6
$\begingroup$

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.

  • 2
    How do you even define $x^\pi$ if x<0? The answer to this question should show you why there's a singularity at 0 without having to take derivatives.2010-09-13

3 Answers 3

7

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.

  • 1
    Planeman: Sure, $\exp(\pi\ln\;x)$ is differentiable at $x=1$.2010-09-13
7

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.

  • 0
    I honestly hadn't considered that issue until you brought it up. It's an excellent point!2010-09-14
4

As a graphical supplement to Jonas's and WWright's answers:

real and imaginary parts of x^\pi

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.

  • 0
    Ah, yes you're right, thanks for the reminder and supplying the additional rigor needed.2010-09-14