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- Newsgroups: sci.math,rec.puzzles
- Path: sparky!uunet!zaphod.mps.ohio-state.edu!caen!hellgate.utah.edu!fcom.cc.utah.edu!mail.physics.utah.edu!freier
- From: freier@mail.physics.utah.edu (rodney james freier)
- Subject: Re: Integral Puzzle (Cute, not Evil)
- Message-ID: <1992Dec26.080020.10668@fcom.cc.utah.edu>
- Sender: news@fcom.cc.utah.edu
- Organization: University of Utah - Physics Department
- References: <1992Dec19.011244.2780@Csli.Stanford.EDU>
- Date: Sat, 26 Dec 92 08:00:20 GMT
- Lines: 26
-
- In article <1992Dec19.011244.2780@Csli.Stanford.EDU>, hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
- |>
- |> Let f be any function such that:
- |> * for 0<=x<=1; 0<=f(x)<= a
- |>
- |> /1
- |> * | f(x)dx = A
- |> /0
- |> where a and A are given constants (obviously, A<=a).
- |>
- |> What are the maximum and minimum values that
- |>
- |> /1 2
- |> | {f(x)} dx
- |> /0
- |>
- |> can take?
- |> The values can be easily induced, but come up with
- |> a simple (cute) proof.
- |>
- |> -Yuzuru Hiraga
-
-
- Eeep. Doesn't the second expression become undefined if we choose f(x) to
- be A*delta(x-.5) or the like? I guess my point is that there may be functions
- for which the first expression is just peachy, but the second becomes pathological.
-