# How to find a function $f(x)$ such that $f(f(x))=\log_ax$?

Is there some method to do this or maybe some method to find a function such that $f(f(x))$ is at least approximately equal to $\log_{a}x$? Maybe the taylor series could be of help. So, we're looking for a function such that $$f(f(x))\approx x-1-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-.........$$

I think such a function can help us extend the definition of superlogarithms. Here's how:

$slog_a x$ is defined as the number of times logarithm with the base $a$ is applied to $x$ to get to 1. But this definition fails when $x$ is not of the form $^na$. So, we're applying $f(x)=\log_a x$ continuously to $x$ in this case until we get 1.

If we define $log_ax$ in a similar manner as the number of times $x$ is divided by $a$ to get to 1, then this definition fails when $x$ is not of the form $a^n$. In this case, we're applying $f(x)=\frac{x}{a}$ continuously to $x$ until we get 1. This definition of $log_ax$ can be extended by looking for functions f, g, h ....such that $f$ such that $f(f(x))=\frac{x}{a}, g(g(g(x)))=\frac{x}{a},$ etc.

These functions are easy to find and are equal to, $f(x)=\frac{x}{a^{\frac{1}{2}}}$, $g(x)=\frac{x}{a^{\frac{1}{3}}}$, etc.

Here's how these functions help in extending this definition of $\log_ax$: Suppose we want to find $log_220$, then first keep dividing $20$ by $2$, i.e. keep applying $f(x)=\frac{x}{2}$ to 20. After four divisions, we get 1.25. We can't divide further by 2 because then the number will get smaller than 1.

Now, keep applying $f(x)=\frac{x}{\sqrt{2}}$ to 1.25 until we again reach a situation in which further division gets us smaller than 1. Let the number of times we we were able to divide by $\sqrt{2}$ be $k_1$. After that, start dividing the resulting number by $2^{\frac{1}{3}}$. Let the number of times we could divide by $2^{\frac{1}{3}}$ be $k_2$. Repeat this process to as much accuracy as you like. Then,

$$log_220=4+\frac{k_1}{2}+\frac{k_2}{3}+...........$$. The 4 in the start of the sum is the number of times we could divide 20 by 2 without getting a number smaller than 1. So, if we can get functions $f$, $g$, $h$,.... such that $f(f(x))=\log_ax$, $g(g(g(x)))=\log_ax$, $h(h(h(h(x))))=log_ax$, etc and calculate the number of times these functions can be applied without getting a number smaller than 1, then I think $slog_ax$ can be calculated by using $$slog_ax=k_1+\frac{k_2}{2}+\frac{k_3}{3}+....$$ where $k_1$ is the number of times log with base $a$ can be applied to $x$ such that the resulting number isn't smaller than 1, $k_2$ is the number of times the function $f(x)$ such that $f(f(x))=log_ax$ can be applied to the number we get after applying $k_1$ logarithms with base $a$ to $x$ without getting a number less than 1, $k_3$ is the number of times the function $g(x)$ such that $g(g(g(x)))=\log_ax$ can be applied to the number we got after the previous step without getting a number less than 1, etc.

• Looks like previous comments were deleted. Are you interested in the Taylor sereis for Kneser's $\ln^{[\circ 0.5]}(x+1)$ function at x=1 to compare with your approximation? $\ln^{[\circ 0.5]}\approx$ 0.498563287941114, and the derivative at x=1 ~= 0.876336132224813 – Sheldon L Feb 8 '17 at 16:16