Revision as of 02:02, 21 March 2015

Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define $$\mathrm{dn \hspace{2pt}} u = \sqrt{1-m\sin^2 \phi} = \sqrt{1-mx^2}.$$

Properties

1. $m \mathrm{sn \hspace{2pt}}^2 u + \mathrm{dn \hspace{2pt}}^2u=1$
2. $\mathrm{dn \hspace{2pt}}(0)=1$
3. $\dfrac{d \phi}{du} = \mathrm{dn \hspace{2pt}}u$
4. $\dfrac{d}{du}\mathrm{sn \hspace{2pt}} u =\mathrm{cn \hspace{2pt}}(u)\mathrm{dn \hspace{2pt}}(u)$

References

Special functions by Leon Hall

Jacobi $\mathrm{cn}$

Jacobi $\mathrm{cd}$

Jacobi $\mathrm{cs}$

Jacobi $\mathrm{dc}$

Jacobi $\mathrm{dn}$

Jacobi $\mathrm{ds}$

Jacobi $\mathrm{nc}$

Jacobi $\mathrm{nd}$

Jacobi $\mathrm{ns}$

Jacobi $\mathrm{sc}$

Jacobi $\mathrm{sd}$

Jacobi $\mathrm{sn}$