# change of modulus

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##### 1: 22.17 Moduli Outside the Interval [0,1]
###### §22.17 Moduli Outside the Interval [0,1]
22.17.6 $\operatorname{sn}\left(z,ik\right)=k_{1}^{\prime}\operatorname{sd}\left(z/k_{1% }^{\prime},k_{1}\right),$
22.17.7 $\operatorname{cn}\left(z,ik\right)=\operatorname{cd}\left(z/k_{1}^{\prime},k_{% 1}\right),$
22.17.8 $\operatorname{dn}\left(z,ik\right)=\operatorname{nd}\left(z/k_{1}^{\prime},k_{% 1}\right).$
For proofs of these results and further information see Walker (2003).
##### 2: 19.7 Connection Formulas
###### §19.7(ii) Change of Modulus and Amplitude
$\Pi\left(\phi,\alpha^{2},k_{1}\right)=k\Pi\left(\beta,k^{2}\alpha^{2},k\right),$ $k_{1}=1/k$, $\sin\beta=k_{1}\sin\phi\leq 1$.
$\kappa=\frac{k}{\sqrt{1+k^{2}}},$
$\kappa^{\prime}=\frac{1}{\sqrt{1+k^{2}}},$
##### 3: 22.11 Fourier and Hyperbolic Series
22.11.1 $\operatorname{sn}\left(z,k\right)=\frac{2\pi}{Kk}\sum_{n=0}^{\infty}\frac{q^{n% +\frac{1}{2}}\sin\left((2n+1)\zeta\right)}{1-q^{2n+1}},$
22.11.5 $\operatorname{sd}\left(z,k\right)=\frac{2\pi}{Kkk^{\prime}}\sum_{n=0}^{\infty}% \frac{(-1)^{n}q^{n+\frac{1}{2}}\sin\left((2n+1)\zeta\right)}{1+q^{2n+1}},$
22.11.6 $\operatorname{nd}\left(z,k\right)=\frac{\pi}{2Kk^{\prime}}+\frac{2\pi}{Kk^{% \prime}}\sum_{n=1}^{\infty}\frac{(-1)^{n}q^{n}\cos\left(2n\zeta\right)}{1+q^{2% n}}.$
22.11.11 $\operatorname{nc}\left(z,k\right)-\frac{\pi}{2Kk^{\prime}}\sec\zeta=-\frac{2% \pi}{Kk^{\prime}}\sum_{n=0}^{\infty}\frac{(-1)^{n}q^{2n+1}\cos\left((2n+1)% \zeta\right)}{1+q^{2n+1}},$
22.11.12 $\operatorname{sc}\left(z,k\right)-\frac{\pi}{2Kk^{\prime}}\tan\zeta=\frac{2\pi% }{Kk^{\prime}}\sum_{n=1}^{\infty}\frac{(-1)^{n}q^{2n}\sin\left(2n\zeta\right)}% {1+q^{2n}}.$
##### 4: 22.2 Definitions
22.2.3 $\zeta=\frac{\pi z}{2K\left(k\right)},$
22.2.4 $\operatorname{sn}\left(z,k\right)=\frac{\theta_{3}\left(0,q\right)}{\theta_{2}% \left(0,q\right)}\frac{\theta_{1}\left(\zeta,q\right)}{\theta_{4}\left(\zeta,q% \right)}=\frac{1}{\operatorname{ns}\left(z,k\right)},$
22.2.5 $\operatorname{cn}\left(z,k\right)=\frac{\theta_{4}\left(0,q\right)}{\theta_{2}% \left(0,q\right)}\frac{\theta_{2}\left(\zeta,q\right)}{\theta_{4}\left(\zeta,q% \right)}=\frac{1}{\operatorname{nc}\left(z,k\right)},$
22.2.6 $\operatorname{dn}\left(z,k\right)=\frac{\theta_{4}\left(0,q\right)}{\theta_{3}% \left(0,q\right)}\frac{\theta_{3}\left(\zeta,q\right)}{\theta_{4}\left(\zeta,q% \right)}=\frac{1}{\operatorname{nd}\left(z,k\right)},$
22.2.7 $\operatorname{sd}\left(z,k\right)=\frac{{\theta_{3}}^{2}\left(0,q\right)}{% \theta_{2}\left(0,q\right)\theta_{4}\left(0,q\right)}\frac{\theta_{1}\left(% \zeta,q\right)}{\theta_{3}\left(\zeta,q\right)}=\frac{1}{\operatorname{ds}% \left(z,k\right)},$
##### 5: 22.7 Landen Transformations
22.7.2 $\operatorname{sn}\left(z,k\right)=\frac{(1+k_{1})\operatorname{sn}\left(z/(1+k% _{1}),k_{1}\right)}{1+k_{1}{\operatorname{sn}}^{2}\left(z/(1+k_{1}),k_{1}% \right)},$
22.7.4 $\operatorname{dn}\left(z,k\right)=\frac{{\operatorname{dn}}^{2}\left(z/(1+k_{1% }),k_{1}\right)-(1-k_{1})}{1+k_{1}-{\operatorname{dn}}^{2}\left(z/(1+k_{1}),k_% {1}\right)}.$
22.7.6 $\operatorname{sn}\left(z,k\right)=\frac{(1+k_{2}^{\prime})\operatorname{sn}% \left(z/(1+k_{2}^{\prime}),k_{2}\right)\operatorname{cn}\left(z/(1+k_{2}^{% \prime}),k_{2}\right)}{\operatorname{dn}\left(z/(1+k_{2}^{\prime}),k_{2}\right% )},$
22.7.7 $\operatorname{cn}\left(z,k\right)=\frac{(1+k_{2}^{\prime})({\operatorname{dn}}% ^{2}\left(z/(1+k_{2}^{\prime}),k_{2}\right)-k_{2}^{\prime})}{k_{2}^{2}% \operatorname{dn}\left(z/(1+k_{2}^{\prime}),k_{2}\right)},$
22.7.8 $\operatorname{dn}\left(z,k\right)=\frac{(1-k_{2}^{\prime})({\operatorname{dn}}% ^{2}\left(z/(1+k_{2}^{\prime}),k_{2}\right)+k_{2}^{\prime})}{k_{2}^{2}% \operatorname{dn}\left(z/(1+k_{2}^{\prime}),k_{2}\right)}.$
##### 6: 22.6 Elementary Identities
###### §22.6(v) Change of Modulus
19.8.14 $2(k^{2}-\alpha^{2})\Pi\left(\phi,\alpha^{2},k\right)=\frac{\omega^{2}-\alpha^{% 2}}{1+k^{\prime}}\Pi\left(\phi_{1},\alpha_{1}^{2},k_{1}\right)+k^{2}F\left(% \phi,k\right)-{(1+k^{\prime})\alpha_{1}^{2}R_{C}\left(c_{1},c_{1}-\alpha_{1}^{% 2}\right)},$
$k_{2}=2\sqrt{k}/(1+k),$
$k_{1}=(1-k^{\prime})/(1+k^{\prime}),$
$\rho=\sqrt{1-(k^{2}/\alpha^{2})},$
$\alpha_{1}^{2}=\alpha^{2}(1+\rho)^{2}/(1+k^{\prime})^{2},$
If (19.36.1) is used instead of its first five terms, then the factor $(3r)^{-1/6}$ in Carlson (1995, (2.2)) is changed to $(3r)^{-1/8}$. For both $R_{D}$ and $R_{J}$ the factor $(r/4)^{-1/6}$ in Carlson (1995, (2.18)) is changed to $(r/5)^{-1/8}$ when the following polynomial of degree 7 (the same for both) is used instead of its first seven terms: …
31.2.8 $\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}\zeta}^{2}}+\left((2\gamma-1)\frac{% \operatorname{cn}\zeta\operatorname{dn}\zeta}{\operatorname{sn}\zeta}-(2\delta% -1)\frac{\operatorname{sn}\zeta\operatorname{dn}\zeta}{\operatorname{cn}\zeta}% -(2\epsilon-1)k^{2}\frac{\operatorname{sn}\zeta\operatorname{cn}\zeta}{% \operatorname{dn}\zeta}\right)\frac{\mathrm{d}w}{\mathrm{d}\zeta}+4k^{2}(% \alpha\beta{\operatorname{sn}}^{2}\zeta-q)w=0.$
In terms of Bessel functions, and with $\xi=\tfrac{2}{3}|x|^{3/2}$,
9.8.11 $\theta\left(x\right)=\tfrac{2}{3}\pi+\operatorname{arctan}\left(Y_{1/3}\left(% \xi\right)/J_{1/3}\left(\xi\right)\right),$