5
$\begingroup$

Let $N \in \mathbb N^*$, $\delta > 0$, $t > 0$ and consider \begin{equation} f(t,\delta, N)= e^{-t/\delta} \sum_{k=0}^{N-1}\frac{(t/\delta)^k}{k!}. \qquad \qquad \qquad (1) \end{equation} Let now $N = \lceil\delta^{-\gamma}\rceil$, with $0 < \gamma < 1$. Can we conclude that \begin{equation} \lim_{\delta \to 0} f(t,\delta,\lceil \delta^{-\gamma}\rceil)=0, \end{equation} for all $t > 0$?

The reasoning is that if $N \to \infty$ (for $\delta$ fixed), then the sum tends to $e^{t/\delta}$ and therefore $f(t,\delta,N)\to 1$. If on the other hand $\delta \to 0$, for $N$ fixed, we have that $f(t,\delta,N)\to 0$ since the exponential goes faster to zero than the polynomial to infinity. In this case, we let $\delta \to 0$ and simultaneously $N \to \infty$, but with a "slower pace". Can we conclude that the limit is zero? If yes, how?

Edit:

After some reasoning I came up with $$ f(t,\delta,N)= \frac{\Gamma(N, t/\delta)}{(N-1)!}, $$ with $\Gamma(\cdot, \cdot)$ the upper incomplete Gamma function. I implemented this in Matlab, and the interesting behaviour is that if $N=\delta^{-1}$, then the limit (1) is $1/2$ if $0 <t \leq 1$ and it is 0 if $t > 1$. If $N=\delta^{-\gamma}$, with $\gamma \in (0, 1)$, the limit is 0, and if $N = \delta^{-\gamma}$, with $\gamma > 1$, then the limit is 1, independently of $t$.

$\endgroup$
9
  • $\begingroup$ my bad, I misread $\endgroup$
    – NHL
    Jan 29 at 15:11
  • 1
    $\begingroup$ Speaking intuitively, the question comes down to identifying the dominant terms in the series for the exponential. If you think about this, the dominant terms are necessarily around $t \delta^{-1}$: until that point, the terms are rapidly growing, and after that they are rapidly decaying. $\endgroup$
    – Ian
    Jan 29 at 15:30
  • $\begingroup$ Sorry Ian but I did not understand your comment. In my intuition (and some computations seem to confirm this) if $\gamma$ is different from $1$ then $t$ does not play a role in the result $\endgroup$
    – G. Gare
    Jan 29 at 15:36
  • 1
    $\begingroup$ If $\gamma$ is different from $1$ and $t$ is different from $0$ then asymptotically you are not cutting off the sum anywhere near $t\delta^{-1}$ and so you either have "all the important terms" ($\gamma>1$) or "none of the important terms" ($0<\gamma<1$). $\endgroup$
    – Ian
    Jan 29 at 15:37
  • $\begingroup$ Ok gotcha. How would you rigorously prove this? $\endgroup$
    – G. Gare
    Jan 29 at 15:39
1
$\begingroup$

For simplicity let $x=t/\delta$ for the moment.

The main point is that $\frac{x^k}{k!}$ is within a polynomial factor of $\left ( \frac{xe}{k} \right )^k$, by Stirling. (It's within a bounded factor of $\frac{(xe)^k}{k^{k+1/2}}$ but this is overkill for this application.)

So for $0<c<1$, the sum of all the terms with $k>x(1+c)$ behaves basically like $\left ( \frac{e}{1+c} \right )^{x(1+c)} \frac{c+1}{c}$. This is the dominant term times $\frac{1}{1-r}$ where $r=\frac{1}{1+c}$ is the common ratio of the bounding geometric series.

Similarly for the first $x(1-c)$ terms you have a bound like $\left ( \frac{e}{1-c} \right )^{x(1-c)} \frac{1}{c}$.

Returning to your setting, if $\gamma>1$ then you exclude only the terms past $x(1+c)$ where $c \to \infty$ as $\delta \to 0$, and so the error goes to zero.

If $\gamma<1$ then you exclude all the terms past $x(1-c)$ for $c \to 1$ as $\delta \to 0$ and so the terms you are including become negligible relative to $e^x$, i.e. the error goes to $1$.

If instead you look at $N \sim C \delta^{-1}$ then the above bounds properly come into play.

$\endgroup$
2
  • $\begingroup$ Thanks. I may be slow, but I'm not grasping it. Could you be a bit more precise? "The error" is $e^{t/\delta} - \sum_{k=N}^{\infty} \cdots$? $\endgroup$
    – G. Gare
    Jan 30 at 17:01
  • $\begingroup$ @G.Gare The error I'm referring to is $1-e^{-t/\delta} \sum \dots$. $\endgroup$
    – Ian
    Jan 30 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.