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 #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..

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