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 Theorem on tetration. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 06/09/2014, 02:47 PM (This post was last modified: 06/09/2014, 03:04 PM by JmsNxn.) Hey everyone. I thought I'd post this theorem, perhaps someone has some uses for it. Theorem: A.) If $F$ is holomorphic for $\Re(z) > -a-\epsilon>-1$ for some $a> 0$ and $F(z) < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ for $0 \le \alpha < \pi/2$ and $\rho \ge 0$. B.) for some $b > 1$ and $n \in \mathbb{N}$ we have $F(\log_b(1+n)) = F(1+n) - 1$ Then, for $\Re(z) > 1-a$ we have $F(\log_b(1+z)) = F(1+z) - 1$ Proof: Well this is rather easy: $\frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(1-\xi)x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(1+n)\frac{(-x)^n}{n!}$ Which follows by cauchy's residue formula and the bounds of F (the gamma function along with x small enough pulls the arc next to our line integral to zero at infinity). For those who don't see, $\G(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + \int_1^\infty e^{-t}t^{z-1}\,dt$ where the right term is entire in z and only contribute asymptotics, observe stirlings asymptotic formula $\G(z) \sim \sqrt{2\pi} z^{z-1/2}e^{-z}$ Therefore this holds. Now observe that by a similar argument: $\frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)F(\log_b(1-\xi))x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(\log_b(1+n))\frac{(-x)^n}{n!}= \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x)$ And of course, by another similar argument: $\frac{1}{2\pi i} \int_{a - i\infty}^{a + i\infty} \G(\xi)(F(1-\xi)-1)x^{-\xi}\,d\xi = \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x)$ Therefore since the kernel of this integral transform is zero (its a modified fourier transform). On the line $[a-i\infty,a+i\infty]$ we have $F(1-\xi) -1 = F(\log_b(1-\xi))$. Therefore since both functions are analytic we get the desired. $\box$ I'm wondering, does anyone see any uses for this? I know with some formal manipulation we can say that, if $G(a_n) = 1+n$ and $G(a_n-1) = \log_b(n+1)$ and $G$ is holo and is invertible which satisfies the bounds above. Then $b^{G(z)} = G(z+1)$ « Next Oldest | Next Newest »

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