I'm stuck in the following series: $$\sum_{n=0}^{+\infty} \frac{1}{n!}\frac{d}{dx^n} \left( f(n-2x) \right) \left|_{x=0} \right.$$ where $f$ is a smooth function. At first glance it resembles a Taylor Type series, but the argument of the function depends on the index $n$ of the sum, and so it's not a Taylor expansion around a fixed point as usual. I have no idea how to treat this sum. Any ideas, tricks or references? Thanks!

  • $\begingroup$ what exactly is the question though? what do you want to accomplish? $\endgroup$ – Paulo Mourão May 24 at 15:10

$$\frac{\mathrm{d}}{\mathrm{d}x^n} f(n-2x) \bigg|_{x=0}=(-2)^nf^{(n)}(n)$$ So the result is not a series expansion as it does not depend on $x$, but a constant solely dependant on the function $f(x)$ used. Although the summation may not converge for example when $f(x)=n!/(-2)^n$, its value is given by $$\sum_{n=0}^\infty \frac{(-2)^nf^{(n)}(n)}{n!}$$

  • $\begingroup$ Thanks Peter. The fact is that, the coeffients of the original series are really horrific; after long calculations I proved that the coeffients are exactly in the form above; my hope was that, thanks to this false similarity to a Taylor series, I could say something on the sum. $\endgroup$ – Papemax89 May 24 at 15:31
  • $\begingroup$ @Papemax89 What was the original series? Perhaps you should post that in the question instead? $\endgroup$ – Peter Foreman May 24 at 15:32

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