# Graph for the curved portion of the equation $y^y=x^x$

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.

• When $x$ is less than the value where the two curves intersect (which I think is $e^{-1}$), the solution is given by the product log function: $y = \frac{\ln(x^x)}{W(\ln(x^x))}$ which is equivalent to $e^{W(\ln(x^x))}$. – Varun Vejalla Sep 21 '20 at 0:49
• @VarunVejalla What is the name of the function W(x)? – KingLogic Sep 21 '20 at 0:57
• It is the Lambert $W$ function also known as the product log function. It is the inverse of $x e^x$. – Varun Vejalla Sep 21 '20 at 1:03
• @VarunVejalla Do you have a graph of the function $y=\frac{\ln(x^x)}{W(\ln(x^x))}$? – KingLogic Sep 21 '20 at 1:06
• It is here, but $y \not = x$ only for $x$ less than some value (I think $e^{-1}$). – Varun Vejalla Sep 21 '20 at 2:15

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 $$y=1-\frac{(e-1)}{\sqrt[3]{e}} x^{2/3}$$

• Thank you very much for the answer. However, I'm not sure if it approximates the entire curve too well, it only approximates the area in the vicinity of $\frac{1}{e}$ well, using the sequence you have given. Therefore, I have upvoted, but will withhold the checkmark. desmos.com/calculator/5m07abeesj – KingLogic Sep 21 '20 at 10:11
• @KingLogic.Let me check again – Claude Leibovici Sep 21 '20 at 10:12
• @KingLogic. Checked ! Look at my edit for values. – Claude Leibovici Sep 21 '20 at 10:26
• @KingLogic. Try $y=1-\frac{(e-1)}{\sqrt[3]{e}} x^{2/3}$ ! – Claude Leibovici Sep 21 '20 at 10:51
• Btw, is there a closed form for its factor $\frac{y^y-x^x}{y-x}?$ – Narasimham Sep 21 '20 at 11:00