How can we compute averages over infinite sets of functions?

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Discussion Overview

The discussion revolves around the concept of computing averages over infinite sets of functions, particularly focusing on the implications of defining what an "average" means in this context. Participants explore theoretical considerations and mathematical definitions related to averaging functions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant notes that the set of all functions is larger than \(2^{\aleph_0}\) and questions how to average over such a set.
  • Another participant challenges the clarity of the term "average," suggesting it lacks a precise mathematical definition.
  • A suggestion is made to "add them all up and then divide by the total number," but this is critiqued for lacking a defined mathematical procedure.
  • Some participants mention that "average" can refer to "expectation" under a probability distribution, while others argue that the concept of averaging over all real numbers is similarly undefined.
  • One participant asserts that the set of all functions does not exist due to its size, proposing that if one considers real-valued functions, the average could be \(f(x) = 0\) because for every function \(f(x)\), there exists a function \(g(x) = -f(x)\).
  • Another participant counters that symmetry does not lead to a unique answer, suggesting that for every function \(h(x)\), there exists a function \(g(x) = 5 - h(x)\), which could imply an average of \(f(x) = 5\).
  • A later reply expresses a tentative preference for \(f(x) = 0\) but acknowledges the need for further consideration of the problem.

Areas of Agreement / Disagreement

Participants express differing views on the definition of "average" and its applicability to infinite sets of functions. There is no consensus on a specific method for computing such an average, and multiple competing ideas are presented.

Contextual Notes

The discussion highlights limitations in defining mathematical operations like "add" and "divide" in the context of infinite sets. The implications of symmetry in determining averages remain unresolved.

cragar
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The set of all functions is larger than [itex]2^{\aleph_0}[/itex].
So let's 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.
 
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Define what you mean by "average". The word is pretty much useless in mathematics.
 
add them all up and then divide by the total number.
 
cragar said:
add them all up and then divide by the total number.

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.
 
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)
 
Dmobb Jr. said:
,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)

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.
 
Stephen Tashi said:
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.

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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