Question

The function `y = x^x` is differentiable at x = 1/e, What is the Taylor series ( if any ) for y, about x = 1/e?

Answer 1

The function `x^x` appears to be a constant, in an infinitesimal) interval `x in (1/e-in, 1/e+in)`.

Explanation:

At x = 1/e, y = `(1/e)^(1/e)`, and this is also transcendental.

It is easy to prove that the derivatives of all orders vanish at x = 1/e.

So, the function should behave as

`y=( 1/e)^(1/e)`

in an infinitesimal neighborhood

`x in(1/e-in, 1/e+in)`

The graph of `y = x^x` is having a multiple (?) -point

contact, with the bracing tangent

`y = (1/e)^(1/e)`,

at `(1/e, (1/e)^(1/e))`.

In my opinion, this contact is transcendental..

.