There is ONE fractional derivative to be used for interpolation, as I've said it. It's nothing more than the Mellin transform.
\[
\frac{d^{-z}}{dw^{-z}}\vartheta(w) = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(w-x)x^{z-1}\,dx\\
\]
Which converges in a vertical strip \(0 \le \Re(z) \le b\). We assume that \(\vartheta\) is entire, and additionally has some kind of \(O(x^{-b})\) growth as \(x \to \infty\). This fractional derivative can be extended to \(-b \le \Re(z)\):
\[
\Gamma(-z)\frac{d^{z}}{dw^{z}}\vartheta(w) = \sum_{n=0}^\infty \vartheta^{(n)}(w) \frac{(-1)^n}{n!(z-n)} + \int_1^\infty \vartheta(w-x)x^{-z-1}\,dx\\
\]
This is absolutely a unique operator, so long as you assume that \(F(z) = \frac{d^{z}}{dw^{z}}\Big{|}_{w=0}\vartheta(w) \) is in the exponential space:
\[
|F(z)| \le Ce^{\rho|\Re(z)| + \tau|\Im(z)|}\,\,\text{for}\,\,\rho,\tau \in \mathbb{R}^+\,\,\tau < \pi/2\\
\]
There then exists a correspondence of functions \(\vartheta\) which produce \(F\)'s. What I have shown, is that certain functions produce \(\vartheta\), which then produce \(F\) again through this formula. It's nothing really more than Ramanujan's Master Theorem, described using Fractional calculus. THIS IS UNIQUE.
NOW, what I think you are getting at, is that there are uncountably many infinite fractional derivatives. This is true. We can iterate the derivative uncountably many ways. BUT! There is only one way to do this such that:
\[
\frac{d^z}{dw^z} e^w = e^w\\
\]
THERE'S ONLY ONE FRACTIONAL DERIVATIVE THAT DOES THIS!
It's known as the Riemann-Liouville, or the Exponential, fractional derivative. It's also known as the Weyl fractional derivative. IT IS UNIQUE.
So it is not arbitrary to interpolate sequences using fractional calculus, if you only use this operator. It's absolutely the exact opposite of arbitrary. It's unique. It either works or it doesn't. And if it works, it's unique.
I don't know what to say, tommy. You've never liked the fractional calculus approach. But you clearly also don't understand it. And I don't know what else to say.
\[
\frac{d^{-z}}{dw^{-z}}\vartheta(w) = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(w-x)x^{z-1}\,dx\\
\]
Which converges in a vertical strip \(0 \le \Re(z) \le b\). We assume that \(\vartheta\) is entire, and additionally has some kind of \(O(x^{-b})\) growth as \(x \to \infty\). This fractional derivative can be extended to \(-b \le \Re(z)\):
\[
\Gamma(-z)\frac{d^{z}}{dw^{z}}\vartheta(w) = \sum_{n=0}^\infty \vartheta^{(n)}(w) \frac{(-1)^n}{n!(z-n)} + \int_1^\infty \vartheta(w-x)x^{-z-1}\,dx\\
\]
This is absolutely a unique operator, so long as you assume that \(F(z) = \frac{d^{z}}{dw^{z}}\Big{|}_{w=0}\vartheta(w) \) is in the exponential space:
\[
|F(z)| \le Ce^{\rho|\Re(z)| + \tau|\Im(z)|}\,\,\text{for}\,\,\rho,\tau \in \mathbb{R}^+\,\,\tau < \pi/2\\
\]
There then exists a correspondence of functions \(\vartheta\) which produce \(F\)'s. What I have shown, is that certain functions produce \(\vartheta\), which then produce \(F\) again through this formula. It's nothing really more than Ramanujan's Master Theorem, described using Fractional calculus. THIS IS UNIQUE.
NOW, what I think you are getting at, is that there are uncountably many infinite fractional derivatives. This is true. We can iterate the derivative uncountably many ways. BUT! There is only one way to do this such that:
\[
\frac{d^z}{dw^z} e^w = e^w\\
\]
THERE'S ONLY ONE FRACTIONAL DERIVATIVE THAT DOES THIS!
It's known as the Riemann-Liouville, or the Exponential, fractional derivative. It's also known as the Weyl fractional derivative. IT IS UNIQUE.
So it is not arbitrary to interpolate sequences using fractional calculus, if you only use this operator. It's absolutely the exact opposite of arbitrary. It's unique. It either works or it doesn't. And if it works, it's unique.
I don't know what to say, tommy. You've never liked the fractional calculus approach. But you clearly also don't understand it. And I don't know what else to say.