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

##### 1 Answer

Aug 27, 2016

The function

#### Explanation:

At x = 1/e, y =

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

So, the function should behave as

in an infinitesimal neighborhood

The graph of

contact, with the bracing tangent

at

In my opinion, this contact is transcendental..

.