I have found a number. Google and OEIS come up blank: 0.696340872970033948754981...

Question

It kind of looks like $\log(2)$, but it isn't.

This is a fixed point of a map $f:\Bbb{R}\to\Bbb{R}$ with $f(x) = (1-x)^{1-x}$. When you iterate the map it misbehaves at $0$ and $1$, so I started at $1/2$. Convergence is geometric, so it's not hard to find many decimals. I'm using Python $3$ with gmpy2.

My question is how do I proceed? I guess I'm putting it out here to see if someone can find some nice formula for this value. Tried posting to reddit but that got nuked real quick.

If this gets nuked for not being a question or unproductive or whatever, please advice how I should post!

Answer 1

Via a substitution this value is also $1-u$ in the following equations $$u^u=1-u$$ or $$u=\frac{\ln(1-u)}{\ln(u)}$$ At the end of the day it's just a number - among uncountably infinitely many others. It's not surprising at all that we don't have a way to express it in terms of a finite number of constants and functions we happen to have given names and notation.

Answer 2

Since you are looking for the fixed point, you want the solution of the equation $(1-x)^{1-x}=x$. This equation does have a solution in $(0,1)$ and can be approximated by numerical methods but this kind of equations rarely have a closed form for the solutions.

Answer 3

Quoting @K.defaoite, "it's just a number" that you can compute with as many decimals as you wish.

You could also consider it as the limit of an infinite sequence of rational numbers. Suppose that you make an expansion to $O(x^{n+1})$, say for example $$f(x)=(1-x)^{1-x}-x=1-2 x+x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{12}+\frac{3 x^6}{40}+\frac{x^7}{120}+\frac{59 x^8}{2520}+O\left(x^{9}\right)$$ Perform a series reversion to get $$x_{(8)}=t+\frac{t^2}{2}+\frac{t^3}{4}+\frac{t^4}{6}+\frac{5 t^5}{24}+\frac{79 t^6}{240}+\frac{727 t^7}{1440}+\frac{30271 t^8}{40320}+O\left(t^9\right)$$ where $t=\frac{1-f(x)}2$.

Make $f(x)=0$, that is to say $t=\frac 12$ to get $$x_{(8)}=\frac{785839}{1146880}\approx 0.685197$$

The first terms of the sequence will be $$\left\{\frac{1}{2},\frac{5}{8},\frac{21}{32},\frac{2}{3},\frac{517}{768},\frac{3473 }{5120},\frac{25151}{36864},\frac{785839}{1146880},\frac{28380857}{41287680},\frac {68280757}{99090432},\cdots\right\}$$

I shall not reproduce the exact values but below are some numerical values $$\left( \begin{array}{cc} n & x_{(n)} \\ 10 & 0.689075177308744 \\ 20 & 0.695101903140602 \\ 30 & 0.696062162951107 \\ 40 & 0.696269787293523 \\ 50 & 0.696321372306707 \\ 60 & 0.696335260924677 \\ 70 & 0.696339202000697 \\ 80 & 0.696340362612384 \\ 90 & 0.696340713976303 \\ 100 & 0.696340822646690 \\ 110 & 0.696340856833953 \\ 120 & 0.696340867739664 \\ 130 & 0.696340871259015 \\ 140 & 0.696340872405876 \\ 150 & 0.696340872782743 \end{array} \right)$$

We could find another sequence of numbers building the $[1,n]$ Padé approximants of $f(x)$ around $x=0$. The first terms would be

$$\left\{ \frac{2}{3} , \frac{2}{3} , \frac{27}{40} , \frac{80}{117} , \frac{130}{189} , \frac{567}{821} , \frac{51723}{74687} , \frac{149374}{215309} , \frac{430618}{619955} , \frac{619955}{891822} \right\} $$