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Does there exist a rational function $f$ and a polynomial in $\mathbb{R}[x]$ so that: $$f(\frac{x^2}{x+1}) = p(x)$$

mtheorylord
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All such functions $f$ must be constant. Here's a proof.

Suppose $f$ and $p$ satisfy the conditions of the problem. We must then have: $$p(-1)=\lim_{x\to-1^+}p(x)=\lim_{x\to-1^+}f\left(\frac{x^2}{x+1}\right)=\lim_{x\to\infty}f\left(\frac{x^2}{x+1}\right)=\lim_{x\to\infty}p(x).$$ Since $p(x)$ tends to a finite value as $x\to\infty$, $p$ 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. $\blacksquare$

ViHdzP
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  • Why the downvote? – ViHdzP Oct 31 '19 at 03:26
  • URL. What an idea! But as it stands f is not defined at x=-1, hence cannot be a polynomial. Define f at x=-1 to be = constant pol. =c , then OK.Hope this makes sense:) – Peter Szilas Oct 31 '19 at 06:30
  • @PeterSzilas You can do pretty much whatever at the point $x=-1$, if we ignore it altogether in the original problem redaction. My solution only really cares about what happens in its neighborhood ;) – ViHdzP Oct 31 '19 at 06:36
  • Fine.The problem : one equates a continuos polynomial with a function that is not defined at x=-1. Assuming a continuos extension of f at x=-1, you show that it is impossible unless f=p=constant. – Peter Szilas Oct 31 '19 at 07:10