在數學 中，雅可比橢圓函數 是由卡爾·雅可比 在1830年左右研究的一類橢圓函數 。這類函數可用於擺 之類的應用問題，並具有與三角函數 相似的性質。
雅可比矩形 雅可比橢圓函數有十二種，各對映到某個矩形的頂點 連線。此諸頂點記作
s
{\displaystyle s\,}
c
{\displaystyle c\,}
d
{\displaystyle d\,}
n
{\displaystyle n\,}
視此矩形為複數 平面的一部分，
s
{\displaystyle s\,}
c
{\displaystyle c\,}
K
,
d
{\displaystyle K,d\,}
K
+
i
K
′
{\displaystyle K+{\rm {i}}K'\,}
n
{\displaystyle n\,}
i
K
′
{\displaystyle {\rm {i}}K'\,}
K
{\displaystyle K\,}
i
K
′
{\displaystyle {\rm {i}}K'\,}
十二個橢圓函數分別記為
s
c
,
s
d
,
s
n
,
c
s
,
c
d
,
c
n
,
d
s
,
d
c
,
d
n
,
n
s
,
n
c
,
n
d
{\displaystyle {\rm {sc}},{\rm {sd}},{\rm {sn}},{\rm {cs}},{\rm {cd}},{\rm {cn}},{\rm {ds}},{\rm {dc}},{\rm {dn}},{\rm {ns}},{\rm {nc}},{\rm {nd}}\,}
p
,
q
{\displaystyle p,q\,}
頂點 ，則函數
p
q
{\displaystyle pq\,}
亞純函數
p
{\displaystyle p\,}
q
{\displaystyle q\,}
p
q
{\displaystyle pq\,}
p
q
→
{\displaystyle {\vec {pq}}}
p
,
q
{\displaystyle p,q\,}
p
{\displaystyle p\,}
p
q
{\displaystyle pq\,}
p
q
{\displaystyle pq\,}
p
{\displaystyle p\,}
q
{\displaystyle q\,}
表列如次：
函數
週期
零點
極點
留數
s
n
(
z
;
k
)
{\displaystyle \mathrm {sn} \,(z;k)}
4
K
,
2
i
K
′
{\displaystyle 4\,K,\ 2\,\mathrm {i} K'}
2
m
K
+
2
n
i
K
′
{\displaystyle 2mK+2\,n\,\mathrm {i} \,K'}
2
m
K
+
(
2
n
+
1
)
i
K
′
{\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}
(
−
1
)
m
1
k
{\displaystyle (-1)^{m}{\frac {1}{k}}}
c
n
(
z
;
k
)
{\displaystyle \mathrm {cn} \,(z;k)}
4
K
,
2
(
K
+
i
K
′
)
{\displaystyle 4\,K,\ 2\,(K+\mathrm {i} K')}
(
2
m
+
1
)
K
+
2
n
i
K
′
{\displaystyle (2m+1)\,K+2\,n\,\mathrm {i} \,K'}
2
m
K
+
(
2
n
+
1
)
i
K
′
{\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}
(
−
1
)
m
+
n
1
i
k
{\displaystyle (-1)^{m+n}{\frac {1}{{\rm {i}}k}}}
d
n
(
z
;
k
)
{\displaystyle \mathrm {dn} \,(z;k)}
2
K
,
4
i
K
′
{\displaystyle 2\,K,\ 4\,\mathrm {i} K'}
(
2
m
+
1
)
K
+
2
(
n
+
1
)
i
K
′
{\displaystyle (2\,m+1)\,K+2\,(n+1)\,\mathrm {i} \,K'}
2
m
K
+
(
2
n
+
1
)
i
K
′
{\displaystyle 2\,mK+(2n+1)\,\mathrm {i} \,K'}
(
−
1
)
n
−
1
i
{\displaystyle (-1)^{n-1}{\rm {i}}\,}
n
{\displaystyle n\,}
m
{\displaystyle m\,}
一般而言，須以平行四邊形代替上述矩形，以考慮更一般的週期。
以上定義略顯抽象，更具體的定義是將之表為某類橢圓積分 (第一類不完全橢圓積分)之逆。設
u
=
∫
0
ϕ
d
θ
1
−
m
sin
2
θ
.
{\displaystyle u=\int _{0}^{\phi }{\frac {d\theta }{\sqrt {1-m\sin ^{2}\theta }}}.}
橢圓正弦函數 sn u 定義為
sn
u
=
sin
ϕ
{\displaystyle \operatorname {sn} \;u=\sin \phi \,}
而橢圓餘弦函數 cn u 定義為
cn
u
=
cos
ϕ
{\displaystyle \operatorname {cn} \;u=\cos \phi }
同理，橢圓德爾塔函數有
dn
u
=
1
−
m
sin
2
ϕ
.
{\displaystyle \operatorname {dn} \;u={\sqrt {1-m\sin ^{2}\phi }}.\,}
這裡的
m
∈
R
{\displaystyle m\in \mathbb {R} }
0
≤
m
≤
1
{\displaystyle 0\leq m\leq 1}
剩下的九種橢圓函數能由這三種構造。
雅可比橢圓函數的反函數可以像三角函數與反三角函數那樣被定義。因為橢圓函數往往是橢圓積分之逆，這些反函數也都可以用勒讓德橢圓積分 來描述。如同反三角函數一樣，雅可比橢圓函數的反函數也是多值的，因此需要支割線。以下是部分反函數的積分表達：
a
r
c
s
n
(
z
,
k
)
=
∫
0
z
d
t
(
1
−
t
2
)
(
1
−
k
2
t
2
)
{\displaystyle \mathrm {arcsn} \,(z,k)=\int _{0}^{z}{\frac {\mathrm {d} t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}}
a
r
c
c
n
(
z
,
k
)
=
∫
z
1
d
t
(
1
−
t
2
)
(
1
−
k
2
+
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})}}}}
a
r
c
d
n
(
z
,
k
)
=
∫
z
1
d
t
(
1
−
t
2
)
(
t
2
+
k
2
−
1
)
{\displaystyle \mathrm {arcdn} \,(z,k)=\int _{z}^{1}{\frac {\mathrm {d} t}{\sqrt {(1-t^{2})(t^{2}+k^{2}-1)}}}}
a
r
c
n
s
(
z
,
k
)
=
∫
0
∞
d
t
(
t
2
−
1
)
(
t
2
−
k
2
)
{\displaystyle \mathrm {arcns} \,(z,k)=\int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {(t^{2}-1)(t^{2}-k^{2})}}}}
a
r
c
n
c
(
z
,
k
)
=
∫
1
z
d
t
(
t
2
−
1
)
(
1
−
k
2
)
(
k
2
+
t
2
)
{\displaystyle \mathrm {arcnc} \,(z,k)=\int _{1}^{z}{\frac {\mathrm {d} t}{\sqrt {(t^{2}-1)(1-k^{2})(k^{2}+t^{2})}}}}
a
r
c
n
d
(
z
,
k
)
=
∫
1
z
d
t
(
t
2
−
1
)
(
1
−
(
1
−
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})}}}}
雅可比橢圓函數也可以用Θ函數 來定義。如果我們把
ϑ
(
0
;
τ
)
{\displaystyle \vartheta (0;\tau )}
ϑ
{\displaystyle \vartheta }
ϑ
01
(
0
;
τ
)
,
ϑ
10
(
0
;
τ
)
,
ϑ
11
(
0
;
τ
)
{\displaystyle \vartheta _{01}(0;\tau ),\vartheta _{10}(0;\tau ),\vartheta _{11}(0;\tau )}
ϑ
01
,
ϑ
10
,
ϑ
11
{\displaystyle \vartheta _{01},\vartheta _{10},\vartheta _{11}}
橢圓模 k 是
k
=
(
ϑ
10
ϑ
)
2
{\displaystyle k=({\vartheta _{10} \over \vartheta })^{2}}
u
=
π
ϑ
2
z
{\displaystyle u=\pi \vartheta ^{2}z}
sn
(
u
;
k
)
=
−
ϑ
ϑ
11
(
z
;
τ
)
ϑ
10
ϑ
01
(
z
;
τ
)
{\displaystyle {\mbox{sn}}(u;k)=-{\vartheta \vartheta _{11}(z;\tau ) \over \vartheta _{10}\vartheta _{01}(z;\tau )}}
cn
(
u
;
k
)
=
ϑ
01
ϑ
10
(
z
;
τ
)
ϑ
10
ϑ
01
(
z
;
τ
)
{\displaystyle {\mbox{cn}}(u;k)={\vartheta _{01}\vartheta _{10}(z;\tau ) \over \vartheta _{10}\vartheta _{01}(z;\tau )}}
dn
(
u
;
k
)
=
ϑ
01
ϑ
(
z
;
τ
)
ϑ
ϑ
01
(
z
;
τ
)
{\displaystyle {\mbox{dn}}(u;k)={\vartheta _{01}\vartheta (z;\tau ) \over \vartheta \vartheta _{01}(z;\tau )}}
cn
2
+
sn
2
=
1
,
{\displaystyle \operatorname {cn} ^{2}+\operatorname {sn} ^{2}=1,\,}
dn
2
+
k
2
sn
2
=
1.
{\displaystyle \operatorname {dn} ^{2}+k^{2}\operatorname {sn} ^{2}=1.\,}
由此可見 (cn,sn,dn) 描出射影空間
P
3
(
C
)
{\displaystyle \mathbb {P} ^{3}(\mathbb {C} )}
曲面 之交，這同構於一條橢圓曲線 。曲線上的群 運算由下列加法公式描述：
cn
(
x
+
y
)
=
cn
(
x
)
cn
(
y
)
−
sn
(
x
)
sn
(
y
)
dn
(
x
)
dn
(
y
)
1
−
k
2
sn
2
(
x
)
sn
2
(
y
)
,
{\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)}},}
sn
(
x
+
y
)
=
sn
(
x
)
cn
(
y
)
dn
(
y
)
+
sn
(
y
)
cn
(
x
)
dn
(
x
)
1
−
k
2
sn
2
(
x
)
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)}},}
dn
(
x
+
y
)
=
dn
(
x
)
dn
(
y
)
−
k
2
sn
(
x
)
sn
(
y
)
cn
(
x
)
cn
(
y
)
1
−
k
2
sn
2
(
x
)
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)}}.}
−
dn
2
(
u
)
+
(
1
−
k
2
)
=
−
k
2
cn
2
(
u
)
=
k
2
sn
2
(
u
)
−
k
2
{\displaystyle -\operatorname {dn} ^{2}(u)+(1-k^{2})=-k^{2}\;\operatorname {cn} ^{2}(u)=k^{2}\;\operatorname {sn} ^{2}(u)-k^{2}}
(
k
2
−
1
)
nd
2
(
u
)
+
(
1
−
k
2
)
=
k
2
(
k
2
−
1
)
sd
2
(
u
)
=
k
2
cd
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}}
(
1
−
k
2
)
sc
2
(
u
)
+
(
1
−
k
2
)
=
(
1
−
k
2
)
nc
2
(
u
)
=
dc
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}}
cs
2
(
u
)
+
(
1
−
k
2
)
=
ds
2
(
u
)
=
ns
2
(
u
)
−
k
2
{\displaystyle \operatorname {cs} ^{2}(u)+(1-k^{2})=\operatorname {ds} ^{2}(u)=\operatorname {ns} ^{2}(u)-k^{2}}
三個基本的雅可比橢圓函數的導數 為：
d
d
z
s
n
(
z
;
k
)
=
c
n
(
z
;
k
)
d
n
(
z
;
k
)
,
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {sn} \,(z;k)=\mathrm {cn} \,(z;k)\,\mathrm {dn} \,(z;k),}
d
d
z
c
n
(
z
;
k
)
=
−
s
n
(
z
;
k
)
d
n
(
z
;
k
)
,
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {cn} \,(z;k)=-\mathrm {sn} \,(z;k)\,\mathrm {dn} \,(z;k),}
d
d
z
d
n
(
z
;
k
)
=
−
k
2
s
n
(
z
;
k
)
c
n
(
z
;
k
)
.
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\,\mathrm {dn} \,(z;k)=-k^{2}\mathrm {sn} \,(z;k)\,\mathrm {cn} \,(z;k).}
根據以上的加法定理，可知它們是以下非線性常微分方程 的解：
s
n
(
x
;
k
)
{\displaystyle \mathrm {sn} \,(x;k)}
d
2
y
d
x
2
+
(
1
+
k
2
)
y
−
2
k
2
y
3
=
0
,
{\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+(1+k^{2})y-2k^{2}y^{3}=0,}
(
d
y
d
x
)
2
=
(
1
−
y
2
)
(
1
−
k
2
y
2
)
{\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(1-y^{2})(1-k^{2}y^{2})}
c
n
(
x
;
k
)
{\displaystyle \mathrm {cn} \,(x;k)}
d
2
y
d
x
2
+
(
1
−
2
k
2
)
y
+
2
k
2
y
3
=
0
,
{\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+(1-2k^{2})y+2k^{2}y^{3}=0,}
(
d
y
d
x
)
2
=
(
1
−
y
2
)
(
1
−
k
2
+
k
2
y
2
)
{\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(1-y^{2})(1-k^{2}+k^{2}y^{2})}
d
n
(
x
;
k
)
{\displaystyle \mathrm {dn} \,(x;k)}
d
2
y
d
x
2
−
(
2
−
k
2
)
y
+
2
y
3
=
0
,
{\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}-(2-k^{2})y+2y^{3}=0,}
(
d
y
d
x
)
2
=
(
y
2
−
1
)
(
1
−
k
2
−
y
2
)
{\displaystyle \left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}=(y^{2}-1)(1-k^{2}-y^{2})}
Abramowitz, Milton; Stegun, Irene A. eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover. 1972. ISBN 0-486-61272-4 . 第16章 （頁面存檔備份 ，存於網際網路檔案館 ）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 
