(*)オイラー多項式の総和

by nomura · 2025年1月23日

オイラー多項式の総和
オイラー多項式\(E_{n}\left(x\right)\)の総和について次が成り立つ。

(1)

\[ x^{n}=E_{n}\left(x\right)+\frac{1}{2}\sum_{k=0}^{n-1}C\left(n,k\right)E_{k}\left(x\right) \]

(2)

\[ \sum_{k=0}^{n}C\left(n,k\right)E_{k}\left(x\right)+E_{n}\left(x\right)=2x^{n} \]

(3)平行移動

\[ E_{n}\left(x+y\right)=\sum_{k=0}^{n}C\left(n,k\right)E_{k}\left(x\right)y^{n-k} \]

(4)乗法定理

\[ E_{n}\left(mx\right)=\begin{cases} m^{n}\sum_{k=0}^{m-1}\left(-1\right)^{k}E_{n}\left(x+\frac{k}{m}\right) & m=1,3,\cdots\\ -\frac{2}{n+1}m^{n}\sum_{k=0}^{m-1}\left(-1\right)^{k}B_{n+1}\left(x+\frac{k}{m}\right) & m=2,4,\cdots \end{cases} \]

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\(B_{n}\left(x\right)\)はベルヌーイ多項式

(1)

\begin{align*} E_{n}\left(x\right)+\frac{1}{2}\sum_{k=0}^{n-1}C\left(n,k\right)E_{k}\left(x\right) & =E_{n}\left(x\right)+\frac{1}{2}\left(\sum_{k=0}^{n}C\left(n,k\right)E_{k}\left(x\right)-C\left(n,n\right)E_{k}\left(x\right)\right)\\ & =\frac{1}{2}\left(E_{n}\left(x+1\right)+E_{k}\left(x\right)\right)\cmt{\because E_{n}\left(x+y\right)=\sum_{k=0}^{n}C\left(n,k\right)E_{k}\left(x\right)y^{n-k}}\\ & =\frac{1}{2}\left(-E_{n}\left(x\right)+2x^{n}+E_{k}\left(x\right)\right)\\ & =x^{n} \end{align*}

(2)

\begin{align*} \sum_{k=0}^{n}C\left(n,k\right)E_{k}\left(x\right)+E_{n}\left(x\right) & =\sum_{k=0}^{n-1}C\left(n,k\right)E_{k}\left(x\right)+C\left(n,n\right)E_{n}\left(x\right)+E_{n}\left(x\right)\\ & =\sum_{k=0}^{n-1}C\left(n,k\right)E_{k}\left(x\right)+C\left(n,n\right)E_{n}\left(x\right)+E_{n}\left(x\right)\\ & =2\left(x^{n}-E_{n}\left(x\right)\right)+E_{n}\left(x\right)+E_{n}\left(x\right)\cmt{\because x^{n}=E_{n}\left(x\right)+\frac{1}{2}\sum_{k=0}^{n-1}C\left(n,k\right)E_{k}\left(x\right)}\\ & =2x^{n} \end{align*}

(3)

\begin{align*} E_{n}\left(x+y\right) & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(x+y-\frac{1}{2}\right)^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\sum_{j=0}^{n-k}C\left(n-k,j\right)\left(x-\frac{1}{2}\right)^{j}y^{n-k-j}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\sum_{j=k}^{n}C\left(n-k,j-k\right)\left(x-\frac{1}{2}\right)^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)\frac{E_{k}}{2^{k}}C\left(n-k,j-k\right)\left(x-\frac{1}{2}\right)^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,j\right)\frac{E_{k}}{2^{k}}C\left(j,k\right)\left(x-\frac{1}{2}\right)^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}C\left(n,j\right)y^{n-j}\sum_{k=0}^{j}\frac{E_{k}}{2^{k}}C\left(j,k\right)\left(x-\frac{1}{2}\right)^{j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)y^{n-j}E_{j}\left(x\right)\\ & =\sum_{j=0}^{n}C\left(n,j\right)E_{j}\left(x\right)y^{n-j} \end{align*}