# Solving the equation: $p^a (1-p)^b = q^a (1-q)^b$

I am currently studying the following equation:

$p^a(1-p)^b=q^a(1-q)^b$

where $p,q \in (0,1)$, and $a,b \in \mathbb{N}$.

I would like to show that the equation is satisfied if and only if $p=q$.

Is it possible to do this in an exact way? I came across this equation when studying dynamical systems, and I don't have much of a background with these sorts of equations.

(Actually, more precisely, I would like to show that

$\sum_k (p^{a_k}(1-p)^{b_k} - q^{a_k}(1-q)^{b_k}) = 0 \Leftrightarrow p = q$

for $p,q \in (0,1)$ and $a_k,b_k \in \mathbb{N}$.)

Thanks

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Sorry, what do you think I should tag it as? – Solaris May 16 '12 at 13:45
Do you have $a\neq b$? Otherwise $(p,1-p)$ is a solution. – Simon Markett May 16 '12 at 13:52

This cannot be quite true. Look at the function $x^a(1-x)^b$. Note that on $[0,1]$ the function increases, then decreases. So it takes on all but one value twice.

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Thanks, I didn't realise that was a feature for all powers. – Solaris May 16 '12 at 21:37
@Solaris: It is the same for all positive $a$ and $b$, they don't even have to be integers. If it is not geometrically clear to you, you can take the derivative. You will find it is positive for a while, then negative. – André Nicolas May 16 '12 at 21:45

It's clear that the function $f(x)=x^a(1-x)^b$ has endpoints $f(0)=0$ and $f(1)=0$. It's also clearly positive, continuous and bounded $(0,1)$, so it must have a maximum $y$-value, attained at say $(u,v)$.

Let $\lambda$ be any height greater than $0$ but less than the maximum height $v$. By the intermediate value theorem (applied twice), there exists $p\in(0,u)$ and $q\in(u,1)$ such that $f(p)=\lambda=f(q)$.

Edit: This reasoning applies just as well to the more general function

$$f(x)=\sum_{k=1}^n x^{a_k}(1-x)^{b_k}.$$

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Consider $f(x) = x^a (1-x)^b$ for natural numbers $a,b > 0$. Then $f'(x) = x^{a-1} (1-x)^{b-1} (a-(a + b)x) > 0$ if and only if $x < \frac{a}{a+b} \in (0,1)$. So $f(x)$ is increasing on $[0, \frac{a}{a+b}]$ and decreasing on $[\frac{a}{a+b}, 1]$, with $f(0) = 0$ and $f(1) = 0$. So for each $0 \leq p \leq \frac{a}{a+b}$ there is a unique $\frac{a}{a+b} \leq q \leq 1$ also satisfying the equation. Only for $p = \frac{a}{a+b}$ you will not find a different value $q$ satisfying the equation.

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