2項変換とベルヌーイ数
2項変換とベルヌーイ数
数列\(\left(a_{n}\right)_{n\in\mathbb{N}_{0}},\left(b_{n}\right)_{n\in\mathbb{N}_{0}}\)があるとき次の変換と逆変換が成り立つ。
\[ a_{n}=\sum_{k=0}^{n}C\left(n,k\right)\frac{b_{n-k}}{k+1} \] \[ b_{n}=\sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} \]
数列\(\left(a_{n}\right)_{n\in\mathbb{N}_{0}},\left(b_{n}\right)_{n\in\mathbb{N}_{0}}\)があるとき次の変換と逆変換が成り立つ。
\[ a_{n}=\sum_{k=0}^{n}C\left(n,k\right)\frac{b_{n-k}}{k+1} \] \[ b_{n}=\sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} \]
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\(B_{n}\)はベルヌーイ数\[
a_{n}=\sum_{k=0}^{n}C\left(n,k\right)\frac{b_{n-k}}{k+1}
\]
であるとき、
\begin{align*} \sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=0}^{n-k}C\left(n-k,j\right)\frac{b_{n-k-j}}{j+1}\\ & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=k}^{n}C\left(n-k,j-k\right)\frac{b_{n-j}}{j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)B_{k}C\left(n-k,j-k\right)\frac{b_{n-j}}{j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)B_{k}C\left(n-k,n-j-k\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)B_{k}C\left(n-k,j\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,j\right)B_{k}C\left(n-j,k\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n,j\right)B_{k}C\left(n-j+1,n-j-k+1\right)b_{j}\\ & =\sum_{j=0}^{n}\frac{C\left(n,j\right)b_{j}}{n-j+1}\sum_{k=0}^{n-j}B_{k}C\left(n-j+1,k\right)\\ & =\sum_{j=0}^{n}\frac{C\left(n,j\right)b_{j}}{n-j+1}\delta_{n,j}\cmt{\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}}\\ & =b_{n} \end{align*} となる。
従って題意は成り立つ。
\[ b_{n}=\sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} \] であるとき、
\begin{align*} \sum_{k=0}^{n}C\left(n,k\right)\frac{b_{n-k}}{k+1} & =\sum_{k=0}^{n}C\left(n,k\right)\frac{1}{k+1}\sum_{j=0}^{n-k}C\left(n-k,j\right)B_{j}a_{n-k-j}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{1}{k+1}\sum_{j=k}^{n}C\left(n-k,j-k\right)B_{j-k}a_{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)\frac{1}{k+1}C\left(n-k,j-k\right)B_{j-k}a_{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)\frac{1}{k+1}C\left(n-k,n-j-k\right)B_{n-j-k}a_{j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}\frac{1}{k+1}C\left(n,k\right)C\left(n-k,j\right)B_{n-j-k}a_{j}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\sum_{k=0}^{n-j}\frac{1}{k+1}C\left(n-j,k\right)B_{n-j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,k+1\right)B_{n-j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,n-j-k+1\right)B_{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,k\right)B_{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\delta_{0,n-j}\cmt{\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}}\\ & =C\left(n,n\right)a_{n}\\ & =a_{n} \end{align*} となるので逆向きも成り立つ。
\begin{align*} \sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=0}^{n-k}C\left(n-k,j\right)\frac{b_{n-k-j}}{j+1}\\ & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=k}^{n}C\left(n-k,j-k\right)\frac{b_{n-j}}{j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)B_{k}C\left(n-k,j-k\right)\frac{b_{n-j}}{j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)B_{k}C\left(n-k,n-j-k\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)B_{k}C\left(n-k,j\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,j\right)B_{k}C\left(n-j,k\right)\frac{b_{j}}{n-j-k+1}\\ & =\sum_{j=0}^{n}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n,j\right)B_{k}C\left(n-j+1,n-j-k+1\right)b_{j}\\ & =\sum_{j=0}^{n}\frac{C\left(n,j\right)b_{j}}{n-j+1}\sum_{k=0}^{n-j}B_{k}C\left(n-j+1,k\right)\\ & =\sum_{j=0}^{n}\frac{C\left(n,j\right)b_{j}}{n-j+1}\delta_{n,j}\cmt{\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}}\\ & =b_{n} \end{align*} となる。
従って題意は成り立つ。
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逆向きは次のようにする。\[ b_{n}=\sum_{k=0}^{n}C\left(n,k\right)B_{k}a_{n-k} \] であるとき、
\begin{align*} \sum_{k=0}^{n}C\left(n,k\right)\frac{b_{n-k}}{k+1} & =\sum_{k=0}^{n}C\left(n,k\right)\frac{1}{k+1}\sum_{j=0}^{n-k}C\left(n-k,j\right)B_{j}a_{n-k-j}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{1}{k+1}\sum_{j=k}^{n}C\left(n-k,j-k\right)B_{j-k}a_{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)\frac{1}{k+1}C\left(n-k,j-k\right)B_{j-k}a_{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}C\left(n,k\right)\frac{1}{k+1}C\left(n-k,n-j-k\right)B_{n-j-k}a_{j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n-j}\frac{1}{k+1}C\left(n,k\right)C\left(n-k,j\right)B_{n-j-k}a_{j}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\sum_{k=0}^{n-j}\frac{1}{k+1}C\left(n-j,k\right)B_{n-j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,k+1\right)B_{n-j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,n-j-k+1\right)B_{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\sum_{k=0}^{n-j}C\left(n-j+1,k\right)B_{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)a_{j}\frac{1}{n-j+1}\delta_{0,n-j}\cmt{\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}}\\ & =C\left(n,n\right)a_{n}\\ & =a_{n} \end{align*} となるので逆向きも成り立つ。
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ベルヌーイ数の定義
\[
\frac{x}{e^{x}-1}=\sum_{k=0}^{\infty}\frac{B_{k}}{k!}x^{k}
\]
奇数ベルヌーイ数
\[
B_{2n-1}=-\frac{1}{2}\delta_{1,n}
\]
ベルヌーイ数の一般項
\[
B_{n}=\sum_{k=0}^{n}\left(-1\right)^{k}k^{n}\sum_{j=k}^{n}\frac{C\left(j,k\right)}{j+1}
\]
(*)ベルヌーイ数の総和と漸化式
\[
\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}
\]