02/10/2023, 03:49 AM
Hey, Caleb; both of your solutions are unintegrable. And there is no interpolation for \(\frac{1}{n!}\). What happens is very different.
For \(A(z)\) to be a FRactional Calculus interpolation of \(a_n\); it must be that:
\[
A(z) = \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{n=0}^\infty a_n \frac{w^n}{n!}\\
\]
End of story; where in this space, the bounds are pretty much \(A(z)\) is holomorphic for at least \(\Re(z) \ge 0\) and satisfies \(A(z) = O(e^{\rho |\Re(z)| + \tau|\Im(z)|})\) for \(0 \le \tau < \pi/2\). The reason, you find the \(1/n!\) in this camp is a much more difficult idea.
\[
\Gamma(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + h(z)\\
\]
Where \(h\) is entire. So the reason you are seeing these factorials, is because you are seeing them as residues of the Gamma function. Not because you are flipping the Gamma function on it's head. I should've added, because I made a mistake, that the integral is in the opposite sign (man I am making too many sign mistakes lately
);
\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}\Gamma(z) \frac{d^{-z}}{dw^{-z}}\Big{|}_{w=0} f(w)\,dz
\]
Whereupon:
\[
f(w) = \sum_{n=0}^\infty f^{(n)}(0) \frac{(-w)^n}{n!}\\
\]
The \(1/n!\) you are seeing are actually the RESIDUES of the Gamma function. Not the gamma function itself.
If you want to instead write:
\[
\Gamma(z)\frac{d^{-z}}{dw^{-z}}\Big{|}_{w=0} f(w) = B(-z)\\
\]
Where \(c_n = \frac{f^{(n)}(0)(-1)^n}{n!}\):
\[
B(-z) = \sum_{n=0}^\infty c_n\frac{(-1)^n}{(n+z)} + h(z)\\
\]
Then the equation reads as:
\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}B(-z)\,dz
\]
And lastly, but not least; we can write it as Ramanujan would; which inspired Carlson:
\[
f(w) = \sum_{n=0}^\infty c_n (-w)^n\\
\]
\[
\int_0^\infty f(w)w^{z-1}\,dw = \frac{\pi C(-z)}{\sin \pi z}\\
\]
And:
\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}\frac{\pi C(-z)}{\sin \pi z}\,dz\\
\]
Where now, we solely ask that \(C(z) = O(e^{\rho|\Re(z)|+\kappa|\Im(z)|})\) but \(0 \le \kappa < \pi\), rather than the half. And \(C(n) = c_n\).
A lot of what's going on under the hood; when nature chooses the Gamma function; is this operation. You find it with Bernoulli numbers, all the way to zeta functions...
For \(A(z)\) to be a FRactional Calculus interpolation of \(a_n\); it must be that:
\[
A(z) = \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{n=0}^\infty a_n \frac{w^n}{n!}\\
\]
End of story; where in this space, the bounds are pretty much \(A(z)\) is holomorphic for at least \(\Re(z) \ge 0\) and satisfies \(A(z) = O(e^{\rho |\Re(z)| + \tau|\Im(z)|})\) for \(0 \le \tau < \pi/2\). The reason, you find the \(1/n!\) in this camp is a much more difficult idea.
\[
\Gamma(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + h(z)\\
\]
Where \(h\) is entire. So the reason you are seeing these factorials, is because you are seeing them as residues of the Gamma function. Not because you are flipping the Gamma function on it's head. I should've added, because I made a mistake, that the integral is in the opposite sign (man I am making too many sign mistakes lately
);\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}\Gamma(z) \frac{d^{-z}}{dw^{-z}}\Big{|}_{w=0} f(w)\,dz
\]
Whereupon:
\[
f(w) = \sum_{n=0}^\infty f^{(n)}(0) \frac{(-w)^n}{n!}\\
\]
The \(1/n!\) you are seeing are actually the RESIDUES of the Gamma function. Not the gamma function itself.
If you want to instead write:
\[
\Gamma(z)\frac{d^{-z}}{dw^{-z}}\Big{|}_{w=0} f(w) = B(-z)\\
\]
Where \(c_n = \frac{f^{(n)}(0)(-1)^n}{n!}\):
\[
B(-z) = \sum_{n=0}^\infty c_n\frac{(-1)^n}{(n+z)} + h(z)\\
\]
Then the equation reads as:
\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}B(-z)\,dz
\]
And lastly, but not least; we can write it as Ramanujan would; which inspired Carlson:
\[
f(w) = \sum_{n=0}^\infty c_n (-w)^n\\
\]
\[
\int_0^\infty f(w)w^{z-1}\,dw = \frac{\pi C(-z)}{\sin \pi z}\\
\]
And:
\[
f(w) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} w^{-z}\frac{\pi C(-z)}{\sin \pi z}\,dz\\
\]
Where now, we solely ask that \(C(z) = O(e^{\rho|\Re(z)|+\kappa|\Im(z)|})\) but \(0 \le \kappa < \pi\), rather than the half. And \(C(n) = c_n\).
A lot of what's going on under the hood; when nature chooses the Gamma function; is this operation. You find it with Bernoulli numbers, all the way to zeta functions...