# A functional equation

• Jul 16th 2009, 11:36 PM
NonCommAlg
A functional equation
Find all polynomials $\displaystyle p(x) \in \mathbb{C}[x]$ which satisfy the identity $\displaystyle p(2x-x^2)=(p(x))^2.$
• Jul 17th 2009, 01:12 AM
TwistedOne151
$\displaystyle 2x-x^2=1-(1-x)^2$, so if $\displaystyle p(x)=(1-x)^n$, we see that
$\displaystyle p(2x-x^2)=[1-(2x-x^2)]^n=(1-2x+x^2)^n=(1-x)^{2n}=(p(x))^2$.

--Kevin C.
• Jul 18th 2009, 09:37 PM
NonCommAlg
Quote:

Originally Posted by TwistedOne151
$\displaystyle 2x-x^2=1-(1-x)^2$, so if $\displaystyle p(x)=(1-x)^n$, we see that
$\displaystyle p(2x-x^2)=[1-(2x-x^2)]^n=(1-2x+x^2)^n=(1-x)^{2n}=(p(x))^2$.

--Kevin C.

$\displaystyle p(x)=0$ is also a solution. the question is that whether or not these are the only possible solutions?
• Jul 30th 2009, 05:11 PM
halbard
OK, let's offer this to the snake and see if it strikes.

Using the functional equation, first show that if $\displaystyle a$ is a root of $\displaystyle p(z)=0$ then so are $\displaystyle 2a-a^2$ and $\displaystyle 1\pm\surd(1-a)$.

Also, since $\displaystyle p(z)=0$ is a polynomial equation it has only a finite number of roots.

Now let $\displaystyle m=\max_{p(z)=0}|1-z|$. By the previous assertion this is well-defined. Aim to show that $\displaystyle m=0$ and therefore that $\displaystyle 1$ is the only possible root of $\displaystyle p(z)=0$.

1. Assume $\displaystyle m>1$. If $\displaystyle z$ is a root with $\displaystyle |1-z|=m$, let $\displaystyle w=2z-z^2$. Then $\displaystyle w$ is a root with $\displaystyle |1-w|=|1-z|^2=m^2>m$, contradicting the choice of $\displaystyle m$.

2. Assume $\displaystyle 0<m<1$. Choose a root $\displaystyle z$ with $\displaystyle |1-z|=m$ and let $\displaystyle w=1\pm\surd(1-z)$. Then $\displaystyle w$ is a root with $\displaystyle |1-w|=\surd|1-z|=\surd m>m$, contradiction.

3. Assume $\displaystyle m=1$. Choose a root $\displaystyle z$ with $\displaystyle |1-z|=1$. If $\displaystyle z=0$ is a root then $\displaystyle z=2$ is also a root. Therefore assume WLOG that $\displaystyle z=1-\mathrm e^{\mathrm i\theta}$ for some $\displaystyle \theta$ in $\displaystyle (0,2\pi)$.

Then $\displaystyle w=1-\surd(1-z)=1-\mathrm e^{\mathrm i\theta/2}$ is a root. Similarly $\displaystyle 1-\mathrm e^{\mathrm i\theta/4}$ is a root, and $\displaystyle 1-\mathrm e^{\mathrm i\theta/8}$, etc.

Continue in this way to show the existence of an infinite number of distinct roots of the polynomial equation $\displaystyle p(z)=0$, another contradiction.

Therefore $\displaystyle m=0$ is the only possibility, and consequently $\displaystyle z=1$ is the only possible root. Therefore write $\displaystyle p(z)=k(1-z)^m$ where $\displaystyle m\in\mathbb N\cup\{0\}$ and $\displaystyle k$ is a constant.

Use the functional equation to get $\displaystyle k=k^2$, and so $\displaystyle k=0$ or $\displaystyle k=1$.