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I have seen posts on the inverse function of $x+e^x$ (which is $y-W(e^y)$), but (out of curiosity) how do we derive an inverse function for $x^n+e^x$ in terms of the Lambert W function? I tried to do what @MarkViola did here but I couldn't.

IV_
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Kamal Saleh
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  • There isn't always a closed formula for an inverse in terms of known functions. Just think of all the functions that are defined to be inverses. Logarithms, nth roots, arctan, etc. The Lambert $W,$ used in your example, is function is a function defined to be n inverse. Sometimes, you can't get an inverse in closed form. – Thomas Andrews Dec 14 '22 at 17:22
  • @ThomasAndrews I edited the OP – Kamal Saleh Dec 14 '22 at 17:24
  • @ThomasAndrews I know that at times I couldn't get closed form solutions but so far there is no proof that this specific function doesn't have an inverse. – Kamal Saleh Dec 14 '22 at 17:25
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    If it's bijective, it has an inverse. It looks easy enough to prove it's bijective...? But the existence of an inverse does not mean you can write this inverse as elementary functions. – kevinkayaks Dec 14 '22 at 17:27
  • Still, if you give me $x$ with some precision, I'll give you $y$ with the same precision, and vice versa. The function is invertible – kevinkayaks Dec 14 '22 at 17:29
  • $x^n+e^x$ is an algebraic expression of more than one algebraically independent monomial ($x^n,e^x$). Therefore it's not in a form that allows to read partial inverses that are elementary functions. We therefore don't know how we can invert the expression. If partial inverses exist that are elementary functions, is a different mathematical problem. – IV_ Dec 15 '22 at 16:41
  • @KamalSaleh For rational $n$, $x^n+e^x$ is an algebraic expression of $x$ and $e^x$. I recently found that the non-existence of partial inverses of such kind of expression that are elementary functions can be proved with help of Ritt's proof for Kepler's equation: https://math.stackexchange.com/questions/4586412/how-can-we-show-that-az-ez-and-a-ln-z-z-have-no-elementary-inverse/4586413#4586413 – IV_ Dec 15 '22 at 17:09
  • Series solution demo with subsitution here – Тyma Gaidash Dec 16 '22 at 02:05
  • I did not close vote, but please try to add more details on your ideas for the inverse function etc otherwise more close votes may occur. – Тyma Gaidash Dec 17 '22 at 12:24
  • @TymaGaidash I couldn't add more details :( – Kamal Saleh Dec 17 '22 at 15:56

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$x^n+e^x$ is not always invertible, for example when $n=2$. Therefore, your expression does not always have an inverse.

Certain values of $n$ will give elementary inverses, for example $n=0$ has the simple inverse $\log(x-1)$, and $n=1$ has $x- W(e^x)$ as its inverse, as can be seen here. It is likely, though I could not prove it, there is no elementary way to create the inverse when $n=3$ using just the $W$ function.

Snared
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    $x^n+e^x$ has a local inverse, not a global inverse. The same is true for $e^x$ and $x+e^x.$ This is common. We still call $\arctan$ an inverse. $x^n+e^x$ has a local inverse near $f(x_0)$ when $f'(x_0)\neq 0.$ Solving: $0=f'(x_0)=nx_0^{n-1}+e^{x_0}:$ when $n>0$ is even, at least, we can solve for $x_0$ using the $W$ function. There is no real $x_0$ when $n$ is odd. So there is a local inverse everywhere when $n$ is odd, and everywhere but one point when $n$. Is even. – Thomas Andrews Dec 14 '22 at 21:50