For $x>0$, define $\tilde f(x) = \sum\limits_{k = 0}^\infty {\frac{{(x - k) ^k }}{{k!}} {\bf 1}(x>k)}$, where ${\bf 1}$ is the indicator function. I know (actually, proved) that $\tilde f(x)$ is asymptotically $a {\rm e}^{bx} $ as $x \to \infty$, for certain $a>0$ and $0<b<1$. So, with $f(x)={\rm e}^x$, we have $\sum\limits_{k = 0}^\infty {f^{(k)} (0)\frac{{(x - k)^k }}{{k!}}{\bf 1}(x > k)} \sim af(bx)$. Are there other (non-trivial, but preferably simple) examples of functions $f$ for which such asymptotic equality holds?

Since the result for $f(x)={\rm e}^x$ is interesting in its own right, it is worth noting that, in fact, $\tilde f(x) \sim \frac{1}{{1 + b}}{\rm e}^{bx}$, where $b = 0.567143...$ is the solution $b \in (0,1)$ of $b{\rm e}^b = 1$. Moreover, the approximation $\tilde f(x) \approx \frac{1}{{1 + b}}{\rm e}^{bx}$ is most impressive, and is valid even for relatively small $x$ values; for example, $\tilde f(5) = 10.875$ while $\frac{1}{{1 + b}}{\rm e}^{5b} \approx 10.87495$ (for large $x$ values the approximation is extremely accurate).

That's what I meant:you need to fix the first formula where you forgot to write the $f^{(k)}(0)$ part. – Thierry Zell Nov 9 '10 at 20:52