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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.

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    $\begingroup$ 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 $e^{W(\ln(x^x))}$. $\endgroup$ – Varun Vejalla Sep 21 '20 at 0:49
  • $\begingroup$ @VarunVejalla What is the name of the function W(x)? $\endgroup$ – KingLogic Sep 21 '20 at 0:57
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    $\begingroup$ It is the Lambert $W$ function also known as the product log function. It is the inverse of $x e^x$. $\endgroup$ – Varun Vejalla Sep 21 '20 at 1:03
  • $\begingroup$ @VarunVejalla Do you have a graph of the function $y=\frac{\ln(x^x)}{W(\ln(x^x))}$? $\endgroup$ – KingLogic Sep 21 '20 at 1:06
  • $\begingroup$ It is here, but $y \not = x$ only for $x$ less than some value (I think $e^{-1}$). $\endgroup$ – Varun Vejalla Sep 21 '20 at 2:15
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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 $$y=1-\frac{(e-1)}{\sqrt[3]{e}} x^{2/3}$$

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  • $\begingroup$ 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 $\endgroup$ – KingLogic Sep 21 '20 at 10:11
  • $\begingroup$ @KingLogic.Let me check again $\endgroup$ – Claude Leibovici Sep 21 '20 at 10:12
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    $\begingroup$ @KingLogic. Checked ! Look at my edit for values. $\endgroup$ – Claude Leibovici Sep 21 '20 at 10:26
  • $\begingroup$ @KingLogic. Try $y=1-\frac{(e-1)}{\sqrt[3]{e}} x^{2/3}$ ! $\endgroup$ – Claude Leibovici Sep 21 '20 at 10:51
  • $\begingroup$ Btw, is there a closed form for its factor $\frac{y^y-x^x}{y-x}?$ $\endgroup$ – Narasimham Sep 21 '20 at 11:00

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