"베버(Weber) 모듈라 함수"의 두 판 사이의 차이
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* [[q-초기하급수(q-hypergeometric series) (통합됨)|q-초기하급수(q-hypergeometric series)]] 의 공식<br><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><br><math>z=q^{1/2}</math> 인 경우<br><math>\prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} (q^{1/2})^n=\sum_{n\geq 0}\frac{q^{n^2/2}}{(1-q)(1-q^2)\cdots(1-q^n)} </math><br><math>\prod_{n=1}^{\infty} (1+q^{2n-1})=\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})} </math><br><math>z=q</math> 인 경우<br><math>\prod_{n=1}^{\infty} (1+q^{n})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}q^n=\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}</math><br> | * [[q-초기하급수(q-hypergeometric series) (통합됨)|q-초기하급수(q-hypergeometric series)]] 의 공식<br><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><br><math>z=q^{1/2}</math> 인 경우<br><math>\prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} (q^{1/2})^n=\sum_{n\geq 0}\frac{q^{n^2/2}}{(1-q)(1-q^2)\cdots(1-q^n)} </math><br><math>\prod_{n=1}^{\infty} (1+q^{2n-1})=\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})} </math><br><math>z=q</math> 인 경우<br><math>\prod_{n=1}^{\infty} (1+q^{n})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}q^n=\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}</math><br> | ||
* 위의 결과로부터 다음을 얻을 수 있다<br><math>\mathfrak{f}(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}</math><br><math>\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}</math><br> | * 위의 결과로부터 다음을 얻을 수 있다<br><math>\mathfrak{f}(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}</math><br><math>\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}</math><br> | ||
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** [http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=webe034 Lehrbuch der Algebra (Volume 3)] (1898). [http://astech.library.cornell.edu/ast/math/additional/Digital-Books.cfm available in print] | ** [http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=webe034 Lehrbuch der Algebra (Volume 3)] (1898). [http://astech.library.cornell.edu/ast/math/additional/Digital-Books.cfm available in print] | ||
** [http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=02910001 Theorie der Abelschen Functionen vom Geschlecht 3] (1876). [http://www.amazon.com/dp/1429704683?tag=corneunivelib-20 available in print] | ** [http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=02910001 Theorie der Abelschen Functionen vom Geschlecht 3] (1876). [http://www.amazon.com/dp/1429704683?tag=corneunivelib-20 available in print] | ||
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2012년 7월 19일 (목) 15:10 판
이 항목의 스프링노트 원문주소
개요
- 베버의 class invariant 라는 이름으로 잘 알려져 있으며, 베버는 Schläfli 함수로 불렀음
- class field theory에서 중요한 역할
정의
\(\mathfrak{f}(\tau)=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})\)
\(\mathfrak{f}_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)
\(\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)
\(\gamma_2(\tau)=\frac{\mathfrak{f}(\tau)^{24}-16}{\mathfrak{f}(\tau)^8}=\sqrt[3]{j(\tau)}\)
\(\gamma_3(\tau)= \frac{(\mathfrak{f}(z)^{24} + 8) (\mathfrak{f}_1(z)^8 - \mathfrak{f}_2(z)^8)}{\mathfrak{f}(z)^8}=\sqrt{j(\tau)-1728}\)
여기서 \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\) 는 데데킨트 에타함수
항등식
- \(\mathfrak{f}_1(2\tau)\mathfrak{f}_2(\tau)=\sqrt2\)
- \(\mathfrak{f}(\tau)\mathfrak{f}_1(\tau)\mathfrak{f}_2(\tau)=\sqrt2\)
- \(\mathfrak{f}(\tau)^8=\mathfrak{f}_1(\tau)^8+\mathfrak{f}_2(\tau)^8\)
모듈라 성질
- \(\mathfrak{f}(\tau+1)=\zeta_{48}^{-1}\mathfrak{f}_1(\tau)\)
- \(\mathfrak{f}_1(\tau+1)=\zeta_{48}^{-1}\mathfrak{f}(\tau)\)
- \(\mathfrak{f}_2(\tau+1)=\zeta_{24}\mathfrak{f}_2(\tau)\)
- \(\mathfrak{f}(-\frac{1}{\tau})=\mathfrak{f}(\tau)\)
- \(\mathfrak{f}_1(-\frac{1}{\tau})=\mathfrak{f}_2(\tau)\)
- \(\mathfrak{f}_2(-\frac{1}{\tau})=\mathfrak{f}_1(\tau)\)
j-invariant 와의 관계
- 타원 모듈라 j-함수 (j-invariant)
- \(\mathfrak{f}(\tau)^{24}\), \(-\mathfrak{f}_1(\tau)^{24}\), \(-\mathfrak{f}_2(\tau)^{24}\)는 \((x-16)^3-j(\tau)x=0\) 의 근이다
special values
- \(\mathfrak{f}(i)^8=4\)
- \(\mathfrak{f}_1(i)^8=2\)
- \(\mathfrak{f}_2(i)^8=2\)
- \(\mathfrak{f}_1(2i)^8=8\)
데데킨트 에타함수와의 관계
\(f(\tau)=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})\)
\(f_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)
\(f_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)
여기서 \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\) 는 데데킨트 에타함수
q-초기하급수와의 관계
- q-초기하급수(q-hypergeometric series) 의 공식
\(\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)
\(z=q^{1/2}\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} (q^{1/2})^n=\sum_{n\geq 0}\frac{q^{n^2/2}}{(1-q)(1-q^2)\cdots(1-q^n)} \)
\(\prod_{n=1}^{\infty} (1+q^{2n-1})=\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})} \)
\(z=q\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}q^n=\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\) - 위의 결과로부터 다음을 얻을 수 있다
\(\mathfrak{f}(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\)
\(\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)
역사
메모
관련된 항목들
- 라마누잔의 class invariants
- 데데킨트 에타함수
- 로저스-라마누잔 연분수와 항등식
- q-초기하급수(q-hypergeometric series)
- 자코비 세타함수
- 타원 모듈라 j-함수 (j-invariant)
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- 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
- Weber's class invariants revisited
- [1]Reinhard Schertz, Journal de théorie des nombres de Bordeaux, 14 no. 1 (2002), p. 325-343
- On The Singular Values Of Weber Modular Functions
- Noriko Yui , Don Zagier, Math. Comp. 66 (1997), 1645-1662
- Weber's Class Invariants
- B. J. Birch, Mathematika 16 (1969)
- http://www.jstor.org/action/doBasicSearch?Query=
관련도서
- 베버의 책
- Elliptische functionen und algebraische zahlen (1891). available in print
- Lehrbuch der Algebra (Volume 1) (1898). available in print
- Lehrbuch der Algebra (Volume 2) (1898). available in print
- Lehrbuch der Algebra (Volume 3) (1898). available in print
- Theorie der Abelschen Functionen vom Geschlecht 3 (1876). available in print