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How can I find the inverse function of $f(x)=x^{x^x}$? Has this inverse function ever been defined / studied? are there any asymptotic expansions?

It would be nice if the inverse of $f(x)$ could be expressed in terms of standard mathematical functions, but it would be enough for me to know even just a few properties. The inverse of $x^{x^x}$ can be linked to the inverse of $(x + a)^x$ and to the inverse of $x e^x +a x$, and many other functions, so knowing its properties can have many applications in solving a wide range of equations

Thanks

edit:

The series expansion of $x^{x^x} $ at $x=0$ can be expressed as:

$$x^{x^x}=x \sum_{j=0}^{\infty} \frac{\left( \ln(x^x)\right)^j}{j!}B_j\left( \ln(x)\right)\tag{1}$$

$B_j(x)$ is the Bell Polynomial.

By the General Leibniz rule we have that the nth derivative of $x^{x^x}$ can be shown as

$$\sum_{k=1}^n \binom{n}{k}f^{(n-k)}(x)P_j^{(k)}(\ln(x))$$

$$f(x)=x^{j+1}\Rightarrow f^{(n-k)}(x)=x^{j+1-n+k} \frac{(j+1)!}{(j+1-n+k)!}$$ $$P_j(\ln(x))=S_{j}^{(1)}\ln(x)^{j+1}+S_{j}^{(2)}\ln(x)^{j+2}+\dots+S_{j}^{(j)}\ln(x)^{j+j}$$ $S_j^{(k)}$ is the Stirling number of the second kind

$$P^{(n)}_j(\ln(x))=\sum_{k=1}^j \ \sum_{r=n-k-j}^{n-1}\frac{(j+k)!}{(j+k-n+r)!}s_{n}^{(n-r)}S_{j}^{(k)}\frac{\ln(x)^{j+k-n+r}}{x^n}$$ $s_j^{(k)}$ is the Stirling number of the first kind

$$P_j^{(n)}(\ln(1))=P_j^{(n)}(0)=\sum_{k=1}^j (j+k)!s_{n}^{(j+k)}S_{j}^{(k)}$$

Therefore the n th derivative of $x^{x^x}$ in $x = 1$ can be expressed as: $$D_n(1)=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^n(j+1)\frac{(h+j)!}{(j+1+k-n)!} \binom{n}{k}s_{n}^{(j+k)}S_{j}^{(k)}$$

With $n$ between $2$ and $10$ we have $D_n(1)={2,9,32,180,954,6524,45016,360144,3023640} $ A179230 obtaining an explicit form for the Taylor expansion coefficients for $x\to 1:$

$$ x^{x^x}=f(x)=x+\sum_{n=2}^{\infty}\frac{D_{n}(1)}{n!}(x-1)^n$$

Since the series has no constant terms, it is possible to express the inverse function $f^{-1} (x)$ by Series Reversion

$$f^{-1}(x)=x-(-1+x)^2+ \frac{1}{2}(-1+x)^3+\frac{7}{6}(-1+x)^4-\frac{17}{4} (-1+x)^5+O(x^6) $$

Patrick Danzi
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    One thing to note is that just because the inverse exists, this does not mean that it can be expressed in terms of elementary functions. – Joe May 27 '21 at 11:24
  • You might be able to write it in terms of Lambert's W function, see: https://math.stackexchange.com/questions/1261825/inverse-function-of-xx – Adam Rubinson May 27 '21 at 11:30
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    @AdamRubinson: no. –  May 27 '21 at 11:33
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    @Joe It would be nice if the inverse of $f(x)$ could be expressed in terms of standard mathematical functions, but it would be enough for me to know even just a few properties. The inverse of $x^{x^x}$ can be linked to the inverse of $(x + a)^x$ and to the inverse of $x e^x +a x$, and many other functions, so knowing its properties can have many applications in solving a wide range of equations – Patrick Danzi May 27 '21 at 11:34
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    @PatrickDanzi: I agree that it would interesting to learn about these properties. That's why I upvoted your question. I'm afraid I do not have the expertise to give you any answers. – Joe May 27 '21 at 11:38
  • You won't find a nice expression for the inverse. But you can determine a lot of its properties. What kind of properties are you looking for? For instance, it is differentiable, increasing, and tends to infinity (but very slowly). – TonyK May 27 '21 at 11:41
  • @TonyK since I believe that this function can be applied in the solution of equations, the purpose of this post is to group the properties of this function, for those who may need it in the future – Patrick Danzi May 27 '21 at 11:45
  • You can know many properties of the inverse just by knowing the properties of $x^{x^x}$ – Vivaan Daga May 27 '21 at 11:48
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    If you take a function like $f(x) = x^\frac{1}{x}$ its inverse is the basic infinite power tower. I believe that there is a known relationship between power towers and their inverses expressed as power towers. It is the type of thing you see in math magazine or American math monthly on a rare occasion. It has some relationships with continued fractions as well I think. – open problem May 27 '21 at 11:52
  • You can sketch its detailed graph , find its derivative in terms of itself, etc. – Vishu May 27 '21 at 12:14
  • How do you define that function , for example for x=3, you can end up with either $3^9$ or $9^3$ – jimjim Jul 05 '21 at 08:14
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    $x^{(x^x)}$ @jimjim – Patrick Danzi Jul 05 '21 at 08:26
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    Has this inverse function ever been defined / studied My guess: it has never been studied, because it has never been found useful for anything. Usually, studying is done of things that are already known to be useful. – GEdgar Jul 05 '21 at 16:12
  • Here is a simpler series expansion of the inverse function. After a series/limit expansion or 2 on the derivatives, we are done – Тyma Gaidash Mar 07 '23 at 18:06
  • series iversion – Max0815 Mar 07 '23 at 18:19
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    With Lambert, you can establish the inverse of $x^{x^{x+1}}$ is $e^{W(W(\ln(x))}$ (restricted to the right domains). Not the same of course. – 2'5 9'2 Mar 07 '23 at 19:14
  • @2'59'2 Interestingly, making one of the W functions into $W_{-1}(x)$ is another solution. It makes one wonder which other branches make the inverse work – Тyma Gaidash Mar 07 '23 at 19:20

2 Answers2

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Basic recursive method:

This is not a closed form, but it actually works. It uses a strategy of solving for a variable recursively. See this graph for proof as the recursive function converges to a vertical or horizontal line as the solution is x=recursive(x). It will be easier to calculate without splitting the bases of the logarithm:

$$x^{x^x}=y=y_0\implies x=A_0(y_0)=\log_x\left(\log_x\left(y_0\right)\right)=A_1(y_0)\implies x=\lim_{n\to\infty}A_n(y_0), A_n= \log_{A_n(y_0)}\left(\log_{A_n(y_0)}\left(y_0\right)\right)\implies A_2(y_0)= \log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(\log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(y_0\right)\right), A_3(y_0)= \log_{\log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(\log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(y_0\right)\right)}\left(\log_{\log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(\log_{\log_x\left(\log_x\left(y_0\right)\right)}\left(y_0\right)\right)}\left(y_0\right)\right),…\implies x=A_\infty (x_0)$$

See graph 2 to see how to solve the the value that gets you y. The graph converges to a line as well like in paragraph 1:

$$x^{x^x}=y, B_0(a)=x=a\implies x=B_1(a)=\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}\implies x=\lim_{n\to \infty} B_n(a), B_{n+1}(a)= \frac{\ln(\ln(a))-\ln\left(\ln\left(B_n(a)\right)\right)}{\ln\left(B_n(a)\right)}\implies x=B_{\infty}(a)\implies B_2(a)= \frac{\ln(\ln(a))-\ln\left(\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{ln\left(x\right)}}\right)\right)}{\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}}\right)}, B_3(a)= \frac{\ln(\ln(a))-\ln\left(\ln\left({ \frac{\ln(\ln(a))-\ln\left(\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}}\right)\right)}{\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}}\right)}}\right)\right)}{\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}}\right)\right)}{\ln\left({\frac{\ln(\ln(a))-\ln\left(\ln\left(x\right)\right)}{\ln\left(x\right)}}\right)}}\right)},…\implies x=B_\infty(a)$$

$\def\srt{\operatorname{srt}}$ Analytic solutions:

With Lagrange reversion, we get:

$$y^{y^y}=z\mathop\iff^{y=e^w} w=\ln(y)e^{-we^w}\implies y=1+\sum_{n=1}^\infty\frac{\ln^n(z)}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}e^{t\left(1-ne^t\right)}\right|_0$$

General Leibniz rule and a Maclaurin series of $e^y$ uses Kronecker $\delta_{m,k}$ and factorial power $a^{(b)}$:

$$\lim_{t\to0}\sum_{m=0}^\infty\sum_{k=0}^{n-1}\frac{(-n)^m}{m!}\binom{n-1}k\left.\frac{d^{n-1}t^m}{dt^{n-1}}\right|_0\left.\frac{d^{n-1}e^{(m+1)t}}{dt^{n-1}}\right|_0=\sum_{m=0}^\infty\sum_{k=0}^{n-1}\binom{n-1}k\frac{(-n)^m m^{(k)}\delta_{m,k}}{(m+1)^{k+1-n}m!}$$

Quantile mechanics (93) to (96) hints at $\delta_{m,k}$ removing the $m$ sum and replacing $m=k$:

$$\sum_{m=0}^\infty\sum_{k=0}^{n-1}\binom{n-1}k\frac{(-n)^m m^{(k)}\delta_{m,k}}{(m+1)^{k+1-n}m!}=\Gamma(n)\sum_{k=0}^{n-1}\frac{(-1)^k(k+1)^{n-k}n^k}{(k+1)!\Gamma(n-k)}$$

Therefore:

$$\boxed{\sqrt[3]z_s=\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}}$$

shown here. Both sums are interchangeable if $k\to\infty$ and the region of convergence is near $|z|=1$. Also, @Dark Malthorp found an integral representation.

Тyma Gaidash
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  • From $y^{y^y}=x$ for $x=0$ we must have $y=0$, but your Lagrange reversion formula yields $y=1$ as $x=0$. – Diger Mar 07 '23 at 21:06
  • Maybe you could add how you applied the reversion formula. – Diger Mar 07 '23 at 21:20
  • Not sure what is going on there, but $x^x \rightarrow 1$ as $x\rightarrow 0$ and so $x^{x^x}$ goes to $0$ as $x\rightarrow 0$. – Diger Mar 08 '23 at 12:41
  • https://www.wolframalpha.com/input?i2d=true&i=Limit%5BPower%5Bx%2CPower%5Bx%2Cx%5D%5D%2Cx-%3E0%5D – Diger Mar 08 '23 at 12:47
  • That looks a lot different. Maybe you meant $\ln^n(x)$ not $x^n$ as you wrote it. – Diger Mar 08 '23 at 12:52
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$$x^{x^x}=y$$ $$e^{\ln(x){e^{\ln(x)x}}}=y$$

The main theorem in [Ritt 1925] implies that the function term $e^{\ln(x){e^{\ln(x)x}}}$ isn't in a form to read if a bijective elementary function $x\mapsto x^{x^x}$ has an elementary inverse or not.

Look for the terms Wexzal and super-root.

related: Solving $x^{x^x}=c$ with Lambert's W function

[Edwards 2020] Edwards, S.: Extension of Algebraic Solutions Using the Lambert W Function. 2020

Fantini, J.; Kloepfer, G.: Wexzal/The coupled Exponent, 1998

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

Galidakis, I.; Weisstein, E. W.: Power Tower. Wolfram MathWorld

Helms, G.: Wexzal"/"coupled exponent" , "Superroot" and a generalized Lambert-W

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

Wikipedia: Tetration - Super-root

IV_
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