathrm{ker}\/}\nolimits$, $\mathop{\mathrm{kei}\/}\nolimits$, $\mathop{N\/}\nolimits$, and $\mathop{\phi\/}\nolimits$, respectively. In place of (10.68.7),

 10.68.15 $\displaystyle\mathop{N_{-\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\phi_{-\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)+\nu\pi.$

§10.68(iii) Asymptotic Expansions for Large Argument

When $\nu$ is fixed, $\mu=4\nu^{2}$, and $x\to\infty$

 10.68.16 $\displaystyle\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\left(1-\frac{\mu-1% }{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)^{2}}{256}\frac{1}{x^{2}}-\frac{(\mu-1)(% \mu^{2}+14\mu-399)}{6144\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(% \frac{1}{x^{4}}\right)\right),$ 10.68.17 $\displaystyle\mathop{\ln\/}\nolimits\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{x}{\sqrt{2}}-\frac{1}{2}\mathop{\ln\/}\nolimits\!\left(2% \pi x\right)-\frac{\mu-1}{8\sqrt{2}}\frac{1}{x}-\frac{(\mu-1)(\mu-25)}{384% \sqrt{2}}\frac{1}{x^{3}}-\frac{(\mu-1)(\mu-13)}{128}\frac{1}{x^{4}}+\mathop{O% \/}\nolimits\!\left(\frac{1}{x^{5}}\right),$ 10.68.18 $\displaystyle\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{x}{\sqrt{2}}+\left(\frac{1}{2}\nu-\frac{1}{8}\right)\pi+% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{\mu-1}{16}\frac{1}{x^{2}}-\frac{(\mu-% 1)(\mu-25)}{384\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(\frac{1}{% x^{5}}\right).$ 10.68.19 $\displaystyle\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\left(1+% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)^{2}}{256}\frac{1}{x^{2}}+% \frac{(\mu-1)(\mu^{2}+14\mu-399)}{6144\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}% \nolimits\!\left(\frac{1}{x^{4}}\right)\right),$ 10.68.20 $\displaystyle\mathop{\ln\/}\nolimits\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=-\frac{x}{\sqrt{2}}+\frac{1}{2}\mathop{\ln\/}\nolimits\!\left(% \frac{\pi}{2x}\right)+\frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)(\mu-25)% }{384\sqrt{2}}\frac{1}{x^{3}}-\frac{(\mu-1)(\mu-13)}{128}\frac{1}{x^{4}}+% \mathop{O\/}\nolimits\!\left(\frac{1}{x^{5}}\right),$ 10.68.21 $\displaystyle\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=-\frac{x}{\sqrt{2}}-\left(\frac{1}{2}\nu+\frac{1}{8}\right)\pi-% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{\mu-1}{16}\frac{1}{x^{2}}+\frac{(\mu-% 1)(\mu-25)}{384\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(\frac{1}{% x^{5}}\right).$

§10.68(iv) Further Properties

Additional properties of the modulus and phase functions are given in Young and Kirk (1964, pp. xi–xv). However, care needs to be exercised with the branches of the phases. Thus this reference gives $\mathop{\phi_{1}\/}\nolimits\!\left(0\right)=\tfrac{5}{4}\pi$ (Eq. (6.10)), and $\lim_{x\to\infty}(\mathop{\phi_{1}\/}\nolimits\!\left(x\right)+(x/\sqrt{2}))=-% \tfrac{5}{8}\pi$ (Eqs. (10.20) and (Eqs. (10.26b)). However, numerical tabulations show that if the second of these equations applies and $\mathop{\phi_{1}\/}\nolimits\!\left(x\right)$ is continuous, then $\mathop{\phi_{1}\/}\nolimits\!\left(0\right)=-\tfrac{3}{4}\pi$; compare Abramowitz and Stegun (1964, p. 433).