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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.

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Are you sure it's infinitely differentiable? –  Qiaochu Yuan Sep 13 '10 at 22:48
    
Well I may be missing something but it would go $\pi * x^(\pi-1)$ then $\pi * (\pi -1)x^(\pi-2)$...I don't see where it would have a gap. –  Planeman Sep 13 '10 at 22:54
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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. –  Jonas Meyer Sep 13 '10 at 23:05

3 Answers 3

up vote 7 down vote accepted

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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Oh, ok. I was forgetting to evaluate the power and therefore I didn't realize that at a certain point it turned negative. Would a function like this have an expansion around something like x=1? –  Planeman Sep 13 '10 at 23:04
    
I think centering it at 1 or any positive number would work fine –  WWright Sep 13 '10 at 23:16
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Planeman: Sure, $\exp(\pi\ln\;x)$ is differentiable at $x=1$. –  J. M. Sep 13 '10 at 23:17

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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I'll add that I agree that WWright's answer is a good one: the fourth derivative going to infinity as x decreases to 0 shows that the function can't be 4 times differentiable in a neighborhood of 0, no matter what definition you give it for negative x. In fact, I didn't prove anything. My intent is in part to show that there are problems even in determining what is being asked. –  Jonas Meyer Sep 14 '10 at 0:00
    
I honestly hadn't considered that issue until you brought it up. It's an excellent point! –  WWright Sep 14 '10 at 0:17

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.

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I agree that the existence of a cut is relevant to the singularity at 0, but I'm not sure I follow your last comment. For example, the function $f$ in my answer can be uniformly approximated by polynomials (with complex coefficients) on each interval $[-\epsilon,\epsilon]$, by Weierstrass's approximation theorem. Also, I think the question was just about the existence of the Maclaurin series, not whether it approximates the function; for that the fact that the 4th derivative doesn't exist suffices. –  Jonas Meyer Sep 14 '10 at 1:35
    
(But it's true that the cut implies that the Maclaurin series couldn't both exist and locally approximate the function.) –  Jonas Meyer Sep 14 '10 at 1:37
    
I'm essentially saying that trying to find a Maclaurin expansion is the same as asking "does the function behave like a polynomial in the vicinity of the point of expansion?" The cut makes $z^\pi$ not behave like a polynomial in the vicinity of $z=0$. Alternatively, we can think of $z^\pi$ as $z^3 z^{\pi-3}$, the product of a (known) smooth function and the "unknown" $z^{\pi-3}$. $z^{\pi-3}$ has a vertical tangent at the origin, and polynomials cannot have vertical tangents. –  J. M. Sep 14 '10 at 1:52
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Perhaps I'm just nitpicking, but I was reacting to what you wrote: unlike on open sets in the plane, on bounded intervals every continuous function, no matter how non-polynomial-like, can be uniformly approximated by polynomials. Approximation by the Maclaurin series is asking for much more than this, namely analyticity, and I believe that this is what you intended. –  Jonas Meyer Sep 14 '10 at 2:00
    
Ah, yes you're right, thanks for the reminder and supplying the additional rigor needed. –  J. M. Sep 14 '10 at 2:19

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