# Can I interpret the exponential of the derivative operator, $e^D$, as infinite shift operators each shifting “infinitesimally”?

To better explain what I mean, an example can be very useful. Consider $e^{i\theta}$. We could express this using the series definition or the limit definition of $e^x$ instead:

$$e^{i\theta} = \lim_{n\to\infty}\left(1+\frac{\theta}{n}i\right)^n$$

If we want to see visually what's going on as $n$ grows we can look at the following GIF from Wikipedia:

$\hskip2in$

Essentially each factor in the product above as $n$ grows has its norm shrink to $1$ and its angle get "infinitesimally" small. The intuitive idea would be that when you multiply all the factors you add all the (very small) angles to the total $\theta$ and multiply all the norms which in the limit gives $1$.

I've considered applying this intuition to the exponential of the derivative operator, which shifts the function it's applied to:

$$e^{cD} f(x)=f(x+c)$$

I've considered rewriting the derivative operator as a limit in terms of the shift operator, and then insert it into the product definition of the exponential ($I$ is the identity in what follows), where I take $c=1$ for simplicity:

$$Df(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\equiv\lim_{h\to0}\frac{S(h)-I}{h} f(x)$$

$$e^{D} = \lim_{n\to\infty}\left(1+\frac{1}{n} \lim_{h\to 0}\frac{S(h)-I}{h} \right)^n$$

The only way I see to get to the exponential being the composition of a lot of "small" shift operators is if we evaluate the double limit along $h=1/n$ (or $n=1/h$ alternatively) in which case the above simplifies and we get $e^{D}=\lim_{n\to\infty}S^n(1/n)=S(1)$ but I cannot think of any way to justify taking that specific path, $h=1/n$, when evaluating the limits.

To sum up, I would like help with two things:

1. Is it conceptually sound to interpret the $e^D$ in the above manner?
2. If it is, can I prove with some rigour the above intuition?

Let's do this instead: $$e^Df = \lim_{n\to\infty}\left(1+\frac Dn\right)^nf.$$ But \begin{align} \left(1+\frac Dn\right)f &= f+\frac1nf', \\ S_{1/n}f &= f + \frac1n f' + O(1/n^2), \end{align} at least when $f''$ is bounded. So each $(1 + D/n)$ operation is like a little shift, and \begin{align} e^Df &= \lim_{n\to\infty}\bigl(S_{1/n}+O(1/n^2)\bigr)^nf \\ &= \lim_{n\to\infty}\bigl((S_{1/n})^n+O(1/n)\bigr) f \\ &= S_1 f. \end{align}
• was is the definition of the operator $[e^D f]$ ?? shouldn't it be : $[e^D f] (x_0) = \sum_{k=0}^\infty \frac{1}{k!}\frac{d^k f}{dx^k}(x_0) = f(x_0+1)$ ?? – reuns Jun 13 '15 at 0:07