19 Aug 08
Originally posted by David113My 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.
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?
The derivative of the function is a little ugly, so I'm going to look for a simplification.
21 Aug 08
6 edits
Originally posted by PBE6For 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.
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.
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.