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From what I know the function $$f(x)=xe^{x}$$ doesn't have an inverse function $f^{-1}$ that can be written in a closed form. However his inverse function $f^{-1}$ does exists and it's "famous": it's the Lambert W function. Wonderful, but then what about finding the inverse function of functions similar to $f$? For example: in the case of the function $$g(x)=\frac{1}{x}e^x$$ turns out that its inverse can be written using the $W$ function: $$g^{-1}(x)=-W\left(-\frac{1}{x}\right)$$ But what about other similar functions? In particular I am interested in the function: $$h(x)=\frac{1}{x^2}e^x$$ what is the inverse of this function? Can we express it via $W$ as we have done for $g(x)$?

Noumeno
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  • did you plot $e^x / x^2$ ? – G Cab Jun 27 '21 at 13:56
  • @GCab Why? I don't care about the plot, I want to know how to write the inverse of it. – Noumeno Jun 27 '21 at 14:00
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    You are talking about inverses of functions a little loosely. Whenever you are looking for the inverse of a function, you should first think about whether or not your function is bijective. If it is not (as in all of your examples here), you should find some domain on which your function is, so that you can consider a restriction of the function and find its inverse. – Stefan Octavian Jun 27 '21 at 14:02
  • when you plot $y=f(x)$ do you know what is the plot of x=f^{-1}(y)$ – G Cab Jun 27 '21 at 14:03
  • I gave a general method for solving equations by Lambert W in my answer to https://math.stackexchange.com/questions/2578139/algebraic-solution-to-natural-logarithm-equations-like-1-xx-ln-x-0/2578259#2578259 – IV_ Jun 27 '21 at 15:17

2 Answers2

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First, let's rewrite the expression: $$ \begin{align} h &= \frac{1}{x^2}e^x \\ \frac{1}{h} &= x^2e^{-x} \\ \frac{1}{\sqrt h} &= \pm x e^{-x/2} \\ \pm\frac{1}{2\sqrt h} &= -\frac{x}{2} e^{-x/2} \\ \end{align} $$

For $k\in\mathbb Z$ define $W_k\colon\mathbb C\to\mathbb C$ so that $z=W_k(z)e^{W_k(z)}$. Now apply the Lambert function on both sides:

$$ \begin{align} W_k(\pm\frac{1}{2\sqrt h}) &= W_k(-\frac{x}{2} e^{-x/2})=-\frac{x}{2} \\ \end{align} \Longrightarrow x=-2W_k(\pm\frac{1}{2\sqrt h}). $$

vitamin d
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If $x>0$ and $y=e^x/x^2$, then $(-x/2)e^{-x/2}=-y^{-1/2}/2$, leading to an expression using Lambert's $W$. The case $x<0$ is handled similarly.

metamorphy
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