"Slater 31"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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| (같은 사용자의 중간 판 2개는 보이지 않습니다) | |||
| 3번째 줄: | 3번째 줄: | ||
* [[Rogers-Selberg identities]]<math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math> | * [[Rogers-Selberg identities]]<math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math> | ||
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<math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math> | <math>C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}</math> | ||
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==Bailey pair 2== | ==Bailey pair 2== | ||
| − | * Use the | + | * Use the following <math>\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}</math> |
* Specialize<math>a=q,d=-q^{\frac{3}{2}},e=\infty</math> | * Specialize<math>a=q,d=-q^{\frac{3}{2}},e=\infty</math> | ||
| − | * Bailey pair<math>\alpha_{0}=1</math>, | + | * 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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| − | ==Bailey | + | ==Bailey pair == |
| − | * Bailey pairs<math>\delta_n=q^{n^2}</math><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</math><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math> | + | * Bailey pairs<math>\delta_n=q^{n^2}</math><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</math><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math> |
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==q-series identity== | ==q-series identity== | ||
| 36번째 줄: | 36번째 줄: | ||
** 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)== | ==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> | ||
| 57번째 줄: | 57번째 줄: | ||
a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1 | a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1 | ||
| − | + | The equation becomes <math>(1-x^4)(1-x^4)=x^4(1-x^2)</math>. | |
This is factorized into <math>(-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)</math> | This is factorized into <math>(-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)</math> | ||
| 69번째 줄: | 69번째 줄: | ||
<math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> | <math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> | ||
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==dilogarithm identity== | ==dilogarithm identity== | ||
| 79번째 줄: | 79번째 줄: | ||
<math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> where <math>\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</math> | <math>7L(\alpha^2)-7L(\alpha)+L(1)=0</math> where <math>\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots</math> | ||
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==related items== | ==related items== | ||
| 89번째 줄: | 89번째 줄: | ||
* [[asymptotic analysis of basic hypergeometric series]] | * [[asymptotic analysis of basic hypergeometric series]] | ||
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[[분류:개인노트]] | [[분류:개인노트]] | ||
[[분류:math and physics]] | [[분류:math and physics]] | ||
[[분류:migrate]] | [[분류:migrate]] | ||
2020년 12월 28일 (월) 05:06 기준 최신판
Note
- Rogers-Selberg identities\(C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)
\(C(q)=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)
Bailey pair 2
- Use the following \(\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\)
- Specialize\(a=q,d=-q^{\frac{3}{2}},e=\infty\)
- Bailey pair\(\alpha_{0}=1\), \(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}\)
Bailey pair
- Bailey pairs\(\delta_n=q^{n^2}\)\(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)
q-series identity
\(\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\frac{(q^{1};q^{7})_{\infty}(q^{6};q^{7})_{\infty}(q^{7};q^{7})_{\infty}}{(q^{2};q^{2})_{\infty}}\)
- Bailey's lemma\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{n^2}}{(q)_{n}}\)\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\)
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
\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)
To apply the above result, we rewrite the equation in the suitable form.
\((-q;q)_{2n+1}=(-q)_{2n}(1+q^{2n+1})=\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})\) (useful techniques in q-series)
\(\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{ (q^{2};q^{2})_{n}(-q;q)_{2n+1}}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}}{(q^{2};q^{2})_{n}\frac{(q^2;q^4)_{n}(q^4;q^4)_{n}}{(q;q^2)_{n}(q^2;q^2)_{n}}(1+q^{2n+1})}=\sum_{n=0}^{\infty}\frac{q^{2n^2+2n}(q;q^2)_{n}}{(q^2;q^4)_{n}(q^4;q^4)_{n}(1+q^{2n+1})}\)
a=4,d1=4,e1=1,d2=4,e2=1,d3=2,e3=-1
The equation becomes \((1-x^4)(1-x^4)=x^4(1-x^2)\).
This is factorized into \((-1+x) (1+x) \left(-1-x^2+2 x^4+x^6\right)\)
So \(\mu^2=\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots\)
\(4(\frac{1}{4}L(1-\alpha^2)+\frac{1}{4}L(1-\alpha^2)-\frac{1}{2}L(1-\alpha))=\frac{1}{14}(\frac{2}{3}\pi^2)\)
\(2L(1-\alpha^2)-2L(1-\alpha)=\frac{\pi^2}{21}\)
\(7L(\alpha^2)-7L(\alpha)+L(1)=0\)
dilogarithm identity
\(7L(\alpha^2)-7L(\alpha)+L(1)=0\) where \(\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots\)