ヤコビの三重積
Jacobi (1829)の原証明は冪級数の操作のみを用いている。まず
Q ( x ) = ( 1 − x 2 ) ( 1 − x 4 ) ( 1 − x 6 ) ⋯ , R ( x , z ) = ( 1 + x z ) ( 1 + x 3 z ) ( 1 + x 5 z ) ⋯ , S ( x , z ) = 1 ( 1 − x z ) ( 1 − x 2 z ) ( 1 − x 3 z ) ⋯ {\displaystyle {\begin{aligned}Q(x)=&(1-x^{2})(1-x^{4})(1-x^{6})\cdots ,\\R(x,z)=&(1+xz)(1+x^{3}z)(1+x^{5}z)\cdots ,\\S(x,z)=&{\frac {1}{(1-xz)(1-x^{2}z)(1-x^{3}z)\cdots }}\end{aligned}}}
とおく。ヤコビの三重積は Q ( x ) R ( x , z − 1 ) R ( x , z ) {\displaystyle Q(x)R(x,z^{-1})R(x,z)} であらわされる。
まず R ( x , z ) {\displaystyle R(x,z)} について、
R ( x , z x 2 ) = ( 1 + x 3 z ) ( 1 + x 5 z ) ( 1 + x 7 z ) ⋯ {\displaystyle R(x,zx^{2})=(1+x^{3}z)(1+x^{5}z)(1+x^{7}z)\cdots }
より
a n ( x ) = x n 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ ( 1 − x 2 n ) {\displaystyle a_{n}(x)={\frac {x^{n^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2n})}}}
が得られる。
次に S ( x , z ) {\displaystyle S(x,z)} について、
S ( x , z x ) = 1 ( 1 − x 2 z ) ( 1 − x 3 z ) ( 1 − x 4 z ) ⋯ = ( 1 − x z ) S ( x , z ) {\displaystyle S(x,zx)={\frac {1}{(1-x^{2}z)(1-x^{3}z)(1-x^{4}z)\cdots }}=(1-xz)S(x,z)}
が成り立つ。そこで
S ( x , z ) = 1 + b 1 ( x ) z 1 − x z + b 2 ( x ) z 2 ( 1 − x z ) ( 1 − x 2 z ) + ⋯ {\displaystyle S(x,z)=1+{\frac {b_{1}(x)z}{1-xz}}+{\frac {b_{2}(x)z^{2}}{(1-xz)(1-x^{2}z)}}+\cdots }
とおくと
a m ( x ) + a m + 1 ( x ) a 1 ( x ) + a m + 2 ( x ) a 2 ( x ) + ⋯ = a m ( x ) ( 1 + x 2 m + 2 ( 1 − x 2 ) ( 1 − x 2 m + 2 ) + x 4 m + 8 ( 1 − x 2 ) ( 1 − x 4 ) ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) + ⋯ ) = a m ( x ) ∑ k = 0 ∞ x 2 k 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ ( 1 − x 2 k ) × x 2 k m ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) ⋯ ( 1 − x 2 ( m + k ) ) {\displaystyle {\begin{aligned}&a_{m}(x)+a_{m+1}(x)a_{1}(x)+a_{m+2}(x)a_{2}(x)+\cdots \\=&a_{m}(x)\left(1+{\frac {x^{2m+2}}{(1-x^{2})(1-x^{2m+2})}}+{\frac {x^{4m+8}}{(1-x^{2})(1-x^{4})(1-x^{2m+2})(1-x^{2m+4})}}+\cdots \right)\\=&a_{m}(x)\sum _{k=0}^{\infty }{\frac {x^{2k^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2k})}}\times {\frac {x^{2km}}{(1-x^{2m+2})(1-x^{2m+4})\cdots (1-x^{2(m+k)})}}\\\end{aligned}}}
に一致するが、これは、上記の S ( x , z ) {\displaystyle S(x,z)} の展開より
R ( x , z ) = ( 1 + x z ) R ( x , z x 2 ) {\displaystyle R(x,z)=(1+xz)R(x,zx^{2})}
が成り立つ。そこで
R ( x , z ) = 1 + a 1 ( x ) z + a 2 ( x ) z 2 + ⋯ {\displaystyle R(x,z)=1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots }
とおくと
( 1 + a 1 ( x ) z + a 2 ( x ) z 2 + ⋯ ) = ( 1 + x z ) ( 1 + a 1 ( x ) x 2 z + a 2 ( x ) x 4 z 2 + ⋯ ) {\displaystyle (1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots )=(1+xz)(1+a_{1}(x)x^{2}z+a_{2}(x)x^{4}z^{2}+\cdots )}
より
a n ( x ) = a n ( x ) x 2 n + a n − 1 ( x ) x 2 n − 1 , {\displaystyle a_{n}(x)=a_{n}(x)x^{2n}+a_{n-1}(x)x^{2n-1},}
つまり
a n ( x ) = a n − 1 ( x ) x 2 n − 1 1 − x 2 n , a 0 ( x ) = 1 {\displaystyle a_{n}(x)={\frac {a_{n-1}(x)x^{2n-1}}{1-x^{2n}}},a_{0}(x)=1}
が成り立つ。この漸化式を解くと