"Slater 31"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
잔글 (찾아 바꾸기 – “<h5>” 문자열을 “==” 문자열로)
imported>Pythagoras0
잔글 (찾아 바꾸기 – “</h5>” 문자열을 “==” 문자열로)
1번째 줄: 1번째 줄:
−
==Note</h5>
+
==Note==
  
 
* [[Rogers-Selberg identities]]<br><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><br>
 
* [[Rogers-Selberg identities]]<br><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><br>
11번째 줄: 11번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 2em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair 2</h5>
+
<h5 style="line-height: 2em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair 2==
  
 
*  Use the following <br><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><br>
 
*  Use the following <br><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><br>
21번째 줄: 21번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair </h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bailey pair ==
  
 
*  Bailey pairs<br><math>\delta_n=q^{n^2}</math><br><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</math><br><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math><br>  <br>
 
*  Bailey pairs<br><math>\delta_n=q^{n^2}</math><br><math>\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}</math><br><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math><br>  <br>
27번째 줄: 27번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">q-series identity</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">q-series identity==
  
 
<math>\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>\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>
40번째 줄: 40번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bethe type equation (cyclotomic equation)</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Bethe type equation (cyclotomic equation)==
  
 
Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
 
Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
75번째 줄: 75번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">dilogarithm identity</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">dilogarithm identity==
  
 
<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>
85번째 줄: 85번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">related items</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">related items==
  
 
* [[asymptotic analysis of basic hypergeometric series]]<br>
 
* [[asymptotic analysis of basic hypergeometric series]]<br>
93번째 줄: 93번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">books</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">books==
  
 
 
 
 
106번째 줄: 106번째 줄:
 
 
 
 
  
−
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
+
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles==
  
 
*   <br>
 
*   <br>

2012년 10월 28일 (일) 15:41 판

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 \(\alpha=\frac{\sec\frac{2\pi}{7}}{2}=0.80194\cdots\)      
related items==    
books==      
articles==
원본 주소 "https://wiki.mathnt.net/index.php?title=Slater_31&oldid=36178"