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Does anybody know whether or not this sum has a closed form? $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!(2^n+1)}$$ I can't get WA to even understand it when I type it in.

For context, the reason I want to know is that I calculated it as the solution to the functional equation $$f(x)+f(2x)=e^x$$

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    $\begingroup$ It's really close to $e^{x/2}$. I'm skeptical there's a nice closed form; Wolfram won't even evaluate the sum in closed form for $x = 1$. $\endgroup$ – user296602 Aug 12 '17 at 20:09
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    $\begingroup$ If I haven't made a mistake (quite possible since I didn't use pen and paper), it's $$\exp(x/2) - \sum_{k = 2}^{\infty} (-1)^k \bigl(\exp (2^{-k}x) - 1\bigr).$$ Wouldn't call that a closed form, though. $\endgroup$ – Daniel Fischer Aug 12 '17 at 20:23
  • $\begingroup$ Perhaps $f(2x)+f(4x)=(f(x)+f(2x))^2$ is obvious but useful. $\endgroup$ – Somos Aug 12 '17 at 20:24
  • $\begingroup$ @Somos Perhaps not useful, but certainly interesting. :D $\endgroup$ – Frpzzd Aug 12 '17 at 20:27
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    $\begingroup$ Not a closed form, but amusingly $f(x)=e^{\frac x2}-e^{\frac x4}+e^{\frac x8}-e^{\frac{x}{16}}+...$ $\endgroup$ – David Quinn Aug 12 '17 at 20:43

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