復習もかねて級数って可愛いよなと圧をかけてみる
あいさつ
んちゃ！
今回は体調不良かつ病みがちなやなさんのためにずんだもんが復習用の記事を書きましたのだ。
- 差分で遊び直す
- なんかとりあえず面白い級数を作ってみたい
- 超幾何関数基礎
- 二項係数を含む総和
差分で遊び直す
$n$変数の多重数列$H:\mathbb{Z}\ni(k_{1},k_{2},...,k_{n})\mapsto H(k_{1},k_{2},...,k_{n})\in\mathbb{C}$に対して以下の演算$\Delta_{m}$を定義する。
\begin{equation}
\Delta_{m}H(k_{1},k_{2},...,k_{n})\coloneqq H(k_{1},k_{2},...,k_{m}+1,...,k_{n})-H(k_{1},k_{2},...,k_{m},...,k_{n})
\end{equation}
これを差分演算子と呼ぶ。
差分演算子$\Delta_{m}$は線形
任意の複素数$\alpha,\beta\in\mathbb{C}$、多重数列$H,I$について以下の様な計算ができるので証明完了。
\begin{eqnarray}
\Delta_{m}\{\alpha H(k_{1},k_{2},...,k_{n})+\beta L(k_{1},k_{2},...,k_{n})\}&=&\{\alpha H(k_{1},k_{2},...,k_{m}+1,...,k_{n})+\beta L(k_{1},k_{2},...,k_{m}+1,...,k_{n})\}\\
&-&\{\alpha H(k_{1},k_{2},...,k_{m},...,k_{n})+\beta L(k_{1},k_{2},...,k_{m},...,k_{n})\}\\
&=&\alpha\{H(k_{1},k_{2},...,k_{m}+1,...,k_{n})-H(k_{1},k_{2},...,k_{m},...,k_{n})\}\\
&+&\beta\{L(k_{1},k_{2},...,k_{m}+1,...,k_{n})-L(k_{1},k_{2},...,k_{m},...,k_{n})\}\\
&=&\alpha \Delta_{m} H(k_{1},k_{2},...,k_{n})+\beta \Delta_{m} I(k_{1},k_{2},...,k_{n})
\end{eqnarray}
任意の多重数列$H$に対して以下の式が成り立つ。
$[\Delta_{m_{1}},\Delta_{m_{2}}]H(k_{1},k_{2},...,k_{n})=0$
略
任意の多重数列$H,I$に対して以下の式が成り立つ。
\begin{equation}
\Delta_{m}H(k_{1},k_{2},...,k_{n})I(k_{1},k_{2},...,k_{n})=H(k_{1},k_{2},...,k_{m}+1,...,k_{n})\Delta_{m}I(k_{1},k_{2},...,k_{n})+\{\Delta_{m}H(k_{1},k_{2},...,k_{n})\}I(k_{1},k_{2},...,k_{n})
\end{equation}
\begin{eqnarray} \Delta_{m}H(k_{1},k_{2},...,k_{n})I(k_{1},k_{2},...,k_{n})&=&H(k_{1},k_{2},...,k_{m}+1,...,k_{n})\{I(k_{1},k_{2},...,k_{m}+1,...,k_{n})-I(k_{1},k_{2},...,k_{m},...,k_{n})\}+\{H(k_{1},k_{2},...,k_{m}+1,...,k_{n})-H(k_{1},k_{2},...,k_{m},...,k_{n})\}I(k_{1},k_{2},...,k_{m},...,k_{n})\\ &=&H(k_{1},k_{2},...,k_{m}+1,...,k_{n})\Delta_{m}I(k_{1},k_{2},...,k_{n})+\{\Delta_{m}H(k_{1},k_{2},...,k_{n})\}I(k_{1},k_{2},...,k_{n}) \end{eqnarray}
任意の多重数列$H$に対して以下の式が成り立つ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\displaystyle F_{1}(k_{1},k_{2},...,k_{n})\coloneqq \Delta_{m_{1}}H(k_{1},k_{2},...,k_{n})\\
\displaystyle F_{2}(k_{1},k_{2},...,k_{n})\coloneqq\Delta_{m_{2}}H(k_{1},k_{2},...,k_{n})\quad(\lim_{k_{m_{1}}\rightarrow\infty}
F_{2}(k_{1},k_{2},...,k_{n})=0)\\
\displaystyle \sum_{k_{m_{1}}=\overline{k_{m_{1}}}}^{\infty}\Delta_{m_{2}}F_{1}(k_{1},k_{2},...,k_{n})=-F_{2}(k_{1},k_{2},...,\overline{k_{m_{1}}},...,k_{n})
\end{array}
\right.
\end{eqnarray}
なんかとりあえず面白い級数を作ってみたい
Step1
$\displaystyle H(k_{1},k_{2})\coloneqq\frac{1}{\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}$とおく。
Step2
\begin{eqnarray} \left\{ \begin{array}{l} \displaystyle F_{1}(k_{1},k_{2})=\Delta_{k_{1}}H(k_{1},k_{2})=-\frac{k_{2}(2k_{1}+k_{2}+2)}{(k_{1}+k_{2}+1)^{2}}H(k_{1},k_{2})\\ \displaystyle F_{2}(k_{1},k_{2})=\Delta_{k_{2}}H(k_{1},k_{2})=-\frac{k_{1}(k_{1}+2k_{2}+2)}{(k_{1}+k_{2}+1)^{2}}H(k_{1},k_{2}) \end{array} \right. \end{eqnarray}
step3
\begin{eqnarray} \Delta_{k_{2}}F_{1}(k_{1},k_{2})&=&\frac{1}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}}\{k_{2}(k_{2}+1)(2k_{1}+k_{2}+2)(2k_{1}+k_{2}+3)-(k_{2}+1)(2k_{1}+k_{2}+3)(k_{1}+k_{2}+1)^{2}+k_{2}(2k_{1}+k_{2}+2)(k_{1}+k_{2}+2)^{2}\}H(k_{1},k_{2})\\ &=&-\frac{(k_{1}+1)^{2}(k_{2}+1)(2k_{1}+k_{2}+3)-k_{2}(2k_{1}+k_{2}+2)(k_{1}+k_{2}+2)^{2}}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}}H(k_{1},k_{2}) \end{eqnarray}
step4
\begin{equation} -\sum_{k_{1}=0}^{\infty}\frac{(k_{1}+1)^{2}(k_{2}+1)(2k_{1}+k_{2}+3)-k_{2}(2k_{1}+k_{2}+2)(k_{1}+k_{2}+2)^{2}}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}=0 \end{equation}
step5
\begin{equation}
\sum_{k_{1}=0}^{\infty}\frac{(k_{1}+1)^{2}(k_{2}+1)(2k_{1}+k_{2}+3)}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}=
\sum_{k_{1}=0}^{\infty}\frac{k_{2}(2k_{1}+k_{2}+2)(k_{1}+k_{2}+2)^{2}}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}
\end{equation}
折角やし一応まとめておこうぜ
$k_{2}\in\mathbb{N}$に対して以下の等式が成り立つ。
\begin{equation}
\sum_{k_{1}=0}^{\infty}\frac{(k_{1}+1)^{2}(k_{2}+1)(2k_{1}+k_{2}+3)}{(k_{1}+k_{2}+2)^{2}(k_{1}+k_{2}+1)^{2}\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}=
\sum_{k_{1}=0}^{\infty}\frac{k_{2}(2k_{1}+k_{2}+2)}{(k_{1}+k_{2}+1)^{2}\begin{pmatrix}k_{1}+k_{2}\\k_{1}\end{pmatrix}^{2}}
\end{equation}
step1
$\displaystyle H(k_{1},k_{2})\coloneqq\frac{(A)_{k_{1}}(B)_{k_{2}}}{(C)_{k_{1}+k_{2}}}$とおく。
step2
\begin{eqnarray} \left\{ \begin{array}{l} F_{1}(k_{1},k_{2})=\Delta_{k_{1}}H(k_{1},k_{2})=\frac{A-C-k_{2}}{C+k_{1}+k_{2}-1}H(k_{1},k_{2})\\ F_{2}(k_{1},k_{2})=\Delta_{k_{2}}H(k_{1},k_{2})=\frac{B-C-k_{1}}{C+k_{1}+k_{2}-1}H(k_{1},k_{2}) \end{array} \right. \end{eqnarray}
step3
\begin{eqnarray} \Delta_{k_{2}}F_{1}(k_{1},k_{2})&=&\{\frac{(A-C-k_{2})(A-C-k_{2}-1)-(k_{1}+A-1)}{(C+k_{1}+k_{2})(C+k_{1}+k_{2}-1)}\}H(k_{1},k_{2}) \end{eqnarray}
step4
\begin{equation}
\sum_{k_{1}=0}^{\infty}\{\frac{(A-C-k_{2})(A-C-k_{2}-1)-(k_{1}+A-1)}{(C+k_{1}+k_{2})(C+k_{1}+k_{2}-1)}\}\frac{(A)_{k_{1}}(B)_{k_{2}}}{(C)_{k_{1}+k_{2}}}=\frac{C-B}{C+k_{2}-1}\frac{(B)_{k_{2}}}{(C)_{k_{2}}}
\end{equation}
これもメモしとこか
\begin{equation} \sum_{k_{1}=0}^{\infty}\{\frac{(A-C-k_{2})(A-C-k_{2}-1)-(k_{1}+A-1)}{(C+k_{1}+k_{2})(C+k_{1}+k_{2}-1)}\}\frac{(A)_{k_{1}}(B)_{k_{2}}}{(C)_{k_{1}+k_{2}}}=\frac{C-B}{C+k_{2}-1}\frac{(B)_{k_{2}}}{(C)_{k_{2}}} \end{equation}
step1
$\displaystyle H(k_{1},k_{2})\coloneqq\frac{(A+1)_{k_{1}}}{\Gamma(k_{1}+\frac{1}{2})(k_{1}+k_{2})}$とおく。
step2
\begin{eqnarray} \left\{ \begin{array}{l} F_{1}(k_{1},k_{2})=\Delta_{k_{1}}H(k_{1},k_{2})=\frac{(A-\frac{1}{2})(k_{1}+k_{2})-(k_{1}+\frac{1}{2})}{(k_{1}+\frac{1}{2})(k_{1}+k_{2}+1)}H(k_{1},k_{2})\\ F_{2}(k_{1},k_{2})=\Delta_{k_{2}}H(k_{1},k_{2})=-\frac{1}{k_{1}+k_{2}+1}H(k_{1},k_{2}) \end{array} \right. \end{eqnarray}
step3
$A=\frac{1}{2}$とおくと
\begin{eqnarray}
\left\{
\begin{array}{l}
F_{1}(k_{1},k_{2})=\Delta_{k_{1}}H(k_{1},k_{2})=-\frac{1}{k_{1}+k_{2}+1}H(k_{1},k_{2})\\
F_{2}(k_{1},k_{2})=\Delta_{k_{2}}H(k_{1},k_{2})=-\frac{1}{k_{1}+k_{2}+1}H(k_{1},k_{2})
\end{array}
\right.
\end{eqnarray}
step4
\begin{eqnarray} \Delta_{k_{2}}F_{1}(k_{1},k_{2})&=&-\frac{2}{(k_{1}+k_{2}+1)(k_{1}+k_{2}+2)}H(k_{1},k_{2})\\ \end{eqnarray}
step5
\begin{equation}
-\sum_{k_{1}=0}^{\infty}\frac{2(\frac{3}{2})_{k_{1}}}{\Gamma(k_{1}+\frac{1}{2})(k_{1}+k_{2})_{3}}=-\frac{1}{k_{2}(k_{2}+1)}\frac{1}{\sqrt{\pi}}
\end{equation}
記録しておこう
自然数$k_{2}\in\mathbb{N}$に対して以下の式が成り立つ。
\begin{equation}
\sum_{k_{1}=0}^{\infty}\frac{(\frac{3}{2})_{k_{1}}}{\Gamma(k_{1}+\frac{1}{2})(k_{1}+k_{2})_{3}}=\frac{1}{2k_{2}(k_{2}+1)}\frac{1}{\sqrt{\pi}}
\end{equation}
超幾何関数基礎
自然数$p,q\in\mathbb{N}$、ベクトル$\vb*{a}=\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{p}\end{pmatrix}\in\mathbb{C}^{p}$そして$\vb*{b}=\begin{pmatrix}b_{1}\\b_{2}\\\vdots\\b_{q}\end{pmatrix}\in\mathbb{C}^{q}$によって定まる以下の様な級数を定める。これを超幾何関数という。
\begin{equation}
{}_{p}F_{q}(\vb*{a},\vb*{b};z)=\sum_{n=0}^{\infty}\frac{(a_{1})_{n}(a_{2})_{n}\cdots(a_{p})_{n}}{(b_{1})_{n}(b_{2})_{n}\cdots(b_{q})_{n}n!}z^{n}
\end{equation}
ベクトル$\vb*{a}=\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{p}\end{pmatrix}\in\mathbb{C}^{p}$そして$\vb*{b}=\begin{pmatrix}b_{1}\\b_{2}\\\vdots\\b_{q}\end{pmatrix}\in\mathbb{C}^{q}$に対して以下の様な記号を定める。
\begin{equation}
\vb*{a}\sharp\vb*{b}\coloneqq\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{p}\\b_{1}\\b_{2}\\\vdots\\b_{q}\end{pmatrix}\in\mathbb{C}^{p+q}
\end{equation}
ベクトル$\vb*{a}=\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{p}\end{pmatrix}\in\mathbb{C}^{p}$そして複素数$c\in\mathbb{C}$に対して以下の記号を定める。
\begin{equation}
\vb*{a}+c=c+\vb*{a}\coloneqq\begin{pmatrix}a_{1}+c\\a_{2}+c\\\vdots\\a_{p}+c\end{pmatrix}\in\mathbb{C}^{p}
\end{equation}
\begin{equation} \frac{d^{N}}{dz^{N}}\{z^{c}{}_{p}F_{q}(\vb*{a},\vb*{b};z)\}=(c-N+1)_{N}z^{c-N}{}_{p+1}F_{q+1}(c+1\sharp\vb*{a},c-N+1\sharp\vb*{b};z) \end{equation}
\begin{eqnarray} \frac{d^{N}}{dz^{N}}\{z^{c}{}_{p}F_{q}(\vb*{a},\vb*{b};z)\}&=&\frac{d^{N}}{dz^{N}}\sum_{n=0}^{\infty}\frac{(a_{1})_{n}(a_{2})_{n}\cdots(a_{p})_{n}}{(b_{1})_{n}(b_{2})_{n}\cdots(b_{q})_{n}n!}z^{c+n}\\ &=&\sum_{n=0}^{\infty}\frac{(a_{1})_{n}(a_{2})_{n}\cdots(a_{p})_{n}}{(b_{1})_{n}(b_{2})_{n}\cdots(b_{q})_{n}n!}(c+n)(c+n-1)\cdots (c+n-N+1)z^{c+n-N}\\ &=&(c-N+1)_{N}z^{c-N}\sum_{n=0}^{\infty}\frac{(c+1)_{n}(a_{1})_{n}(a_{2})_{n}\cdots(a_{p})_{n}}{(c-N+1)_{n}(b_{1})_{n}(b_{2})_{n}\cdots(b_{q})_{n}n!}z^{n}\\ &=&(c-N+1)_{N}z^{c-N}{}_{p+1}F_{q+1}(c+1\sharp\vb*{a},c-N+1\sharp\vb*{b};z) \end{eqnarray}
上記の式を書き直しておこう
\begin{equation} {}_{p+1}F_{q+1}(c+N\sharp\vb*{a},c\sharp\vb*{b};z)=\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(a_{1})_{k}(a_{2})_{k}\cdots(a_{p})_{k}}{(c)_{k}(b_{1})_{k}(b_{2})_{k}\cdots(b_{q})_{k}}{}_{p}F_{q}(\vb*{a}+k,\vb*{b}+k;z) \end{equation}
[1]
\begin{eqnarray}
{}_{p+1}F_{q+1}(c+1\sharp\vb*{a},c-N+1\sharp\vb*{b};z)&=&\frac{1}{(c-N+1)_{N}z^{c-N}}\frac{d^{N}}{dz^{N}}\{z^{c}{}_{p}F_{q}(\vb*{a},\vb*{b};z)\}\\
&=&\frac{1}{(c-N+1)_{N}z^{c-N}}\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{\Gamma(c+1)}{\Gamma(c+1-N+k)}z^{c-N+k}\frac{d^{k}}{dz^{k}}{}_{p}F_{q}(\vb*{a},\vb*{b};z)\\
&=&\frac{1}{(c-N+1)_{N}z^{c-N}}\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{\Gamma(c+1)}{\Gamma(c+1-N+k)}\frac{(a_{1})_{k}(a_{2})_{k}\cdots(a_{p})_{k}}{(b_{1})_{k}(b_{2})_{k}\cdots(b_{q})_{k}}z^{c-N}{}_{p}F_{q}(\vb*{a}+k,\vb*{b}+k;z)\\
&=&\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(a_{1})_{k}(a_{2})_{k}\cdots(a_{p})_{k}}{(c-N+1)_{k}(b_{1})_{k}(b_{2})_{k}\cdots(b_{q})_{k}}{}_{p}F_{q}(\vb*{a}+k,\vb*{b}+k;z)
\end{eqnarray}
[2]
\begin{equation}
{}_{p+1}F_{q+1}(c+N\sharp\vb*{a},c\sharp\vb*{b};z)=\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(a_{1})_{k}(a_{2})_{k}\cdots(a_{p})_{k}}{(c)_{k}(b_{1})_{k}(b_{2})_{k}\cdots(b_{q})_{k}}{}_{p}F_{q}(\vb*{a}+k,\vb*{b}+k;z)
\end{equation}
$(1-z)^{-(\lambda+k)}=\sum_{n=0}^{\infty}\frac{(\lambda+k)_{n}}{n!}z^{n}={}_{2}F_{1}(\begin{pmatrix}\lambda+k\\1+k\end{pmatrix},1+k;z)$より以下の式が成り立つ。
\begin{equation}
\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}n!}z^{n}=\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(\lambda)_{k}}{(c)_{k}}(1-z)^{-(\lambda+k)}
\end{equation}
\begin{eqnarray} {}_{3}F_{2}(\begin{pmatrix}c+N\\\lambda\\1\end{pmatrix},\begin{pmatrix}c\\1\end{pmatrix};z)&=&\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}n!}z^{n}\\ &=&\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(\lambda)_{k}}{(c)_{k}}(1-z)^{-(\lambda+k)} \end{eqnarray}
任意の自然数$N\in\mathbb{N}$そして複素数$\mu\in\mathbb{C}(Re(\mu)\gt\lambda+N+1)$に対して以下の式が成り立つ。
\begin{equation}
\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}(\mu+1)_{n}}
=\sum_{k=0}^{N}\begin{pmatrix}N\\k\end{pmatrix}\frac{(\lambda)_{k}}{(c)_{k}(\mu-\lambda-k+1)}
\end{equation}
\begin{eqnarray} \int_{0}^{1}(1-z)^{\mu}\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}n!}z^{n}dz&=&\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}n!}\frac{\Gamma(n+1)\Gamma(\mu+1)}{\Gamma(n+\mu+2)}\\ &=&\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}(\mu+n+1)(\mu+n)\cdots(\mu+1)}\\ &=&\sum_{n=0}^{\infty}\frac{(c+N)_{n}(\lambda)_{n}}{(c)_{n}(\mu+1)_{n}}\\ &=&\sum_{k=0}^{\infty}\begin{pmatrix}N\\k\end{pmatrix}\frac{(\lambda)_{k}}{(c)_{k}}\int_{0}^{1}(1-z)^{\mu-\lambda-k}dz\\ &=&\sum_{k=0}^{\infty}\begin{pmatrix}N\\k\end{pmatrix}\frac{(\lambda)_{k}}{(c)_{k}(\mu-\lambda-k+1)} \end{eqnarray}
二項係数を含む総和
$R(k)\in\mathbb{C}(k)(\forall n\in\mathbb{N}:\lim_{k\rightarrow n}(z-k)R(k)=0)$に対して以下の式が成り立つ。
\begin{equation}
\sum_{k=1}^{\infty}\frac{R(k)}{k\begin{pmatrix}ak\\k\end{pmatrix}}=\int_{0}^{1}\sum_{k=1}^{\infty}R(k)x^{k-1}(1-x)^{(a-1)k}dx
\end{equation}
\begin{eqnarray} \sum_{k=1}^{\infty}\frac{R(k)}{k\begin{pmatrix}ak\\k\end{pmatrix}}&=&\sum_{k=1}^{\infty}R(k)\frac{(k-1)!\{(a-1)k\}!}{(ak)!}\\ &=&\sum_{k=1}^{\infty}R(k)\frac{\Gamma(k)\Gamma((a-1)k+1)}{\Gamma(ak+1)}\\ &=&\sum_{k=1}^{\infty}R(k)\int_{0}^{1}x^{k-1}(1-x)^{(a-1)k}dx\\ &=&\int_{0}^{1}\sum_{k=1}^{\infty}R(k)x^{k-1}(1-x)^{(a-1)k}dx \end{eqnarray}
$F(z)\coloneqq \sum_{n=0}^{\infty}f_{n}z^{n}=\sum_{k=1}^{\infty}R(k)z^{k-1}(1-z)^{(a-1)k}$とすると以下の式が成り立つ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\displaystyle f_{n}=\sum_{m=0}^{n}(-1)^{n-m}\frac{\Gamma(am+a-m)}{\Gamma(am+a-n)(n-m)!}R(m+1)\quad(a\not\in \mathbb{N})\\
\displaystyle f_{n}=\sum_{m=\max\{0,\lfloor\frac{n-a+1}{a}\rfloor\}}^{n}(-1)^{n-m}\begin{pmatrix}(a-1)m+a-1\\n-m\end{pmatrix}R(m+1)\quad(a\in\mathbb{N})
\end{array}
\right.
\end{eqnarray}
[1]$a\not\in\mathbb{N}_{0}$に対して以下の式が成り立つ。
\begin{eqnarray}
F(z)&=&\sum_{n=0}^{\infty}f_{n}z^{n}\\
&=&\sum_{k=1}^{\infty}R(k)z^{k-1}(1-z)^{(a-1)k}\\
&=&\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}(-1)^{l}\frac{\Gamma((a-1)k+a)}{\Gamma((a-1)k+a-l)l!}R(k+1)z^{k+l}\\
&=&\sum_{n=0}^{\infty}z^{n}\sum_{m=0}^{n}(-1)^{n-m}\frac{\Gamma((a-1)m+a)}{\Gamma((a-1)m+a-n+m)(n-m)!}R(m+1)\\
&=&\sum_{n=0}^{\infty}z^{n}\sum_{m=0}^{n}(-1)^{n-m}\frac{\Gamma((a-1)m+a)}{\Gamma(am+a-n)(n-m)!}R(m+1)
\end{eqnarray}
[2]$a\in\mathbb{N}$の場合
\begin{eqnarray}
F(z)&=&\sum_{n=0}^{\infty}f_{n}z^{n}\\
&=&\sum_{k=1}^{\infty}R(k)z^{k-1}(1-z)^{(a-1)k}\\
&=&\sum_{k=1}^{\infty}\sum_{l=0}^{\infty}(-1)^{l}R(k)\begin{pmatrix}(a-1)k\\l\end{pmatrix}z^{k+l-1}\\
&=&\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}(-1)^{l}R(k+1)\begin{pmatrix}(a-1)k+a-1\\l\end{pmatrix}z^{k+l}\\
&=&\sum_{n=0}^{\infty}z^{n}\sum_{m=0}^{n}(-1)^{n-m}\begin{pmatrix}(a-1)m+a-1\\n-m\end{pmatrix}R(m+1)\\
&=&\sum_{n=0}^{\infty}z^{n}\sum_{m=\max\{0,\lceil\frac{n-a+1}{a}\rceil\}}^{n}(-1)^{n-m}\begin{pmatrix}(a-1)m+a-1\\n-m\end{pmatrix}R(m+1)
\end{eqnarray}
ただし$\forall k\in\mathbb{N}:\begin{pmatrix}n\\n+k\end{pmatrix}\coloneqq0$とした。
先に数列$\{f_{n}\}_{n\in\mathbb{N}_{0}}\subset\mathbb{C}$が定まっている場合は以下の様になる。
\begin{eqnarray}
\left\{
\begin{array}{l}
R(1)=f_{0}\\
\displaystyle R(n+1)=f_{n}-\sum_{m=0}^{n-1}(-1)^{n-m}\frac{\Gamma((a-1)m+a)}{\Gamma(am+a-n)(n-m)!}R(m+1)\quad(a\not\in\mathbb{N})\\
\displaystyle R(n+1)=f_{n}-\sum_{m=\max\{0,\lceil\frac{n-a+1}{a}\rceil\}}^{n-1}(-1)^{n-m}\begin{pmatrix}(a-1)m+a-1\\n-m\end{pmatrix}R(m+1)\quad(n\in\mathbb{N})
\end{array}
\right.
\end{eqnarray}
略
$\displaystyle f_{0}\coloneqq0,f_{n}\coloneqq\frac{1}{n}\quad(n\in\mathbb{N}),a=2$とする。この時以下の値を求めよ。
- $\displaystyle R(n)\quad(n\in\mathbb{N})$←未解決
- $\displaystyle \sum_{n=1}^{\infty}\frac{R(n)}{n\begin{pmatrix}2n\\n\end{pmatrix}}$
[1]
\begin{eqnarray}
\left\{
\begin{array}{l}
\displaystyle R(1)=0\\
\displaystyle R(n+1)=\frac{1}{n}-\sum_{m=\lceil\frac{n-a+1}{a}\rceil\}}^{n-1}(-1)^{n-m}\begin{pmatrix}m+1\\n-m\end{pmatrix}R(m+1)\quad(n\in\mathbb{N})
\end{array}
\right.
\end{eqnarray}
[2]
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{R(n)}{n\begin{pmatrix}2n\\n\end{pmatrix}}&=&-\int_{0}^{1}\log{(1-x)}dx\\
&=&(1-x)\log{(1-x)}|_{0}^{1}+\int_{0}^{1}dx\\
&=&1
\end{eqnarray}
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