Taylor series about $x = e$ of $x^y = y^x$

Question

I was wondering how I could find the Taylor series expansion about $x = e$ of the branch of $x^y = y^x$ that isn't $y = x$.

I know for a general Taylor series, you are supposed to find the $n$th derivatives. When I use implicit differentiation, I get $$\frac{dy}{dx}=\frac{\ln\left(y\right)-\frac{y}{x}}{\ln\left(x\right)-\frac{x}{y}}$$

Looking at the graph and considering that it is symmetric about $y = x$, the first derivative should be $-1$. However, if I plug in $(x,y) = (e,e)$, I get an indeterminate form of $0/0$. If I used L'Hopital's on this, I would have to calculate the derivative of $y$, which just results in the same problem. I'm guessing the problem lies in the fact that $(e,e)$ is on both branches of the function.

So, how exactly can I find the Taylor series expansion about $x = e$ of $x^y = y^x$ if I can't even find the first derivative mathematically?

This function is seen in this graph

Answer 1

The solution you're looking for is $$y = -{\frac {x}{\ln \left( x \right) }{\rm W} \left(-{\frac {\ln \left( x \right) }{x}}\right)}$$ where $W$ is the Lambert W function, and you use the "$-1$" branch for $0 < x \le e$ and the "$0$" branch for $x \ge e$.

The first few terms of the series (according to Maple) are

$$\eqalign{({\rm e}&-(x-{\rm e})+{\frac {5\,{{\rm e}^{-1}}}{3}} \left( x-{\rm e} \right) ^{2}-{\frac {25\,{{\rm e}^{-2}}}{9}} \left( x-{\rm e} \right) ^{3}+{\frac {1243\,{{\rm e}^{-3}}}{270}} \left( x-{\rm e} \right) ^{4}-{\frac {1229\,{{\rm e}^{-4}}}{162}} \left( x-{\rm e} \right) ^{5}\cr &+{\frac {14107\,{{\rm e}^{-5}}}{1134}} \left( x-{\rm e} \right) ^{6}-{\frac {575927\,{{\rm e}^{-6}}}{28350}} \left( x-{\rm e} \right) ^{7}+{\frac {4217764\,{{\rm e}^{-7}}}{127575}} \left( x-{ \rm e} \right) ^{8}-{\frac {1408003\,{{\rm e}^{-8}}}{26244}} \left( x- {\rm e} \right) ^{9}\cr &+{\frac {18804662561\,{{\rm e}^{-9}}}{216513000}} \left( x-{\rm e} \right) ^{10} }$$