"자코비 삼중곱(Jacobi triple product)"의 두 판 사이의 차이
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(피타고라스님이 이 페이지의 이름을 자코비 삼중곱(triple product)로 바꾸었습니다.) |
Pythagoras0 (토론 | 기여) |
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| (사용자 2명의 중간 판 13개는 보이지 않습니다) | |||
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| − | < | + | ==개요== |
| + | * 세타함수의 삼중곱 | ||
| + | :<math>\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)</math> | ||
| + | * <math>z=1</math> 인 경우 | ||
| + | :<math>\sum_{n=-\infty}^\infty q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + q^{2m-1}\right)^2</math> | ||
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| − | + | ;증명 | |
| + | [[Q-초기하급수(q-hypergeometric series)와 양자미적분학(q-calculus)|q-초기하급수(q-hypergeometric series)]]의 다음 등식을 활용 | ||
| + | :<math>\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</math> | ||
| + | :<math>\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</math> | ||
| − | < | + | :<math>\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})}</math> |
| − | + | '''[Andrews65] '''참조 ■ | |
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| − | + | ==또다른 형태== | |
| − | ( | + | :<math>\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)</math> |
| − | + | :<math>\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</math> | |
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| − | + | ==특별한 경우== | |
| + | :<math>\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)</math> | ||
| − | + | :<math>\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)</math> | |
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| − | + | ==예== | |
| − | + | * [[오일러의 오각수정리(pentagonal number theorem)]]:<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> | |
| + | * [[로저스-라마누잔 항등식]] | ||
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| − | + | ==역사== | |
| + | * [[수학사 연표]] | ||
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| − | + | ==메모== | |
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| − | + | ==관련된 항목들== | |
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| − | + | ==매스매티카 파일 및 계산 리소스== | |
| − | * | + | * https://docs.google.com/file/d/0B8XXo8Tve1cxSEM4UjZmckVwbFk/edit |
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| − | + | ==관련논문== | |
| − | + | * '''[Andrews65]'''[http://www.jstor.org/stable/2033875 Shorter Notes: A Simple Proof of Jacobi's Triple Product Identity] | |
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| − | * '''[Andrews65]'''[http://www.jstor.org/stable/2033875 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 | ** George E. Andrews, Proceedings of the American Mathematical Society, Vol. 16, No. 2 (Apr., 1965), pp. 333-334 | ||
| − | * [http://www.jstor.org/stable/2320552 An Easy Proof of the Triple-Product Identity] | + | * [http://www.jstor.org/stable/2320552 An Easy Proof of the Triple-Product Identity] |
| − | ** John A. Ewell, | + | ** John A. Ewell, <cite style="line-height: 2em;">[http://www.jstor.org/action/showPublication?journalCode=amermathmont The American Mathematical Monthly]</cite>, Vol. 88, No. 4 (Apr., 1981), pp. 270-272 |
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| − | + | [[분류:q-급수]] | |
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2014년 10월 30일 (목) 06:07 기준 최신판
개요
- 세타함수의 삼중곱
\[\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\]
- \(z=1\) 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)의 다음 등식을 활용 \[\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\] \[\prod_{n=0}^{\infty}\frac{1}{1+zq^n}=\sum_{n\geq 0}\frac{(-1)^n}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\]
\[\prod_{m=0}^\infty \left( 1 + zq^{2m+1}\right)=\sum_{n\geq 0}\frac{q^nz^n}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\]
[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
역사
메모
관련된 항목들
매스매티카 파일 및 계산 리소스
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