After seeing a discussion about graphs of the relationship $x^x + y^y = r^r$, it got me interested in attempting to see what the graphical appearance of ${^{\infty}x}+{^{\infty}y}={^{\infty}r}$ would look like. The first step I did was use the relationship of ${^{\infty}n}=-\frac{W(-\ln(n))}{\ln(n)}$ to give me the equation of: $$-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $$

Where the range of possible real values for $r$ is $e^{−e} < r < e^{1/e}$. Using this I attempted to define the Lambert W function inside of Desmos using $$\text{}$$ $$W\left(x\right)=-\frac{2}{\pi}\int_{0}^{\pi}\frac{\sin\left(\frac{t}{2}\right)\left(\sin\left(\frac{3t}{2}\right)+e^{\cos\left(t\right)}x\sin\left(\frac{5t}{2}-\sin\left(t\right)\right)\right)}{1+e^{2\cos\left(t\right)}x^{2}+2e^{\cos\left(t\right)}x\cos\left(t-\sin\left(t\right)\right)}dt\left\{-\frac{1}{e}<x<e\right\}$$ $$\text{}$$

and from there try to graph $-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $, in the hopes of being able to get a rough idea of what value of $r$ gives the greatest arc length. However, Desmos was extremely laggy and any change in the value of $r$ would take over a minute to be reflected by the graph.

This would take far too long to reasonably do so instead I decided to try to find a way to define a new function as being the arc length of the equation $-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $, get the derivative of said new function and solve for when the derivative equals zero to get the value of $r$ that produces the maximum arc length.

However, at this point I encountered another problem. The only way I know how to calculate the length of a curve is by using the formula $\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx$ for carteasian coordinates and $\int_{\theta_1}^{\theta_2} \sqrt{r^2+(\frac{dr}{d\theta})^2} d\theta$ for polar coordinates, but the prior isn't useful without rearranging for $y$ to be the subject and the latter isn't useful as I'm unsure of how to rearrange for $r$ after converting this particular equation to polar coordinates.

So, to summarize, I would very much appreciate if someone could help by telling me if there's a way to make $y$ the subject in $-\frac{W(-\ln(x))}{\ln(x)} -\frac{W(-\ln(y))}{\ln(y)} = -\frac{W(-\ln(r))}{\ln(r)} $ or if there's an alternate method I could use to evaluate the arc length of this curve.

  • 3
    $\begingroup$ what do you mean by $^\infty x$ ? $\endgroup$ – Jean Marie Dec 7 '19 at 9:34
  • $\begingroup$ @Jean Tetration is used to help express repeated exponents. $^{n}a$ means to repeatedly raise a to the power of itself ($a^{{a}^{a}}$ and so on) until the total number of $a$'s equals $n$. So $^{4}a$=a^a^a^a. $\endgroup$ – Aussie Mathematician Dec 7 '19 at 10:04
  • $\begingroup$ The inverse of $x=W(y)$ is $y=x e^x$ if that helps. $\endgroup$ – Gottfried Helms Dec 10 '19 at 7:26

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