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Find a general solution for this recurrence relation: $$f(n) = 2f(n-1) + \frac{(-1)^n}{n!}$$ when $f(0) = 0, f(1) = 1$ EDIT: n >= 2

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closed as off-topic by Abcd, Martín Vacas Vignolo, A. Goodier, Paul Frost, José Carlos Santos Dec 30 '18 at 15:51

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  • $\begingroup$ What have you tried? $\endgroup$ – MisterRiemann Dec 30 '18 at 11:04
  • $\begingroup$ I have tried to use generating functions, but it just didn't go well... $\endgroup$ – Robo Yonuomaro Dec 30 '18 at 11:07
  • $\begingroup$ Iz is $$a_n=c_1 2^{n-1}+\frac{2^n \left(\frac{\Gamma \left(n+1,-\frac{1}{2}\right)}{\Gamma (n+1)}-\sqrt{e}\right)}{\sqrt{e}}$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '18 at 11:12
  • $\begingroup$ @Dr.SonnhardGraubner. This is pure magics ! $\endgroup$ – Claude Leibovici Dec 30 '18 at 14:25
  • $\begingroup$ Expanding this into an answer would be appreciated. $\endgroup$ – marty cohen Dec 30 '18 at 14:26
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Remark: If $f(0)=0$, then $f(1)$ must be equal to $-1$, not $1$ as claimed. [Just plug $n=1$ in the recursion]

Multiply by $z^n$ and sum over $n\geq 1$ $$ \sum_{n\geq 1}f(n)z^n=2\sum_{n\geq 1}f(n-1)z^n+\sum_{n\geq 1}\frac{(-1)^n}{n!}z^n $$ and define $\sum_{n\geq 0}f(n)z^n=G(z)$. Hence $$ G(z)-f(0)=2zG(z)-e^{-z} \left(e^z-1\right)\ . $$ Using $f(0)=0$, this leads to $$ G(z)=\frac{ 1-e^{-z}}{2z-1}\ . $$ The general term $f(n)$ of the recursion can be obtained from Cauchy's formula. Can you proceed from here?

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  • $\begingroup$ You are right about that, I forgot to mention that n >= 2. How can it be solved with this? $\endgroup$ – Robo Yonuomaro Dec 30 '18 at 15:57

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