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Zero to the zeroth power

  1. Feb 26, 2015 #1
    Suppose we wish to represent ##f(x)## as a power series:

    $$f(x) = \sum_{k=0}^{∞} a_k x^k = a_0 x^0 + a_1 x + a_2 x^2 + .....$$

    How is it that ##f(0) = a_0## if ##x^0 = 1## only for nonzero ##x##?
     
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  3. Feb 26, 2015 #2

    Orodruin

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    You just have to be mindful of your definitions. In this case ##x^0## should really be interpreted as 1. It is simply the only way of doing it here, you could also just pull the term out of the sum. If it was not interpreted as one your function would not even be differentiable.
     
  4. Feb 26, 2015 #3

    mathman

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    Writing f(x) as a power series implies that f(x) is analytic (and therefore continuous) within its circle of convergence, so f(0)=a0.
     
    Last edited: Feb 27, 2015
  5. Feb 27, 2015 #4

    micromass

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    Many authors use the convention that ##0^0 = 1##, and there are many good reasons. For example, Knuth (who invented LaTeX) wrote in his (very beautiful) book "concrete mathematics:

    Some textbooks leave the quantity ##0^0## undefined, because the functions ##0^x## and ##x^0## have different limiting values when ##x## decreases to ##0##. But this is a mistake. We must define ##x^0=1## for all ##x## , if the binomial theorem is to be valid when ##x=0## , ##y=0## , and/or ##x=-y## . The theorem is too important to be arbitrarily restricted! By contrast, the function ##0^x## is quite unimportant.

    On the other hand, there are also many mathematicians leaving it undefined. The resolution I take is that ##0^0## is ##1## if the exponent is only allowed to be integers. So ##a^n## where ##n\in \mathbb{Z}## only. If the exponent is allowed to be more general real numbers, then it's best to leave it undefined.

    That said, if you choose to follow the convention of ##0^0 = 1##, then there is nothing wrong with that as long as you're consistent.
     
  6. Mar 20, 2015 #5
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