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

Aug 27, 2016

The function ${x}^{x}$ appears to be a constant, in an infinitesimal) interval $x \in \left(\frac{1}{e} - \in , \frac{1}{e} + \in\right)$.

#### Explanation:

At x = 1/e, y = ${\left(\frac{1}{e}\right)}^{\frac{1}{e}}$, 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

$y = {\left(\frac{1}{e}\right)}^{\frac{1}{e}}$

in an infinitesimal neighborhood

$x \in \left(\frac{1}{e} - \in , \frac{1}{e} + \in\right)$

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

contact, with the bracing tangent

$y = {\left(\frac{1}{e}\right)}^{\frac{1}{e}}$,

at $\left(\frac{1}{e} , {\left(\frac{1}{e}\right)}^{\frac{1}{e}}\right)$.

In my opinion, this contact is transcendental..

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