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Questions

$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$

Also obeys (see background for argument):

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} dk = \frac{\partial S(n,x)}{\partial x}$$

Can this be used in the Borel summation sense for divergent series? If so, when can it be used for analytical continuity (convergence issues)? Is it useful(intuitively I feel it should be more powerful than Borel summation)? In the following heuristic sense:

$$ \kappa = \sum_{n=1}^\infty a_n = \sum_n a_n \frac{\frac{\partial S(n,x)}{\partial x}}{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} } $$

Using the L.H.S of the first equation:

$$ \kappa = \frac{1}{2 \pi i} \sum_n a_n \frac{\oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} dk }{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} } $$

Swapping order of summation and contour integral:

$$ \kappa =^! \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_n \frac{ a_n \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} }{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} }dk $$

How can I make this rigorous?

Background

I've been recently studying the following series:

$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$

Where the $n$'th term is raising the $x$ exponentially $n$ number of times.

$$ b_n(x) = \underbrace{e^{e^{e^{\dots}x}}}_{\text{$n$ times exponentially raised}} $$

$n$ number of times.

Hence, we notice:

$$ e^{S(r,x)} = \frac{\partial b_{r+1}(x)}{\partial x}$$

Summing both sides and defining $S(0,x) \equiv 0$:

$$ \sum_{r=0}^n e^{S(r,x)} = \sum_{r=1}^{n+1} \frac{\partial b_{r}(x)}{\partial x} $$

Hence, we get:

$$ \sum_{r=0}^n e^{S(r,x)} = \frac{\partial S(n+1,x)}{\partial x}$$

Rewriting the R.H.S using complex analysis as a contour integral over the whole complex plane:

$$\frac{1}{2 \pi i} \oint \sum_{r=0}^n \frac{e^{S(k,x)}}{k-r}dk = \frac{\partial S(n+1,x)}{\partial x}$$

Taking $e^{S(k,x)}$ common:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_{r=0}^n \frac{1}{k-r}dk = \frac{\partial S(n+1,x)}{\partial x}$$

Further using $d \ln x = dx/x$

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_{r=0}^n d \ln({k-r}) = \frac{\partial S(n+1,x)}{\partial x}$$

Rewriting as factorial:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{(k)!}{(k-n-1)!})}{\partial k} dk = \frac{\partial S(n+1,x)}{\partial x}$$

Analytically continuing $k!$ using the gamma function:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n-1)} dt})}{\partial k} dk = \frac{\partial S(n+1,x)}{\partial x}$$

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    Reason for downvote? – More Anonymous Jul 01 '18 at 13:46
  • Hmm, assume $t$ is a fixpoint (thus $t$ is complex) then $S(x)|_{x=t}$ should possibly equal $t+t+t+t+..$ and then $t \zeta(0)$ . Is there any chance with your ansatz to say something about this? – Gottfried Helms May 11 '20 at 09:31
  • In extension of the idea in the previous comment: I'm trying to find a periodical point of period $2$ where the two points $(t_1,t_2)$ have alternate signs, such that the series $S(x)$ becomes a Cesaro-summable alternating series $S(t_1)=t_1+t_2+t_1+...$- and then to see, what your method gives for such an $x=t_1$ or $x=t_2$. Unfortunately the found pairs of $2$-periodic points so far have not alternate signs so far, only one pair has (at least) conjugate values (the imagniary compnent has alternate signs) – Gottfried Helms May 11 '20 at 10:02

1 Answers1

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(This is no answer but continuation of my earlier comments, the answer-box is used because of the picture that I want to show).

Here is a small image on the existence and some relations of period-2 -points, in the hope to get an idea how to find period-2 points of alternate signs, following the rule:
$$e^{p_1} \to p_2 ; e^{p_2} \to p_1; e^{p_1} \to p_2; \cdots \\ S(p_1) = p_1 + p_2 + p_1 + p_2 + \cdots = ? $$

picture Observations: there are "a lot of" (infinitely many) pairs of 2-periodic points. They occur even in subsets, each which can be described as sequence of pairs (here colored with different colours, pairs of one subset in the same color).
The subsets with color yellow, green and blue have sequences of pairs which converge to 1-periodic (fixed-) points!

But I still didn't find pairs which alternating signs in the real components...