a^(-k)及びa^(-kx)の無限級数

\(a^{-k}\)及び\(a^{-kx}\)の無限級数

\(a^{-k}\)の無限級数

$$\begin{eqnarray} S=\sum_{k=0}^{\infty}a^{-k} &=&\sum_{k=0}^{\infty}\left(a^{-1}\right)^k \;\ldots\;a\in\mathbb{R},\;\left|a\right|^{-1}=\frac{1}{\left|a\right|}\lt1 \\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a}\right)^k \\&=&\sum_{k=0}^{\infty}\frac{1}{a^{k}} \\&=&\frac{1}{a^0}+\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots \\&=&1+\frac{1}{a}+\frac{1}{a\cdot a}+\frac{1}{a\cdot a^{2}}+\cdots \\&=&1+\frac{1}{a}+\frac{1}{a}\frac{1}{a}+\frac{1}{a}\frac{1}{a^{2}}+\cdots \\&=&1+\frac{1}{a}\left(1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right) \\&=&1+\frac{1}{a}S \\S-\frac{1}{a}S&=&1 \\S\left(1-\frac{1}{a}\right)&=&1 \\S&=&\frac{1}{1-\frac{1}{a}} \\&=&\frac{1}{\frac{a-1}{a}} \\&=&\frac{a}{a-1} \end{eqnarray}$$

\(a^{-kx}\)の無限級数

$$\begin{eqnarray} S=\sum_{k=0}^{\infty}a^{-kx}&=&\sum_{k=0}^{\infty}\left(a^{-x}\right)^k \;\ldots\;a,x\in\mathbb{R},\;|a|^{-x}=\frac{1}{|a|^{x}}\lt1 \\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a^{x}}\right)^k \\&=&\sum_{k=0}^{\infty}\frac{1}{a^{kx}} \\&=&\frac{1}{a^{0x}}+\frac{1}{a^x}+\frac{1}{a^{2x}}+\frac{1}{a^{3x}}+\cdots \\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}a^{x}}+\frac{1}{a^{x}a^{2x}}+\cdots \\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{2x}}+\cdots \\&=&1+\frac{1}{a^{x}}\left(1+\frac{1}{a^{x}}+\frac{1}{a^{2x}}+\cdots\right) \\&=&1+\frac{1}{a^{x}}S \\S-\frac{1}{a^x}S&=&1 \\S\left(1-\frac{1}{a^x}\right)&=&1 \\S&=&\frac{1}{1-\frac{1}{a^x}} \\&=&\frac{1}{\frac{a^x-1}{a^x}} \\&=&\frac{a^x}{a^x-1} \end{eqnarray}$$

偶数の二重階乗(double factorial / semifactorial)の逆数(reciprocal)の和(無限級数(infinite series))

$$\begin{array}{rcl} \frac{1}{0!!}+\frac{1}{2!!}+\frac{1}{4!!}+\cdots &=&\displaystyle \sum_{k=0}^{\infty}\frac{1}{2^k k!} \\&&\;\dots\;n!!(二重階乗) \neq (n!)! (=階乗凾数の二回反復),\;偶数nの二重階乗n!!=\prod_{k=0}^{\frac{n}{2}}(2k),偶数2k(k\geq0)の二重階乗(2k)!!=2^kk! \\&=&\displaystyle \sum_{k=0}^{\infty}\frac{\left(\frac{1}{2}\right)^k}{k!} \\&=&\displaystyle \mathrm{e}^{\frac{1}{2}} \,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{k=0}^{\infty}\frac{x^k}{k!}= \mathrm{e}^x}\\ &=&\displaystyle \sqrt{\mathrm{e}}\\ \end{array}$$

階乗(factorial)の逆数(reciprocal)の和(無限級数(infinite series))

$$\begin{array}{rcl} \displaystyle f(x) &=& \displaystyle \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\,\dotso\,a点まわりのテイラー展開\\ &=& \displaystyle \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb\\ \end{array}$$

\(e^x\)のマクローリン展開

$$\begin{array}{rcl} \displaystyle \mathrm{e}^x　 &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}\left\{\left(\frac{\mathrm{d}^k}{\mathrm{d}x^k}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^k \,\dotso\,0点まわりのテイラー展開(Taylor\,series)=マクローリン展開(Maclaurin\,expansion)\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\frac{\mathrm{d}^0}{\mathrm{d}x^0}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\frac{\mathrm{d}^1}{\mathrm{d}x^1}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\,1\,x^0 \displaystyle +\frac{1}{1!}\,1\,x^1 \displaystyle +\frac{1}{2!}\,1\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}x^k\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!}&=&\displaystyle \mathrm{e}^x\\ \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}=\sum_{k=0}^{\infty}\frac{1^k}{k!}&=&\displaystyle \mathrm{e}^1=\mathrm{e}\,\dotso\,x=1\\ \end{array}$$

\(e^{Cx}\)のマクローリン展開

$$\begin{array}{rcl} \displaystyle \mathrm{e}^{Cx}　 &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}\left\{\left(\frac{\mathrm{d}^k}{\mathrm{d}x^k}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^k \,\dotso\,0点まわりのテイラー展開(Taylor\,series)=マクローリン展開(Maclaurin\,expansion)\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\frac{\mathrm{d}^0}{\mathrm{d}x^0}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\frac{\mathrm{d}^1}{\mathrm{d}x^1}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(C\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(C^2\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\,1\,x^0 \displaystyle +\frac{1}{1!}\,C\,x^1 \displaystyle +\frac{1}{2!}\,C^2\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \frac{1}{0!}\,C^0\,x^0 \displaystyle +\frac{1}{1!}\,C^1\,x^1 \displaystyle +\frac{1}{2!}\,C^2\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{(Cx)^k}{k!}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \sum_{k=0}^{\infty}\frac{(Cx)^k}{k!} &=& \mathrm{e}^{Cx}\\ \end{array}$$