Editing Jacobi elliptic functions (section)

==Identities==
===Half angle formula===
<math display="block">\operatorname{sn}\left(\frac{u}{2},m\right)=\pm\sqrt{\frac{1-\operatorname{cn}(u,m)}{1+\operatorname{dn}(u,m)}}</math>
<math display="block">\operatorname{cn}\left(\frac{u}{2},m\right)=\pm\sqrt{\frac{\operatorname{cn}(u,m)+\operatorname{dn}(u,m)}{1+\operatorname{dn}(u,m)}}</math>
<math display="block">\operatorname{cn}\left(\frac{u}{2},m\right)=\pm\sqrt{\frac{m'+\operatorname{dn}(u,m)+m\operatorname{cn}(u,m)}{1+\operatorname{dn}(u,m)}}</math>

===K formulas===
'''Half K formula'''

<math display="block">\operatorname{sn}\left[\tfrac{1}{2}K(k); k\right] = \frac{\sqrt{2}}{\sqrt{1+k} + \sqrt{1-k}} </math>

<math display="block">\operatorname{cn}\left[\tfrac{1}{2}K(k); k\right] = \frac{\sqrt{2}\,\sqrt[4]{1-k^2}}{\sqrt{1+k} + \sqrt{1-k}} </math>

<math display="block"> \operatorname{dn}\left[\tfrac{1}{2}K(k); k\right] = \sqrt[4]{1-k^2} </math>
	
'''Third K formula'''
	
:<math>\operatorname{sn}\left[\frac{1}{3}K\left(\frac{x^3}{\sqrt{x^6+1}+1}\right);\frac{x^3}{\sqrt{x^6+1}+1}\right] = 
	
\frac{\sqrt{2\sqrt{x^4-x^2+1}-x^2+2}+\sqrt{x^2+1}-1}{\sqrt{2\sqrt{x^4-x^2+1}-x^2+2}+\sqrt{x^2+1}+1} </math>
	
To get ''x''<sup>3</sup>, we take the tangent of twice the arctangent of the modulus.
	
Also this equation leads to the sn-value of the third of ''K'':
	
:<math>k^2s^4-2k^2s^3+2s-1 = 0 </math>
	
:<math>s = \operatorname{sn}\left[\tfrac{1}{3}K(k); k\right] </math>
	
These equations lead to the other values of the Jacobi-Functions:
	
:<math>\operatorname{cn}\left[\tfrac{2}{3}K(k); k\right] = 1 - \operatorname{sn}\left[\tfrac{1}{3}K(k); k\right] </math>
	
:<math>\operatorname{dn}\left[\tfrac{2}{3}K(k); k\right] = 1/\operatorname{sn}\left[\tfrac{1}{3}K(k); k\right] - 1 </math>
	
'''Fifth K formula'''
	
Following equation has following solution:
	
:<math>4k^2x^6+8k^2x^5+2(1-k^2)^2x-(1-k^2)^2 = 0 </math>
	
:<math>x = \frac{1}{2}-\frac{1}{2}k^2\operatorname{sn}\left[\tfrac{2}{5}K(k); k\right]^2 \operatorname{sn}\left[\tfrac{4}{5}K(k); k\right]^2 = \frac{\operatorname{sn}\left[\frac{4}{5}K(k); k\right]^2-\operatorname{sn}\left[\frac{2}{5}K(k); k\right]^2}{2\operatorname{sn}\left[\frac{2}{5}K(k); k\right]\operatorname{sn}\left[\frac{4}{5}K(k); k\right]} </math>
	
To get the sn-values, we put the solution x into following expressions:
	
:<math>\operatorname{sn}\left[\tfrac{2}{5}K(k); k\right] = (1 + k^2)^{-1/2}\sqrt{2(1-x-x^2)(x^2+1-x\sqrt{x^2+1})} </math>
	
:<math>\operatorname{sn}\left[\tfrac{4}{5}K(k); k\right] = (1 + k^2)^{-1/2}\sqrt{2(1-x-x^2)(x^2+1+x\sqrt{x^2+1})} </math>

=== Relations between squares of the functions ===
Relations between squares of the functions can be derived from two basic relationships (Arguments (''u'',''m'') suppressed):
<math display="block">\operatorname{cn}^2+\operatorname{sn}^2=1</math>
<math display="block">\operatorname{cn}^2+m' \operatorname{sn}^2=\operatorname{dn}^2</math>
where ''m&nbsp;+&nbsp;m'&nbsp;''=&nbsp;1. Multiplying by any function of the form ''nq'' yields more general equations:

<math display="block">\operatorname{cq}^2+\operatorname{sq}^2=\operatorname{nq}^2</math>
<math display="block">\operatorname{cq}^2{}+m' \operatorname{sq}^2=\operatorname{dq}^2</math>

With ''q''&nbsp;=&nbsp;''d'', these correspond trigonometrically to the equations for the unit circle (<math>x^2+y^2=r^2</math>) and the unit ellipse  (<math>x^2{}+m' y^2=1</math>), with ''x''&nbsp;=&nbsp;''cd'', ''y''&nbsp;=&nbsp;''sd'' and ''r''&nbsp;=&nbsp;''nd''. Using the multiplication rule, other relationships may be derived. For example:

<math display="block">
-\operatorname{dn}^2{}+m'= -m\operatorname{cn}^2 = m\operatorname{sn}^2-m
</math>

<math display="block">
-m'\operatorname{nd}^2{}+m'= -mm'\operatorname{sd}^2 = m\operatorname{cd}^2-m
</math>

<math display="block">
m'\operatorname{sc}^2{}+m'= m'\operatorname{nc}^2 = \operatorname{dc}^2-m
</math>

<math display="block">
\operatorname{cs}^2{}+m'=\operatorname{ds}^2=\operatorname{ns}^2-m
</math>

=== Addition theorems ===
The functions satisfy the two square relations (dependence on ''m'' suppressed)
<math display="block">\operatorname{cn}^2(u) + \operatorname{sn}^2(u) = 1,\,</math>

<math display="block">\operatorname{dn}^2(u) + m \operatorname{sn}^2(u) = 1.\,</math>

From this we see that (cn, sn, dn) parametrizes an [[elliptic curve]] which is the intersection of the two [[quadric]]s defined by the above two equations. We now may define a group law for points on this curve by the addition formulas for the Jacobi functions<ref name="DLMF22" />

<math display="block">
\begin{align}
\operatorname{cn}(x+y) & =
{\operatorname{cn}(x) \operatorname{cn}(y)
- \operatorname{sn}(x) \operatorname{sn}(y) \operatorname{dn}(x) \operatorname{dn}(y)
\over {1 - m \operatorname{sn}^2 (x) \operatorname{sn}^2 (y)}}, \\[8pt]
\operatorname{sn}(x+y) & =
{\operatorname{sn}(x) \operatorname{cn}(y) \operatorname{dn}(y) +
\operatorname{sn}(y) \operatorname{cn}(x) \operatorname{dn}(x)
\over {1 - m \operatorname{sn}^2 (x) \operatorname{sn}^2 (y)}}, \\[8pt]
\operatorname{dn}(x+y) & =
{\operatorname{dn}(x) \operatorname{dn}(y) - m \operatorname{sn}(x) \operatorname{sn}(y) \operatorname{cn}(x) \operatorname{cn}(y)
\over {1 - m \operatorname{sn}^2 (x) \operatorname{sn}^2 (y)}}.
\end{align}
</math>

The Jacobi epsilon and zn functions satisfy a quasi-addition theorem:
<math display="block">\begin{align}\mathcal{E}(x+y,m)&=\mathcal{E}(x,m)+\mathcal{E}(y,m)-m\operatorname{sn}(x,m)\operatorname{sn}(y,m)\operatorname{sn}(x+y,m),\\
\operatorname{zn}(x+y,m)&=\operatorname{zn}(x,m)+\operatorname{zn}(y,m)-m\operatorname{sn}(x,m)\operatorname{sn}(y,m)\operatorname{sn}(x+y,m).\end{align}</math>

Double angle formulae can be easily derived from the above equations by setting ''x''&nbsp;=&nbsp;''y''.<ref name="DLMF22" /> Half angle formulae<ref name="WolframJE" /><ref name="DLMF22" /> are all of the form:

<math display="block">\operatorname{pq}(\tfrac{1}{2}u,m)^2 = f_{\mathrm p}/f_{\mathrm q}</math>

where:
<math display="block">f_{\mathrm c} = \operatorname{cn}(u,m)+\operatorname{dn}(u,m)</math>
<math display="block">f_{\mathrm s} = 1-\operatorname{cn}(u,m)</math>
<math display="block">f_{\mathrm n} = 1+\operatorname{dn}(u,m)</math>
<math display="block">f_{\mathrm d} = (1+\operatorname{dn}(u,m))-m(1-\operatorname{cn}(u,m))</math>