# Is there any meaning to this “Super Derivative” operation I invented?

Does anyone know anything about the following “super-derivative” operation? I just made this up so I don’t know where to look, but it appears to have very meaningful properties. An answer to this question could be a reference and explanation, or known similar idea/name, or just any interesting properties or corollaries you can see from the definition here? Is there perhaps a better definition than the one I am using? What is your intuition for what the operator is doing (i.e. is it still in any sense a gradient)? Is there a way to separate the log part out, or remove it? Or is that an essential feature?

Definition: I’m using the word “super-derivative” but that is a made-up name. Define the “super-derivative”, operator $$SαxS_x^{\alpha}$$, about $$α\alpha$$, using the derivative type limit equation on the fractional derivative operator $$DαxD_x^\alpha$$
$$Sαx=limh→0Dα+hx−Dαxh S_x^{\alpha} = \lim_{h \to 0} \frac{D^{\alpha+h}_x-D^{\alpha}_x}{h}$$
then for a function
$$S_x^{\alpha} f(x) = \lim_{h \to 0} \frac{D^{\alpha+h}_xf(x)-D^{\alpha}_x f(x)}{h} S_x^{\alpha} f(x) = \lim_{h \to 0} \frac{D^{\alpha+h}_xf(x)-D^{\alpha}_x f(x)}{h}$$
for example, the [Riemann-Liouville, see appendix] fractional derivative of a power function is
$$D_x^\alpha x^k = \frac{\Gamma(k+1)}{\Gamma(k-\alpha+1)}x^{k-\alpha} D_x^\alpha x^k = \frac{\Gamma(k+1)}{\Gamma(k-\alpha+1)}x^{k-\alpha}$$
and apparently
$$S_x^{\alpha} x^k = \frac{\Gamma (k+1) x^{k-\alpha} (\psi ^{(0)}(-\alpha+k+1) – \log (x))}{\Gamma (-\alpha+k+1)} = (\psi ^{(0)}(-\alpha+k+1) – \log (x)) D_x^\alpha x^k S_x^{\alpha} x^k = \frac{\Gamma (k+1) x^{k-\alpha} (\psi ^{(0)}(-\alpha+k+1) - \log (x))}{\Gamma (-\alpha+k+1)} = (\psi ^{(0)}(-\alpha+k+1) - \log (x)) D_x^\alpha x^k$$
a nice example of this, the super-derivative of $$xx$$ at $$\alpha=1\alpha=1$$ is $$-\gamma – \log(x)-\gamma - \log(x)$$, which turns up commonly. I’m wondering if this could be used to describe the series expansions of certain functions that have log or $$\gamma\gamma$$ terms, e.g. BesselK functions, or the Gamma function.

Potential relation to Bessel functions: For example, a fundamental function with this kind of series, (the inverse Mellin transform of $$\Gamma(s)^2\Gamma(s)^2$$), is $$2 K_0(2 \sqrt{x})2 K_0(2 \sqrt{x})$$ with
$$2 K_0(2 \sqrt{x}) = (-\log (x)-2 \gamma )+x (-\log (x)-2 \gamma +2)+\frac{1}{4} x^2 (-\log (x)-2 \gamma +3)+\\ +\frac{1}{108} x^3 (-3 \log (x)-6 \gamma +11)+\frac{x^4 (-6 \log (x)-12 \gamma +25)}{3456}+O\left(x^5\right) 2 K_0(2 \sqrt{x}) = (-\log (x)-2 \gamma )+x (-\log (x)-2 \gamma +2)+\frac{1}{4} x^2 (-\log (x)-2 \gamma +3)+\\ +\frac{1}{108} x^3 (-3 \log (x)-6 \gamma +11)+\frac{x^4 (-6 \log (x)-12 \gamma +25)}{3456}+O\left(x^5\right)$$
in the end, taking the super-derivative of polynomials and matching coefficients we find
$$S_x^1[2 \sqrt{x}I_1(2\sqrt{x})] + I_0(2 \sqrt{x})\log(x) = 2K_0(2 \sqrt{x}) S_x^1[2 \sqrt{x}I_1(2\sqrt{x})] + I_0(2 \sqrt{x})\log(x) = 2K_0(2 \sqrt{x})$$
which can also potentially be written in terms of linear operators as
$$[2 S_x x D_x + \log(x)]I_0(2 \sqrt{x}) = 2K_0(2 \sqrt{x}) [2 S_x x D_x + \log(x)]I_0(2 \sqrt{x}) = 2K_0(2 \sqrt{x})$$
likewise
$$[2 S_x x D_x – \log(x)]J_0(2 \sqrt{x}) = \pi Y_0(2 \sqrt{x}) [2 S_x x D_x - \log(x)]J_0(2 \sqrt{x}) = \pi Y_0(2 \sqrt{x})$$
I like this because it’s similar to an eigensystem, but the eigenfunctions swap over.

Gamma Function: We can potentially define higher-order derivatives, for example
$$(S_x^{\alpha})^2 = \lim_{h \to 0} \frac{D^{\alpha+h}_x-2 D^{\alpha}_x + D^{\alpha-h}_x}{h^2} (S_x^{\alpha})^2 = \lim_{h \to 0} \frac{D^{\alpha+h}_x-2 D^{\alpha}_x + D^{\alpha-h}_x}{h^2}$$
and
$$(S_x^{\alpha})^3 = \lim_{h \to 0} \frac{D^{\alpha+3h}_x-3 D^{\alpha+2h}_x + 3 D^{\alpha+h}_x – D^{\alpha}_x}{h^3} (S_x^{\alpha})^3 = \lim_{h \to 0} \frac{D^{\alpha+3h}_x-3 D^{\alpha+2h}_x + 3 D^{\alpha+h}_x - D^{\alpha}_x}{h^3}$$

this would be needed if there was any hope of explaining the series
$$\Gamma(x) = \frac{1}{x}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) x+\frac{1}{6} x^2 \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)+ \\+\frac{1}{24} x^3 \left(\gamma ^4+\gamma ^2 \pi ^2+\frac{3 \pi ^4}{20}-4 \gamma \psi ^{(2)}(1)\right)+O\left(x^4\right) \Gamma(x) = \frac{1}{x}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) x+\frac{1}{6} x^2 \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)+ \\+\frac{1}{24} x^3 \left(\gamma ^4+\gamma ^2 \pi ^2+\frac{3 \pi ^4}{20}-4 \gamma \psi ^{(2)}(1)\right)+O\left(x^4\right)$$
using the ‘super-derivative’. This appears to be
$$\Gamma(x) = [(S^1_x)^0 x]_{x=1} x^{-1} + [(S^1_x)^1 x]_{x=1} x + \frac{1}{2}[(S^1_x)^2 x]_{x=1} x^2 + \frac{1}{6} [(S^1_x)^3 x]_{x=1} x^3 + \cdots \Gamma(x) = [(S^1_x)^0 x]_{x=1} x^{-1} + [(S^1_x)^1 x]_{x=1} x + \frac{1}{2}[(S^1_x)^2 x]_{x=1} x^2 + \frac{1}{6} [(S^1_x)^3 x]_{x=1} x^3 + \cdots$$
so one could postulate
$$\Gamma(x) = \frac{1}{x}\sum_{k=0}^\infty \frac{1}{k!}[(S^1_x)^k x]_{x=1} x^{k} \Gamma(x) = \frac{1}{x}\sum_{k=0}^\infty \frac{1}{k!}[(S^1_x)^k x]_{x=1} x^{k}$$
which I think is quite beautiful.

Appendix: I used the following definition for the fractional derivative:
$$D_x^\alpha f(x) = \frac{1}{\Gamma(-\alpha)}\int_0^x (x-t)^{-\alpha-1} f(t) \; dt D_x^\alpha f(x) = \frac{1}{\Gamma(-\alpha)}\int_0^x (x-t)^{-\alpha-1} f(t) \; dt$$
implemented for example by the Wolfram Mathematica code found here

FractionalD[\[Alpha]_, f_, x_, opts___] :=
Integrate[(x - t)^(-\[Alpha] - 1) (f /. x -> t), {t, 0, x},
opts, GenerateConditions -> False]/Gamma[-\[Alpha]]

FractionalD[\[Alpha]_?Positive, f_, x_, opts___] :=  Module[
{m = Ceiling[\[Alpha]]},
If[\[Alpha] \[Element] Integers,
D[f, {x, \[Alpha]}],
D[FractionalD[-(m - \[Alpha]), f, x, opts], {x, m}]
]
]


I’m happy to hear more about other definitions for the fractional operators, and whether they are more suitable.

I’ve thought about this for a few days now, I didn’t originally intend to answer my own question but it seems best to write this as an answer rather than add to the question. I think there is nice interpretation in the following:
$$f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h} f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h}$$
also consider the Abel shift operator
$$e^{h D_x}f(x) = f(x+h) e^{h D_x}f(x) = f(x+h)$$
from the limit form of the derivative we have (in the sense of an operator)
$$D_x = \lim_{h \to 0} \frac{e^{h D_x}-e^{0 D_x}}{h} = \lim_{h \to 0} \frac{e^{h D_x}-1}{h} D_x = \lim_{h \to 0} \frac{e^{h D_x}-e^{0 D_x}}{h} = \lim_{h \to 0} \frac{e^{h D_x}-1}{h}$$
now we can also manipulate the first equation to get
$$\log f(x) = \lim_{h \to 0} \frac{f^h(x)-1}{h} \log f(x) = \lim_{h \to 0} \frac{f^h(x)-1}{h}$$
so by (a very fuzzy) extrapolation, we might have
$$\log(D_x) = \lim_{h \to 0} \frac{D_x^h-1}{h} \log(D_x) = \lim_{h \to 0} \frac{D_x^h-1}{h}$$
and applying that to a function we now get
$$\log(D_x) f(x) = \lim_{h \to 0} \frac{D_x^h f(x)-f(x)}{h} \log(D_x) f(x) = \lim_{h \to 0} \frac{D_x^h f(x)-f(x)}{h}$$
which is the $$\alpha = 0\alpha = 0$$ case of the ‘super-derivative’. So one interpretation of this case is the logarithm of the derivative?
If we apply the log-derivative to a fractional derivative then we have
$$\log(D_x) D^\alpha_x f(x) = \lim_{h \to 0} \frac{D_x^h D^\alpha_x f(x)-D^\alpha_x f(x)}{h} \log(D_x) D^\alpha_x f(x) = \lim_{h \to 0} \frac{D_x^h D^\alpha_x f(x)-D^\alpha_x f(x)}{h}$$
there might be a question of the validity of $$D_x^h D^\alpha_x = D_x^{\alpha+h}D_x^h D^\alpha_x = D_x^{\alpha+h}$$ which I believe isn’t always true for fractional derivatives.

This interpretation would explain the $$\log(x)\log(x)$$ type terms arising in the series above. I’d be interested to see if anyone has any comments on this? I’d love to see other similar interpretations or developments on this. What are the eigenfunctions for the $$\log D_x\log D_x$$ operator for example? Can we form meaningful differential equations?

Edit: For some functions I have tried we do have the expected property
$$n \log(D_x) f(x) = \log(D_x^n) f(x) n \log(D_x) f(x) = \log(D_x^n) f(x)$$
with
$$\log(D_x^n) f(x) = \lim_{h \to 0} \frac{D_x^{n h} f(x)-f(x)}{h} \log(D_x^n) f(x) = \lim_{h \to 0} \frac{D_x^{n h} f(x)-f(x)}{h}$$