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- Newsgroups: sci.math
- Path: sparky!uunet!caen!umeecs!quip.eecs.umich.edu!kanad
- From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
- Subject: Re: Integral Puzzle (SPOILER)
- Message-ID: <1992Dec20.183116.23990@zip.eecs.umich.edu>
- Sender: news@zip.eecs.umich.edu (Mr. News)
- Organization: University of Michigan EECS Dept., Ann Arbor
- References: <1992Dec19.011244.2780@Csli.Stanford.EDU> <Dec.20.11.00.28.1992.16004@romulus.rutgers.edu>
- Date: Sun, 20 Dec 1992 18:31:16 GMT
- Lines: 44
-
- In article <Dec.20.11.00.28.1992.16004@romulus.rutgers.edu> clong@romulus.rutgers.edu (Chris Long) writes:
- >In article <1992Dec19.011244.2780@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?
- >
- >Let I = \int_0^1 f(x)^2 dx; the fact that I >= A^2 follows immediately
- >from the Cauchy-Schwarz inequality. For the other (more interesting)
- >direction, consider that if f(x) is non-negative, then so is a-f(x),
- >and so (f(x)+a-f(x))^2 >= f(x)^2+(a-f(x))^2, hence \int_0^1 a^2 dx >=
- >\int_0^1 f(x)^2 dx + \int_0^1 (a-f(x))^2 dx ==> a^2 >= I + a^2 - 2aA
- >+ I ==> I <= aA. To finish, such functions exist. For the first case,
- >consider f(x)=A. For the second, consider f(x)=a for 0 <= x <= A/a,
- >f(x)=0 elsewhere.
- >--
- >Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
- >
- >"S.B., with an I.Q. of 161, failed to complete his course of study,
- >running away instead with his professor's wife."
- >H. J. Eysenck, _Know Your Own I.Q._
-
- An easier way to prove that the integral is <= aA is as follows :
-
- f(x)^2 = f(x) * f(x) <= a * f(x), since f(x) <= a for 0<=x<=1;
- hence :
-
- /1 2 /1
- | {f(x)} dx <= | a * f(x) dx = aA
- /0 /0
-
- Kanad
-