Alternating Exponential Generating Functions

SPECIES

$\displaystyle e^{-x} = \sum_{n \geq 0} (-1)^n\,\dfrac{x^n}{n!} \\ = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} - \dfrac{x^5}{5!} + \dfrac{x^6}{6!} \pm \cdots$

$\displaystyle \dfrac{1}{e^x\!-\!1} = \sum_{n \geq 0} B_n^{-}\,\dfrac{x^{n-1}}{n!} \\ = \dfrac{1}{x} - \dfrac{1}{2} + \dfrac{1}{6}\,\dfrac{x}{2!} - \dfrac{1}{30}\,\dfrac{x^3}{4!} + \dfrac{1}{42}\,\dfrac{x^5}{6!} - \dfrac{1}{30}\,\dfrac{x^7}{8!} \pm \cdots$

$\displaystyle \ln(1\!+\!e^x) = \sum_{n \geq 0} \eta(1\!-\!n)\,\dfrac{x^n}{n!} \\ = \ln 2 + \dfrac{1}{2}\,x + \dfrac{1}{4}\,\dfrac{x^2}{2!} - \dfrac{1}{8}\,\dfrac{x^4}{4!} + \dfrac{1}{4}\,\dfrac{x^6}{6!} - \dfrac{17}{16}\,\dfrac{x^8}{8!} \pm \cdots$

$\displaystyle \dfrac{e^x}{1\!+\!e^x} = \dfrac{1}{1\!+\!e^{-x}} = \sum_{n \geq 0} \eta(-n)\,\dfrac{x^n}{n!} \\ = \dfrac{1}{2} + \dfrac{1}{4}\,x - \dfrac{1}{8}\,\dfrac{x^3}{3!} + \dfrac{1}{4}\,\dfrac{x^5}{5!} - \dfrac{17}{16}\,\dfrac{x^7}{7!} + \dfrac{31}{4}\,\dfrac{x^9}{9!} \pm \cdots$

$\displaystyle \dfrac{e^x}{(1\!+\!e^x)^2} = \dfrac{e^{-x}}{(1\!+\!e^{-x})^2} = \sum_{n \geq 0} \eta(-n\!-\!1)\,\dfrac{x^n}{n!} \\ = \dfrac{1}{4} - \dfrac{1}{8}\,\dfrac{x^2}{2!} + \dfrac{1}{4}\,\dfrac{x^4}{4!} - \dfrac{17}{16}\,\dfrac{x^6}{6!} + \dfrac{31}{4}\,\dfrac{x^8}{8!} - \dfrac{691}{8}\,\dfrac{x^{10}}{10!} \pm \cdots$

$\displaystyle \dfrac{e^x(1\!-\!e^x)}{(1\!+\!e^x)^3} = -\dfrac{e^{-x}(1\!-\!e^{-x})}{(1\!+\!e^{-x})^3} \\ = \dfrac{e^x}{(1\!+\!e^x)^3} \, \sum_{k=0}^1 (-1)^k\,A_2(k)\,e^{kx} = \sum_{n \geq 0} \eta(-n\!-\!2)\,\dfrac{x^n}{n!}$

$\displaystyle \dfrac{e^x(1\!-\!4e^x\!+\!e^{2x})}{(1\!+\!e^x)^4} = \dfrac{e^{-x}(1\!-\!4e^{-x}\!+\!e^{-2x})}{(1\!+\!e^{-x})^4} \\ = \dfrac{e^x}{(1\!+\!e^x)^4} \, \sum_{k=0}^2 (-1)^k\,A_3(k)\,e^{kx} = \sum_{n \geq 0} \eta(-n\!-\!3)\,\dfrac{x^n}{n!}$

$\displaystyle \dfrac{e^x(1\!-\!11e^x\!+\!11e^{2x}\!-\!e^{3x})}{(1\!+\!e^x)^5} = -\dfrac{e^{-x}(1\!-\!11e^{-x}\!+\!11e^{-2x}\!-\!e^{-3x})}{(1\!+\!e^{-x})^5} \\ = \dfrac{e^x}{(1\!+\!e^x)^5} \, \sum_{k=0}^3 (-1)^k\,A_4(k)\,e^{kx} = \sum_{n \geq 0} \eta(-n\!-\!4)\,\dfrac{x^n}{n!}$

$\displaystyle \dfrac{e^x(1\!-\!26e^x\!+\!66e^{2x}\!-\!26e^{3x}\!+\!e^{4x})}{(1\!+\!e^x)^6} = \dfrac{e^{-x}(1\!-\!26e^{-x}\!+\!66e^{-2x}\!-\!26e^{-3x}\!+\!e^{-4x})}{(1\!+\!e^{-x})^6} \\ = \dfrac{e^x}{(1\!+\!e^x)^6} \, \sum_{k=0}^4 (-1)^k\,A_5(k)\,e^{kx} = \sum_{n \geq 0} \eta(-n\!-\!5)\,\dfrac{x^n}{n!}$

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