Jacobi elliptic functions: Difference between revisions

Revision as of 20:05, 11 August 2024

Revision as of 18:07, 12 August 2024

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===Amplitude transformations===

In the following, the second variable is suppressed and is equal to <math>m</math>:

:<math>\operatorname{am}\left(\sqrt{m'}u,-\frac{m}{m'}\right)=\frac{\pi}{2}-\operatorname{am}(K-u,m),\quad u\in\mathbb{R},\, 0<m<1,</math>

:<math>\operatorname{am}(u,m')=-2\arctan\left(i\tan \frac{\operatorname{am}(iu,m)}{2}\right),\quad \left|\operatorname{Re}u\right|<K',\, \left|\operatorname{Im}u\right|<K,\, 0<m<1,</math>

:<math>\sin(\operatorname{am}\left(u+v)+\sqrtoperatorname{mam}(u,-v))=\frac{12\operatorname{sn}u\operatorname{mcn}u\right)=\arcsin(\sqrtoperatorname{dn}v}{1-m\operatorname{sn}^2u\operatorname{sn}(u^2v},m)),\quad -2K<u<2K,\, 0<m<1</math>

:<math>\cos(\operatorname{am}(u,m'+v)=-2\arctan\left(i\tan \frac{\operatorname{am}(iu,mu-v)}{2}\right),=\quad dfrac{\left|operatorname{cn}^2v-\operatorname{Resn}u^2v\right|<K',\, \left|operatorname{dn}^2u}{1-m\operatorname{Imsn}u^2u\right|<K,\, 0<m<1,operatorname{sn}^2v}.</math>

where both identities are valid for all <math>u,v,m\in\mathbb{C}</math> such that both sides are well-defined.

With

:<math>m_1=\left(\frac{1-\sqrt{m'}}{1+\sqrt{m'}}\right)^2,</math>

we have

:<math>\cos (\operatorname{am}(u,m)+\operatorname{am}(K-u,m))=-\operatorname{sn}((1-\sqrt{m'})u,1/m_1),</math>

:<math>\sin(\operatorname{am}(\sqrt{m'}u,-m/m')+\operatorname{am}((1-\sqrt{m'})u,1/m_1))=\operatorname{sn}(u,m),</math>

:<math>\sin(\operatorname{am}\left((1+\sqrt{m'})u,-m_1)+\fracoperatorname{mam}((1-\sqrt{m'}\right)u,1/m_1))=\frac{\pi}{sin(2}-\operatorname{am}(K-u,m),\quad u\in\mathbb{R},\, 0<m<1,)</math>

where all the identities are valid for all <math>u,m\in\mathbb{C}</math> such that both sides are well-defined.

==The Jacobi hyperbola==