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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

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    $\begingroup$ In the theory of distributions, the derivative is commonly expressed as a convolution of the function with the derivative of the Dirac delta, i.e. the original of $s$. $\endgroup$ – Yves Daoust May 10 '16 at 7:56
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    $\begingroup$ @YvesDaoust Thank you for your comment, after looking into it, I was essentially trying to derive the distributional derivative haha! Looks like I need to learn this(: $\endgroup$ – R. Rankin May 10 '16 at 23:17
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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.::

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  • $\begingroup$ Apologies sums should be large pi (multiplications) which will turn into a sum, will fix in the morning (: $\endgroup$ – R. Rankin May 10 '16 at 7:43

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