A WZ PROOF OF RAMANUJAN’S FORMULA FOR π
수학자 아키크메스와 라마누잔의 π에 관한 계산 방법을 논하는 학술논문입니다. 이 논문은 WZ 증명을 통해 라마누잔이 발견한 무한급수식 공식 중 하나를 증명하고 있습니다. 이 공식은 다음과 같습니다.
π = 2∞∑k=0(1/4)^k(1/2)^k(3/4)^k \* (1103 + 26390k)(1/99)^(4k+2)
논문에서 WZ 증명 방법을 사용하여 이 공식의 적분형인 다음 공식을 증명합니다.
∞∑k=0(-1)^k(4k+1)(1/2)^3k / k!^3 = 2/π
이 논문은 수학자들 사이에서 중요한 계산 방법을 제공하며, WZ 증명의 원리와 라마누잔의 무한급수식 공식에 대한 이해를 높일 것입니다.
영어 요약 시작:
This paper presents a WZ proof for Ramanujan's formula for π. The formula is an example of a non-terminating hypergeometric series identity, which was discovered by Ramanujan. The paper uses the WZ method to prove that
π = 2∞∑k=0(1/4)^k(1/2)^k(3/4)^k \* (1103 + 26390k)(1/99)^(4k+2)
is equivalent to another formula for π, which is given by:
∞∑k=0(-1)^k(4k+1)(1/2)^3k / k!^3 = 2/π
The WZ proof method is used to prove the latter formula, and the paper shows that the two formulas are equivalent. This result provides a new understanding of Ramanujan's formula for π and highlights the power of the WZ method in proving non-terminating hypergeometric series identities.
영어 요약 끝
A WZ PROOF OF RAMANUJAN’S FORMULA FOR π
arXiv:math/9306213v1 [math.CO] 3 Jun 1993A WZ PROOF OF RAMANUJAN’S FORMULA FOR πShalosh B. EKHAD1 and Doron ZEILBERGER1Dedicated to Archimedes on his 2300th birthdayArchimedes computed π very accurately. Much later, Ramanujan discovered several infinite seriesfor 1/π that enables one to compute π even more accurately.
The most impressive one is([Ra]):((a)k denotes, as usual, a(a + 1)...(a + k −1). )1π = 2√2∞Xk=0(1/4)k(1/2)k(3/4)kk!3(1103 + 26390k)(1/99)4k+2.
(1)This formula is an example of a non-terminating hypergeometric series identity. Many times, non-terminating series are either limiting cases or ”analytic continuations” of terminating identities,which are now known to be routinely provable by computer.
[WZ].While we do not know of a terminating generalization of (1), we do know how to give a WZ proofof another formula for π, also given by Ramanujan[Ra], and included in his famous letter to Hardy.This formula is:2π =∞Xk=0(−1)k(4k + 1)(1/2)3kk!3. (2)The terminating version, that we will prove isΓ(3/2 + n)Γ(3/2)Γ(n + 1) =∞Xk=0(−1)k(4k + 1) (1/2)2k(−n)kk!2(3/2 + n)k.(3)To prove it for all positive integers n, we call the summand divided by the left side F(n, k), andcleverly constructG(n, k) :=(2k + 1)2(2n + 2k + 3)(4k + 1)F(n, k) ,with the motive that F(n + 1, k) −F(n, k) = G(n, k) −G(n, k −1) (check!
), and summing this lastidentity w.r.t k shows that Pk F(n, k) ≡Constant, which is seen to be 1, by plugging in n = 0.This proves (3). To deduce (2), we ”plug” in n = −1/2, which is legitimate in view of Carlson’stheorem [Ba].REFERENCES[Ba] W.N.
Bailey, “Generalized Hypergeometric Series”, (Cambridge Univ. Press, 1935), p. 39.1 Department of Mathematics, Temple University, Philadelphia, PA 19122.
[ekhad,zeilberg]@math.temple.edu ;http://www.math.temple.edu/~ [ekhad,zeilberg].The work of the second author was supported in part by theNSF. This paper was published in p.107-108 of ‘Geometry, Analysis, and Mechanics’, ed.
by J. M. Rassias, WorldScientific, Singapore 1994.1
[Ra] K.S. Rao, in “Srinivasa Ramanujan”, ed.
K.R. Nagarajan and T. Soundararajan, (MACMIL-LAN INDIA, Madras, 1988).
[WZ] H. S. Wilf and D. Zeilberger, Amer. Math.Soc.
B3 (1990) 147.2
출처: arXiv:9306.213 • 원문 보기