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Sort of a sorting problem (Posted on 2006-08-22) Difficulty: 5 of 5
Sort the set of functions f:R→R, with R being the set of real numbers. This means you have to find an order "«" that lets you compare any two pairs of unequal functions f and g; unequal means, f(x)≠g(x) for at least one x.

More precisely, these are the requirements for the order "«" you are challenged to find:

1. If f≠g, either f«g or g«f.
2. If f«g and g«h then f«h

You might be tempted to declare f«g when f(x)<g(x) for all x but that would of course fail because e.g. f(x)=x and g(x)=-x would not be comparable with respect to your order.

For a much, much easier challenge, start by finding an order for all continuous functions.

See The Solution Submitted by JLo    
Rating: 4.3333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): For continuous functions... | Comment 6 of 14 |
(In reply to re(2): For continuous functions... by Eric)

Eric:

I don't think your idea in comment 4 works either.

Consider:   Define functions
 
  f(x) = 0 except between d and 2d
        = sine(2pi(x-d)/d) between d and 2d

  g(x) = -f(x)


The integral of f(x) and g(x) is 0 for any region that includes all of or none of [d,2d].   The integral of |f(x)| and |g(x)| are equal for any region that includes all of or none of [d,2d].

These functions is indistinguishable if the integral is evaluated between 0 and d, -d, d/2, 2d, -d/2, -2d, d/4, 4d, -d/4, -4d, d/8, 8d, etc.

  Posted by Steve Herman on 2006-08-24 01:45:52

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