2項係数が0になるとき
2項係数が0になるとき
\[ \forall m,n\in\mathbb{Z},\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\Leftrightarrow C\left(m,n\right)=0 \]
\[ \forall m,n\in\mathbb{Z},\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\Leftrightarrow C\left(m,n\right)=0 \]
\(\Rightarrow\)
\(0\leq m<n\)のとき
\begin{align*} C\left(m,n\right) & =\frac{P\left(m,n\right)}{n!}\\ & =\frac{m!}{n!}P\left(0,n-m\right)\\ & =\frac{m!}{n!}0P\left(-1,n-m-1\right)\\ & =\frac{m!}{n!}0\left(-1\right)^{n-m-1}Q\left(1,n-m-1\right)\\ & =0\left(-1\right)^{n-m-1}\frac{m!}{n!}\left(n-m-1\right)!\\ & =0 \end{align*}\(n<0\leq m\)のとき
\begin{align*} C\left(m,n\right) & =\frac{P\left(m,n\right)}{n!}\\ & =\frac{1}{n!P\left(m-n,-n\right)}\\ & =\frac{1}{n!Q\left(m+1,-n\right)}\\ & =0 \end{align*}\(m<n<0\)のとき、
\begin{align*} C\left(m,n\right) & =\frac{P\left(m,n\right)}{n!}\\ & =\frac{\left(-1\right)^{n}Q\left(-m,n\right)}{n!}\\ & =\frac{\left(-1\right)^{n}}{n!P\left(-m-1,-n\right)}\\ & =\frac{\left(-1\right)^{n}}{n!Q\left(-m+n,-n\right)}\\ & =0 \end{align*}-
これらより、\(\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\Rightarrow C\left(m,n\right)=0\)となる。\(\Leftarrow\)
\begin{align*} \lnot\left\{ \left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\right\} & \Leftrightarrow\lnot\left(0\leq m<n\right)\land\lnot\left(n<0\leq m\right)\land\lnot\left(m<n<0\right)\\ & \Leftrightarrow\left(n\leq m\lor m<0\right)\land\left(m<0\lor0\leq n\right)\land\left(0\leq n\lor n\leq m\right)\\ & \Leftrightarrow\left(n\leq m<0\right)\lor\left(0\leq n\leq m\right)\lor\left(m<0\land0\leq n\right) \end{align*}\(n\leq m<0\)のとき
\begin{align*} C\left(m,n\right) & =\frac{m!}{n!\left(m-n\right)!}\\ & =\left(-1\right)^{m-n}\frac{\Gamma\left(-n\right)}{\Gamma\left(-m\right)\left(m-n\right)!}\\ & =\left(-1\right)^{m-n}\frac{\left(-n-1\right)!}{\left(-m-1\right)!\left(m-n\right)!}\\ & =\left(-1\right)^{m-n}C\left(-n-1,-m-1\right)\\ & \ne0 \end{align*}\(0\leq n\leq m\)のとき
\begin{align*} C\left(m,n\right) & =\frac{m!}{n!\left(m-n\right)!}\\ & \ne0 \end{align*}\(m<0\land0\leq n\)のとき
\begin{align*} C\left(m,n\right) & =\frac{P\left(m,n\right)}{n!}\\ & =\frac{\left(-1\right)^{n}Q\left(-m,n\right)}{n!}\\ & \ne0 \end{align*}-
これより、\(\lnot\left\{ \left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\right\} \Rightarrow C\left(m,n\right)\ne0\)が成り立つ。故に対偶をとると、\(C\left(m,n\right)=0\Rightarrow\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\)が成り立つ。
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\(\Rightarrow\)も\(\Leftarrow\)も成り立つので与式は成り立つ。ページ情報
| タイトル | 2項係数が0になるとき |
| URL | https://www.nomuramath.com/vdi853rk/ |
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ディクソンの等式
\[
\sum_{k=-a}^{a}(-1)^{k}C(a+b,a+k)C(b+c,b+k)C(c+a,c+k)=\frac{(a+b+c)!}{a!b!c!}
\]
パスカルの法則の一般形
\[
C\left(x+n,y+n\right)=\sum_{k=0}^{n}C\left(n,k\right)C\left(x,y+k\right)
\]
2項係数の2乗和
\[
\sum_{j=0}^{m}C^{2}(m,j)=C(2m,m)
\]
2項係数の微分
\[
\frac{d}{dx}C(x,y) =C(x,y)\left(\psi(1+x)-\psi(1+x-y)\right)
\]