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.
Choose some small delta, d. Then sort all functions such that
f(x)<<g(x) whenever the integral from 0 to d of f(x)dx is less
than the integral of g(x)dx. For cases where these integrals are
equivalent, sort further by the integral from 0 to d/2, then 0 to d/4
etc. If f(x)==g(x) in this interval we move on to the interval -d
to 0, continually halving (-d/2,0), (-d/4,0) etc. if f(x)==g(x) over
(-d,d) we move on to (d,2d), then (-2d,-d) until we find some
discrepancy in the integrals. Eventually we will locate the
difference between the functions (because they are continuous).
Discontinuous functions don't seem so easy.
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Posted by Eric
on 2006-08-22 13:25:47 |