"타원곡선의 주기"의 두 판 사이의 차이

수학노트
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* [[타원곡선의 주기]]<br>
  
 
 
 
 
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">개요</h5>
 
 
 
 
  
 
 
 
 
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 <br> 타원곡선 <math>y^2=(x-e_1)(x-e_2)(x-e_3)</math>의 주기는 <math>\mathbb{Z}\</math><br><math>\omega_1=2\int_{e_1}^{\infty}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}</math><br><math>\omega_2=2\int_{e_2}^{e_1}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}</math><br>
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*  타원곡선 <math>y^2=(x-e_1)(x-e_2)(x-e_3)</math>의 주기는 두 복소수 <math>\omega_1,\omega_2</math>에 의해 생성되는 2차원 격자 <math>\Lambda=\{m_1\omega_1+m_2\omega_2)|m_1,m_2\in\mathbb{Z}\}</math>이다<br><math>\omega_1=2\int_{e_1}^{\infty}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}</math><br><math>\omega_2=2\int_{e_2}^{e_1}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}</math><br>
  
 
 
 
 
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">예</h5>
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<h5 style="line-height: 2em; margin: 0px;">예</h5>
  
 
*  타원곡선 <math>y^2=x^3-x</math>의 경우 ([[타원곡선 y²=x³-x|타원곡선 y^2=x^3-x]] 에서 가져옴)<br><math>e_1=1, e_2=0, e_3=-1</math>로 두자<br><math>\omega_1=2\int_{1}^{\infty}\frac{dx}{\sqrt{x^3-x}}=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots</math><br><math>\omega_2=2i\int_0^1\frac{dx}{\sqrt{x-x^3}}=2i\int_{\infty}^1\frac{-dy}{\sqrt{y^3-y}}=i\omega_{1}</math><br>
 
*  타원곡선 <math>y^2=x^3-x</math>의 경우 ([[타원곡선 y²=x³-x|타원곡선 y^2=x^3-x]] 에서 가져옴)<br><math>e_1=1, e_2=0, e_3=-1</math>로 두자<br><math>\omega_1=2\int_{1}^{\infty}\frac{dx}{\sqrt{x^3-x}}=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots</math><br><math>\omega_2=2i\int_0^1\frac{dx}{\sqrt{x-x^3}}=2i\int_{\infty}^1\frac{-dy}{\sqrt{y^3-y}}=i\omega_{1}</math><br>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">1종타원적분과의 관계</h5>
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<h5 style="line-height: 2em; margin: 0px;">1종타원적분과의 관계</h5>
  
 
* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br>
 
* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br>
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<math>\int_{c}^{b}\frac{dt}{\sqrt{(a-t)(b-t)(t-c)}}=\frac{2}{\sqrt{a-c}}K(\sqrt{\frac{b-c}{a-c}})</math>
 
<math>\int_{c}^{b}\frac{dt}{\sqrt{(a-t)(b-t)(t-c)}}=\frac{2}{\sqrt{a-c}}K(\sqrt{\frac{b-c}{a-c}})</math>
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<math>K(k)=\frac{1}{2}\int_{0}^{\infty} \frac{dx}{\sqrt{x (x^2 - (4k^2-2)x + 1)}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math>
 
<math>K(k)=\frac{1}{2}\int_{0}^{\infty} \frac{dx}{\sqrt{x (x^2 - (4k^2-2)x + 1)}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math>
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">재미있는 사실</h5>
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">역사</h5>
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">관련된 항목들</h5>
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* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br>
 
* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br>
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">수학용어번역</h5>
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
 
* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
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* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
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* http://www.wolframalpha.com/input/?i=
 
* http://www.wolframalpha.com/input/?i=
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
 
** http://www.research.att.com/~njas/sequences/?q=
 
** http://www.research.att.com/~njas/sequences/?q=
  
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* http://www.jstor.org/action/doBasicSearch?Query=
 
* http://www.jstor.org/action/doBasicSearch?Query=
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">관련도서</h5>
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*  도서내검색<br>
 
*  도서내검색<br>
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*  네이버 뉴스 검색 (키워드 수정)<br>
 
*  네이버 뉴스 검색 (키워드 수정)<br>
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*  구글 블로그 검색<br>
 
*  구글 블로그 검색<br>

2011년 3월 14일 (월) 19:18 판

이 항목의 스프링노트 원문주소

 

 

개요

 

 

정의
  • 타원곡선 \(y^2=(x-e_1)(x-e_2)(x-e_3)\)의 주기는 두 복소수 \(\omega_1,\omega_2\)에 의해 생성되는 2차원 격자 \(\Lambda=\{m_1\omega_1+m_2\omega_2)|m_1,m_2\in\mathbb{Z}\}\)이다
    \(\omega_1=2\int_{e_1}^{\infty}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}\)
    \(\omega_2=2\int_{e_2}^{e_1}\frac{dx}{\sqrt{(x-e_1)(x-e_2)(x-e_3)}}\)

 

 

  • 타원곡선 \(y^2=x^3-x\)의 경우 (타원곡선 y^2=x^3-x 에서 가져옴)
    \(e_1=1, e_2=0, e_3=-1\)로 두자
    \(\omega_1=2\int_{1}^{\infty}\frac{dx}{\sqrt{x^3-x}}=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots\)
    \(\omega_2=2i\int_0^1\frac{dx}{\sqrt{x-x^3}}=2i\int_{\infty}^1\frac{-dy}{\sqrt{y^3-y}}=i\omega_{1}\)

 

 

 

1종타원적분과의 관계

 

 

1종완전타원적분과 타원곡선의 주기1

\(\int_{c}^{b}\frac{dt}{\sqrt{(a-t)(b-t)(t-c)}}=\frac{2}{\sqrt{a-c}}K(\sqrt{\frac{b-c}{a-c}})\)

 

 

일종완전타원적분과 타원곡선의 주기2

\(K(k)=\frac{1}{2}\int_{0}^{\infty} \frac{dx}{\sqrt{x (x^2 - (4k^2-2)x + 1)}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx\)

(증명)

\(k=\cos \alpha\) 로 두자.

\(K(\cos\alpha)=\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1 - \cos^2 \alpha \sin^2 \theta }}\)

\(=\int_{0}^{1} \frac{2dt}{\sqrt{t^4 - 2(2\cos^2 \alpha - 1)t^2 + 1}}\) (\(t =\tan (\theta/2) \)로 치환)

\(=\int_{1}^{\infty} \frac{2dx}{\sqrt{x^4 - 2x^2 \cos 2\alpha + 1}}\) (\(x=\frac{1}{t}\) 로 치환)

\(=\int_{1}^{\infty} \frac{du}{\sqrt{u (u^2 - 2u \cos 2\alpha + 1)}}\) (\(u=x^2\)로 치환)

 

한편,  \(u=\frac{1}{v}\) 치환을 통하여

\(\int_{1}^{\infty} \frac{du}{\sqrt{u (u^2 - 2u \cos 2\alpha + 1)}}=\int_{0}^{1} \frac{dv}{\sqrt{v (v^2 - 2v \cos 2\alpha + 1)}}\)

임을 보일 수 있으므로,

\(2K(\cos\alpha)=\int_{0}^{\infty} \frac{du}{\sqrt{u (u^2 - 2u \cos 2\alpha + 1)}}\)

\(4K(\cos\alpha)\)는 타원곡선 \(y^2=x(x-e^{2i\alpha})(x-e^{-2i\alpha})=x(x^2 - 2x \cos 2\alpha + 1)}\)의 주기임을 알 수 있다. ■

 

 

 

재미있는 사실

 

 

 

역사

 

 

 

메모

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문

 

 

관련도서

 

 

관련기사

 

 

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