q超幾何級数によるJacobiの六平方和定理, 八平方和定理の証明

Baileyの${}_6{\psi }_6$和公式
$$\begin{array}{rl}& {}_6{\psi }_6[\begin{array}{c}\sqrt{a}q,-\sqrt{a}q,b,c,d,e\\ \sqrt{a},-\sqrt{a},aq/b,aq/c,aq/d,aq/e\end{array};\frac{a^{2}q}{bcde}]\\ & =\frac{(aq,aq/bc,aq/bd,aq/be,aq/cd,aq/ce,aq/de,q,q/a;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,q/b,q/c,q/d,q/e,a^{2}q/bcde;q{)}_{\mathrm{\infty }}}\end{array}$$
において, $b=c=d=-1,e\mapsto \mathrm{\infty }$とすると,
$$\begin{array}{rl}\frac{(q,q/a;q{)}_{\mathrm{\infty }}(aq;q{)}_{\mathrm{\infty }}^4}{(-q,-aq;q{)}_{\mathrm{\infty }}^3}& =\sum _{n\in \mathbb{Z}}\frac{1-a{q}^{2n}}{1-a}\frac{(-1;q{)}_n^3}{(-aq;q{)}_n^3}{a}^{2n}{q}^{(\genfrac{}{}{0ex}{}{n+1}{2})}\\ & =1+\sum _{0<n}\frac{1-a{q}^{2n}}{1-a}\frac{(-1;q{)}_n^3}{(-aq;q{)}_n^3}{a}^{2n}{q}^{(\genfrac{}{}{0ex}{}{n+1}{2})}\\ & +\sum _{0<n}\frac{1-a{q}^{-2n}}{1-a}\frac{(-1;q{)}_{-n}^3}{(-aq;q{)}_{-n}^3}{a}^{-2n}{q}^{(\genfrac{}{}{0ex}{}{1-n}{2})}\\ & =1+\sum _{0<n}\frac{q^{(\genfrac{}{}{0ex}{}{n+1}{2})}}{1-a}((1-a{q}^{2n})a^{2n}\frac{(-1;q{)}_n^3}{(-aq;q{)}_n^3}+(q^{2n}-a)a^n\frac{(-1/a;q{)}_n^3}{(-q;q{)}_n^3})\end{array}$$
ここで, $a\to 1$とすると,
$$\begin{array}{rl}& \frac{d}{da}{((1-a{q}^{2n})a^{2n}\frac{(-1;q{)}_n^3}{(-aq;q{)}_n^3}+(q^{2n}-a)a^n\frac{(-1/a;q{)}_n^3}{(-q;q{)}_n^3})|}_{a=1}\\ & =(2n-(2n+1)q^{2n}+n{q}^{2n}-(n+1))\frac{(-1;q{)}_n^3}{(-q;q{)}_n^3}\\ & -3(1-q^{2n})\frac{(-1;q{)}_n^3}{(-q;q{)}_n^3}\sum _{k=1}^n\frac{q^k}{1+q^k}+3(1-q^{2n})\frac{(-1;q{)}_n^3}{(-q;q{)}_n^3}\sum _{k=0}^{n-1}\frac{q^k}{1+q^k}\\ & =\frac{8(n-1-(n+1)q^{2n})}{(1+q^n{)}^3}+\frac{12(1-q^{2n})}{(1+q^n{)}^3}(1-\frac{2q^n}{1+q^n})\\ & =\frac{8(n-1-(n+1)q^{2n})+12(1-q^n{)}^2}{(1+q^n{)}^3}\\ & =\frac{4(2n+1-6q^n-(2n-1)q^{2n})}{(1+q^n{)}^3}\end{array}$$
であるから,
$$\begin{array}{rl}{\left(\frac{(q;q{)}_{\mathrm{\infty }}}{(-q;q{)}_{\mathrm{\infty }}}\right)}^6& =1+4\sum _{0<n}\frac{q^{(\genfrac{}{}{0ex}{}{n+1}{2})}}{(1+q^n{)}^3}(6q^n+(2n-1)q^{2n}-(2n+1))\end{array}$$
を得る. ここで,
$$\begin{array}{rl}& \sum _{0<n}\frac{q^{(\genfrac{}{}{0ex}{}{n+1}{2})}}{(1+q^n{)}^3}(6q^n+(2n-1)q^{2n}-(2n+1))\\ & =\sum _{0<n}\sum _{0\le k}(-1{)}^{k-1}{q}^{(\genfrac{}{}{0ex}{}{n+1}{2})+nk}(6(\genfrac{}{}{0ex}{}{k+1}{2})-(2n-1)(\genfrac{}{}{0ex}{}{k}{2})+(2n+1)(\genfrac{}{}{0ex}{}{k+2}{2}))\\ & =\sum _{0<n}\sum _{0\le k}(-1{)}^{k-1}{q}^{\frac{n(n+2k+1)}{2}}((n+2k+1{)}^2-n^2)\end{array}$$
ここで, $n$の偶奇で分けることによって,
$$\begin{array}{rl}& \sum _{0<n}\sum _{0\le k}(-1{)}^{k-1}{q}^{\frac{n(n+2k+1)}{2}}((n+2k+1{)}^2-n^2)\\ & =\sum _{0<n}\sum _{0\le k}(-1{)}^{k-1}{q}^{n(2n+2k+1)}((2n+2k+1{)}^2-(2n{)}^2)\\ & +\sum _{0<n}\sum _{0\le k}(-1{)}^{k-1}{q}^{(2n-1)(n+k)}((2n+2k{)}^2-(2n-1{)}^2)\\ & =\sum _{0<n\le k}(-1{)}^{n+k-1}{q}^{n(2k+1)}((2k+1{)}^2-(2n{)}^2)\\ & +\sum _{0<n\le k}(-1{)}^{n+k-1}{q}^{(2n-1)k}((2k{)}^2-(2n-1{)}^2)\\ & =\sum _{0<n\le k}(-1{)}^{n+k-1}{q}^{n(2k+1)}((2k+1{)}^2-(2n{)}^2)\\ & +\sum _{0\le k<n}(-1{)}^{n+k-1}{q}^{n(2k+1)}((2k+1{)}^2-(2n{)}^2),(n\mapsto k+1,k\mapsto n)\\ & =\sum _{0<n,0\le k}(-1{)}^{n+k-1}{q}^{n(2k+1)}((2k+1{)}^2-(2n{)}^2)\\ & =\sum _{0<n,0\le k}(-1{)}^{n+k-1}{q}^{n(2k+1)}(2k+1{)}^2-\sum _{0<n,0\le k}(-1{)}^{n+k-1}{q}^{n(2k+1)}(2n{)}^2\\ & =\sum _{0\le k}\frac{(-1{)}^k(2k+1{)}^2{q}^{2k+1}}{1+q^{2k+1}}+\sum _{0<n}\frac{(-1{)}^n(2n{)}^2{q}^n}{1+q^{2n}}\end{array}$$
よって,
$$\begin{array}{rl}{\left(\sum _{n\in \mathbb{Z}}{q}^{n^2}\right)}^6& ={\left(\frac{(-q;-q{)}_{\mathrm{\infty }}}{(q;-q{)}_{\mathrm{\infty }}}\right)}^6\\ & =1+16\sum _{0<n}\frac{n^2{q}^n}{1+q^{2n}}-4\sum _{0\le n}\frac{(-1{)}^n(2n+1{)}^2{q}^{2n+1}}{1-q^{2n+1}}\end{array}$$
を得る.

Baileyの${}_6{\psi }_6$和公式
$$\begin{array}{rl}& {}_6{\psi }_6[\begin{array}{c}\sqrt{a}q,-\sqrt{a}q,b,c,d,e\\ \sqrt{a},-\sqrt{a},aq/b,aq/c,aq/d,aq/e\end{array};\frac{a^{2}q}{bcde}]\\ & =\frac{(aq,aq/bc,aq/bd,aq/be,aq/cd,aq/ce,aq/de,q,q/a;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,q/b,q/c,q/d,q/e,a^{2}q/bcde;q{)}_{\mathrm{\infty }}}\end{array}$$
において, $b=c=d=e=-1$とすると,
$$\begin{array}{rl}\frac{(q,q/a;q{)}_{\mathrm{\infty }}(aq;q{)}_{\mathrm{\infty }}^7}{(a^{2}q;q{)}_{\mathrm{\infty }}(-q,-aq;q{)}_{\mathrm{\infty }}^8}& =\sum _{n\in \mathbb{Z}}\frac{1-a{q}^{2n}}{1-a}\frac{(-1;q{)}_n^4}{(-aq;q{)}_n^4}(a^{2}q{)}^n\\ & =1+\sum _{0<n}\frac{1-a{q}^{2n}}{1-a}\frac{(-1;q{)}_n^4}{(-aq;q{)}_n^4}(a^{2}q{)}^n\\ & +\sum _{0<n}\frac{1-a{q}^{-2n}}{1-a}\frac{(-1;q{)}_{-n}^4}{(-aq;q{)}_{-n}^4}(a^{2}q{)}^{-n}\\ & =1+\sum _{0<n}\frac{a^{2n}{q}^n}{1-a}((1-a{q}^{2n})\frac{(-1;q{)}_n^4}{(-aq;q{)}_n^4}+(q^{2n}-a)\frac{(-1/a;q{)}_n^4}{(-q;q{)}_n^4})\end{array}$$
ここで, $a\to 1$とすると,
$$\begin{array}{rl}& \frac{d}{da}{((1-a{q}^{2n})\frac{(-1;q{)}_n^4}{(-aq;q{)}_n^4}+(q^{2n}-a)\frac{(-1/a;q{)}_n^4}{(-q;q{)}_n^4})|}_{a=1}\\ & =-(1+q^{2n})\frac{(-1;q{)}_n^4}{(-q;q{)}_n^4}-4\frac{(-1;q{)}_n^4}{(-q;q{)}_n^4}(1-q^{2n})(\sum _{k=1}^n\frac{q^k}{1+q^k}-\sum _{k=0}^{n-1}\frac{q^k}{1+q^k})\\ & =\frac{16}{(1+q^n{)}^4}(-1-q^{2n}+2(1-q^n{)}^2)\\ & =\frac{16(1-4q^n+q^{2n})}{(1+q^n{)}^4}\end{array}$$
より,
$$\begin{array}{rl}{\left(\frac{(q;q{)}_{\mathrm{\infty }}}{(-q;q{)}_{\mathrm{\infty }}}\right)}^8& =1+16\sum _{0<n}\frac{q^n(1-4q^n+q^{2n})}{(1+q^n{)}^4}\end{array}$$
を得る. ここで,
$$\begin{array}{rl}\sum _{0<n}\frac{q^n(1-4q^n+q^{2n})}{(1+q^n{)}^4}& =\sum _{0<n,k}(-1{)}^k(4(\genfrac{}{}{0ex}{}{k+1}{3})+(\genfrac{}{}{0ex}{}{k}{3})+(\genfrac{}{}{0ex}{}{k+2}{3}))q^{nk}\\ & =\sum _{0<n,k}(-1{)}^k{k}^3{q}^{nk}\\ & =\sum _{0<k}\frac{(-1{)}^k{k}^3{q}^k}{1-q^k}\end{array}$$
となる.　よって,
$$\begin{array}{rl}{\left(\sum _{n\in \mathbb{Z}}{q}^{n^2}\right)}^8& ={\left(\frac{(-q;-q{)}_{\mathrm{\infty }}}{(q;-q{)}_{\mathrm{\infty }}}\right)}^8\\ & =1+16\sum _{n=1}^{\mathrm{\infty }}\frac{n^3{q}^n}{1-(-q{)}^n}\end{array}$$
を得る.