TITLE: Exponential of the differential operator
QUESTION [19 upvotes]: I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and
$$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$$
(ref. Dan Piponi's answer)
So I was curious as to whether the differential equation
$$\frac{df(t)}{dt} = e^Df(t) = f(t+1)$$
has any solutions apart from $f = 0$?

REPLY [9 votes]: Let us elaborate on the differential operator $ {\rm e}^{aD} $. It is defined by the power series $ \sum_{k=0}^{\infty}\frac{D^k}{k!} $. We would like to know its effect on the function $f(x)$. First let's see its effect on the function $x^m$. 
$$ {\rm e}^{ aD } x^m =  \sum_{k=0}^{\infty} \frac{a^kD^k}{k!} x^m = \sum_{k=0}^{\infty}\frac{\Gamma(m+1)}{\Gamma(k-m+1)k!} a^k x^{m-k} = \sum_{k=0}^{\infty}\frac{m!}{(k-m)!k!} a^k x^{m-k}$$
$$= \sum_{k=0}^{\infty}{m \choose k}  a^k x^{m-k} = \sum_{k=0}^{m}{m \choose k} a^k x^{m-k} = (x+a)^m\,, $$
by binomial theorem. That means the differential operator works as a shift operator that shift the argument by $a$.
Now, we can use this result to see the effect on a function $f(x)$. If the function $f(x)$ has a power series at $x=0$, then
$$ {\rm e}^{aD} \sum_{k=0}^{\infty} f_k x^k = \sum_{k=0}^{\infty} f_k {\rm e}^{aD} x^k = \sum_{k=0}^{\infty} f_k (x+a)^k = f(x+a)\,.$$
So, the general effect of the differential operator ${\rm e}^{aD}$ on the function $f(x)$ is nothing but shifts the argument by $a$. In your case $a=1$.