雅可比橢圓函數

介紹

• ${\displaystyle p\,}$  是單零點，${\displaystyle q\,}$  是單極點。
• ${\displaystyle pq\,}$ ${\displaystyle {\vec {pq}}}$  方向的週期等於 ${\displaystyle p,q\,}$  距離的兩倍。對另兩個從 ${\displaystyle p\,}$ 出發的方向，${\displaystyle pq\,}$ 亦滿足同樣性質。
• ${\displaystyle pq\,}$  在頂點 ${\displaystyle p\,}$  ${\displaystyle q\,}$  的展式首項係數均為一。

${\displaystyle \mathrm {sn} \,(z;k)}$  ${\displaystyle 4\,K,\ 2\,\mathrm {i} K'}$  ${\displaystyle 2mK+2\,n\,\mathrm {i} \,K'}$  ${\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}$  ${\displaystyle (-1)^{m}{\frac {1}{k}}}$
${\displaystyle \mathrm {cn} \,(z;k)}$  ${\displaystyle 4\,K,\ 2\,(K+\mathrm {i} K')}$  ${\displaystyle (2m+1)\,K+2\,n\,\mathrm {i} \,K'}$  ${\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}$  ${\displaystyle (-1)^{m+n}{\frac {1}{{\rm {i}}k}}}$
${\displaystyle \mathrm {dn} \,(z;k)}$  ${\displaystyle 2\,K,\ 4\,\mathrm {i} K'}$  ${\displaystyle (2\,m+1)\,K+2\,(n+1)\,\mathrm {i} \,K'}$  ${\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}$  ${\displaystyle (-1)^{n-1}{\rm {i}}\,}$
${\displaystyle n\,}$ ${\displaystyle m\,}$  是整數

表為橢圓積分之逆

${\displaystyle u=\int _{0}^{\phi }{\frac {d\theta }{\sqrt {1-m\sin ^{2}\theta }}}.}$

${\displaystyle \operatorname {sn} \;u=\sin \phi \,}$

${\displaystyle \operatorname {cn} \;u=\cos \phi }$

${\displaystyle \operatorname {dn} \;u={\sqrt {1-m\sin ^{2}\phi }}.\,}$

反函數

• ${\displaystyle \mathrm {arcsn} \,(z,k)=\int _{0}^{z}{\frac {\mathrm {d} t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}}$
• ${\displaystyle \mathrm {arccn} \,(z,k)=\int _{z}^{1}{\frac {\mathrm {d} t}{\sqrt {(1-t^{2})(1-k^{2}+k^{2}t^{2})}}}}$
• ${\displaystyle \mathrm {arcdn} \,(z,k)=\int _{z}^{1}{\frac {\mathrm {d} t}{\sqrt {(1-t^{2})(t^{2}+k^{2}-1)}}}}$
• ${\displaystyle \mathrm {arcns} \,(z,k)=\int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {(t^{2}-1)(t^{2}-k^{2})}}}}$
• ${\displaystyle \mathrm {arcnc} \,(z,k)=\int _{1}^{z}{\frac {\mathrm {d} t}{\sqrt {(t^{2}-1)(1-k^{2})(k^{2}+t^{2})}}}}$
• ${\displaystyle \mathrm {arcnd} \,(z,k)=\int _{1}^{z}{\frac {\mathrm {d} t}{\sqrt {(t^{2}-1)(1-(1-k^{2})t^{2})}}}}$

用Θ函數来定义

${\displaystyle {\mbox{sn}}(u;k)=-{\vartheta \vartheta _{11}(z;\tau ) \over \vartheta _{10}\vartheta _{01}(z;\tau )}}$
${\displaystyle {\mbox{cn}}(u;k)={\vartheta _{01}\vartheta _{10}(z;\tau ) \over \vartheta _{10}\vartheta _{01}(z;\tau )}}$
${\displaystyle {\mbox{dn}}(u;k)={\vartheta _{01}\vartheta (z;\tau ) \over \vartheta \vartheta _{01}(z;\tau )}}$

加法定理

${\displaystyle \operatorname {cn} ^{2}+\operatorname {sn} ^{2}=1,\,}$
${\displaystyle \operatorname {dn} ^{2}+k^{2}\operatorname {sn} ^{2}=1.\,}$

${\displaystyle \operatorname {cn} (x+y)={\operatorname {cn} (x)\;\operatorname {cn} (y)-\operatorname {sn} (x)\;\operatorname {sn} (y)\;\operatorname {dn} (x)\;\operatorname {dn} (y) \over {1-k^{2}\;\operatorname {sn} ^{2}(x)\;\operatorname {sn} ^{2}(y)}},}$
${\displaystyle \operatorname {sn} (x+y)={\operatorname {sn} (x)\;\operatorname {cn} (y)\;\operatorname {dn} (y)+\operatorname {sn} (y)\;\operatorname {cn} (x)\;\operatorname {dn} (x) \over {1-k^{2}\;\operatorname {sn} ^{2}(x)\;\operatorname {sn} ^{2}(y)}},}$
${\displaystyle \operatorname {dn} (x+y)={\operatorname {dn} (x)\;\operatorname {dn} (y)-k^{2}\;\operatorname {sn} (x)\;\operatorname {sn} (y)\;\operatorname {cn} (x)\;\operatorname {cn} (y) \over {1-k^{2}\;\operatorname {sn} ^{2}(x)\;\operatorname {sn} ^{2}(y)}}.}$

函数的平方之间的关系

${\displaystyle -\operatorname {dn} ^{2}(u)+(1-k^{2})=-k^{2}\;\operatorname {cn} ^{2}(u)=k^{2}\;\operatorname {sn} ^{2}(u)-k^{2}}$
${\displaystyle (k^{2}-1)\;\operatorname {nd} ^{2}(u)+(1-k^{2})=k^{2}(k^{2}-1)\;\operatorname {sd} ^{2}(u)=k^{2}\;\operatorname {cd} ^{2}(u)-k^{2}}$
${\displaystyle (1-k^{2})\;\operatorname {sc} ^{2}(u)+(1-k^{2})=(1-k^{2})\;\operatorname {nc} ^{2}(u)=\operatorname {dc} ^{2}(u)-k^{2}}$
${\displaystyle \operatorname {cs} ^{2}(u)+(1-k^{2})=\operatorname {ds} ^{2}(u)=\operatorname {ns} ^{2}(u)-k^{2}}$

常微分方程的解

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {sn} \,(z;k)=\mathrm {cn} \,(z;k)\,\mathrm {dn} \,(z;k),}$
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {cn} \,(z;k)=-\mathrm {sn} \,(z;k)\,\mathrm {dn} \,(z;k),}$
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {dn} \,(z;k)=-k^{2}\mathrm {sn} \,(z;k)\,\mathrm {cn} \,(z;k).}$

• ${\displaystyle \mathrm {sn} \,(x;k)}$ 是微分方程${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+(1+k^{2})y-2k^{2}y^{3}=0,}$ ${\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(1-y^{2})(1-k^{2}y^{2})}$ 的解；
• ${\displaystyle \mathrm {cn} \,(x;k)}$ 是微分方程${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+(1-2k^{2})y+2k^{2}y^{3}=0,}$ ${\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(1-y^{2})(1-k^{2}+k^{2}y^{2})}$ 的解；
• ${\displaystyle \mathrm {dn} \,(x;k)}$ 是微分方程${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}-(2-k^{2})y+2y^{3}=0,}$ ${\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(y^{2}-1)(1-k^{2}-y^{2})}$ 的解。

文獻

• Naum Illyich Akhiezer, Elements of the Theory of Elliptic Functions, (1970) Moscow, translated into English as AMS Translations of Mathematical Monographs Volume 79 (1990) AMS, Rhode Island ISBN 0-8218-4532-2
• E. T. Whittaker and G. N. Watson A Course of Modern Analysis, (1940, 1996) Cambridge University Press. ISBN 0-521-58807-3