广义二项级数 / 广义指数级数
SalomeJLQ
2024-06-15 14:55:54
复合方程
$$
\begin{aligned}
\mathcal B_t(z)^{1-t}-\mathcal B_t(z)^{-t}&=z,\\
\mathcal E_t(z)^{-t}\log\mathcal E_t(z)&=z.
\end{aligned}
$$
当 $t=1$ 时此下标可略去不写。
其系数显然能用 Lagrange 反演求出。记 $F(z):=\mathcal B_t(z)-1$，则 $F(z)=z(1+F(z))^t$，由 Lagrange 反演
$$
\begin{aligned}
[z^n]\mathcal B_t(z)^r&=[z^n](1+F(z))^r\\
&=\frac{1}{n}\left[z^{n-1}\right]\left((1+z)^{r}\right)'(1+z)^{tn}\\
&=\frac{k}{n}\binom{tn+r-1}{n-1}=\frac{r}{tn+r}\binom{tn+r}{n}.
\end{aligned}
$$
同理，设 $f(z)=\log\mathcal E_t(z)$，有 $f(z)=z\exp(tf(z))$，于是
$$
\begin{aligned}
[z^n]\mathcal E_t(z)^r&=[z^n]\exp(rf(z))\\
&=\frac{1}{n}\left[z^{n-1}\right]\exp(rz)'\exp(tnz)\\
&=\frac{r}{n}\frac{(tn+r)^{n-1}}{(n-1)!}.
\end{aligned}
$$
$\bf Theorem\;2.\quad$广义二项级数 $\mathcal B_t(z)$ 与广义指数级数 $\mathcal E_t(z)$ 满足
$$
\begin{aligned}
\mathcal B_t(z)^r&=\sum_{n=0}^{\infty}\binom{tn+r}{n}\frac{r}{tn+r}z^n,\\
\mathcal E_t(z)^r&=\sum_{n=0}^{\infty}r(tn+r)^{n-1}\frac{z^n}{n!}.
\end{aligned}
$$
$Proof.\quad$如上，进行 Lagrange 反演。$\square
对此二函数我们有基本的结论，其一是
\mathcal B_t(z)\mathcal B_{1-t}(-z)=1.
这是因为
\begin{aligned}
\mathcal B_{1-t}(-z)^{-1}&=\sum_{n=0}^{\infty}\binom{(1-t)n-1}{n}\frac{(-1)^{n+1}z^n}{(1-t)n-1}\\
&=\sum_{n=0}^{\infty}\binom{tn}{n}\frac{z^n}{tn-n+1}\\
&=\mathcal B_t(z).
\end{aligned}
其二是
\mathcal E_t(z)^t=\mathcal E(tz)=\sum_{n=0}^{\infty}\frac{(n+1)^{n-1}t^nz^n}{n!}.
微分形式及其推论
$$
\begin{aligned}
\mathcal B_t'(z)&=\frac{\mathcal B_t(z)^t}{1-t+t\mathcal B_t(z)^{-1}},\\
\mathcal E_t'(z)&=\frac{\mathcal E_t(z)^{t+1}}{1-tz\mathcal E_t(z)^t}.
\end{aligned}
$$
$Proof.\quad$对复合方程两端微分。$\square
$$
\begin{aligned}
\frac{\mathcal B_t(z)^{r}}{1-t+t\mathcal B_t(z)^{-1}}&=\sum_{n=0}^{\infty}\binom{tn+r}{n}z^n,\\
\frac{\mathcal E_t(z)^r}{1-tz\mathcal E_t(z)^t}&=\sum_{n=0}^{\infty}(tn+r)^n\frac{z^n}{n!}.
\end{aligned}
$$
$Proof.\quad$考察 $(\mathcal B_t(z)^r)'$ 及 $(\mathcal E_t(z)^r)$ 即可。对于前者
$$
\begin{aligned}
\frac{r\mathcal B_t(z)^{t+r-1}}{1-t+t\mathcal B_t(z)^{-1}}&=r\mathcal B_t(z)^{r-1}\mathcal B_t'(z)=(\mathcal B_t(z)^r)'\\
&=\sum_{n=1}^{\infty}n\binom{tn+r}{n}\frac{r}{tn+r}z^{n-1}\\
&=r\sum_{n=0}^{\infty}\binom{tn+(t+r-1)}{n}z^n,
\end{aligned}
$$
故结论成立。对于广义指数级数则是
$$
\begin{aligned}
\mathcal E_t'(z)&=\frac{r\mathcal E_t(z)^{t+r}}{1-tz\mathcal E_t(z)^t}=\sum_{n=1}^{\infty}r\frac{(tn+r)^{n-1}}{(n-1)!}z^{n-1}.\quad\square
\end{aligned}
$$
### 一般卷积恒等式
分别用 $\mathcal B_t(z)^r,\frac{r\mathcal B_t(z)^{r}}{1-t+t\mathcal B_t(z)^{-1}},\mathcal E_t(z)^r,\frac{\mathcal E_t(z)^r}{1-tz\mathcal E_t(z)^t}$ 互相做乘积可以得到四个卷积恒等式：
$$
\begin{aligned}
\frac{\mathcal B_t(z)^{r+s}}{1-t+t\mathcal B_t(z)^{-1}}=\mathcal B_t(z)^r\frac{\mathcal B_t(z)^{s}}{1-t+t\mathcal B_t(z)^{-1}}\iff\quad\;\;\quad&\\
\sum_{k=0}^n\binom{tk+r}{k}\binom{tn-tk+s}{n-k}\frac{r}{rk+r}&=\binom{tn+r+s}{n},\\\\
\mathcal B_t(z)^{r+s}=\mathcal B_t(z)^r\mathcal B_t(z)^s\iff\qquad\qquad\qquad\qquad\quad\;\;\quad&\\
\sum_{k=0}^n\binom{tk+r}{k}\binom{tn-tk+s}{n-k}\frac{r}{rk+r}\frac{s}{tn-tk+s}&=\binom{tn+r+s}{n}\frac{r+s}{tn+r+s},\\\\
\frac{\mathcal E_t(z)^{r+s}}{1-tz\mathcal E_t(z)^t}=\mathcal E_t(z)^r\frac{\mathcal E_t(z)^s}{1-tz\mathcal E_t(z)^t}\iff\qquad\qquad\quad\;\,\quad\\
\sum_{k=0}^n\binom{n}{k}(tk+r)^k(tn-tk+s)^{n-k}\frac{r}{rk+r}&=(tn+r+s)^n,\\\\
\mathcal E_t(z)^{r+s}=\mathcal E_t(z)^r\mathcal E_t(z)^s\iff\qquad\qquad\qquad\qquad\quad\quad\quad&\\
\sum_{k=0}^n\binom{n}{k}(tk+r)^k(tn-tk+s)^{n-k}\frac{r}{rk+r}\frac{s}{tn-tk+s}&=(tn+r+s)^n\frac{r+s}{tn+r+s}.
\end{aligned}
$$
## Catalan 级数与求和
$\mathcal B_2(z)$ 是 Catalan 数的 OGF（其满足 $\mathcal B_2(z)=1+\mathcal B_2(z)^2$，这正是 Catalan 数的组合意义之一）。根据我们之前得到的关系，$\mathcal B_{-1}(z)$ 与之满足关系
$$
\mathcal B_{-1}(z)=\mathcal B_2(-z)^{-1}.
$$
除了常数项外，$\mathcal B_{-1}(z)$ 的系数无非是带符号的 Catalan 数，即
$$
\mathcal B_{-1}(z)=1+z\mathcal B_2(-z),
$$
具体地，根据 $\rm Theorem2/4$ 有以下定理。
$\bf Theorem\;5.\quad$以上二个广义二项级数符合以下展式：
$$
\begin{aligned}
\mathcal B_2(z)=\sum_{k=0}^{\infty}\binom{2k}{k}\frac{z^k}{1+k}&=\sum_{k=0}^{\infty}\binom{2k+1}{k}\frac{z^k}{1+2k}=\frac{1-\sqrt{1-4z}}{2z},\\
\mathcal B_{-1}(z)=\sum_{k=0}^{\infty}\binom{1-k}{k}\frac{z^k}{1-k}&=\sum_{k=0}^{\infty}\binom{2k-1}{k}\frac{(-z)^k}{1-2k}=\frac{1+\sqrt{1+4z}}{2},\\
\mathcal B_2(z)^r&=\sum_{k=0}^{\infty}\binom{2k+r}{k}\frac{r}{2k+r}z^k,\\
\mathcal B_{-1}(z)^r&=\sum_{k=0}^{\infty}\binom{r-k}{k}\frac{r}{r-k}z^k,\\
\frac{\mathcal B_2(z)^r}{\sqrt{1-4z}}&=\sum_{k=0}^{\infty}\binom{2k+r}{k}z^k,\\
\frac{\mathcal B_{-1}(z)^{r+1}}{\sqrt{1+4z}}&=\sum_{k=0}^{\infty}\binom{r-k}{k}z^k.\\
\end{aligned}
$$
$Proof.\quad$前两行乃从复合方程中直接求解；后四行为 $\rm Theorem2/4$ 中直接代入所得。$\square
最后，观察相关系数可以得到一下定理。
$$
\begin{aligned}
\sum_{k=0}^n\binom{n-k}{k}z^k&=\frac{\mathcal B_{-1}(z)^{n+1}-(-z)^{n+1}\mathcal B_{2}(-z)^{n+1}}{\sqrt{1+4z}}\\
&=\frac{1}{\sqrt{1+4z}}\left(\left(\frac{1+\sqrt{1+4z}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{1+4z}}{2}\right)^{n+1}\right),\\
\sum_{k=0}^{n-1}\binom{n-k}{k}\frac{n}{n-k}z^k&=\mathcal B_{-1}(z)^{n}+(-z)^{n}\mathcal B_{2}(-z)^{n}\\
&=\left(\frac{1+\sqrt{1+4z}}{2}\right)^{n}+\left(\frac{1-\sqrt{1+4z}}{2}\right)^{n}.
\end{aligned}
$$
$Proof.\quad$当 $k>n$ 时有
$$
\begin{aligned}
\left[z^k\right]\frac{(-z)^{n+1}\mathcal B_{2}(-z)^{n+1}}{\sqrt{1+4z}}&=(-1)^{n+1}\big[z^{k-n-1}\big]\frac{\mathcal B_{2}(-z)^{n+1}}{\sqrt{1+4z}}\\
&=(-1)^{n+1}(-1)^{k-n-1}\big[z^{k-n-1}\big]\frac{\mathcal B_{2}(z)^{n+1}}{\sqrt{1+4z}}\\
&=(-1)^k\binom{2(k-n-1)+n+1}{k-n-1}=\binom{n-k}{k}\\
&=\left[z^k\right]\frac{\mathcal B_{-1}(z)^{n+1}}{\sqrt{1+4z}},\\\\
\big[z^k\big](-z)^n\mathcal B_{2}(-z)^n&=(-1)^k\big[z^{n-k}\big]\mathcal B_{2}(z)^n\\
&=(-1)^k\binom{2(k-n)+n}{k-n}\frac{n}{2(k-n)+n}\\
&=\frac{n}{2k-n}\binom{n-k-1}{k}=\frac{(n-k)!(n-2k)}{k!(n-2k)!(n-k)}\frac{n}{2k-n}\\
&=-\binom{n-k}{k}\frac{n}{n-k}=-\big[z^{k}\big]\mathcal B_{-1}(z)^n.
\end{aligned}
$$
因而大于 $n$ 的项都被抵消。故和式成立。$\square