A functional equation

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

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

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

--Kevin C.

$p(x)=0$ 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 $a$ is a root of $p(z)=0$ then so are $2a-a^2$ and $1\pm\surd(1-a)$.

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

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

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

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

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

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

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

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

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