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Advanced Math/Extrema of f(x) = ln(x) * ln(1-x)

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Question
I wonder how one can give a rigorous argument about the uniqueness of extremum of the function above.

This question arises when minimizing the probability of false positives,
p(k) = (1 - [1 - 1/m]^(k*n))^k /approx (1 - e^(-(k*n)/m))^k ,
of Bloom filters; see http://en.wikipedia.org/wiki/Bloom_filter for background.

All of the interenet resources dealing with the subject I could find, e.g.,
http://www.cs.princeton.edu/courses/archive/spr05/cos598E/bib/bloom_filters.pdf
and
http://www.cs.umd.edu/class/spring2011/cmsc818k/Lectures/bloom-filters.pdf,
seem to brush the question of uniqueness aside.

Thanks!

Answer
The function f(x) = ln(x) ln(1-x) has derivative:

f'(x) = log(1-x)/x - log(x)/(1-x)

It is clear that f'(1/2) = 0, and in fact, f' is strictly decreasing on [0,1], so it can only be f'(x)=0 at one place (at x=1/2). You know f' is decreasing because:

f''(x) = - [ (x^2-1)ln(1-x) + x(2-2x+xln(x)) ] / x^2 (x+1)^2

You can quickly decide/verify for yourself that:

(x^2-1)ln(1-x)>0

2-2x+xln(x)>0

for all 0<x<1. This implies f'' is always negative, i.e. f' is always decreasing, i.e. f' can only be zero once on (0,1).

Because f(x) is differentiable on the interval (0,1) and continuous on [0,1], its extreme values can only occur at 0, 1, or places where f'(x)=0. This is only true when x=1/2.

f(0)=0 and f(1)=0, while f(1/2) is clearly not zero (it is ln(2)^2) so this is the unique maximum.

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