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.
Explanation of Maclaurin Series of $x^\pi$
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2How 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
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.
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1Planeman: Sure, $\exp(\pi\ln\;x)$ is differentiable at $x=1$. – 2010-09-13
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.
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0I honestly hadn't considered that issue until you brought it up. It's an excellent point! – 2010-09-14
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.
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0Ah, yes you're right, thanks for the reminder and supplying the additional rigor needed. – 2010-09-14