Question

What is the inverse fun of `y=x^x`?

Answer 1

For `x > 0` the real valued function is given by:

`y = e^(W_0(ln x))`

More generally:

`y = e^(W_n(ln x + 2kpi i))" "` for any `k, n in ZZ`

Explanation:

Given:

`y = x^x`

This question is essentially asking us to solve for `x` in terms of `y`.

This cannot be done with elementary functions, but we can use a family of functions called the Lambert `W` function, that provide the inverse to:

`w = ze^z`

namely:

`z = W_n(w)`

where `n in ZZ` is the index of the branch, with `W_0` being the principal branch, providing the unique real valued solution when `w > 0`.

Putting `x = e^t` in our given equation we have:

`y = x^x = (e^t)^(e^t) = e^(t e^t)`

So taking logs, we have:

`ln y = t e^t`

or more generally for complex solutions:

`ln y + 2kpi i = t e^t" "` for any `k in ZZ`

So using the Lambert `W` function, we have:

`t = W_n (ln y)`

or more generally

`t = W_n (ln y + 2kpi i)`

Then taking the exponent of both sides:

`x = e^t = e^(W_n(ln y))`

or more generally:

`x = e^t = e^(W_n(ln y + 2kpi i))`

So the inverse of:

`f(x) = x^x`

is:

`f^(-1)(y) = e^(W_n(ln y))`

or more generally:

`f^(-1)(y) = e^(W_n(ln y + 2kpi i))`

or if you prefer in terms of `x` and `y`, swap them in the expression we found to get:

`y = e^(W_n(ln x))`

or more generally:

`y = e^(W_n(ln x + 2kpi i))`

This is actually a family of functions indexed by branch number `n` rather than a single function.

For `x > 0` we can choose the real valued principal branch and write:

`y = e^(W_0(ln x))`