2項変換と交代2項変換の母関数

by nomura · 2024年12月11日

2項変換と交代2項変換の母関数
2項変換と交代2項変換について、母関数は次のようになります

2項変換
\[ b_{n}=\sum_{k=0}^{n}C\left(n,k\right)a_{k} \]

(1)通常型母関数

\[ \sum_{k=0}^{\infty}b_{k}x^{k}=\frac{1}{1-x}\sum_{k=0}^{\infty}a_{k}\left(\frac{x}{1-x}\right)^{k} \]

(2)指数型母関数

\[ \sum_{k=0}^{\infty}b_{k}\frac{x^{k}}{k!}=e^{x}\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{k!} \]
交代2項変換
\[ b_{n}=\sum_{k=0}^{n}\left(-1\right)^{k}C\left(n,k\right)a_{k} \]

(3)通常型母関数

\[ \sum_{k=0}^{\infty}b_{k}x^{k}=\frac{1}{1-x}\sum_{k=0}^{\infty}a_{k}\left(\frac{-x}{1-x}\right)^{k} \]

(4)指数型母関数

\[ \sum_{j=0}^{\infty}b_{k}\frac{x^{k}}{k!}=e^{x}\sum_{k=0}^{\infty}a_{k}\frac{\left(-x\right)^{k}}{k!} \]

交代2項変換の母関数は、2項変換の母関数で\(a_{k}\rightarrow\left(-1\right)^{k}a_{k}\)とすれば導出できます。

(1)

\begin{align*} \sum_{j=0}^{\infty}b_{j}x^{j} & =\sum_{j=0}^{\infty}\sum_{k=0}^{j}C\left(j,k\right)a_{k}x^{j}\\ & =\sum_{k=0}^{\infty}a_{k}\sum_{j=k}^{\infty}C\left(j,k\right)x^{j}\\ & =\sum_{k=0}^{\infty}a_{k}x^{k}\left(1-x\right)^{-\left(k+1\right)}\\ & =\frac{1}{1-x}\sum_{k=0}^{\infty}a_{k}\left(\frac{x}{1-x}\right)^{k} \end{align*}

(2)

\begin{align*} \sum_{j=0}^{\infty}b_{j}\frac{x^{j}}{j!} & =\sum_{j=0}^{\infty}\sum_{k=0}^{j}C\left(j,k\right)a_{k}\frac{x^{j}}{j!}\\ & =\sum_{k=0}^{\infty}a_{k}\sum_{j=k}^{\infty}C\left(j,k\right)\frac{x^{j}}{j!}\\ & =\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{k!}e^{x}\\ & =e^{x}\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{k!} \end{align*}

(3)

\begin{align*} \sum_{j=0}^{\infty}b_{j}x^{j} & =\sum_{j=0}^{\infty}\sum_{k=0}^{j}\left(-1\right)^{k}C\left(j,k\right)a_{k}x^{j}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}a_{k}\sum_{j=k}^{\infty}C\left(j,k\right)x^{j}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}a_{k}x^{k}\left(1-x\right)^{-\left(k+1\right)}\\ & =\frac{1}{1-x}\sum_{k=0}^{\infty}a_{k}\left(\frac{-x}{1-x}\right)^{k} \end{align*}

(4)

\begin{align*} \sum_{j=0}^{\infty}b_{j}\frac{x^{j}}{j!} & =\sum_{j=0}^{\infty}\sum_{k=0}^{j}\left(-1\right)^{k}C\left(j,k\right)a_{k}\frac{x^{j}}{j!}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}a_{k}\sum_{j=k}^{\infty}C\left(j,k\right)\frac{x^{j}}{j!}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}a_{k}\frac{x^{k}}{k!}e^{x}\\ & =e^{x}\sum_{k=0}^{\infty}a_{k}\frac{\left(-x\right)^{k}}{k!} \end{align*}