Subspace Inversion
Elegia
2021-11-19 19:20:12
(这里采取符号 $(\omega;q)=(1-\omega) (1-\omega q)\cdots(1-\omega q^{n-1})$)
设 $f,g$ 满足
$$
g_n = \sum_k \binom n k_q f_k
$$
那就是
$$
\frac{g_n}{(q;q)_n} = \sum_k \frac{f_k}{(q;q)_k} \frac{1}{(q;q)_{n-k}}
$$
若要进行反演,只需计算乘法逆
$$
\left(\sum_n \frac{x^n}{(q;q)_n}\right)^{-1}
$$
不妨记括号内为 $\zeta(x)$,注意到
$$
\zeta(x)-\zeta(qx) = \sum_n \frac{(1-q^n)x^n}{(q;q)_n}
= x \zeta(x)
$$
因此
$$
\zeta(x)= \frac{\zeta(qx)}{1-x} = \frac{\zeta(q^2x)}{(1-x)(1-qx)} = \cdots = \frac1{(x;q)_{\infty}}
$$
故有
$$
\zeta(x)^{-1} = (x;q)_{\infty} = (1-x)\zeta(qx)^{-1}
$$
解得
$$
\zeta(x) = \sum_n \frac{(-1)^n q^{n(n-1)/2}}{(q;q)_n}x^n
$$
因此,我们得到了子空间反演系数
$$
f_n = \sum_k (-1)^k \binom n k_q q^{k(k-1)/2} g_{n-k}
$$