# §29.5 Special Cases and Limiting Forms

 29.5.1 $a^{m}_{\nu}\left(0\right)=b^{m}_{\nu}\left(0\right)=m^{2},$
 29.5.2 $\mathit{Ec}^{0}_{\nu}\left(z,0\right)=2^{-\frac{1}{2}},$ ⓘ Symbols: $\mathit{Ec}^{\NVar{m}}_{\NVar{\nu}}\left(\NVar{z},\NVar{k^{2}}\right)$: Lamé function, $z$: complex variable and $\nu$: real parameter Permalink: http://dlmf.nist.gov/29.5.E2 Encodings: TeX, pMML, png See also: Annotations for §29.5 and Ch.29
 29.5.3 $\displaystyle\mathit{Ec}^{m}_{\nu}\left(z,0\right)$ $\displaystyle=\cos\left(m(\tfrac{1}{2}\pi-z)\right),$ $m\geq 1$, $\displaystyle\mathit{Es}^{m}_{\nu}\left(z,0\right)$ $\displaystyle=\sin\left(m(\tfrac{1}{2}\pi-z)\right),$ $m\geq 1$.

Let $\mu=\max{(\nu-m,0)}$. Then

 29.5.4 $\lim_{k\to 1-}a^{m}_{\nu}\left(k^{2}\right)=\lim_{k\to 1-}b^{m+1}_{\nu}\left(k% ^{2}\right)=\nu(\nu+1)-\mu^{2},$
 29.5.5 ${\lim_{k\to 1-}\frac{\mathit{Ec}^{m}_{\nu}\left(z,k^{2}\right)}{\mathit{Ec}^{m% }_{\nu}\left(0,k^{2}\right)}=\lim_{k\to 1-}\frac{\mathit{Es}^{m+1}_{\nu}\left(% z,k^{2}\right)}{\mathit{Es}^{m+1}_{\nu}\left(0,k^{2}\right)}}=\frac{1}{(\cosh z% )^{\mu}}F\left({\tfrac{1}{2}\mu-\tfrac{1}{2}\nu,\tfrac{1}{2}\mu+\tfrac{1}{2}% \nu+\tfrac{1}{2}\atop\tfrac{1}{2}};{\tanh}^{2}z\right),$ $m$ even,
 29.5.6 $\lim_{k\to 1-}\frac{\mathit{Ec}^{m}_{\nu}\left(z,k^{2}\right)}{\left.\ifrac{% \mathrm{d}\mathit{Ec}^{m}_{\nu}\left(z,k^{2}\right)}{\mathrm{d}z}\right|_{z=0}% }=\lim_{k\to 1-}\frac{\mathit{Es}^{m+1}_{\nu}\left(z,k^{2}\right)}{\left.% \ifrac{\mathrm{d}\mathit{Es}^{m+1}_{\nu}\left(z,k^{2}\right)}{\mathrm{d}z}% \right|_{z=0}}=\frac{\tanh z}{(\cosh z)^{\mu}}F\left({\tfrac{1}{2}\mu-\tfrac{1% }{2}\nu+\tfrac{1}{2},\tfrac{1}{2}\mu+\tfrac{1}{2}\nu+1\atop\tfrac{3}{2}};{% \tanh}^{2}z\right),$ $m$ odd,

where $F$ is the hypergeometric function; see §15.2(i).

If $k\to 0+$ and $\nu\to\infty$ in such a way that $k^{2}\nu(\nu+1)=4\theta$ (a positive constant), then

 29.5.7 $\displaystyle\lim\mathit{Ec}^{m}_{\nu}\left(z,k^{2}\right)$ $\displaystyle=\operatorname{ce}_{m}\left(\tfrac{1}{2}\pi-z,\theta\right),$ $\displaystyle\lim\mathit{Es}^{m}_{\nu}\left(z,k^{2}\right)$ $\displaystyle=\operatorname{se}_{m}\left(\tfrac{1}{2}\pi-z,\theta\right),$

where $\operatorname{ce}_{m}\left(z,\theta\right)$ and $\operatorname{se}_{m}\left(z,\theta\right)$ are Mathieu functions; see §28.2(vi).