Derivatives and integrals, normal and fractional, and their explanations and relations Assuming, naively, that one acquires the nth derivative of a function by repeatedly differentiating and finding a pattern. Thus one gets $f^{(n)}(x)=g(x,n)$. I have a few questions about this situation.
The first derivative at a point is the slope of the line tangent to the curve at the point. The second derivative measures the concavity of the graph of a function. Are there corresponding "explanations" for higher order derivatives?
The first integral measures the area between the curve and the x-axis. Are there similar "explanations" for higher integrals?
If one has the function $g(x,n)$, does it make sense to put a non-integer real number or imaginary number on the place of $n$? I know there exists a rigorous field of fractional calculus, but I know next to nothing about it.
If one has  $g(x,n)$, when will $g(x,0)=f(x)$? How about when will $g(x,-1)=\int f(x) dx ?$
If one replaces $n$ in $g(x,n)$ with a fraction, one will get a proper function. What is this function, what does it "mean", does it have an "explanation" similar to first or second derivative?
 A: Firstly, with higher order derivatives, we have a thing known as Taylor's theorem, and using your notation, Taylor's theorem says that for some $|x-x_0|<r$, we have
$$f(x)=\sum_{n=0}^\infty\frac{g(x,n)}{n!}(x-x_0)^n$$
You could imagine it takes into consideration that each higher order derivative has an increasing less affect near the point $x_0$, which is accounted by the $n!$ in the denominator, while the $(x-x_0)^n$ accounts for the dominating behavior of higher order derivatives as we get farther from $x_0$.
To answer the second question, let us first define $g(x,-n)$ to be the $n$th anti-derivative of $f(x)$.  We have a formula that you might want to say is the above, except that it has the power to consider negative $n$ values, and hence, negative values in $g(x,n)$.  Naively put, we almost have
$$f(x)\stackrel?=\sum_{n=-\infty}^\infty\frac{g(x,n)}{n!}(x-x_0)^n$$
However, this is a confusing thing because it does not take into account the affects of constants of integration.  Instead, we have
$$f(x)=\sum_{n=-\infty}^\infty a_n(x-x_0)^n$$
$$a_n= \frac1{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz.$$
Which, whenever $n\in\mathbb N$, reduces down to $g(x,n)$.  On the contrary, this, as one may realize, can lead to a good generalized fractional derivative, that being the Riemann-Louisville definition.  It is given as follows:
$$g(x,n)=\frac1{\Gamma(\lceil n\rceil-n)}\frac{d^{\lceil n\rceil}}{dx^{\lceil n\rceil}}\int_c^xf(y)(x-y)^{\lceil n\rceil-n-1}dy$$
If $n<0$ and $k=-n$, it can be more simply rewritten as
$$g(x,k)=\frac1{\Gamma(k)}\int_c^xf(y)(x-y)^{k-1}dy$$
Interestingly, this may be proven to be true by induction.  Also, the lower bound to the integral, $c$, sort of represents a constant of integration.
As to how I interpret it, it is merely something that satisfies the following basic operations:
$D^aD^bf(x)=D^{a+b}f(x)$
$D^1f(x)=f'(x)$
$\lim_{a\to b}D^af(x)=D^bf(x)$
though I haven't much idea how else I would interpret it.
