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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).

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Do you have an upper bound on $a$? I presume also that $a$ and $b$ for your general question are not the same as for the case $f(x) = e^x$... –  David Roberts Nov 9 '10 at 5:42
    
Your formula for $\tilde{f}$ does not have an $f$ in it. –  Thierry Zell Nov 9 '10 at 6:02
    
@David: Yes, a and b for the general question are not necessarily the same as for the case $f(x)=e^x$. As for your first question, I don't have an upper bound on $a$ for the general case, but I can evaluate it (as well as $b$) for the case $f(x)=e^x$. –  Shai Covo Nov 9 '10 at 7:16
    
@Thierry: The formula for $\tilde f$ has $f$ in it, if you write it as $\tilde f(x) = \sum\nolimits_{k = 0}^\infty {f^{(k)} (0)\frac{{(x - k)^k }}{{k!}}{\bf 1}(x > k)}$. –  Shai Covo Nov 9 '10 at 7:30
    
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

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