Explanation of Maclaurin Series of $x^\pi$ 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. 
 A: Consider what happens for higher powered derivatives. $\frac{d^4}{dx^4}x^{\pi}=(\pi)(\pi -1)(\pi -2)(\pi -3)x^{\pi -4}$ You can think of this as $c \frac{1}{x^{k}}$ for some positive real number k which is not defined at 0.
A: I decided to elaborate on my comment above.
What is $x^\pi$ if $x$ is negative?  Trying to approximate with $x^r$ for rational $r$ is problematic, depending on which rational numbers you use to approximate; should the answer be positive, negative, imaginary?  Perhaps more reasonable would be to take $x^\pi=e^{\pi \log(x)}$ for some suitably chosen branch of the logarithm on $(-\infty,0)$, the most common choice being $\log x=\ln(-x)+\pi i$.  Your function then becomes 
 $$
   f(x) = \left\{
     \begin{array}{lr}
       e^{\pi \ln x} & : x>0 \\
       0 & : x=0 \\
       e^{i\pi^2}e^{\pi \ln(-x)} & :x<0
     \end{array}
   \right.
$$
In particular, it is not real-valued.  You can still go ahead and try to take its derivatives, but this piecewise representation may make it less surprising that there is going to be a singularity at zero.
A: 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.
