슬레이터 14

개요

• 로저스-라마누잔 항등식 의 하나\[\sum_{n=0}^{\infty}\frac{q^{n(n+1)}}{ (q)_{n}}=\frac{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}\]
• 슬레이터 18 는 또다른 로저스-라마누잔 항등식

항등식의 분류

• 슬레이터 목록 (Slater's list)
• B(2)

켤레 베일리 쌍의 유도

• q-가우스 합 에서 얻어진 다음 결과를 이용\[\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\], \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)\[\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}\]
• 다음의 특수한 경우 \(x=q,y\to\infty, z\to\infty\).
• 얻어진 켤레 베일리 쌍 (relative to 1)\[\delta_n=q^{n^2}\]\[\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\]

베일리 쌍의 유도

• 다음을 이용 [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
• 다음의 특수한 경우\[a=q,c\to\infty,d\to\infty\]
• 얻어진 베일리 쌍 (relative to 1)\[\alpha_{0}=1\], \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{3}{2}n}+q^{-\frac{3}{2}n})\)\[\beta_n=\frac{q^n}{(q)_{n}}\]\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q)_{n-r}(q)_{n+r}}=\frac{q^n}{(q)_{n}}\]

베일리 쌍

• 베일리 쌍과 켤레 베일리 쌍\[\delta_n=q^{n^2}\]\[\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\]\[\alpha_{0}=1\], \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{3}{2}n}+q^{-\frac{3}{2}n})\)\[\beta_n=\frac{q^n}{(q)_{n}}\]

q-series 항등식

• 항등식\[\sum_{n=0}^{\infty}\frac{q^{n(n+1)}}{ (q)_{n}}=\frac{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}\]

• 베일리 쌍(Bailey pair)과 베일리 보조정리\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\]\[\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{n^2+n}}{(q)_{n}}\]\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{5n^2+3n}{2}}+q^{\frac{5n^2-3n}{2}})}{(q)_{\infty}}=\frac{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}\]

• The On-Line Encyclopedia of Integer Sequences
  • http://www.research.att.com/~njas/sequences/?q=

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=2,d=1,e=1

\((1-x)^{1}=x^{2}\).

\(x=\frac{\sqrt{5}-1}{2}\)

\(4L(\frac{3-\sqrt{5}}{2})=\frac{2}{5}(\frac{2}{3}\pi^2)=\frac{4}{15}\pi^2\)

다이로그 항등식

\(L(\frac{3-\sqrt{5}}{2})=\frac{1}{15}\pi^2\)

\(L(\frac{\sqrt{5}-1}{2})=\frac{1}{10}\pi^2\)

메모

• B(3) type
• 베일리 쌍 relative to q\[\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\], \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\) Take \(x=q^2,y\to\infty, z\to\infty\).\[\delta_n=q^{n^2+n}\]\[\gamma_n=\frac{q^{n^2+n}}{(q^2;q)_{\infty}}\]\[\alpha_{0}=1\], \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2+\frac{1}{2}n}\frac{(1-q^{2n +1})}{(1-q)}\)\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\beta_n=\frac{1}{(q)_{n}}\]