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Rational function and polynomial

Prove that there does not exist a rational function f with real coefficients such that

    \[f \left ( \frac{x^2}{x+1} \right ) = \mathrm{P}(x)\]

where \mathrm{P}(x) \in \mathbb{R}[x] is a non constant polynomial.

Solution

Since polynomials are defined on x=-1 we have that

    \begin{align*} \mathbb{R} \ni \mathrm{P}(-1) & =\lim_{x \rightarrow -1^+} \mathrm{P}(x) \\ &=\lim_{x \rightarrow -1^+}f \left(\frac{x^2}{x+1} \right) \\ &=\lim_{x \rightarrow \infty} f\left(\frac{x^2}{x+1}\right) \\ &=\lim_{x \rightarrow \infty} \mathrm{P}(x) \end{align*}

Since \mathrm{P} tends to a finite value as x \rightarrow \infty it must be a constant polynomial. In particular, f must be constant in the range of \frac{x^2}{x+1} which is an infinite set, implying that f must also be constant. This proves what we wanted.

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