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Thread: A functional equation

  1. #1
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    A functional equation

    Find all polynomials $\displaystyle p(x) \in \mathbb{C}[x]$ which satisfy the identity $\displaystyle p(2x-x^2)=(p(x))^2.$
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  2. #2
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    $\displaystyle 2x-x^2=1-(1-x)^2$, 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.
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  3. #3
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    Quote Originally Posted by TwistedOne151 View Post
    $\displaystyle 2x-x^2=1-(1-x)^2$, 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?
    Last edited by NonCommAlg; Jul 22nd 2009 at 03:12 AM.
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  4. #4
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    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 $\displaystyle k=1$.
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