I was considering the notion of taylor series which posit that the sum
$$ \sum_{i=0}^{\infty} \frac{1}{i!} a_ix^i $$
Where:
$$ a_i = \frac{d^if}{dx^i}_{x= a} $$
Converge to the function f in a disc of radius $0 \le R \le \infty$
What is was considering was the fractional derivative such as
$$ \frac{d^{\frac{1}{2}}f}{dx^{\frac{1}{2}}} $$
Which is well defined using
https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration
For polynomials.
I wanted to use the half derivative to build taylor-series like representations for functions involving in this case terms of the form $x^{\frac{i}{2}} $ as opposed to just whole valued powers but it is not immediately clear how to do this. If I naively write:
$$ \sum_{i=0}^{\infty} \frac{1}{\Gamma \left( \frac{i}{2}+1\right)}a_ix^\frac{i}{2} $$
Where:
$$ a_i = \frac{d^\frac{i}{2}f}{dx^\frac{i}{2}}_{x= a} $$
Then it must be the case that
$$ \sum_{i=0}^{\infty} \frac{1}{\Gamma \left( i + \frac{1}{2}\right) }a_{2i}x^{i+\frac{1}{2}} = 0 $$
Over the same disc with radius of convergence R, that the clean Taylor series converged over, or else Taylor's theorem wouldn't be correct.
This statement follows from the fact that
$$ \frac{d^{\frac{1}{2}}}{dx^\frac{1}{2}} \frac{d^{\frac{1}{2}}f}{dx^\frac{1}{2}} = \frac{df}{dx} $$
Thus the series just mentioned contains EVERY Taylor term in it already along with a bunch of extra fractional power terms. IF (and i'm not sure if this is true) the series above too converges to f, for some region G that overlaps with the region of radius R mentioned above. Then it must be the case that all the fractional power terms must converge to 0.
So that leaves me mystified since then I'm no longer sure how to generate proper series using the fractional derivative.