Golden Exponent? Tetration

Question

I read somebody say “golden exponent $ x^x=x+1 $” now I didn’t understand what he meant but it really fascinated me thinking about some type of tetration version of the golden number. A number with a special or awesome property (not necessarily linked to the golden number). Now that I think about the equation mentioned $ x^x=x+1 $ that does look like an interesting equation for x to satisfy. Anyways, what does everybody think about whether such an analog for tetration exists or not?

Answer 1

Given that $0^0=1$ let $x=0$, then $x^x=x+1$ reduces to $0^0=0+1=1$, so $x=0$.

Answer 2

Consider that you look for the non trivial zero's of function $$f(x)=x^x-x-1$$ $$f'(x)=x^x (\log (x)+1)-1\qquad \text{and} \qquad f''(x)=x^{x-1}+x^x (\log (x)+1)^2$$ The first derivative cancels at $x=1$ and $f(1)=-1$ with $f''(1)=2$. So, we have a minimum and two roots (one of them being the trivial $x=0$.

If you do not want to use a root-finding method, expand $f(x)$ as a series around $x=1$ and obtain $$f(x)=-1+\sum_{n=1}^\infty a_n(x-1)^{n+1}$$ where the first coefficients are $$\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{3}{40},-\frac{1}{120},\frac{59}{2520 },-\frac{71}{5040},\frac{131}{10080},-\frac{53}{5040},\frac{179063}{19958400},\cdots\right\}$$

Now, using series reversion, we obtain $$x=1+\sum_{n=1}^p b_n\,t^n \qquad \text{where}\qquad t=\sqrt{f(x)+1}$$ the first $b_n$ being $$\left\{1,-\frac{1}{4},-\frac{1}{96},\frac{1}{12},-\frac{7621}{92160},\frac{173}{2880},- \frac{243979}{6881280},\frac{901}{60480},\frac{33853717}{118908518400},\cdots\right\}$$

Making $f(x)=0$ that is to say $t=1$, we then have the estimate $$x=\frac{211659504277}{118908518400}\sim 1.78002$$ while the "exact" solution given by Newton method would be $1.77678$.