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Define a function $\mathscr E(x)$ on the interval $(e,\infty)$ implicitly via $$y=\mathscr E(x)~\Leftrightarrow~x^y=y^x,\quad x\neq y.$$ Show that $\frac{x+2}{x-1}$ approximates $\mathscr E(x)$ to an accuracy of $<0.05~$ for all $~x>e$.

The transcendental nature of $\mathscr E$ is the only thing that has this problem seemingly out of my reach; if a hint is all I may need please just drop one of those! I don't remember exactly how I originally got the value $0.05$, but I did so (clumsily) using Desmos' graphing calculator.

Supware
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  • I played with this years ago and found a parameterization for $x$ and $y$ in terms of one variable, but cannot recover it. I believe I made the guess that we could write $x=a^b,y=b^a$, so the equation becomes $(a^b)^{(b^a)}=(b^a)^{(a^b)}$ or $a^{b^{a+1}}=b^{a^{b+1}}$ and found $a$ and $b$ could be rational functions of the parameter. – Ross Millikan Mar 11 '19 at 00:04

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For the given error, you may be able to just compute. Define $\mathscr F(x) = \mathscr E^{-1}(x)$. We can find $\mathscr F(1.05) \approx 99, \mathscr F(1.1) \approx 43.5$ Then for $x \gt 99, 1 \lt \mathscr E(x) \lt 1.05, 1 \lt \frac {x+2}{x-1}=1+\frac 3{x-1}\lt 1.030303\ldots$ and the inequality is satisfied. Compute $\mathscr F(x)$ on steps of $0.025$ from $1.05$ up to $e$ and verify that the two expressions are close.

Ross Millikan
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