# Tetration Forum

Full Version: Theorem on tetration.
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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 $b^{G(z)} = G(z+1)$