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Does 0^0 equal 1, or is it undefined?

  1. May 15, 2007 #1
    Does 0^0 equal one, or is it undefined?
  2. jcsd
  3. May 16, 2007 #2


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    It's an indeterminate form. (It's many-valued; not the same as undefined.)

    - Warren
  4. May 16, 2007 #3
    Is the limit of sin(x), as x approaches infinity, also "indeterminate"?
  5. May 16, 2007 #4
    It does not exist because the values oscillate between 1 and -1.
  6. May 16, 2007 #5
    There are three interpretations (each one depends on context):

    0^0 = 1, or indeterminate form, or undefined

    The first is the set-theoretic interpretation. Justification?

    Consider f:A->B. The set of all such functions is denoted B^A. In this context, {}^{} would represent the set of
    functions f:{}->{}, and 0^0 would represent the number of functions in this set. There is only one such function.
    (This is not the only justification for this particular interpretation, i.e., 0^0 = 1.)

    It's a strange post. A cursory preliminary investigation would turn up something like the following:

    Last edited: May 16, 2007
  7. May 16, 2007 #6
    In combinatorics it's usually convention to use 0^0=1. This makes some problems easier (less book-keeping).

    Plot y^x and x^y and see what you get as y->0. The answer clearly depends on the direction of approach.
  8. May 16, 2007 #7

    D H

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    Just to add one very important interpretation of when 0^0 is defined to be 1: Power series.

    In power series, the form 0^0 must be taken to mean 1. If not, we would have to write [itex]\exp(x) = 1 + \sum_{n=1}^{\infty} x^n/n![/itex] rather than [itex]\exp(x) = \sum_{n=0}^{\infty} x^n/n![/itex]
  9. May 16, 2007 #8

    Gib Z

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    And for the binomial theorem:

    [tex](1+x)^n = \sum_{k = 0}^n \binom{n}{k} x^k[/tex]

    Its not valid for x=0 except when defined 0^0 = 1
  10. May 18, 2007 #9
    We really only say indeterminate when we're talking about limits. It's either undefined, or defined arbitrarily depending on context.
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