Difference between revisions of "Jacobi sn"
From specialfunctionswiki
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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 | 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 | ||
$$\mathrm{sn \hspace{2pt}}u = \sin \phi = x.$$ | $$\mathrm{sn \hspace{2pt}}u = \sin \phi = x.$$ | ||
| + | |||
| + | <div align="center"> | ||
| + | <gallery> | ||
| + | File:Domcoljacobisn.png|[[Domain coloring]] of $\mathrm{sn}$ corresponding to $m=0.8$. | ||
| + | </gallery> | ||
| + | </div> | ||
=Properties= | =Properties= | ||
Revision as of 02:01, 21 August 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 $$\mathrm{sn \hspace{2pt}}u = \sin \phi = x.$$
Domain coloring of $\mathrm{sn}$ corresponding to $m=0.8$.
Properties
- $\mathrm{sn \hspace{2pt}}^2u+\mathrm{cn \hspace{2pt}}^2u=1$
- $\mathrm{sn \hspace{2pt}}(0)=0$
- $m \mathrm{sn \hspace{2pt}}^2 u + \mathrm{dn \hspace{2pt}}^2u=1$
- $\mathrm{sn \hspace{2pt}}$ is an odd function
- $\dfrac{d}{du}\mathrm{sn \hspace{2pt}} u =\mathrm{cn \hspace{2pt}}(u)\mathrm{dn \hspace{2pt}}(u)$
References
Special functions by Leon Hall
