The solution for x in this the equation $(\cfrac1x)^{\cfrac1x}=1-\cfrac1x$ is approximately 3.29316628741186103150802829...
Other forms of this equation which have the same solution include :
$y^{y}=1-y$ , $\cfrac1y$ = x
(In which y is equal to 0.3036591270299660... for an extended decimal form try this link and expand the solution :
https://www.wolframalpha.com/input/?i=y%5Ey%3D1-y
$x^\cfrac {x-1} x - x = -1$ (This is the first equation Wolfram Math Alpha gave me the actual approximate answer)
This number can also be described as $1/$ root of $x^x + x -1 = -1$
I find this problem very similar to this one where the same constants are used : number pair's in the self-root function $f(x) = x^{1/x}$
Related Equations that I think are also interesting :
$(\cfrac1a)^{\cfrac1a}=1+\cfrac1a$ (This one is about 0.562817451524631909792003735...)
$(\cfrac1b)^{\cfrac1b}=b+\cfrac1b$ (Approx. 0.609902981394622279977041843...)
$(\cfrac1c)^{\cfrac1c}=c-\cfrac1c$ (c ≈ 1.45799331624361272150026003...)
Is it possible to find an exact form for this solution and is there any significance to equations/solutions like this?
(If you want an extended decimal form for the 1st equation go to this link and expand the 'Numerical Solution') https://www.wolframalpha.com/input/?i=x%5E%28%28x+-+1%29%2Fx%29-x%3D-1
There is no analytical solution for the equation $$y^y=1-y$$ even using special functions.
We can make decent approximation considering that we look for the zero of function $$f(y)=y\log(y)-\log(1-y)$$
Using Taylor expansion, we have $$f(y)=y (\log (y)+1)+\frac{y^2}{2}+O\left(y^3\right)$$ Ignoring the higher order terms, the non-trivial solution is $$y_0=2 W\left(\frac{1}{2 e}\right)$$ where $W(.)$ is Lambert function.
Now, iterating using Newton method, we have $$y_{n+1}=\frac{(y_n-2) y_n+(y_n-1) \log (1-y_n)}{(y_n-2)+(y_n-1) \log (y_n)}$$ This can be considered as an explicit recurrence equation which, as shown below, converges very fast. $$\left( \begin{array}{cc} n & y_n \\ 0 & \color{red}{0.3}14369902967628019173913392043 \\ 1 & \color{red}{0.303}893759613688751450583356220 \\ 2 & \color{red}{0.303659}245373552388724380071295 \\ 3 & \color{red}{0.303659127029996}192264680381756 \\ 4 & \color{red}{0.30365912702996605124501895}3168 \\ 5 & \color{red}{0.303659127029966051245018951213} \end{array} \right)$$