Diferentiation and differential operators

[Jhenrique]

If the gradient of f is equal to differential of f wrt s: $$\vec{\nabla}f=\frac{df}{d\vec{s}}$$ so, what is the curl of f and the gradient of f in terms of fractional differentiation?

 

[Matterwave]

I'm not sure what you mean by "fractional differentiation" (wikipedia has a definition that seems very different than what you're talking about here), but these three operations are all related to the exterior derivative (if that's what you were talking about) $\bf{d}$.

The gradient
$$\nabla f = (\bf{d} f)^\sharp$$
The Curl
$$\nabla\times \bf{f}=[\star(\bf{d}\bf{f}^\flat)]^\sharp$$
The Divergence
$$\nabla\cdot \bf{f}=\star\bf{d}(\star \bf{f}^\flat)$$

Edit: I have no idea why the f is boldfaced in the gradient formula...and I don't know how to fix it. It should be non-boldfaced as it's a functinon.

 

[micromass]

Edit: I have no idea why the f is boldfaced in the gradient formula...and I don't know how to fix it. It should be non-boldfaced as it's a functinon.

Hmm, this is weird. It looks like a bug, thanks for bringing this to our attention.

 

[Matterwave]

Oh I should have explained my notation. The sharp and flat signs mean the musical isomorphisms which, given a metric, maps a one form to a vector and vice versa. The star is the Hodge dual operator which, again being metric dependent, maps a (k)-form to a (n-k) form where n is the dimension of the manifold.

 

[Jhenrique]

I'm not sure what you mean by "fractional differentiation" (wikipedia has a definition that seems very different than what you're talking about here), but these three operations are all related to the exterior derivative (if that's what you were talking about) $\bf{d}$.

Fractional differentiation means, for me, express a derivative in the form of a fraction.

 

[Matterwave]

Oh, so you mean like...

$$\nabla\cdot\vec{f}=\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$$

?

 

[Jhenrique]

Oh, so you mean like...

$$\nabla\cdot\vec{f}=\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$$

?

Yes and not. Yes because you used fraction, and not because your fraction is a scalar.

A Hessian of f, for example, can be write like:

$$Hf = \frac{d^2f}{d\vec{r}^2}$$

 

[Mark44]

Fractional differentiation means, for me, express a derivative in the form of a fraction.

Then your definition is at odds with how the term is already used, which has nothing to do with the derivative appearing in the form of a fraction. See this article on fractional derivatives.