I read that $e^\mathrm{D}f(x)=f(x+1)$, were $\mathrm{D}$ is the differential operator and $f(x)$ is an analytical function.
I tried writing out the definitions as I'd expected them to be (I haven't read much about operator theory):
$$e^\mathrm{D}=\sum_{n=0}^\infty \frac{\mathrm{D}^n}{n!}$$ and $$f(x)=\sum_{m=0}^\infty \frac{\mathrm{D}^m f(x)|_{x=0}}{m!},$$ so $$e^\mathrm{D}f(x)=\sum_{n=0}^\infty \frac{\mathrm{D}^n}{k!}\left(\sum_{m=0}^\infty \frac{\mathrm{D}^m f(x)|_{x=0}}{m!}\right).$$
How do I proceed from here?
Thanks.
Call $\epsilon = 1 / n$, for some integer $n$, then
$$ e^D = (e^{\epsilon D})^n \tag{1} $$
Now, note that for large $n$
$$ f(x + \epsilon) \approx f(x) + \epsilon f'(x) \tag{2} $$
therefore
$$ e^{\epsilon D}f(x) = \left(1 + \epsilon D + \cdots\right)f(x) \approx f(x) + \epsilon Df(x) = f(x) + \epsilon f'(x) \stackrel{(2)}{=} f(x + \epsilon) \tag{3} $$
Going back to Eq. (1)
$$ e^Df(x) = \lim_{n\to \infty}(e^{\epsilon D})^nf(x) = \lim_{n\to\infty}f(x + n\epsilon ) = f(x + 1) $$