# Tetration Forum

Full Version: An asymptotic expansion for \phi
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Hey everyone, so I've recently been trying to streamline some of my proofs in my paper. It's a bit of a clusterfuck at the moment; fairly disorganized. But I thought I'd point out an interesting asymptotic expansion I've been fiddling with.

If we define,

$
-h_n(x) = x - \log(x+1) + \log \log(x+2) +...+ (-1)^n\log^{\circ n}(x+n)\\
$

Then,

$
\lim_{x\to\infty} \frac{\phi(t+\pi i) - h_n(x)}{\log^{\circ n}(x+n)} = (-1)^{n}\\
$

Now this comes from the initial statement that, Sheldon noticed which I thought to be too good to be true,

$
\frac{\phi(x+\pi i)}{x} \to (-1)\\
$

This has me asking if there are asymptotic expansions elsewhere. And it led me to an algorithm which may help future work on $\phi$.

Suppose that,

$
\psi_y(t) = e^{-iy}\phi(t+iy)\\
\lim_{t\to\infty} \frac{\psi_y(t)}{A(t)} = 1\\
$

Then,

$
\frac{\psi_y(t+1)}{A(t+1)} = \frac{e^{t + e^{iy}\psi_y(t)}}{A(t+1)} \to 1\\
$

So we should expect,

$
t + e^{iy}\psi_y(t) = \log(A(t+1)) + o(\log(A(t+1)))\\
$

Which we can iterate to get successive asymptotic expansions. It may not look as nice for the case $y=\pi$ but I think this may be helpful for iterations. Where we can begin to  approximate successive iterations better. And we'll get something like a log asymptotic series.

If we continue with an assumption of superexponential,

$
\log(A(t+1))/A(t) \to \lambda \in\mathbb{C}\\
$

Then,

$
\psi_y(t) = \lambda e^{-iy}A(t) -e^{-iy}t-o(A(t))\\
$

And then,

$
\frac{\psi_y(t+2)}{A(t+2)} = \frac{e^{t+1+e^{iy}e^{t+e^{iy}\psi_y(t)}}}{A(t+2)} \to 1\\
$

meaning,

$
t+1+e^{iy}e^{t+e^{iy}\psi_y(t)} \sim \log A(t+2)\\
t+1+e^{iy}e^{\lambda A(t) + o(A(t))} \sim \log A(t+2)\\

$

This may be a nicer way of thinking about growth. I'm not sure yet. But with super exponentials the asymptotics behave properly. It should also say that exponential asymptotic growth is impossible. It's either dead in the water growth, or super-exponential growth; which means it's in for a penny or in for a dime.
ACK,

So this result is only true if $\phi(t+\pi i) / t \to -1$. The correct statement without this is,

$
\psi_m(t,x) = \Omega_{j=1}^m e^{t-j-x}\bullet x\\
$

Then,

$
\psi_m(t+m,h_m(t)) = t+m\\
$

Where since $h_m\to\infty$ we really can say much, unless $|h_m(t-m)|< M$ is bounded fixed $t$ and $m>0$. Which, is doubtful.

So damn close.