Difference between revisions of "Jacobi cn"

From specialfunctionswiki
Jump to: navigation, search
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
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
 
$$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$
 
$$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$
 +
 +
<div align="center">
 +
<gallery>
 +
File:Complexjacobicn,m=0.8plot.png|[[Domain coloring]] of $\mathrm{cn}$ with $m=0.8$.
 +
</gallery>
 +
</div>
  
 
=Properties=
 
=Properties=
Line 11: Line 17:
 
[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]
  
<center>{{:Jacobi elliptic functions footer}}</center>
+
{{:Jacobi elliptic functions footer}}
 +
 
 +
[[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 $$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$

Properties

  1. $\mathrm{sn \hspace{2pt}}^2u+\mathrm{cn \hspace{2pt}}^2u=1$
  2. $\mathrm{cn \hspace{2pt}}(0)=1$
  3. $\mathrm{cn \hspace{2pt}}$ is an even function
  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 Elliptic Functions