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Disclaimer: I do not know that much math, so there might be obvious mistakes that I make in this question.

As far as I can tell, taking the derivative of $x^{x^x}$ infinitely is possible. However, this does not mean a Taylor series is possible. So what I am asking is, is it possible. I see 5 possible outcomes:

  1. It is possible, and the answer is known/easy enough to figure out
  2. It has been proven to be possible, but we don’t know what it is yet
  3. It has been proven to be impossible, and cannot be done
  4. It may or not be possible, we don’t know yet
  5. Other

So, which one of these is it?

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  • $\begingroup$ This might be what you're looking for math.stackexchange.com/questions/2322635/… $\endgroup$ – Joey May 8 at 2:37
  • $\begingroup$ @Joey that is for x^x, my question is about x^x^x, so they are not duplicates (as far as I can tell, I might be wrong). If there is some way to correlate the two that would fit nicely in an answer :) $\endgroup$ – Ekadh Singh May 8 at 2:40
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    $\begingroup$ They are not duplicates, I just thought it would help to look at. Try looking up Taylor expansion on Wolfram alpha, it lets you try out different functions, and you can try $x^{x^x}$ at centered at different points. $\endgroup$ – Joey May 8 at 2:41
  • $\begingroup$ Do you have a specific center of expansion in mind? $\endgroup$ – Lubin May 8 at 3:14
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It is possible and it is the same for all tetrations. In this case, around $x=0$, it would be $$x^{x^x}=x+\sum_{n=2}^p \frac {t^n}{(n-1)!} P_n(t)\, x^{n}+O(x^{p+1})\qquad \text{with} \qquad t=\log(x)$$ and the first polynomials are $$\left( \begin{array}{cc} 2 & 1 \\ 3 & t+1 \\ 4 & t^2+3 t+1 \\ 5 & t^3+6 t^2+7 t+1 \\ 6 & t^4+10 t^3+25 t^2+15 t+1 \\ 7 & t^5+15 t^4+65 t^3+90 t^2+31 t+1 \\ 8 & t^6+21 t^5+140 t^4+350 t^3+301 t^2+63 t+1 \end{array} \right)$$ where you can notice interesting patterns.

Have a look here for the next one.

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  • $\begingroup$ Nice ! ..What is the interval of convergence of the series u presented ? $\endgroup$ – Balaji sb May 8 at 4:58
  • $\begingroup$ While this is good mathematics, it is technically not a Taylor expansion, since it is an expression in both $x$ and $\log x$. $\endgroup$ – Paul Sinclair May 8 at 14:41
  • $\begingroup$ @PaulSinclair. At the same time, I agree and disagree. $\endgroup$ – Claude Leibovici May 8 at 14:43
  • $\begingroup$ Well, arguing over nomenclature is rarely if ever of any use. I certainly do not dispute the value of the expression. $\endgroup$ – Paul Sinclair May 8 at 14:50
  • $\begingroup$ @PaulSinclair. In fact, for me, it is a real debate (which started in my mind 65+ years ago) $\endgroup$ – Claude Leibovici May 8 at 14:52

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