［整理］Jacobi三乘积恒等式的一个简单证明

Jacobi恒等式就是

$prod_{n=0}^infty (1-x^{2n+2})(1+x^{2n+1}z)(1+x^{2n+1}z^{-1})=sum_{n=-infty}^{infty}x^{n^2}z^n ag{A}$

下面给出一个证明。

先证两个引理：

$prod_{n=0}^infty(1+x^n z)=sum_{n=0}^inftyfrac{x^{n(n-1)/2}z^n}{(1-x)(1-x^2)dots(1-x^n)} ag{E1}$ 与 $prod_{n=0}^infty(1+x^n z)^{-1}=sum_{n=0}^inftyfrac{(-1)^nz^n}{(1-x)(1-x^2)dots(1-x^n)}. ag{E2}$

需要；后者还需要。不妨设

$F(x, z)=prod_{n=0}^{infty} (1+x^nz)=sum_{n=0}^{infty}a_nz^n. ag{E1-1}$

由乘积形式，显然有

写成求和形式，就是

$sum_{n=0}^{infty}a_nz^n=(1+z)sum_{n=0}^{infty}a_nx^nz^n,$

也就是递推关系

$egin{align} a_n&=a_nx^n+a_{n-1}x^{n-1}\ a_n&=frac{x^{n-1}}{1-x^n}a_{n-1}. end{align}$

然后累乘就可以了。第二个等式同理。

接下来，我们开始证明。

$egin{align} prod_{n=0}^infty (1+x^{2n+1}z)&=prod_{n=0}^infty (1+x^{2n}(xz))\ {color{blue}{ ext{(E1)}}}&=sum_{n=0}^infty frac{x^{n(n-1)}(xz)^n}{(1-x^2)cdots(1-x^{2n})}\ &=sum_{n=0}^infty frac{x^{n^2}z^n}{(1-x^2)cdots(1-x^{2n})}\ &=sum_{n=0}^infty x^{n^2}z^ncdotfrac{prod_{j=0}^infty(1-x^{2n+2+2j})}{prod_{j=0}^infty(1-x^{2j+2})}\ &=frac{1}{prod_{j=0}^infty(1-x^{2j+2})} sum_{n=0}^infty x^{n^2}z^ncdot{prod_{j=0}^infty(1-x^{2n+2+2j})} end{align}$

注意到当，后面的因子中一定会出现一项，因此可以放心地把求和下限放在负无穷。

接下来我们看右边的连乘式： $egin{align} &{prod_{j=0}^infty(1-x^{2n+2+2j})}\ =&{prod_{j=0}^infty(1+x^{2j}(-x^{2n+2}))}\ =&sum_{m=0}^inftyfrac{x^{m(m-1)}(-x^{2n+2})^m}{(1-x^2)dots(1-x^{2m})}\ =&sum_{m=0}^inftyfrac{x^{m^2+m+2nm}(-1)^m}{(1-x^2)dots(1-x^{2m})} end{align}$

再代回去：

$egin{align} &frac{1}{prod_{j=0}^infty(1-x^{2j+2})} sum_{n=-infty}^infty x^{n^2}z^ncdotsum_{m=0}^inftyfrac{x^{m^2+m+2nm}(-1)^m}{(1-x^2)dots(1-x^{2m})} end{align}$

交换求和顺序，把 $x^{m^2+2nm}, x^{n^2}$ 凑成指数上的完全平方，并且创造一对 $z^m z^{-m}$ ，分别放进前面和后面：

$egin{align} &frac{1}{prod_{j=0}^infty(1-x^{2j+2})} sum_{n=-infty}^infty color{blue}{x^{n^2}}color{green}{z^n}cdotsum_{m=0}^inftyfrac{x^{color{blue}{m^2}+color{violet}m+color{blue}{2nm}}color{violet}{z^{-m}}color{green}{z^m}(-1)^m}{(1-x^2)dots(1-x^{2m})}\ =&frac{1}{prod_{j=0}^infty(1-x^{2j+2})}sum_{m=0}^infty frac{(-1)^mcolor{violet}{(xz^{-1})^m}}{(1-x^2)cdots(1-x^{2m})}sum_{n=-infty}^inftycolor{blue}{x^{(m+n)^2}color{green}{z^{m+n}}}\ color{blue}{ ext{(E2) }}=&frac{1}{prod_{j=0}^infty(1-x^{2j+2})}prod_{j=0}^infty (1+x^{2j+1}z^{-1})^{-1}sum_{k=-infty}^infty x^{k^2}z^k end{align}$