"자코비 삼중곱(Jacobi triple product)"의 두 판 사이의 차이
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− | * [[오일러의 오각수정리(pentagonal number theorem)]]<br><math>\sum _{m=-\infty }^{\infty } (-1)^mq^{\frac{3}{2}m^2\pm \frac{1}{2}m} = \prod _{n=1}^{\infty } \left(1-q^{3 n}\right)\left(1-q^{3n-2}\right)\left(1-q^{3n-1}\right)</math><br> | + | * [[오일러의 오각수정리(pentagonal number theorem)]]<br><math>\sum _{m=-\infty }^{\infty } (-1)^mq^{\frac{3}{2}m^2\pm \frac{1}{2}m} = \prod _{n=1}^{\infty } \left(1-q^{3 n}\right)\left(1-q^{3n-2}\right)\left(1-q^{3n-1}\right)=\prod _{n=1}^{\infty } \left(1-q^{n}\right)</math><br> |
+ | * [[로저스-라마누잔 항등식]]<br> | ||
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2011년 6월 19일 (일) 10:21 판
이 항목의 스프링노트 원문주소
개요
\(\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\)
\(z=1\) 인 경우
\(\sum_{n=-\infty}^\infty q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + q^{2m-1}\right)^2\)
(증명)
q-초기하급수(q-hypergeometric series)
\(\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)
\(\prod_{n=0}^{\infty}\frac{1}{1+zq^n}=\sum_{n\geq 0}\frac{(-1)^n}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)
를 활용
\(\prod_{m=0}^\infty \left( 1 + zq^{2m+1}\right)=\sum_{n\geq 0}\frac{q^nz^n}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\)
[Andrews65] 참조 ■
또다른 형태
\(\sum _{n=-\infty }^{\infty } (-1)^na^nq^{n(n-1)/2}=\prod _{n=1}^{\infty } \left(1-aq^{n-1}\right)\left(1-a^{-1}q^n\right)\left(1-q^n\right)\)
\(\prod _{n=1}^{\infty } \left(1-x^{2n}\right)\left(1+x^{2n-1}Z\right)\left(1+x^{2n-1}Z^{-1}\text{}\text{}\right)=\sum _{m=-\infty }^{\infty } x^{m^2}Z^m\)
특별한 경우
\(\sum _{m=-\infty }^{\infty } (-1)^mq^{a m^2\pm b m +c}=q^c\prod _{n=1}^{\infty } \left(1-q^{2a n}\right)\left(1-q^{2a n-a+b}\right)\left(1-q^{2a n-a-b}\right)\)
\(\sum _{m=-\infty }^{\infty } q^{a m^2\pm b m +c}=q^c\prod _{n=1}^{\infty } \left(1-q^{2a n}\right)\left(1+q^{2a n-a+b}\right)\left(1+q^{2a n-a-b}\right)\)
예
- 오일러의 오각수정리(pentagonal number theorem)
\(\sum _{m=-\infty }^{\infty } (-1)^mq^{\frac{3}{2}m^2\pm \frac{1}{2}m} = \prod _{n=1}^{\infty } \left(1-q^{3 n}\right)\left(1-q^{3n-2}\right)\left(1-q^{3n-1}\right)=\prod _{n=1}^{\infty } \left(1-q^{n}\right)\) - 로저스-라마누잔 항등식
재미있는 사실
- Math Overflow http://mathoverflow.net/search?q=
- 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=
역사
- http://www.google.com/search?hl=en&tbs=tl:1&q=
- Earliest Known Uses of Some of the Words of Mathematics
- Earliest Uses of Various Mathematical Symbols
- 수학사연표
메모
관련된 항목들
관련논문
- [Andrews65]Shorter Notes: A Simple Proof of Jacobi's Triple Product Identity
- George E. Andrews, Proceedings of the American Mathematical Society, Vol. 16, No. 2 (Apr., 1965), pp. 333-334
- An Easy Proof of the Triple-Product Identity
- John A. Ewell, The American Mathematical Monthly, Vol. 88, No. 4 (Apr., 1981), pp. 270-272
- http://www.jstor.org/action/doBasicSearch?Query=
- http://www.ams.org/mathscinet
- http://dx.doi.org/