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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 $\displaystyle f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right )$ exists and that must be of the form $\displaystyle f(x)=1 \pm x^n$. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be $\displaystyle f(x)=1 \pm x^n$
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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 $\displaystyle f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right )$ exists and that must be of the form $\displaystyle f(x)=1 \pm x^n$. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be $\displaystyle f(x)=1 \pm x^n$
    $\displaystyle f(x)=\frac{f(\frac{1}{x})}{f(\frac{1}{x})-1}$

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

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

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


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

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

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

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

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