[2014] Beyond Gamma and Barnes-G tommy1729 Ultimate Fellow Posts: 1,906 Threads: 409 Joined: Feb 2009 12/28/2014, 05:04 PM Most of you here are familiar with the Gamma function , Barnes-G function , the K-function , the double gamma function etc. But in the spirit of my generalized distributive law / commutative assosiative hyperoperators I was wondering about f(z+1) = f(z)^ln(z) It makes sense afterall : f1(z+1) = z + f1(z) leads to the triangular numbers. f2(z+1) = z f2(z) leads to the gamma function. f3(z+1) = f3(z)^ln(z) Notice f_n(z) = exp^[n]( ln^[n]f_n(z) + ln^[n](z) ) So is there an integral representation for f3 ? How does it look like ? Analogues of Bohr-Mullerup etc ? Hence the connection to the hyperoperators. regards tommy1729 MphLee Long Time Fellow Posts: 374 Threads: 30 Joined: May 2013 12/28/2014, 05:48 PM (This post was last modified: 12/29/2014, 03:06 PM by MphLee.) Mhh maybe I have made some errors but the solution of the eqation $f(x+1)=f(x)^{ln(x)}$ seems to have some problems in the first values... that should make it not determinate for natural values Let assume that $f(0)=\beta\gt 1$ then $f(1)=\beta^{ln(0)}=\beta^{-\infty}=0$ because of the limit $f(2)=0^{ln(1)}=0^0$ is not determinate (multivalued?) I'm not sure how to continue in this case... maybe we can start by fixing the value of $f(2)=\gamma$ and continue with the recursion $f(3)=\gamma^{ln(2)}$ $f(4)=\gamma^{ln(2)ln(3)}$ ... $f(2+n+1)=\gamma^{{\Pi}_{i=2}^{2+n} ln(i)}$ So actually the solution is of the form for some $\gamma$ $f(z+1)=\gamma^{{\Pi}_{i=2}^{z} ln(i)}$? ------ Btw I see the relation with the Bennet's commutative Hyperoperations (denote it by $a\odot^e_i b=a\odot_i b$)! $f_i(z+1)=f_i(z)\odot_i z$ This is like a kind of "Hyper-factorial" that uses the Bennet's Hyperops.. -------------- Edit: I think that the correct forumla is something of the form Given $f(z_0)=\alpha \neq 0$ or $\neq 1$ $f(z_0+n+1)=\exp_\alpha ({\Pi}_{i=0}^{n} ln(z_0+i))=\alpha^{{\Pi}_{i=0}^{n} ln(z_0+i)}$ MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ « Next Oldest | Next Newest »

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