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Is there a non-constant polynomial $p$ with positive coefficients such that function $ x \mapsto p(x^2) - p(x)$ is decreasing on the interval $[1,+\infty)$?

This was a question from the last year's math analysis exam. The answer was the following:

For $1\le x \lt y$ and $n \in \Bbb N $ we have $y^{2n}-x^{2n} = (y^n+x^n)(y^n-x^n) > y^n - x^n$, so $ x \mapsto p(x^2)-p(x)$ is strictly increasing function on $[1,+\infty)$, for every $n \in \Bbb N$. Because polynomial p has positive coefficients and is not constant, such function does not exist.

I have only recently started this class so I have hard times wrapping my head around this. Why is $y^{2n} - x^{2n}$ used to prove that the function is increasing? Why does this ($y^{2n}-x^{2n} > y^n - x^n$) mean that the function is increasing? Does it mean that $p(y^{2n})-p(x^{2n}) > p(y)-p(x)$ so that means $p(y^{2n}) - p(y)- (p(x^{2n})- p(x))>0$ or something like this?

If somebody could just explain the answer, it would be much appreciated!

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    $\begingroup$ The answer is a bit unclear, but it is suggesting that you first try the case when $p(x)=x^n$ and look at $p(y^2)-p(y)- (p(x^2)-p(x))$ for $y>x$. Their argument shows that this is positive, so the function $p(x^2)-p(x)$ is increasing. Now you use the fact that if $f$ and $g$ are increasing and $a,b\geq 0$ then $af+bg$ is increasing too to get the result for any polynomial $p$ with positive coefficients. $\endgroup$ – Matthew Towers Nov 8 '17 at 20:27
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Say $p(x)=ax^n+...$ and $a\ne 0$, $n\geq 1$. Then $$f(x) = p(x^2)-p(x)=ax^{2n}+...$$ Since $f(x)\sim ax^{2n}$ for $x\to \infty$ we see that $a<0$ since it is decreasing. So it is impossible.

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  • $\begingroup$ Why is this down voted? Is it to simple argument to be true? $\endgroup$ – Aqua Nov 8 '17 at 20:49
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    $\begingroup$ This argument is perfectly valid for polynomials. To elaborate just a bit further, $f(x) = ax^{2n} + g(x)$ where $g$ is a polynomial of degree at most $2n-1$. Then $f'(x) = 2nax^{2n-1}(1 + \frac{g'(x)}{x^{2n-1}})$. Since $\frac{g'(x)}{x^{2n-1}} \to 0$ as $x \to \infty$, we obtain $f'(x) > 0$ for sufficiently large $x$. This is true as long as the leading coefficient of $p$ is positive. The answer given to the OP is (a) complicated and (b) uses assumptions that may be relaxed, hence it is less than optimal. $\endgroup$ – Hans Engler Nov 8 '17 at 20:57
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Okay so I think I got it, but I'm still not sure.

Let's say that $f(x)= p(x^2)-p(x)$. If this function is decreasing on the interval $[1, +\infty)$, that means that if $1\le x < y$, then $f(x) > f(y)$, that is, $f(x)-(fy) >0$.

$$f(x)-f(y) >0 $$ $$p(x^2)-p(x) -p(y^2)+p(y)>0 $$ $$a_nx^{2n}+...+a_1x^2 + a_0 -a_nx^n-...-a_1x-a_0 - (a_ny^{2n}+...+a_1y^2 + a_0 -a_ny^n-...-a_1y-a_0)>0 $$ $$a_n(x^{2n}-y^{2n}-(x^n-y^n))+...+a_1(x^2-y^2-(x-y))>0$$

Then I just prove that $x^{2n}-y^{2n}-(x^n-y^n)<0$, for $1\le x < y$. And, because $a_i$ are all positive, this function cannot be decreasing.

I'm pretty sure that this is the proof from the answer. Still, thanks for help, everybody!

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