"슬레이터 2"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 2명의 중간 판 15개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==개요== | |
| − | * [[ | + | * 항등식:<math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math> |
| + | * [[베버(Weber) 모듈라 함수]] 의 하나:<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> | ||
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| − | + | ==항등식의 분류== | |
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* [[슬레이터 목록 (Slater's list)]] | * [[슬레이터 목록 (Slater's list)]] | ||
| − | * E(3) | + | * E(3) |
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| − | + | ==켤레 베일리 쌍의 유도== | |
| − | * | + | * [[q-가우스 합]] 에서 얻어진 다음 결과를 이용:<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>, <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>:<math>\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}</math> |
| − | * | + | * 위의 결과에 다음을 이용:<math>x=q^2, y=-q, z\to\infty</math>. |
| − | * | + | * 켤레 베일리 쌍:<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math>:<math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math> |
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| − | + | ==베일리 쌍의 유도== | |
| − | * Use the | + | * Use the following '''[Slater51] '''(4.1):<math>\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}</math> |
| − | * Specialize | + | * Specialize:<math>a=q,c=-q,d=\infty</math> |
| − | * Bailey pair | + | * Bailey pair:<math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math>:<math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}</math> |
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| − | + | ==베일리 쌍== | |
<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math> | <math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math> | ||
| 55번째 줄: | 47번째 줄: | ||
<math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math> | <math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math> | ||
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| − | + | ==q-series 항등식== | |
| − | <math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n= | + | <math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math> |
| − | * [[베일리 쌍(Bailey pair)과 베일리 보조정리]] | + | * [[베일리 쌍(Bailey pair)과 베일리 보조정리]]:<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math>:<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math>:<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math> ([[오일러의 오각수정리(pentagonal number theorem)]] 가 사용됨) |
| − | * [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] | + | * [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] |
** http://www.research.att.com/~njas/sequences/?q= | ** http://www.research.att.com/~njas/sequences/?q= | ||
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| − | + | ==베테 타입 방정식 (cyclotomic equation)== | |
| − | + | Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ | |
\prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>. | \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>. | ||
| − | + | Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math> has a unique root <math>0<\mu<1</math>. We get | |
<math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | <math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | ||
| 83번째 줄: | 75번째 줄: | ||
a=1,d=1,e=1 | a=1,d=1,e=1 | ||
| − | + | The equation becomes <math>1-x=x</math>. | |
<math>4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2</math> | <math>4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2</math> | ||
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| − | + | ==다이로그 항등식== | |
<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | <math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | ||
| + | [[분류:슬레이터 목록]] | ||
| + | [[분류:q-급수]] | ||
2020년 12월 28일 (월) 03:39 기준 최신판
개요
- 항등식\[\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\]
- 베버(Weber) 모듈라 함수 의 하나\[\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\]
항등식의 분류
켤레 베일리 쌍의 유도
- q-가우스 합 에서 얻어진 다음 결과를 이용\[\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\], \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)\[\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}\]
- 위의 결과에 다음을 이용\[x=q^2, y=-q, z\to\infty\].
- 켤레 베일리 쌍\[\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\]\[\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\]
베일리 쌍의 유도
- Use the following [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
- Specialize\[a=q,c=-q,d=\infty\]
- Bailey pair\[\alpha_{0}=1\], \(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}\]
베일리 쌍
\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)
\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)
\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)
\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)
q-series 항등식
\(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)
- 베일리 쌍(Bailey pair)과 베일리 보조정리\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\]\[\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\]\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\] (오일러의 오각수정리(pentagonal number theorem) 가 사용됨)
베테 타입 방정식 (cyclotomic equation)
Let \(\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}\).
Then \(\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a\) has a unique root \(0<\mu<1\). We get
\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)
a=1,d=1,e=1
The equation becomes \(1-x=x\).
\(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)
다이로그 항등식
\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)