ベルヌーイ多項式の級数表示
ベルヌーイ多項式の級数表示
\[ B_{n}\left(x\right)=\sum_{k=0}^{n}\frac{1}{k+1}\sum_{j=0}^{k}\left(-1\right)^{j}C\left(k,j\right)\left(x+j\right)^{n} \]
\[ B_{n}\left(x\right)=\sum_{k=0}^{n}\frac{1}{k+1}\sum_{j=0}^{k}\left(-1\right)^{j}C\left(k,j\right)\left(x+j\right)^{n} \]
-
\(B_{n}\left(x\right)\)はベルヌーイ多項式第2種スターリング数を最初から展開するともう一度第2種スターリング数に戻すことになるので手間がかかります。
\begin{align*} B_{n}\left(x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}x^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\left(\sum_{j=0}^{k}\left(-1\right)^{j}j^{k}\sum_{m=j}^{k}\frac{C\left(m,j\right)}{m+1}\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}C\left(n,k\right)\left(-1\right)^{j}j^{k}\sum_{m=j}^{k}\frac{C\left(m,j\right)}{m+1}x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{m=j}^{n}\sum_{k=m}^{n}C\left(n,k\right)\left(-1\right)^{j}j^{k}\frac{C\left(m,j\right)}{m+1}x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{m=j}^{n}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=m}^{n}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=m}^{n}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(\sum_{k=0}^{n}C\left(n,k\right)j^{k}x^{n-k}+\sum_{k=0}^{m-1}C\left(n,k\right)j^{k}x^{n-k}\right)\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=0}^{m-1}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{k=0}^{m-1}\left(-1\right)^{m}\frac{C\left(n,k\right)m!}{m+1}x^{n-k}\frac{1}{m!}\sum_{j=0}^{m}\left(-1\right)^{m-j}C\left(m,j\right)j^{k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{k=0}^{m-1}\left(-1\right)^{m}\frac{C\left(n,k\right)m!}{m+1}x^{n-k}S_{2}\left(k,m\right)\\ & =\sum_{m=0}^{n}\frac{1}{m+1}\sum_{j=0}^{m}\left(-1\right)^{j}C\left(m,j\right)\left(x+j\right)^{n} \end{align*}
\begin{align*} B_{n}\left(x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}x^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\left(\sum_{j=0}^{k}\left(-1\right)^{j}j^{k}\sum_{m=j}^{k}\frac{C\left(m,j\right)}{m+1}\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}C\left(n,k\right)\left(-1\right)^{j}j^{k}\sum_{m=j}^{k}\frac{C\left(m,j\right)}{m+1}x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{m=j}^{n}\sum_{k=m}^{n}C\left(n,k\right)\left(-1\right)^{j}j^{k}\frac{C\left(m,j\right)}{m+1}x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{m=j}^{n}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=m}^{n}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=m}^{n}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(\sum_{k=0}^{n}C\left(n,k\right)j^{k}x^{n-k}+\sum_{k=0}^{m-1}C\left(n,k\right)j^{k}x^{n-k}\right)\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\sum_{k=0}^{m-1}C\left(n,k\right)j^{k}x^{n-k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{k=0}^{m-1}\left(-1\right)^{m}\frac{C\left(n,k\right)m!}{m+1}x^{n-k}\frac{1}{m!}\sum_{j=0}^{m}\left(-1\right)^{m-j}C\left(m,j\right)j^{k}\\ & =\sum_{m=0}^{n}\sum_{j=0}^{m}\left(-1\right)^{j}\frac{C\left(m,j\right)}{m+1}\left(x+j\right)^{n}+\sum_{m=0}^{n}\sum_{k=0}^{m-1}\left(-1\right)^{m}\frac{C\left(n,k\right)m!}{m+1}x^{n-k}S_{2}\left(k,m\right)\\ & =\sum_{m=0}^{n}\frac{1}{m+1}\sum_{j=0}^{m}\left(-1\right)^{j}C\left(m,j\right)\left(x+j\right)^{n} \end{align*}
ベルヌーイ数\(B_{k}\)は第2種スターリング数\(S_{2}\left(k,j\right)\)を用いて、
\[ B_{k}=\sum_{j=0}^{k}\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right) \] で表され、第2種スターリング数は、
\[ S_{2}\left(k,j\right)=\frac{1}{j!}\sum_{m=0}^{j}\left(-1\right)^{j-m}C\left(j,m\right)m^{k} \] であるので、
\begin{align*} B_{n}\left(x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}x^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\left(\sum_{j=0}^{k}\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}\left(\frac{1}{j!}\sum_{m=0}^{j}\left(-1\right)^{j-m}C\left(j,m\right)m^{k}\right)x^{n-k}\\ & =\sum_{j=0}^{n}\frac{1}{j+1}\sum_{m=0}^{j}\left(-1\right)^{m}C\left(j,m\right)\sum_{k=0}^{n}C\left(n,k\right)m^{k}x^{n-k}\\ & =\sum_{j=0}^{n}\frac{1}{j+1}\sum_{m=0}^{j}\left(-1\right)^{m}C\left(j,m\right)\left(x+m\right)^{n} \end{align*} となり与式が成り立つ。
\[ B_{k}=\sum_{j=0}^{k}\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right) \] で表され、第2種スターリング数は、
\[ S_{2}\left(k,j\right)=\frac{1}{j!}\sum_{m=0}^{j}\left(-1\right)^{j-m}C\left(j,m\right)m^{k} \] であるので、
\begin{align*} B_{n}\left(x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}x^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\left(\sum_{j=0}^{k}\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}S_{2}\left(k,j\right)x^{n-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{j}\frac{j!}{j+1}\left(\frac{1}{j!}\sum_{m=0}^{j}\left(-1\right)^{j-m}C\left(j,m\right)m^{k}\right)x^{n-k}\\ & =\sum_{j=0}^{n}\frac{1}{j+1}\sum_{m=0}^{j}\left(-1\right)^{m}C\left(j,m\right)\sum_{k=0}^{n}C\left(n,k\right)m^{k}x^{n-k}\\ & =\sum_{j=0}^{n}\frac{1}{j+1}\sum_{m=0}^{j}\left(-1\right)^{m}C\left(j,m\right)\left(x+m\right)^{n} \end{align*} となり与式が成り立つ。
ページ情報
| タイトル | ベルヌーイ多項式の級数表示 |
| URL | https://www.nomuramath.com/d87a5mie/ |
| SNSボタン |
ベルヌーイ多項式の定義
\[
B_{n}\left(x\right)=\sum_{k=0}^{n}C\left(n,k\right)B_{k}x^{n-k}
\]
(*)ベルヌーイ多項式の総和
\[
\sum_{j=0}^{n}C\left(n,j\right)B_{j}\left(x\right)=\left(-1\right)^{n}B_{n}\left(-x\right)
\]
ベルヌーイ多項式の微分表示
\[
B_{n}\left(x\right)=\frac{D}{e^{D}-1}x^{n}
\]
(*)ベルヌーイ多項式の特殊値
\[
B_{n}\left(0\right)=B_{n}
\]