# Is taking a Taylor expansion of $x^{x^x}$ possible? [closed]

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?

• This might be what you're looking for math.stackexchange.com/questions/2322635/…
– Joey
May 8, 2021 at 2:37
• @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 :) May 8, 2021 at 2:40
• 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.
– Joey
May 8, 2021 at 2:41
• Do you have a specific center of expansion in mind? May 8, 2021 at 3:14

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.
• While this is good mathematics, it is technically not a Taylor expansion, since it is an expression in both $x$ and $\log x$. May 8, 2021 at 14:41