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Notation

We employ the following notation:

$$ a_1 = e^{x} $$

$$ a_2 = e^{e^{x}} $$

$$ a_3 = e^{e^{e^{x}}} $$

$$ a_n = e^{\vdots^{e^{x}}}$$

We also use define $c$ by:

$$ e^c =c$$

Motivation

Let us try to define $a_{1.5}$

Possible method?

Now, consider $a_n$:

$$ a_n = a_n \implies \lim_{x \to c} a_n = c$$

Differentiating:

$$ a_n' = a_1 a_2 \dots a_n \implies \lim_{x \to c} a_n' = c^n $$

Again differentiating:

$$ a_n'' = a_n'(1+ a_1' + a_2' + \dots + a_{n-1}') \implies \lim_{x \to c} a_n'' = c^n(1 + c + c^2 + \dots c^{n-1}) = \frac{c^n - c^{2n}}{1 - c} $$

Again differentiating:

$$ a_n''' = a_n''(1+ a_1' + a_2' + \dots + a_{n-1}') + a_n'( a_1'' + a_2'' + \dots + a_{n-1}'')$$ $$ \implies \lim_{x \to c} a_n''' = \frac{c^n(1-c^n)^2}{(1 - c)^2} + c^n \frac{(c - c^{n})}{(1 - c)^2} + \frac{c^n(c^2 - c^{2n})}{(1 - c)(1-c^2)}$$

Now, using Taylor expansion:

$$ a_n(z) = \lim_{z \to c} a_n + (z-c) \lim_{z \to c} a_n' + \frac{(z-c)^2}{2!} \lim_{z \to c} a_n'' + \frac{(z-c)^3}{3!} \lim_{z \to c} a_n''' + \dots$$

Notice, by substituting the differentiated versions we can analytically continue the R.H.S for $n$:

$$ a_n(z) = c+ (z-c) c^n + \frac{(z-c)^2}{2!} \frac{c^n - c^{2n}}{1 - c} + \dots$$

For example $a_{1.5}(z)$ is:

$$ a_{1.5}(z) = c + (z-c) c^{1.5} + \frac{(z-c)^2}{2!} \frac{c^{1.5} - c^{3}}{1 - c} + \dots$$

Note: since $c$ is a complex number one has freedom of choice in $\sqrt c$

Question

Is there a general formula for $a_n$ where $n$ can take non-integer values?

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  • $\begingroup$ Should the first term in your Taylor series be $c$ rather than $a_n$? $\endgroup$
    – kccu
    Dec 30, 2019 at 17:58
  • $\begingroup$ @kccu sorry forgot the limit $\endgroup$ Dec 30, 2019 at 18:00
  • $\begingroup$ en.wikipedia.org/wiki/Abel_equation $\endgroup$
    – Max
    Mar 3, 2020 at 13:26
  • $\begingroup$ @Grottfried Helms mind sharing a link of this mechanism? $\endgroup$ May 24, 2020 at 6:04
  • $\begingroup$ Doesn't look this somehow like the explicite representation by the Schroeder-mechanism? If I got this right, then the problem is that the fixpoint $c$ is complex and fractional powers of $c$ are not simple to handle, if usable at all. Let $1/h$ be some fractional value for the iteration-height, then it is relevant whether we write $(c^{1/h})^k$ in the power series or $(c^k){1/h})$ or $c^{k/h})$. There is also an approach of H. Kneser to avoid the complex-valued power series and arrive at a real-to-real fractional iteration at least. $\endgroup$ May 24, 2020 at 6:05

1 Answer 1

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Wikipedia reports:

It has now been proven[16] that there exists a unique function F which is a solution of the equation F(z + 1) = exp(F(z)) and satisfies the additional conditions that F(0) = 1 and F(z) approaches the fixed points of the logarithm (roughly 0.318 ± 1.337i) as z approaches ±i∞ and that F is holomorphic in the whole complex z-plane, except the part of the real axis at z ≤ −2. This proof confirms a previous conjecture.[17] The complex map of this function is shown in the figure at right. The proof also works for other bases besides e, as long as the base is bigger than $e^{\frac {1}{e}}$. The complex double precision approximation of this function is available online.[citation needed]

The above applies directly for tetration to the base $e$. However, for a given initial value $x$ we can use this solution to render the inverse function, the superlogarithm. For real $x$ we would choose a real superlogarithm $s>-2$, which will be unique because the interpolated tetration function is monotonic; then add the desired height $h$ of $e$'s and evaluate the tetration function at $s+h$.

References 16 and 17 from the above:

16. W. Paulsen and S. Cowgill (March 2017). "Solving $F(z+1)=b^{F(z)}$ in the complex plane" (PDF). Advances in Computational Mathematics. 43: 1–22. https://doi.org/10.1007%2Fs10444-017-9524-1

17. D. Kouznetsov (July 2009). "Solution of $F(z+1)=\exp(F(z))$ in complex z-plane" (PDF). Mathematics of Computation. 78 (267): 1647–1670. https://doi.org/10.1090%2FS0025-5718-09-02188-7.

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  • $\begingroup$ That discusses the nature of e^e^...^e, which isn't quite what was asked for ;p $\endgroup$ Dec 31, 2019 at 22:30
  • $\begingroup$ See the edit above. $\endgroup$ Dec 31, 2019 at 23:18
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    $\begingroup$ Though imo doesn't really discuss what the OP was doing but /shrug I guess $\endgroup$ Dec 31, 2019 at 23:22

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