# Truncated Taylor series of the exponential

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 $$0 and it is 0 if $$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$$.

– NHL
Jan 29 at 15:11
• 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.
– Ian
Jan 29 at 15:30
• 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 Jan 29 at 15:36
• 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" ($0<\gamma<1$).
– Ian
Jan 29 at 15:37
• Ok gotcha. How would you rigorously prove this? Jan 29 at 15:39

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 $$0, the sum of all the terms with $$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.

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