2項係数の母関数
2項係数の母関数
\[ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j,k)x^{j}y^{k}=(1-x-xy)^{-1} \]
\[ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j+k,k)x^{j}y^{k}=(1-x-y)^{-1} \]
(1)通常型母関数
\[ \sum_{k=0}^{\infty}C(x,k)t^{k}=(1+t)^{x} \](2)通常型母関数
\[ \sum_{k=0}^{\infty}C(x+k,k)t^{k}=(1-t)^{-(x+1)} \](3)通常型母関数
\[ \sum_{x=y}^{\infty}C(x,y)t^{x}=t^{y}(1-t)^{-(y+1)} \](4)指数型母関数
\[ \sum_{k=0}^{\infty}C(x,k)\frac{t^{k}}{k!}=F\left(-x;1;-t\right) \](5)指数型母関数
\[ \sum_{k=0}^{\infty}C(x+k,k)\frac{t^{k}}{k!}=F\left(x+1;1;t\right) \](6)指数型母関数
\[ \sum_{x=y}^{\infty}C(x,y)\frac{t^{x}}{x!}=\frac{t^{y}}{y!}e^{t} \](7)通常型2変数母関数
\(x+xy<1\)とする。\[ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j,k)x^{j}y^{k}=(1-x-xy)^{-1} \]
(8)通常型2変数母関数
\(x<1-y\)とする。\[ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j+k,k)x^{j}y^{k}=(1-x-y)^{-1} \]
(9)指数型2変数母関数
\[ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j+k,k)\frac{x^{j}y^{k}}{(j+k)!}=e^{x+y} \](1)
\begin{align*} \sum_{k=0}^{\infty}C(x,k)t^{k} & =\sum_{k=0}^{\infty}C(x,k)t^{k}1^{x-k}\\ & =(1+t)^{x} \end{align*}(2)
\begin{align*} \sum_{k=0}^{\infty}C(x+k,k)t^{k} & =\sum_{k=0}^{\infty}P\left(x+k,k\right)\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}Q\left(x+1,k\right)\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}P\left(-\left(x+1\right),k\right)\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}C\left(-\left(x+1\right),k\right)\left(-t\right)^{k}\\ & =\left(1-t\right)^{-\left(x+1\right)} \end{align*}(3)
\begin{align*} \sum_{x=y}^{\infty}C(x,y)t^{x} & =t^{y}\sum_{s=0}^{\infty}C\left(s+y,y\right)t^{s}\cmt{x=s+y}\\ & =t^{y}\sum_{s=0}^{\infty}P\left(y+s,s\right)\frac{t^{s}}{s!}\\ & =t^{y}\sum_{s=0}^{\infty}Q\left(y+1,s\right)\frac{t^{s}}{s!}\\ & =t^{y}\sum_{s=0}^{\infty}\left(-1\right)^{s}P\left(-\left(y+1\right),s\right)\frac{t^{s}}{s!}\\ & =t^{y}\sum_{s=0}^{\infty}C\left(-\left(y+1\right),s\right)\left(-t\right)^{s}\\ & =t^{y}(1-t)^{-(y+1)} \end{align*}(4)
\begin{align*} \sum_{k=0}^{\infty}C(x,k)\frac{t^{k}}{k!} & =\sum_{k=0}^{\infty}\frac{P(x,k)}{k!}\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}Q\left(-x,k\right)}{k!}\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\frac{Q\left(-x,k\right)}{k!}\frac{\left(-t\right)^{k}}{k!}\\ & =F\left(-x;1;-t\right) \end{align*}(5)
\begin{align*} \sum_{k=0}^{\infty}C(x+k,k)\frac{t^{k}}{k!} & =\sum_{k=0}^{\infty}\frac{P\left(x+k,k\right)}{k!}\frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\frac{Q\left(x+1,k\right)}{k!}\frac{t^{k}}{k!}\\ & =F\left(x+1;1;t\right) \end{align*}(6)
\begin{align*} \sum_{x=y}^{\infty}C(x,y)\frac{t^{x}}{x!} & =\frac{1}{y!}\sum_{x=y}^{\infty}\frac{t^{x}}{\left(x-y\right)!}\\ & =\frac{t^{y}}{y!}\sum_{x=0}^{\infty}\frac{t^{x}}{x!}\\ & =\frac{t^{y}}{y!}e^{t} \end{align*}(7)
\begin{align*} \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j,k)x^{j}y^{k} & =\sum_{j=0}^{\infty}(1+y)^{j}x^{j}\\ & =\sum_{j=0}^{\infty}(x+xy)^{j}\\ & =(1-x-xy)^{-1} \end{align*}(8)
\begin{align*} \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j+k,k)x^{j}y^{k} & =\sum_{j=0}^{\infty}(1-y)^{-(j+1)}x^{j}\\ & =(1-y)^{-1}\sum_{j=0}^{\infty}\left(\frac{x}{1-y}\right){}^{j}\\ & =(1-y)^{-1}\frac{1}{1-\frac{x}{1-y}}\\ & =(1-x-y)^{-1} \end{align*}(9)
\begin{align*} \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C(j+k,k)\frac{x^{j}y^{k}}{(j+k)!} & =\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{x^{j}y^{k}}{j!k!}\\ & =e^{x}e^{y}\\ & =e^{x+y} \end{align*}ページ情報
タイトル | 2項係数の母関数 |
URL | https://www.nomuramath.com/c659v9x1/ |
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負の整数の2項係数
\[
C\left(-m,-n\right)=\left(-1\right)^{m-n}C\left(n-1,m-1\right)
\]
2項係数の微分
\[
\frac{d}{dx}C(x,y) =C(x,y)\left(\psi(1+x)-\psi(1+x-y)\right)
\]
ファンデルモンドの畳み込み定理と第1引数の畳み込み
\[
\sum_{j=0}^{k}C(x,j)C(y,k-j)=C(x+y,k)
\]
飛び飛びの2項定理
\[
\sum_{k=0}^{\infty}C\left(n,2k\right)a^{2k}b^{n-2k}=\frac{1}{2}\left\{ \left(a+b\right)^{n}+\left(-a+b\right)^{n}\right\}
\]