# 雅可比橢圓函數

## 介紹

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

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

## 表為橢圓積分之逆

$u=\int_0^\phi \frac{d\theta} {\sqrt {1-m \sin^2 \theta}}.$

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

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

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

## 反函數

• $\mathrm{Arcsn}\,(x,k) = \int_0^x \frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}$
• $\mathrm{Arccn}\,(x,k) =\int_x^1 \frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2+k^2t^2)}}$

## 用Theta函数来定义

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

## 加法定理

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

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

## 函数的平方之间的关系

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

## 常微分方程的解

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

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

## 文獻

• 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
• 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