"타원적분"의 두 판 사이의 차이
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">일종타원적분과 이종타원적분</h5> | <h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">일종타원적분과 이종타원적분</h5> | ||
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* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br><math>K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math><br><math>K(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)</math><br> | * [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br><math>K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math><br><math>K(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)</math><br> | ||
* [[제2종타원적분 E (complete elliptic integral of the second kind)]]<br><math>E(k)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2 \theta} d\theta =\int_{0}^{1}\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx=\int_{0}^{1}\frac{1-k^2x^2}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math><br><math>E(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},-\frac{1}{2};1;k^2)</math><br> | * [[제2종타원적분 E (complete elliptic integral of the second kind)]]<br><math>E(k)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2 \theta} d\theta =\int_{0}^{1}\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx=\int_{0}^{1}\frac{1-k^2x^2}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx</math><br><math>E(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},-\frac{1}{2};1;k^2)</math><br> | ||
| − | * [[초기하급수(Hypergeometric series)]] | + | * <math>\,_2F_1(a,b;c;z)</math>는 [[초기하급수(Hypergeometric series)]] |
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* [http://www.springerlink.com/content/b365w3511067g184/ In Search of the "Birthday" of Elliptic Functions - Bit by bit, the discoverers decided what it was they had discovered.]<br> | * [http://www.springerlink.com/content/b365w3511067g184/ In Search of the "Birthday" of Elliptic Functions - Bit by bit, the discoverers decided what it was they had discovered.]<br> | ||
| − | ** Rice, Adrian, 48-57, The Mathematical Intelligencer, Volume 30, Number 2 | + | ** Rice, Adrian, 48-57, The Mathematical Intelligencer, Volume 30, Number 2, 2008년 3월<br> |
* [http://www.ams.org/bull/2007-44-04/S0273-0979-07-01178-0/home.html Euler and algebraic geometry]<br> | * [http://www.ams.org/bull/2007-44-04/S0273-0979-07-01178-0/home.html Euler and algebraic geometry]<br> | ||
** Burt Totaro, Bull. Amer. Math. Soc. 44 (2007), 541-559.<br> | ** Burt Totaro, Bull. Amer. Math. Soc. 44 (2007), 541-559.<br> | ||
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* [http://www.springerlink.com/content/911pnwauaeggxk13/ The story of Landen, the hyperbola and the ellipse]<br> | * [http://www.springerlink.com/content/911pnwauaeggxk13/ The story of Landen, the hyperbola and the ellipse]<br> | ||
** Elemente der Mathematik, Volume 57, Number 1 / 2002년 2월 | ** Elemente der Mathematik, Volume 57, Number 1 / 2002년 2월 | ||
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* [http://www.jstor.org/stable/2687483 Three Fermat Trails to Elliptic Curves]<br> | * [http://www.jstor.org/stable/2687483 Three Fermat Trails to Elliptic Curves]<br> | ||
2010년 8월 22일 (일) 08:58 판
이 항목의 스프링노트 원문주소
개요
- 먼저 타원적분 입문 참조
- \(R(x,y)\)는 \(x,y\)의 유리함수이고, \(y^2\)은 \(x\)의 3차 또는 4차식
\(\int R(x,\sqrt{ax^3+bx^2+cx+d}) \,dx\) 또는
\(\int R(x,\sqrt{ax^4+bx^3+cx^2+dx+e}) \,dx\)
타원 둘레의 길이
- 역사적으로 타원 둘레의 길이를 구하는 적분에서 그 이름이 기원함.
- 타원 \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)의 둘레의 길이는 \(4aE(k)\) 로 주어짐.
\(k=\sqrt{1-\frac{b^2}{a^2}}\)
\(E(k)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2 \theta} d\theta =\int_{0}^{1}\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx=\int_{0}^{1}\frac{1-k^2x^2}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx\)
정의
- 일반적으로 다음과 같은 형태로 주어지는 적분을 타원적분이라 부름
\(\int R(x,y)\,dx\)
여기서 \(R(x,y)\)는 \(x,y\)의 유리함수, \(y^2\)= 중근을 갖지 않는 \(x\)의 3차식 또는 4차식.
- 예를 들자면,
\(\int \frac{dx}{\sqrt{1-x^4}}\)
\(\int \frac{1-k^2x^2}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx\)
일종타원적분과 이종타원적분
-
- 제1종타원적분 K (complete elliptic integral of the first kind)
\(K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}}=\int_0^1\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx\)
\(K(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)\) - 제2종타원적분 E (complete elliptic integral of the second kind)
\(E(k)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2 \theta} d\theta =\int_{0}^{1}\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx=\int_{0}^{1}\frac{1-k^2x^2}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx\)
\(E(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},-\frac{1}{2};1;k^2)\) - \(\,_2F_1(a,b;c;z)\)는 초기하급수(Hypergeometric series)
르장드르의 항등식
- 일종타원적분과 이종타원적분 사이에는 다음과 같은 관계가 성립
\(E(k)K'(k)+E'(k)K(k)-K(k)K'(k)=\frac{\pi}{2}\)
또는 \(\theta+\phi=\frac{\pi}{2}\) 에 대하여
\(E(\sin\theta)K(\sin\phi)+E(\sin\phi)K(\sin\theta)-K(\sin\theta)K(\sin\phi)=\frac{\pi}{2}\)
- 특별히 다음과 같은 관계가 성립함
\(2K(\frac{1}{\sqrt{2}})E(\frac{1}{\sqrt{2}})-K(\frac{1}{\sqrt{2}})^2=\frac{\pi}{2}\)
AGM과 파이값의 계산에 응용
덧셈공식
- 파그나노의 공식
\(\int_0^x{\frac{1}{\sqrt{1-x^4}}}dx+\int_0^y{\frac{1}{\sqrt{1-x^4}}}dx = \int_0^{A(x,y)}{\frac{1}{\sqrt{1-x^4}}}dx\)
여기서 \(A(x,y)=\frac{x\sqrt{1-y^4}+y\sqrt{1-x^4}}{1+x^2y^2}\) - 오일러의 일반화
\(p(x)=1+mx^2+nx^4\)일 때,
\(\int_0^x{\frac{1}{\sqrt{p(x)}}}dx+\int_0^y{\frac{1}{\sqrt{p(x)}}}dx = \int_0^{B(x,y)}{\frac{1}{\sqrt{p(x)}}}dx\)
여기서 \(B(x,y)=\frac{x\sqrt{p(y)}+y\sqrt{p(x)}}{1-nx^2y^2}\)
메모
하위페이지
관련된 항목들
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/Elliptic_integral
- http://en.wikipedia.org/wiki/
- http://www.wolframalpha.com/input/?i=
- NIST Digital Library of Mathematical Functions
- The On-Line Encyclopedia of Integer Sequences
관련논문
- In Search of the "Birthday" of Elliptic Functions - Bit by bit, the discoverers decided what it was they had discovered.
- Rice, Adrian, 48-57, The Mathematical Intelligencer, Volume 30, Number 2, 2008년 3월
- Rice, Adrian, 48-57, The Mathematical Intelligencer, Volume 30, Number 2, 2008년 3월
- Euler and algebraic geometry
- Burt Totaro, Bull. Amer. Math. Soc. 44 (2007), 541-559.
- Burt Totaro, Bull. Amer. Math. Soc. 44 (2007), 541-559.
- The Lemniscate and Fagnano's Contributions to Elliptic Integrals
- AYOUB R
- A property of Euler's elastic curve
- The story of Landen, the hyperbola and the ellipse
- Elemente der Mathematik, Volume 57, Number 1 / 2002년 2월
- Three Fermat Trails to Elliptic Curves
- Ezra Brown, The College Mathematics Journal, Vol. 31, No. 3 (May, 2000), pp. 162-172
- Elliptic Curves
- John Stillwell, The American Mathematical Monthly, Vol. 102, No. 9 (Nov., 1995), pp. 831-837
- Abel's Theorem on the Lemniscate
- Michael Rosen, The American Mathematical Monthly, Vol. 88, No. 6 (Jun. - Jul., 1981), pp. 387-395
- Three Fermat Trails to Elliptic Curves
- Ezra Brown, The College Mathematics Journal, Vol. 31, No. 3 (May, 2000), pp. 162-172
- Elliptic Curves
- John Stillwell, The American Mathematical Monthly, Vol. 102, No. 9 (Nov., 1995), pp. 831-837
- Abel's Theorem on the Lemniscate
- Michael Rosen, The American Mathematical Monthly, Vol. 88, No. 6 (Jun. - Jul., 1981), pp. 387-395
관련도서 및 추천도서
- Elliptic functions and elliptic integrals
- Viktor Prasolov, Yuri Solovyev
- Pi and the AGM
- Jonathan M. Borwein, Peter B. Borwein
- 도서내검색
- 도서검색
관련기사
- 네이버 뉴스 검색 (키워드 수정)
블로그
- 삼각치환에서 타원적분으로
- 피타고라스의 창, 2009-8-19
- 피타고라스의 창, 2009-8-19
- 구글 블로그 검색
- 네이버 오늘의과학
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- Mathematical Moments from the AMS
- BetterExplained