If the derivative is written as shifts, can you relate it to the laplace/fourier tranform? I was wondering if there is a way to write the derivative as an exponential? This might sound crazy at first, but I recently came across this formula for the Taylor expansion in three dimensions:
$$\phi(\vec{x}+\vec{a})=\sum_{n=0}^{\infty}\frac{(\vec{a}\cdotp\vec{\nabla_{x}})^{n}}{n!}\Phi(\vec{x})=e^{\vec{a}\cdotp\vec{\nabla}}\Phi(\vec{x})$$
where, the hat arrows denote vectors. Found in: [http://www2.ph.ed.ac.uk/~rhorsley/SI09-10_t+f/lec09.pdf||http://www2.ph.ed.ac.uk/~rhorsley/SI09-10_t+f/lec09.pdf] I'd never seen a Taylor expansion written as an exponential. 
From a physics perspective, I think of it as propagating $\phi$
  from $x$
  to $x+a$
 . 
Anyway, I was wondering if you can write the derivative (or directional derivative) in a manner such as (for a one dimensional case):
$$\frac{\partial\Phi}{\partial x}=Lim_{\triangle x\rightarrow0},\frac{\Phi(x+\triangle x)-\Phi(x)}{\triangle x}=Lim_{\triangle x\rightarrow0}\frac{(e^{\triangle x\frac{\partial}{\partial x}}-1)}{\triangle x}\Phi(x)$$
$$=\frac{(e^{\partial x\frac{\partial}{\partial x}}-1)}{\partial x}\Phi(x)$$???
If correct, that implies:
$$\Longrightarrow\partial\Phi=(e^{\partial x\frac{\partial}{\partial x}}-1)\Phi(x)$$
And that made me wonder:
$$\Phi(\frac{\partial}{\partial x})=\int\partial\Phi=\int(e^{\partial x\frac{\partial}{\partial x}}-1)\Phi(x)$$
Of course this isn't correct as there's no differential under the latter integral, but it got me thinking. Since there are a ton (ok infinite) number of ways to write the derivative as a limit, I was playing with some and writing them as an integral as above, and all I can think is how much they remind me of laplace and fourier transforms (especially the latter if we move to general minkowski space). like in a fourier transform we have $\phi(x)\longrightarrow\tilde{\phi}(k)$
  where k is a wavenumber, whereas in this treatment we have (when actually done correctly) $\phi(x)\rightarrow\tilde{\phi}(\frac{\partial}{\partial x})$
 but $\frac{\partial}{\partial x}$
  can be related to a wavenumber. 
Anyway, can anyone point me in the right direction (a book perhaps) to learn more about this type of thing??? It would be immensely useful in solving a problem I'm working on.
thank you
 A: While this isn't a Full answer, it's gotten me started so I thought I'd share it:
(Note that for a general space or spacetime, $\nabla$
  will be the covariant derivative) 
If we're now considering a closed space of constant curvature R, things get more interesting and we have a periodicity requirement (this pertains to my personal problem, for the forum, let's just consider a periodic function) 
Consider now a function with with period R such that:
$$f(\overrightarrow{x}+\overrightarrow{R})=f(\overrightarrow{x})=e^{\overrightarrow{R}\cdot\nabla}f(\overrightarrow{x})=(\prod_{n}^{\infty}e^{\overrightarrow{nR}\cdot\nabla})f(\overrightarrow{x})$$
Which we may rewrite as:
$$f(\overrightarrow{x}+\overrightarrow{R})=e^{\overrightarrow{x}\cdot\nabla}f(\overrightarrow{R})=f(\overrightarrow{x})$$
since $\nabla$
  is a vector field of the tangent space, one can write it in terms of it's dual vector, which follow the relations (with $\overrightarrow{A}$
  tangent vector, and $\overrightarrow{a}$
  dual vector):
$$\overrightarrow{a}=\star\overrightarrow{A}\quad and\quad\overrightarrow{A}=\star a$$
where \star
  is the Hodge star operator (see https://en.wikipedia.org/wiki/Hodge_dual), which in three dimensions yields
$$\overrightarrow{A}=i\overrightarrow{a}\qquad and\qquad\overrightarrow{a}=-i\overrightarrow{A}$$
Then we obtain, denoting the dual vector k:
$$e^{\overrightarrow{R}\cdot\nabla}f(\overrightarrow{R})\Rightarrow\sum_{n}^{\infty}e^{\overrightarrow{R}\cdot i\overrightarrow{k_{n}}}f(\overrightarrow{x})=f(\overrightarrow{x})$$
Such that:$$k_{n}=\frac{2\pi n}{R}$$
we can write (in one dimension for simplicity):
$$f(\overrightarrow{x})=\sum_{n}^{\infty}e^{\overrightarrow{x}\cdot i\overrightarrow{k_{n}}}f(\overrightarrow{R})=\sum_{n}^{\infty}(cos(k_{n}x)+isin(k_{n}x))f(R)$$
One can write f(x) as a series expansion from this point, solving for the coefficients via orthogonality relations and then probably do something like take the limit as $R\rightarrow\infty$
  making k a continuous variable while simultaneously getting an integral via the consequent Riemann sum. This should be the Fourier transform. Note also that the inverse would necessarily have a negative sign from the hodge dual relations (it's worth learning about the Hodge duals if you haven't, I'm new to it myself) Sorry I'm tired.::
