TITLE: Explanation of Maclaurin Series of $x^\pi$
QUESTION [6 upvotes]: I am reviewing Calc $2$ material and I came across a problem which asked me to explain why $x^\pi$  does not have a Taylor Series expansion around $x=0$. To me it seems that it would have an expansion but it would just be $0$, so maybe it's not a suitable expansion. It doesn't have any holes and it is infinitely differentiable so I don't know why it couldn't have an expansion.

REPLY [4 votes]: As a graphical supplement to Jonas's and WWright's answers:

This is a plot of the real and imaginary parts of $(x+iy)^\pi$ in the complex plane. Note the cut running across the negative real axis. This cut is precisely the reason why you cannot have a Maclaurin expansion; polynomials cannot exhibit cuts, and a Maclaurin expansion amounts to approximating your function with a sequence of polynomials.