08/11/2022, 12:07 AM
(08/10/2022, 11:07 AM)Leo.W Wrote: This extension problem can be solved striaghtly if you'd take analytic methods, or if you happen to know some higher-level functions.
First we define a function as a semi-solution:\(f(z)=\frac{\Gamma(z+1)^2}{z}\), or you can call it "a bounce of height z" if you wish, then we extend this hyper bouncing or whatever-you-like-to-call factorial \(z\Lambda=f(1)f(2)f(3)\dots f(z)=\prod_{i=1}^{z}{f(i)}\)
Before so long we have already studied summations and products and their relationship, thus we transform the function by this equation \(\log(z\Lambda)=\log(f(1))+\log(f(2))+\log(f(3))+\dots+\log(f(z))=\sum_{i=1}^{z}{\log(f(i))}\)
It happened that since \(\log(f(i))=2\log(\Gamma(i+1))-\log(i)\) (not considering multivalued-ity), this sum can be written in closed form with special functions:
\[\log(z\Lambda)=\sum_{i=1}^{z}{\log(f(i))}=2\log(G(z+2))-\log(\Gamma(z+1))\]
where \(G\) denotes Barnes' G function, or more explicitly it's represented by \(G(z)=\Gamma(z)^{z-1}e^{\frac{z\log(2\pi)+z(1-z)}{2}-\phi^{(-2)}(z)}\) where \(\phi^{(-2)}(z)\) is polygamma function and defined originally by the recurrence \(G(z+1)=G(z)\Gamma(z)\). And all used functions are "entire". This can be more presented with antiderivatives and derivatives of Zeta function, if desired.
Yes, that's the bouncing Factorial. Thanks for doing the work I was too lazy to do, lol.
But the bouncing HYPER factorial. Get outta here, that's next century's problem, lmao.

