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Posers and Puzzles

Posers and Puzzles

  1. 19 Aug '08 15:45
    Which value of x between 0 and 1 maximizes x^m*(1-x)^n+x^n*(1-x)^m, where m and n are positive integers?
  2. Standard member PBE6
    Bananarama
    19 Aug '08 20:40
    Originally posted by David113
    Which value of x between 0 and 1 maximizes x^m*(1-x)^n+x^n*(1-x)^m, where m and n are positive integers?
    My first thought was to say that since this function is symmetrical with respect to x and (1-x), the answer would either be an extreme point or the mid-point. The extreme values x=0 and x=1 make the function f(x) equal to 0. However, after graphing this function using various values of m and n, I found that certain combinations create two maximums equally spaced from the centre.

    The derivative of the function is a little ugly, so I'm going to look for a simplification.
  3. Standard member TheMaster37
    Kupikupopo!
    21 Aug '08 12:40 / 6 edits
    Originally posted by PBE6
    My first thought was to say that since this function is symmetrical with respect to x and (1-x), the answer would either be an extreme point or the mid-point. The extreme values x=0 and x=1 make the function f(x) equal to 0. However, after graphing this function using various values of m and n, I found that certain combinations create two maximums equally spa ...[text shortened]...

    The derivative of the function is a little ugly, so I'm going to look for a simplification.
    For x=0 and x=1 the function (let's call it F) is at a minimum, so for the rest of the train of thoughts I will assume x is strictly between 0 and 1.

    Let x = 1/2 - y. Let m be the larger of m and n, let m - n = a

    Then F reduces to

    (1/4 - y^2)^n * ( (1/2 - y)^a + (1/2 + y)^a ) with (1/4 - y^2) > 0

    I'm not going to type out F' here, but from it you can see that either
    y = 0 => x = 1/2
    or (substituting x in again)
    n(1-2x)(x^a + (1-x)^a) - ax(1-x)(x^(a-1) - (1-x)^(a-1)) = 0

    At this point I REALLY hope I didn't make any mistakes.