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Thread: Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

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    Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

    My teacher said that the polynomial function such that f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right ) exists and that must be of the form f(x)=1 \pm x^n. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be f(x)=1 \pm x^n
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  2. #2
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    Re: Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

    Quote Originally Posted by NameIsHidden View Post
    My teacher said that the polynomial function such that f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right ) exists and that must be of the form f(x)=1 \pm x^n. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be f(x)=1 \pm x^n
    f(x)=\frac{f(\frac{1}{x})}{f(\frac{1}{x})-1}

    f(\tfrac{1}{x})=\frac{f(x)}{f(x)-1}

    \begin{align*}f(x)\cdot f(\tfrac{1}{x}) &= \frac{f(\frac{1}{x})}{f(\frac{1}{x})-1} \cdot \frac{f(x)}{f(x)-1}  \\                                                      &= \frac{1}{f(\tfrac{1}{x})-1} \cdot \frac{1}{f(x)-1}\end{align*}

    Note that
    f(\tfrac{1}{x})-1 = \frac{f(x)-[f(x)-1]}{f(x)-1} = \frac{1}{f(x)-1}
    so that
    \frac{1}{f(\tfrac{1}{x})-1} = f(x)-1.


    So,
    f(x)\cdot f(\tfrac{1}{x}) = [f(x)-1] \cdot \frac{1}{f(x)-1} = 1.

    Now let
    g(x)=f(x)-1
    and thus
     g(\tfrac{1}{x}) = f(\tfrac{1}{x}) -1.

    Finally,
    g(x)\cdot g(\tfrac{1}{x}) = [f(x)-1] \cdot [f(\tfrac{1}{x})-1] = \frac{f(x)-1}{f(x)-1} = 1.

    (Of course g(x)\cdot g(x^{-1}) = 1!)

    If g(x) is polynomial, then g(x) = \pm x^n \implies f(x)=1\pm x^n.
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