"Slater 37"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 3명의 중간 판 14개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==Note== | |
| − | * not checked | + | * not checked |
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| − | + | ==type of identity== | |
* [[Slater list|Slater's list]] | * [[Slater list|Slater's list]] | ||
| − | * I(17) | + | * I(17) |
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| − | + | ==Bailey pair 1== | |
| − | * Use the folloing | + | * Use the folloing<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> |
| − | * Specialize | + | * Specialize<math>x=q^{3}, y=-q, z\to\infty</math>. |
| − | * | + | * Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math> |
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| − | + | ==Bailey pair 2== | |
| − | * Use the | + | * Use the following '''[Slater52-1] '''(4.2) |
| − | * Specialize | + | * Specialize<math>a=q^{2},d=q^2,e=q</math> |
| − | * Bailey pair | + | * Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</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^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math> |
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| − | + | ==Bailey pair == | |
| − | * Bailey pairs | + | * Bailey pairs <math>\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}</math><math>\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)</math> <math>\alpha_{0}=1</math>, <math>\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)</math>,<math>\alpha_{2n+1}=0</math><math>\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}</math> |
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| − | + | ==q-series identity== | |
<math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math> | <math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math> | ||
| − | * [[Bailey pair and lemma|Bailey's lemma]] | + | * [[Bailey pair and lemma|Bailey's lemma]]<math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{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> ([http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)] was used to verify this)<math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math> |
| − | * [http://www.research.att.com/ | + | * [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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| − | + | ==Bethe type equation (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> | ||
| − | + | ||
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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| − | + | ==dilogarithm identity== | |
<math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | <math>L(\frac{1}{2})=\frac{1}{12}\pi^2</math> | ||
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| − | + | ==related items== | |
| − | * [[asymptotic analysis of basic hypergeometric series]] | + | * [[asymptotic analysis of basic hypergeometric series]] |
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| − | + | ==articles== | |
| + | * [http://www.combinatorics.org/Surveys/ds15.pdf Rogers-Ramanujan-Slater Type identities] | ||
| + | ** McLaughlin, 2008 | ||
| + | * [http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type] | ||
| + | ** Slater, L. J. (1952), Proceedings of the London Mathematical Society. Second Series 54: 147–167 | ||
| + | * [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series] | ||
| + | ** Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475 | ||
| − | + | [[분류:개인노트]] | |
| − | + | [[분류:math and physics]] | |
| − | + | [[분류:migrate]] | |
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2020년 12월 28일 (월) 04:15 기준 최신판
Note
- not checked
type of identity
- Slater's list
- I(17)
Bailey pair 1
- Use the folloing\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)
- Specialize\(x=q^{3}, y=-q, z\to\infty\).
- Bailey pair\(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)\(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\)
Bailey pair 2
- Use the following [Slater52-1] (4.2)
- Specialize\(a=q^{2},d=q^2,e=q\)
- Bailey pair\(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{3})_{n-r}(q)_{n+r}}=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)
Bailey pair
- Bailey pairs \(\delta_n=(-q)_{n}q^{\frac{n(n+3)}{2}}\)\(\gamma_n=\frac{(-q^2)_{\infty}}{(q^3)_{\infty}}q^{\frac{n(n+3)}{2}}(1+q)\) \(\alpha_{0}=1\), \(\alpha_{2n}=(-1)^{n}q^{n(2n+1)}(1-q^{2n+1})/(1-q)\),\(\alpha_{2n+1}=0\)\(\beta_n=\frac{(q^2,q^2)_{n}}{(q)_{n}(q^2)_{n}(q^3,q^2)_{n}}\)
q-series identity
\(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)
- Bailey's lemma\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\) (오일러의 오각수정리(pentagonal number theorem) was used to verify this)\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)
Bethe type equation (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 \(1-x=x\).
\(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)
dilogarithm identity
\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)
articles
- Rogers-Ramanujan-Slater Type identities
- McLaughlin, 2008
- Further identities of the Rogers-Ramanujan type
- Slater, L. J. (1952), Proceedings of the London Mathematical Society. Second Series 54: 147–167
- A New Proof of Rogers's Transformations of Infinite Series
- Slater, L. J. (1952), Proc. London Math. Soc. 1951 s2-53: 460-475