inverse function of $f(x)=x^{x^x}$

Question

I have easily found the inverse of $f(x)=x^x$ using the following: $$y=x^x\Rightarrow \ln(y)=x\ln(x)\Rightarrow W(\ln(y))=\ln(x)\therefore x=e^{W(\ln(y))}$$ However I am struggling to do the same for $f(x)=x^{x^x}$, this is what I have tried: $$x=y^{y^y}$$ $$\ln(x)=y^y\ln(y)$$ and if: $u=y^y$ then: $$\frac{ue^u}{e^{W(ln(u))}}=\ln(x)$$ now by letting $\ln(u)=ve^v$ we get: $$e^{ve^v}v=\ln(x)\Rightarrow \ln(v)+ve^v=\ln(\ln(x))$$ However I cannot see an easy way of getting from here or even if this is the correct way to approach it, as I have tried several methods and they have all failed

The closest progress I have made is if i define a function $g(x)=\ln(x)+xe^x$ then $G(x)=g^{-1}(x)$ then I get:

$$y=\exp\left(W\left(G(\ln(\ln(x)))+\ln(G(\ln(\ln(x)))\right)\right)$$ However this is a fairly ugly expression and I still have no way of defining $G(x)$

I don't but I just feel there should be some way of expessing it — Henry Lee, Jun 11 '20 at 01:34
Then you might be disappointed. There are many functions one can write down which do not have an easy inverse (including things like $x\mapsto xe^x$, which is why the product log function needs to be defined in the first place.) — YiFan Tey, Jun 11 '20 at 01:35
Ah I see, but the productlog function has an infinite series representation so could the same not be done here? — Henry Lee, Jun 11 '20 at 01:37
Yes, you're right that it could be possible to find a series expansion, although I'm not sure how to go about that directly. — YiFan Tey, Jun 11 '20 at 01:38
Yeah this is not my area of knowledge either really so we could struggle — Henry Lee, Jun 11 '20 at 01:39
@HenryLee what you are looking for are the super root functions. I’m almost sure there isn’t a concise form for these for $n\neq 2$ or $\infty$. See here: https://en.wikipedia.org/wiki/Tetration#Super-root — Bcpicao, Jun 11 '20 at 01:59
I’m sorry to say Wolfie found nothing, even though it is able to solve $y=x^x$ — gen-ℤ ready to perish, Jun 11 '20 at 05:44
Solving $y=x^x$ using the Lambert function $W$ just means: Giving the inverse function a particular name. In the same way you could call the inverse function of $f$ Lee function, and give it the name $L$. — Christian Blatter, Jul 01 '20 at 08:46
See here, here, and here. One can see numerical computations of the inverse here and here. — Simply Beautiful Art, Jul 01 '20 at 12:10

The_Sympathizer · Accepted Answer · 2020-07-01T19:29:58.093

If anything, I'd say perhaps we should flip this around. The functions

$$f_1(x) := x, f_2(x) := x^x, f_3(x) := x^{x^x}, \cdots$$

form a logical progression, and so too their inverses. Hence, it "seems" like that instead of scratching our head regarding how to understand the inverse of $f_3$ as a Lambert W-function because $f_2$ can be given as such, we should perhaps be more inclined to think of the Lambert W-function as the one that needs special expressing - in terms of the inverse of $f_2$! And indeed you can do that: call the $n$th inverse term from the above a "$n$-th order tetrational root", or $\mathrm{trt}_n(x)$, so that $f_2^{-1} = \mathrm{trt}_2$. Then we have, just as

$$\mathrm{trt}_2(x) = e^{W(\ln(x))}$$

which is what gets you wondering, that we can turn the tables to get

$$W(x) = \ln(\mathrm{trt}_2(e^x))$$

and so the equation that defines $W(x)$ is just seen to be a funny case involving 2nd-order tetrational roots, so when you ask to express $\mathrm{trt}_3(x)$ in terms of $W(x)$, you are essentially asking to express $\mathrm{trt}_3(x)$ in terms of $\mathrm{trt}_2(x)$, and if you think about that, that would be kind of like asking "how can I express a cube root $\sqrt[3]{\cdots}$ using square roots (and arithmetic) $\sqrt{\cdots}$ only?" which clearly isn't something you can do - so why should that change here? Moreover, if we put things this way, you see we probably wouldn't think we needed to; it's just that the sequence of "tetrational roots" is not one that is usually put in your algebraic toolkit.

Moreover, your identity involving what you call "$G(x)$" would seem to show what would be the more "correct" relation between $\mathrm{trt}_3$ and $W$: while the equation that defines $W$, i.e.

$$xe^x = a$$

will be solved using $\mathrm{trt}_2$, the equation

$$xe^x + \ln x = a$$

is the analogous one you would solve using $\mathrm{trt}_3$. If you like, it would define a sort of "higher order" $W$-function.

Ah I see, Thank you for the great explanation! – Henry Lee Jul 01 '20 at 17:30 — Henry Lee, Jul 01 '20 at 17:30

inverse function of $f(x)=x^{x^x}$

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