For the equation $y^y=x^x$, I know that one solution is the line $y=x$ (for $x > 0$), and is shown in this graph here: $y^y=x^x$. However, when I see that graph, I also see a curve that goes from $(0, 1)$ to $1, 0$. Is there an equation (i.e. analytical solution) for just that curve?
I played around with equations, and have discovered that equations in the form $y=\frac{1}{x+a}-a$ kind of fit, but not really.
For example, $y=\frac{1}{x+.62}-0.62$ is close, but not really.
I am a high school student and am taking Pre-Calculus, and so my knowledge of advanced functions are limited. However, I do welcome more complicated functions.
Best Answer
Assuming that you want a "simple" and quite accurate function to represent the curved part of function $$y=\frac{\log(x^x)}{W(\log(x^x))}\qquad \text{for} \qquad 0 \leq x \leq \frac 1e$$ hoping that you do not require too much accuracy for small values of $x$, you could use the series expansion
$$y=\frac 1 e \sum_{n=0}^p (-1)^n a_n\,(ex-1)^n $$ where the first coefficients make the sequence $$\left\{1,1,\frac{1}{3},\frac{1}{9},\frac{17}{270},\frac{31}{810},\frac{151}{5670}, \frac{547}{28350},\frac{7541}{510300},\frac{763}{65610},\frac{14281213}{151559100 0}\right\}$$
Edit
Some numerical results $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ 0.000 & 0.966579 & 1.000000 \\ 0.025 & 0.895605 & 0.902904 \\ 0.050 & 0.833657 & 0.835955 \\ 0.075 & 0.778631 & 0.779374 \\ 0.100 & 0.729009 & 0.729241 \\ 0.125 & 0.683692 & 0.683760 \\ 0.150 & 0.641883 & 0.641902 \\ 0.175 & 0.602997 & 0.603001 \\ 0.200 & 0.566596 & 0.566597 \\ 0.225 & 0.532350 & 0.532350 \\ 0.250 & 0.500000 & 0.500000 \\ 0.275 & 0.469345 & 0.469345 \\ 0.300 & 0.440221 & 0.440221 \\ 0.325 & 0.412494 & 0.412494 \\ 0.350 & 0.386053 & 0.386053 \end{array} \right)$$
If you want a "super simple" approximation use $$y=1-\frac{(e-1)}{\sqrt[3]{e}} x^{2/3}$$