product tetration

[Alex Zuma 2025]

product tetration is the superfunction of \(f(x)=xa^{x}\).
it uses the lambert w function (also known as the product logarithm) to go in the negative direction.
product tetration requires three variables and is denoted as: \(\text{prodtet}(a,b,c)\)
product tetration can be interpolated through lagrange polynomials for extension to real numbers.
and it has this recursive property:
\(\text{prodtet}(a,b,c)= f(\text{prodtet}(a,b,c-1))\)
and this base case:
\(\text{prodtet}(a,b,0)= b\)
more details are given on this desmos graph: https://www.desmos.com/calculator/cnvdx5ce1r

[RaeesHarris]

product tetration is the superfunction of \(f(x)=xa^{x}\).
it uses the lambert w function (also known as the product logarithm) to go in the negative direction.
product tetration requires three variables and is denoted as: \(\text{prodtet}(a,b,c)\)
product tetration can be interpolated through lagrange polynomials for extension to real numbers.
and it has this recursive property:
\(\text{prodtet}(a,b,c)= f(\text{prodtet}(a,b,c-1))\)
and this base case:
\(\text{prodtet}(a,b,0)= b\)
more details are given on this desmos graph: https://www.desmos.com/calculator/cnvdx5ce1r

Hello Alex Zuma,

Instead of Interpolating the function using Lagrange Interpolation. You can instead build a local analytical continuation, and then do some processes to get an approximation for the extension of product tetration. I already made a Desmos Project approximating product tetration(specifically prod(a,1,c)  for the real plane only): https://www.desmos.com/calculator/r5xpcxkol1 (edit: improved version with my new lambert approximation: https://www.desmos.com/calculator/l31kmb12ed )

Let me explain how it works, basically the core function for upwards recursion is \( f\left(x\right)=xa^{x} \). We can actually define the nth functional iteration of f(x) using product tetration:

\( \operatorname{f}^{\circ n}(x) = \operatorname{prodtet}(a, b, \operatorname{prodtet}^{\circ (-1)}(a,b,x) + n )  \)

Using this defintion we can work out a local continuation of product tetration:

\( \operatorname{prodtet}(a,b,c) \approx\ L+\left[\frac{d}{dx}f^{o\left(c\right)}\left(x\right)\right]_{x=L}\cdot\left(b-L\right)+\frac{\left[\frac{d^{2}}{dx^{2}}f^{o\left(c\right)}\left(x\right)\right]_{x=L}}{2!}\cdot\left(b-L\right)^{2}+\frac{\left[\frac{d^{3}}{dx^{3}}f^{o\left(c\right)}\left(x\right)\right]_{x=L}}{3!}\cdot\left(b-L\right)^{3}...\)

Where L is the fixed point \( f(L)=L \)(for product tetration L would be 0). This only solves for a local continuation of product tetration, and so you have to use some other numerical methods to extend it to the entirety of the complex plane(like Analytical continuation along a path).

We can simplify the defintion of product tetration(to compute this I had to use an old project since f'(L)=1 which is a parabolic fixed point so i cant use standard recursion: https://www.programiz.com/online-compiler/2vpMmBtpJaDCK):

\( \operatorname{prodtet}(a,b,c) \approx\ b+c\ln\left(a\right)b^{2}+\frac{c\left(2\cdot c-1\right)\ln\left(a\right)^{2}}{2!}b^{3}+\frac{2c\left(12c^{2}-15c+5\right)\ln\left(a\right)^{3}}{4!}b^{4}... \)

This is what I used in my approximation of product tetration(i used 12 derivatives). 

Anyways, I hypothesise that this is unique wherever it converges, or close to unique since based on the principal-branch-normalization, there is only one way to get the local analytical continuation(I think) using this method, and hypothetically a unique analytical continuation would have to fulfill the derivatives I got from the nth functional iteration of f(x). However I cant prove it due to my lack of formal mathematical knowledge, and so the task of proving uniqueness for this method is left as an excercise to whoever is reading this post.

I hope this graph helps you! If you have any questions, feel free to ask!