I've been trying for days now to find a closed form for the coefficients of the power series about $x=0$ of the function $$ f(x)=\exp\left(r^2\frac{x(n-2)-x^2(n-1)+x^n}{(x-1)^2}\right), $$ but I always end up with an infinite series at best. Can it be done at all?
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Here is the wolframalpha output for taylor series around $0$; $\exp(((n-2) r^2 x+(n-3) r^2 x^2+(n-4) r^2 x^3+(n-5) r^2 x^4+(n-6) r^2 x^5+O(x^6))+x^n (r^2+2 r^2 x+3 r^2 x^2+4 r^2 x^3+5 r^2 x^4+6 r^2 x^5+O(x^6)))$ – UserX Sep 22 '14 at 12:34
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Is it really $x^n$ in the numerator? – Jean-Claude Arbaut Sep 22 '14 at 12:42
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Yup, it's really the same $n$ that's in the other places... – Ziofil Sep 22 '14 at 12:43
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Just an idea, but a bit too long for a comment. You have
$$\frac{f'(x)}{f(x)}=\frac{r^2}{(x-1)^3}\left((n-2)x^n-nx^{n-1}+nx-(n-2)\right)$$
So $f$ is a solution of the differential equation
$$(x-1)^3y'=r^2\left((n-2)x^n-nx^{n-1}+nx-(n-2)\right)y$$
Now you can try to plug the series
$$y=\sum_{k=0}^{\infty}a_kx^k$$
And then find a relation between coefficients. See for example on Wikipedia, Power series solution of differential equations
Jean-Claude Arbaut
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This method is really nice, but I end up with a recurrence relation which connects $a_{k-2}$, $a_{k-1}$, $a_k$, $a_{k+1}$, $a_{k-n}$ and $a_{k-n+1}$. The only solvable one is for $n=0$, but we already know the answer to that..! – Ziofil Sep 22 '14 at 14:02
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@Ziofil Not sure you can do better than that. In what context did you get to expand this in series? – Jean-Claude Arbaut Sep 22 '14 at 14:04
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The function that I wrote is a generating function, I wished there was a closed form for the sequence that it represents. I'm happy that at least I have the GF, though. – Ziofil Sep 22 '14 at 18:16