The "generator" of Taylor Series It's since a bit of time that I'm wondering if to ask this or not, but eventually I will, hoping it's not a stupid question.
Let's write the Taylor Series for a general function $f(x)$ around $x_0$:
$$\sum_{k = 0}^{+\infty} \frac{f^{k}(x_0)}{k!}(x-x_0)^k$$
Now let's call 
$$f^{k}(x_0) = \frac{d^k}{dx^k} f(x)\big|_{x = x_0} = D^k_{x_o} f(x)$$
The Taylor Series expression now reads
$$\sum_{k = 0}^{+\infty}\frac{D^k_{x_0}f(x)}{k!}(x-x_0)^k$$
Which I can write as
$$\sum_{k = 0}^{+\infty} \frac{(D_{x_0}\cdot (x-x_0))^k}{k!}\ f(x)$$
But wait: this is nothing but (operatorially speaking)
$$\large e^{(x-x_0)\ D_{x_0}}f(x)$$
That is
$$f(x_0) + (x-x_0)D_{x_0}f(x) + \frac{1}{2}(x-x_0)^2 D^2_{x_0}f(x) + \ldots$$
which brings me to conclude that Taylor Series is nothing but the series of the exponential of the derivative of a function around a point, applied to the initial function $f(x)$.


*

*Is that correct?

*Moreover, since I came to this result, what meaning (if any) could have to take another "generator" (I beg your pardon for the illicit use of this word) like $\sin$ or $\sinh$ or whatever?
Thank you so much!
 A: This can be given a sense in the theory of strongly continuous one parameter semi-groups of bounded operators in Banach and Hilbert spaces. Consider the translation operator
$$(T_t f)(x) = f(x + t)$$
This is for example an isometry in $L^2({\mathbb R})$, but other spaces are possible. $T_t$ is a one parameter group of bounded operators, that is to say
$$T_t T_s = T_{t+s}\qquad T_0 = I$$
This group is generated by an unbounded closed operator with a dense domain, defined by
$$A f = \lim_\limits{t\to 0} \frac{1}{t}(T_t f - T_0 f)$$
when the limit exists. Here the limit exists if $f\in H^1({\mathbb R})$ and
$$A f = \frac{d }{d x} f$$
Then the group can be identified with the exponential of the unbounded generator, that is to say
$$T_t = e^{t A} = e^{t\frac{d}{d x}}$$
Or, the translation operator is the exponential of the derivation operator.
For suitably smooth function, you can then justify the Taylor series by expanding the exponential series
$$f(\cdot + t) = T_t f = \sum_0^\infty \frac{t^k}{k!} \left(\frac{d}{d x}\right)^k f$$
