Jacobi dn

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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[edit]

  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[edit]

Special functions by Leon Hall

Jacobi Elliptic Functions