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Math Help - Explanation for Maclaurin Series of X^pi

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    Explanation for Maclaurin Series of X^pi

    I am confused as how to explain this. Take the function  f(x) = x^{\pi} . I am fairly certain this does not have a series expansion but is it just because it is already a polynomial? In that case is it a Maclaurin series itself?
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    It's not a polynomial because the power of x is not a nonnegative integer.

    I would think that a MacLaurin series is inappropriate here because if you centre the polynomial around x = 0, you'll find that your constants all end up being 0, until you get into negative powers of x in which case you end up with them having zero denominators.

    I would try a Taylor series centred around x = 1.
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    I am not trying to solve this for any practical purpose. The question specifically asks "Why does this function not posses a Taylor series expansion at x=0?"
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    Quote Originally Posted by Planeman View Post
    I am not trying to solve this for any practical purpose. The question specifically asks "Why does this function not posses a Taylor series expansion at x=0?"
    In that case, it is because when you try to evaluate the coefficients, they all end up being 0 or \frac{\textrm{something}}{0}.
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