I've got a recursive equation of the form

$$ x_{n+1} - x_{n} = \frac{(-1)^n}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}(x_0-x_1)$$ for $n \geq 2$. We can assume $x_0$ and $x_1$ are just real numbers/

I am having trouble getting a closed from of this thing. I attempted to get a generating function, using $$f(y)=\sum_{n \geq 0} x_k y^k $$, but it gets pretty nasty, and you cannot easily extract the coefficients to this function. Anyone have any ideas? I was sort of hoping that a decent looking generating function would exist.

Following the suggestions given, I was able to get a generating function of the form

$$f(x) = \frac{a}{1-x} + \frac{x(x_0-x_1)\exp{\left(\frac{-x}{2}\right)}}{1-x}$$ Letting $b=2x_0-x_1$ and $a=x_0$ in the above expression gives me the expression i was looking for. Now to find a closed form for the coefficients..

Mathematica gives me an answer of $\frac{(a+b)(\sqrt{e}-1)}{\sqrt{e}}$ for the limiting behavior of these coefficients.

  • $\begingroup$ This loos like the partial sum of the exponential : $\sum \frac{(\frac{-1}{2})^n}{n!}$. But I guess you knew it. $\endgroup$ – Pierre Alvarez Nov 10 '14 at 7:47
  • $\begingroup$ The denominator is simply $2\,n!$. $\endgroup$ – mvw Nov 10 '14 at 7:49
  • $\begingroup$ @mvw I feel so silly for not seeing that. $\endgroup$ – DaveNine Nov 10 '14 at 8:05
  • $\begingroup$ @user187373 thanks for the correction $\endgroup$ – mvw Nov 10 '14 at 10:41

You can express these partial sums in terms of the incomplete Gamma function:

$$ \sum_{j=0}^n \dfrac{(-1/2)^j}{j!} = \dfrac{\Gamma(n+1,-1/2)}{n! \sqrt{e}}$$


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