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Does anyone recognise this Taylor series expansion? It is similar to that of $\exp(x)$, but not quite:

$$ 1 - \frac{1}{2!}x + \frac{1}{3!}x^2 - \frac{1}{4!}x^3 + \frac{1}{5!}x^4 - \frac{1}{6!}x^5 + \ldots $$

Is this a well-known function? Thank you very much in advance.

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    $\begingroup$ How about $$x-\frac{x^2}{2!}+\frac{x^3}{3!}-\cdots?$$ $\endgroup$ – Lord Shark the Unknown Mar 12 '18 at 18:08
  • $\begingroup$ Thanks a lot Lord Shark the Unknown, looks like user284331 used this as a step in his derivation. $\endgroup$ – user36247 Mar 12 '18 at 18:25
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\begin{align*} S(x)&=1-\dfrac{1}{2!}x+\dfrac{1}{3!}x^{2}-\cdots\\ &=\dfrac{1}{x}\left(x-\dfrac{1}{2!}x^{2}+\dfrac{1}{3!}x^{3}-\cdots\right)\\ &=-\dfrac{1}{x}\left(-x+\dfrac{1}{2!}x^{2}-\dfrac{1}{3!}x^{3}+\cdots\right)\\ &=-\dfrac{1}{x}(e^{-x}-1). \end{align*}

And define $S(0)=1$.

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  • $\begingroup$ Thank you very much, user284331. $\endgroup$ – user36247 Mar 12 '18 at 18:25

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