Lambert W Function: Solving for the Inverse of x^x = y

[arpon]

If, $f(x)=x^x$, then, f^-1(x)=?

 

[guysensei1]

I don't think there's a closed form expression for that.

 

[jedishrfu]

Here's an old discussion on x^x from PF

https://www.physicsforums.com/threads/x-x-does-this-function-have-a-name.70230/

and here;

http://www.faculty.fairfield.edu/jmac/ther/tower.htm

 

[JJacquelin]

x^x=y
x=exp(W(ln(y)))
W is the LambertW function
http://mathworld.wolfram.com/LambertW-Function.html

 

[HallsofIvy]

Nicely done! I misread your answer at first and thought you had it wrong but saw that you are correct after working it out for myself. The Lambert W function is inverse to f(x)= xe^x but taking the logarithm of both sides of x^x= y gives xln(x)= ln(y) not xe^x= y.

Instead, once you have xln(x)= ln(y), let u= ln(x). Of course, then, x= e^{ln(x)}= e^u so the equation becomes
ue^u= ln(y), u= ln(x)= W(ln(y)) so that, as you say, x= exp(W(ln(y)).