타원적분

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Pythagoras0 (토론 | 기여)님의 2022년 9월 16일 (금) 03:19 판
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개요

  • 먼저 타원적분론 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



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

\[2K(\frac{1}{\sqrt{2}})E(\frac{1}{\sqrt{2}})-K(\frac{1}{\sqrt{2}})^2=\frac{\pi}{2}\]

산술기하평균함수(AGM)와 파이값의 계산에 응용



덧셈공식

\[\int_0^x{\frac{1}{\sqrt{1-x^4}}}dx+\int_0^y{\frac{1}{\sqrt{1-x^4}}}dx = \int_0^{A(x,y)}{\frac{1}{\sqrt{1-x^4}}}dx\] 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 \[B(x,y)=\frac{x\sqrt{p(y)}+y\sqrt{p(x)}}{1-nx^2y^2}\]



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말뭉치

  1. For instance, while the arc length of a circle is given as a simple function of the parameter, computing the arc length of an ellipse requires an elliptic integral.[1]
  2. An elliptic integral is written when the parameter is used, when the elliptic modulus is used, and when the modular angle is used.[1]
  3. Additional insight into the theory of the elliptic integral may be gained through the study of the Schwarz–Christoffel mapping.[2]
  4. These arguments are expressed in a variety of different but equivalent ways (they give the same elliptic integral).[2]
  5. This is referred to as the incomplete Legendre elliptic integral.[3]
  6. The integral is also called Legendres form for the elliptic integral of the first kind.[4]
  7. = 0 (1+2)122 , 0 < < 1, 0, also called Legendres form for the elliptic integral of the third kind.[4]
  8. The upper limit x in the Jacobi form of the elliptic integral of the first kind is related to the upper limit in the Legendre form by = sin .[4]
  9. returns values of the complete elliptic integral E(K).[5]
  10. returns values of the complete elliptic integral E(M).[5]
  11. returns values of the complete elliptic integral F(K).[5]
  12. returns values of the complete elliptic integral F(M).[5]
  13. Any elliptic integral can be expressed as a sum of elementary functions and linear combinations of canonical elliptic integrals of the first, second and third kinds.[6]
  14. TOMS577, a C++ library which evaluates Carlson's elliptic integral functions RC, RD, RF and RJ.[7]
  15. The point here is simply show one of the uses of the elliptic integral of the first kind.[8]
  16. However, the complete elliptic integral is taken from 0 to 90 degrees, but since that is exactly 1/4 over the total arc length due to symmetry.[8]
  17. Complete elliptic integral of the second kind Math.[9]
  18. Complete elliptic integral of the first kind.[10]
  19. Complete elliptic integral of the third kind.[10]
  20. Cubic and sesquiplicate modular transformations of an elliptic integral from the equal-mass Dalitz plot are proven and used extensively.[11]
  21. dy 1 + a cos x + b cos y (1) Keywords: Double elliptic integral, hypergeometric function 1 .[12]
  22. v i X r a SHARP DOUBLE INEQUALITY FOR COMPLETE ELLIPTIC INTEGRAL OF THE FIRST KIND QI BAO r2 sin2 t)1/2dt is known as the Abstract.[13]
  23. Abstract We prove a simple relation for a special case of Carlsons elliptic integral RD.[14]
  24. Let K be the complete elliptic integral of the rst kind.[15]