Difference between revisions of "Jacobi cn"
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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 | ||
$$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$ | $$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$ | ||
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| + | <div align="center"> | ||
| + | <gallery> | ||
| + | File:Complexjacobicn,m=0.8plot.png|[[Domain coloring]] of $\mathrm{cn}$ with $m=0.8$. | ||
| + | </gallery> | ||
| + | </div> | ||
=Properties= | =Properties= | ||
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[http://web.mst.edu/~lmhall/SPFNS/spfns.pdf Special functions by Leon Hall] | [http://web.mst.edu/~lmhall/SPFNS/spfns.pdf Special functions by Leon Hall] | ||
| − | + | {{:Jacobi elliptic functions footer}} | |
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| + | [[Category:SpecialFunction]] | ||
Latest revision as of 19:06, 5 July 2016
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 $$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$
Domain coloring of $\mathrm{cn}$ with $m=0.8$.
Properties
- $\mathrm{sn \hspace{2pt}}^2u+\mathrm{cn \hspace{2pt}}^2u=1$
- $\mathrm{cn \hspace{2pt}}(0)=1$
- $\mathrm{cn \hspace{2pt}}$ is an even 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
