Fractional Integration tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 02/09/2023, 12:46 PM (02/09/2023, 08:09 AM)Caleb Wrote: (02/08/2023, 06:03 AM)JmsNxn Wrote: These are questions I have asked myself for 12 years   STUDY RAMANUJAN'S MASTER THEOREM AND THE MELLIN TRANSFORM! The answer is so banal, that you may not like it, but I'll give a shot. I'll start with, why does math choose $\Gamma(z+1)$ from $n!$. Because: $\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(z) x^{-z} = e^{-x}\\$ And: $\int_0^\infty e^{-x} x^{z-1} \, dx = \Gamma(z)\\$ Where Ramanujan showed in a rough handed way; where $H$ is a linear operator (or a function): $\int_0^\infty e^{-Hx} x^{z-1}\,dx = \Gamma(z) H(z)\\$ Where then: $e^{-Hx} = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(z) H(z) x^{-z}\,dz\\$ And the ONLY SOLUTIONS WHICH WORK ARE THE ONES WHICH HAVE CONVERGENT INTEGRALS! This relates deeply to $\sin(\pi z)$ and Carlson's theorem. Carlson's theorem is actually, in an historical perspective, a manner of justifying Ramanujan's Master Theorem in a much much much broader sense. He wanted to justify things Ramanujan just took for granted, lmao. In perfect Ramanujan form When I think about this some more, I think  Quote:"And the ONLY SOLUTIONS WHICH WORK ARE THE ONES WHICH HAVE CONVERGENT INTEGRALS!" Is not a principal which seems to work in general. If $\Gamma$ is defined in terms of unique thing that has convergent integrals, then $\frac{1}{n!}$ grows pretty large in the imaginary direction, but $\frac{1}{n! + 2\sin(\pi z)}$ is quite small and so it will lead to convergent integrals. Are there some other conditions you have in mind in general?  To be more particular, if I have a specific sequence $a_n$ on the integers, what conditions are you proposing determines the unique analytical continuation? Or perhaps, if I have a sequence $a_n$ and an analytical continuation $A(z)$, what specific conditions determine if this is the right analytical continuation? In the case of $a_n = \frac{1}{n!}$, what do your conditions say is the right answer? (Or, perhaps is more information needed besides just the sequence to determine the right analytical continuation?) There is no unique bijective association with a sequence and a function that solves all matters. You could ask it under conditions and arrive at uniqueness conditions such as Carlsons theorem or other theorems. Or you could ask the function to be entire. Or to have a certain integral transform like a Bernstein function. But it is like interpolation , there is no unique function. And hence no unique analytic continuation because it depends on what type of function you associate. And there is no unique fractional integral or integration constant for the same reasons. In general we use properties like the y the derivative of exp(x) is exp(x) and from there we can define fractional integration. Series expansion and distributive property are then key. Your questions appear to me to be too general to give a full or better answer. But maybe I missed something in your questions ? Regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 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... « Next Oldest | Next Newest »

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