ファウルハーバー公式(冪乗和公式)
ファウルハーバー公式
自然数\(j\)の冪乗和は以下の式で表される。
\(B_{n}\left(x\right)\)はベルヌーイ多項式
自然数\(j\)の冪乗和は以下の式で表される。
(1)
\[ \sum_{j=1}^{n}j^{m}=\frac{1}{m+1}\left(B_{m+1}\left(n+1\right)-B_{m+1}\left(1\right)\right) \](2)
\[ \sum_{j=1}^{n}j^{m}=\frac{1}{m+1}\sum_{j=0}^{m}\left(-1\right)^{j}C\left(m+1,j\right)B_{j}n^{m-j+1} \]-
\(B_{n}\)はベルヌーイ数\(B_{n}\left(x\right)\)はベルヌーイ多項式
(1)
ベルヌーイ多項式の関係\[ B_{m+1}\left(j+1\right)-B_{m+1}\left(j\right)=\left(m+1\right)j^{m} \] より、
\begin{align*} \sum_{j=1}^{n}j^{m} & =\frac{1}{m+1}\sum_{j=1}^{n}\left\{ B_{m+1}\left(j+1\right)-B_{m+1}\left(j\right)\right\} \\ & =\frac{1}{m+1}\left(B_{m+1}\left(n+1\right)-B_{m+1}\left(1\right)\right) \end{align*} となるので与式は成り立つ。
(1)-2
オイラー和公式より、\begin{align*} \sum_{j=1}^{n}j^{m} & =\sum_{k=0}^{N}\left[\frac{B_{k}}{k!}\frac{d^{k-1}}{dx^{k-1}}\left(x^{m}\right)\right]_{x=1}^{x=n+1}+R_{N}\\ & =\sum_{k=0}^{N}\left[\frac{B_{k}}{k!}P\left(m,k-1\right)x^{m-k+1}\right]_{1}^{n+1}+R_{N}\\ & =\frac{1}{m+1}\sum_{k=0}^{N}\left[B_{k}C\left(m+1,k\right)x^{m-k+1}\right]_{1}^{n+1}+R_{N}\qquad,\qquad N=m+1\text{とする}\\ & =\frac{1}{m+1}\sum_{k=0}^{m+1}\left[B_{k}C\left(m+1,k\right)x^{m-k+1}\right]_{1}^{n+1}+R_{m+1}\\ & =\frac{1}{m+1}\sum_{k=0}^{m+1}\left[B_{k}C\left(m+1,k\right)\left\{ \left(n+1\right)^{m-k+1}-1\right\} \right]\qquad,\qquad B_{n}(x)=\sum_{k=0}^{n}C\left(n,k\right)B_{n-k}x^{k}\\ & =\frac{1}{m+1}\left(B_{m+1}\left(n+1\right)-B_{m+1}\left(1\right)\right) \end{align*} ただし、
\begin{align*} R_{N} & =\left(-1\right)^{N+1}\int_{0}^{n+1}\frac{B_{N}\left(x-\left\lfloor x\right\rfloor \right)}{N!}\left(\frac{d^{N}}{dx^{N}}x^{m}\right)dx\\ & =\left(-1\right)^{N+1}\int_{0}^{n+1}B_{N}\left(x-\left\lfloor x\right\rfloor \right)C\left(m,N\right)x^{m-N}dx \end{align*} である。
(2)
\begin{align*} \sum_{k=1}^{n}k^{m} & =\left[\sum_{k=1}^{n}\frac{d^{m}}{dx^{m}}e^{kx}\right]_{x=0}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(e^{x}\frac{1-e^{nx}}{1-e^{x}}\right)\right]_{x=0}\\ \\ & =\left[\frac{d^{m}}{dx^{m}}\left(\frac{1-e^{nx}}{e^{-x}-1}\right)\right]_{x=0}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(\frac{-x}{e^{-x}-1}\cdot\frac{e^{nx}-1}{x}\right)\right]_{x=0}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(\sum_{k=0}^{\infty}\frac{B_{k}}{k!}\left(-x\right)^{k}\cdot\frac{1}{x}\sum_{j=1}^{\infty}\frac{\left(nx\right)^{j}}{j!}\right)\right]_{x=0}\cmt{\because\frac{x}{e^{x}-1}=\sum_{k=0}^{\infty}\frac{B_{k}}{k!}x^{k}}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{B_{k}}{k!}x^{k}\cdot\frac{1}{x}\sum_{j=0}^{\infty}\frac{\left(nx\right)^{j+1}}{\left(j+1\right)!}\right)\right]_{x=0}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\left(-1\right)^{k}\frac{B_{k}n^{j+1}x^{k+j}}{k!\left(j+1\right)!}\right)\right]_{x=0}\\ & =\left[\frac{d^{m}}{dx^{m}}\left(\sum_{j=0}^{\infty}\sum_{k=0}^{j}\left(-1\right)^{k}\frac{B_{k}n^{j-k+1}x^{j}}{k!\left(j-k+1\right)!}\right)\right]_{x=0}\cmt{j\rightarrow j'-k',k\rightarrow k'}\\ & =\left[\sum_{j=0}^{\infty}\sum_{k=0}^{j}\left(-1\right)^{k}\frac{B_{k}n^{j-k+1}P\left(j,m\right)x^{j-m}}{k!\left(j-k+1\right)!}\right]_{x=0}\\ & =\sum_{k=0}^{\infty}\sum_{j=k}^{\infty}\left(-1\right)^{k}\frac{B_{k}n^{j-k+1}P\left(j,m\right)\delta_{j,m}}{k!\left(j-k+1\right)!}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{B_{k}n^{m-k+1}P\left(m,m\right)H_{1}\left(m-k\right)}{k!\left(m-k+1\right)!}\\ & =\sum_{k=0}^{m}\left(-1\right)^{k}\frac{B_{k}n^{m-k+1}m!}{k!\left(m-k+1\right)!}\\ & =\frac{1}{m+1}\sum_{k=0}^{m}\left(-1\right)^{k}\frac{B_{k}n^{m-k+1}\left(m+1\right)!}{k!\left(m-k+1\right)!}\\ & =\frac{1}{m+1}\sum_{k=0}^{m}\left(-1\right)^{k}C\left(m+1,k\right)B_{k}n^{m-k+1} \end{align*}ページ情報
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ウォリス積分の値
\[
\int_{0}^{\frac{\pi}{2}}\sin^{2m}\theta d\theta=\frac{C(2m,m)}{4^{m}}\frac{\pi}{2}
\]
2重根号
\[
\sqrt{a\pm|b|\sqrt{c}}=\frac{\sqrt{2}}{2}\left(\sqrt{a+\sqrt{a^{2}-b^{2}c}}\pm\sqrt{a-\sqrt{a^{2}-b^{2}c}}\right)
\]
ラクランジュの未定乗数法
\[
F\left(x_{1},\cdots,x_{n},\lambda_{1,}\cdots,\lambda_{m}\right)=f\left(x_{1},\cdots,x_{n}\right)-\sum_{k=1}^{m}\lambda_{k}g_{k}\left(x_{1},\cdots,x_{n}\right)
\]
ベルヌーイ数とリーマンゼータ関数
\[
B_{2n}=(-1)^{n+1}\frac{2(2n)!}{(2\pi)^{2n}}\zeta(2n)
\]