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Math Help - A functional equation

  1. #1
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    A functional equation

    Find all polynomials p(x) \in \mathbb{C}[x] which satisfy the identity p(2x-x^2)=(p(x))^2.
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by TwistedOne151 View Post
    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?
    Last edited by NonCommAlg; July 22nd 2009 at 04:12 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 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<m<1. 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.
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