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Averages over infinite sets

  1. Jun 3, 2013 #1
    The set of all functions is larger than [itex] 2^{\aleph_0} [/itex].
    So lets say I wanted to average over all functions over some given region. that was
    larger than [itex] 2^{\aleph_0} [/itex] how would I do that.
  2. jcsd
  3. Jun 3, 2013 #2


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    Define what you mean by "average". The word is pretty much useless in mathematics.
  4. Jun 3, 2013 #3
    add them all up and then divide by the total number.
  5. Jun 3, 2013 #4

    Stephen Tashi

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    That doesn't define a particular mathematical procedure unless you can define "add" and "divide" in your context.

    The problem of defining an "average" of all the real numbers seems conceptually simpler and I don't know of any useful definition for such an average.

    Some people use the term "average" to mean "expectation". If you have a particular probability distribution on the set of all real numbers, the "expectation" of that distribution is defined. Some people use the term "average" to mean the "mean value" of a finite sample of data or the "expectation" of a probability distribution.
  6. Jun 3, 2013 #5
    Technically the set of all functions does not exist. It is to big to be a set.

    If you mean real valued functions, while the above post is completely correct, I think the answer you want is f(x) = 0. If you sum every function, I believe you will get 0 because for every f(x) there exists g(x) = -f(x)
  7. Jun 3, 2013 #6

    Stephen Tashi

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    I don't think symmetry directs us to a particular answer. For every function h(x) = there is a function g(x) = 5 - h(x) so by the same reasoning, the answer would be the function f(x) = 5.
  8. Jun 3, 2013 #7
    Good point. I am pretty sure 0 is still going to be the best answer but I have to consider the problem more thoroughly.

    Edit: The answer will probably be 0 assuming there is some sort of reasonable answer at all.
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