# Inverse function of $y=\frac{\ln(x+1)}{\ln x}$

I've been wondering for a while if it's possible to find the inverse function of $y=\frac{\ln(x+1)}{\ln x}$ over the reals. This is the same as finding the positive real root of $x^y-x-1$. I realize that it's impossible with elementary functions, but is it possible with transcendental ones? There's certainly a pattern.

EDIT: I asked this question on MO, and I was given means to arrive at an answer, but I don't really understand it at all. I was told to go to college. Maybe some of you folks have already been, and can help explain this a little bit.

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Have you looked at the Lambert-W function as a possible route (I am not sure how practical it actually is, but might be worth a look and see the examples they give on this page - maybe example 2)? en.wikipedia.org/wiki/Lambert_W_function –  Amzoti Dec 12 '12 at 22:11
I have, but I never got anywhere with it. Mathematica couldn't either, even though it solved some related problems like $x=\frac{ln(x+y)}{ln(y)}$ using the Lambert-W function. –  hombre Dec 12 '12 at 22:15
This question involves a mixture of powers, not a mixture of exponentials with powers, so I feel Lambert W won't be of much use here. –  anon Dec 13 '12 at 22:48
Maybe I'm completely wrong because everyone is arguing with the Lampert function but can't we go this way : \begin{align} &y = \frac{\log(x+1)}{\log(x)} \\ \Leftrightarrow & \log(x) y = \log(x+1)\, \text{ //now use exp()} \\ \Leftrightarrow & \exp(\log(x) y) = \exp(\log(x+1)) \\ \Leftrightarrow & x \exp(y) = x+1 \\ \Leftrightarrow & x (\exp(y)-1)=1 \\ \Leftrightarrow & x=\frac{1}{\exp(y)-1} \end{align} So the inverse can be expressed in elementary functions. –  sonystarmap Dec 17 '12 at 19:03
@macydanim: It's not quite that simple. $\exp(\log(x)y) \neq x\exp(y)$. Instead, $\exp(\log(x)y) = x^y$. –  Scott Caldwell Dec 17 '12 at 19:19

Well, as you know, I can't actually give you a nice solution to put a box around, but I can give you some information about the function $f(x)$ satisfying $f(x)^x+f(x)+1=0$.

There is a pleasant, if not closed-form, asymptotic expansion for $f(x)$ for large $x$ in polynomials of $s=\log 2$:

\begin{align} f(x)=1+\frac sx+&\frac1{2x^2}\left(s+s^2\right)+\frac1{24x^3}\left(6 s+15 s^2+4 s^3\right)+ \\ +&\frac1{48x^4} \left(6 s+27 s^2+18 s^3+2 s^4\right)+\cdots \end{align}

This was obtained by series inversion on the series for $\dfrac{\log(x+1)}{\log x}$ about $x=1$. More specifically, letting $y=x-1$, the series for

\begin{align} \frac1{f^{-1}(x)} & =\frac{\log x}{\log(x+1)}=\frac{\log(y+1)}{\log(y+2)}=\frac{\log(y+1)}{\log(y/2+1)+s} \\ & =\frac{-\sum_1^\infty(-y)^n/n}{s-\sum_1^\infty(-y/2)^n/n}:=\sum_1^\infty a_ny^n \end{align}

can be solved for $a_n$ using the Cauchy product formula $\sum_0^\infty \alpha_ny^n\sum_0^\infty \beta_ny^n=\sum_{n=0}^\infty[\sum_{k=0}^n\alpha_k\beta_{n-k}]y^n$ to get

$$a_n=\frac{(-1)^{n+1}}{sn}+\sum_{k=1}^{n-1}\frac{(-2)^{-k}a_{n-k}}{sk}$$

(which yields coefficients in $\mathbb Q[1/s]$) and if the inverse of this is $f(x)=\sum_0^\infty b_nx^{-n}$, then

\begin{align} y+1 & =f(f^{-1}(y+1))=\sum_{n=0}^\infty b_n\bigg(\sum_{k=1}^\infty a_ky^k\bigg)^n=\sum_{n=0}^\infty b_n\bigg(\sum_{k=1}^\infty a_k(x+1)^k\bigg)^n \\ & =b_0+(a_1b_1)y+(a_2b_1+a_1^2b_2)y^2+(a_3b_1+2a_1a_2b_2+a_1^3b_3)y^3+\cdots. \end{align}

Unfortunately, the last sum has no nice closed form (although it can be done, using Faà di Bruno's formula), and the solution can only be written recursively anyway, so I'll stop here. This equation is supposed to be solved component-by-component to get the terms in the formula, so $1=b_0$, $1=a_1b_1\Rightarrow b_1=s$, $0=a_2b_1+a_1^2b_2\Rightarrow b_2=(s+s^2)/2$, etc. This is the source of the series quoted above.

The series is convergent for $|x|>1$. Here is a plot of the function, in the complex plane:

Red means near $1$, yellow-green is near $i$, blue is near $-i$, white is near $\infty$, black is near $0$. And here is a plot of the above asymptotic expansion, up to the $x^{-12}$ term:

A basic approximation of the function around $x=0$ is given by

$$x=\frac{\log(y+1)}{\log y}=\frac y{\log y}+O(xy)\Rightarrow -\frac1x=-\frac yxe^{-y/x}=y\Rightarrow y=-xW\Big(\!-\frac1x\!\Big),$$

where $W(x)$ is the Lambert W-function. The asymptotics of this function near $\infty$ are quite complex, and there is no proper power series, or even a generalized asymptotic series (involving terms other than $x^k$ for $k\in\mathbb Z$), because the truncated series, with error term $O\left(\big(\frac{\log\log x}{\log x}\big)^m(\log x)^{-k}\right)$, is never $O(1/x)$. The "basic" solution $y=-xW(-1/x)$ is however accurate to $O(x^2)$ (or so, proof or revision to come).

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Thank you! This is informative. I don't want to interrupt you if you were gonna get to this, but can the series be represented in sigma notation? And can that be generalized to $f(x)^x-af(x)-b$? –  hombre Dec 19 '12 at 19:39
Unfortunately, not that I can tell (well, at least not with one sigma. I'll get a 3 sigma version up shortly.) Graphs are harder for the generalized version, but the asymptotic form can be adapted. –  Mario Carneiro Dec 19 '12 at 23:51
Okay, so much for three sigma. Turns out that the closed-form equations for series division and reversion are rather more complex than I recalled, so you'll have to live with a doubly-recursively defined series, and a sum over another sum of variable dimension (i.e. infinite numbers of $\sum$s), which is not very pretty. At least it's easily computable, to low order. –  Mario Carneiro Dec 21 '12 at 8:43

I may not be leading you anywhere, or I'm maybe saying something that you have checked already, but I learned that a function needs to be injective and surjective (hence bijective) to be an inversible one. (Sorry for any grammar/english mistake)

As a start off point, i'm not looking very hard at the function, but i see that if x goes to 1, there will be a gap in there, thus not being injective. If that shows up to be true, then your equation wouldn't be inversible for real numbers.

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I hear you, there is a gap on the interval [0,1], but I'm pretty sure that we can just be picky with our domain for the inverse function. For example, $x^2$ is surjective but not injective, so we just change the domain of its inverse, $\sqrt x$, when we're working with reals. So I think the domain of our inverse would be something like $x>1$ or $x<0$. And once we get an idea of what that is, maybe we can generalize it to complex numbers or something. –  hombre Dec 18 '12 at 13:39
Well, If you go to complex, then you can probably have the desired inverse, to which extent I won't be able to help with, sorry. But what you said about the inverse of $x^2$ is quite true, as you change the domain to make the function bijective, thusly inversible. As such, you might want to make the function work for the domain $(1,\infty)$ or $(0,1)$ . –  FelipeM. Dec 18 '12 at 15:03
Actually, I just went with that to the Wolfram Alpha site. And, you might be surprisingly correct: this is what is shown. Although the function may not be expressed via an equation, the graph is quite elegant. –  FelipeM. Dec 18 '12 at 15:16
OK, I'm getting quite entertained by this. I tried something, not sure if I'm right, but... Going the normal way to do an inverse, substitute all the instances of $x$ by $y$ and of $y$ by $x$: $y=ln(x+1)/ln(x)$ $\to$ $x=ln(y+1)/ln(y)$ Assuming $y>0$, $xln(y) = ln(y+1)$ $ln(y^x)=ln(y+1)$ taking out the $ln$ part from both sides, $y^x = y+1$ Which gives out a graph without the negative parts, which would be a part of the desired inverse. Note that I'm not an expert, so maybe I butchered math in 2 or 3 parts of this :P –  FelipeM. Dec 18 '12 at 15:39
Yeah, you're spot on. The original point of finding this inverse was to extend the idea of the golden ratio, the positive number $x$ such that $x^2=x+1$, to any power. The reason I stated it in this form is because when working with reals, the natural logarithm is only defined for positive numbers. I don't want to find EVERY root of $x^y-x-1$, just the positive real one. –  hombre Dec 18 '12 at 16:22