A doubt regarding polynomials

Question

Suppose $F(x)$ is a polynomial having coefficients in the field of reals ( $\mathbb{R}[X]$ ) . $F(x)$ has the property that $y \geq 0 \implies F(y)$ is nonzero.( $y$ is a real number ) Can we find a polynomial $T(x) \in \mathbb{R}[X]$ so that $T(x) \cdot F(x)$ has all its coefficients non-negative ?

Answer 1

It is not difficult to see that it is enough to prove this for the case $$F(x)=x^2+ax+b$$ where $F$ is irreducible, i.e. $a^2-4b<0$. So consider $$(x^2+ax+b)\sum_{i=0}^\infty c_ix^i$$ We want to show that we can choose the $c_i$ to be equal to $0$ for $i>N$ for some $N$, and still have all non-negative coefficients. Now: $$c_0b\geq 0\Leftrightarrow c_0\geq 0$$ we set, to make things easier $c_0=1$. Then the next condition reads $$c_0a+c_1b\geq 0\Leftrightarrow c_1\geq -\frac{a}{b}$$ Again, to make things simple, set $c_1=-\frac{a}{b}$. The next condition reads $$c_2b+c_1a+c_0\geq 0\Leftrightarrow c_2\geq \frac{1}{b}(\frac{a^2}{b}-1)=\frac{a^2}{b^2}-\frac{1}{b}$$ Now if $\frac{a^2}{b^2}-\frac{1}{b}\leq 0$ we are done, since we can set $c_2=c_3=\dots=0$. If not, we set $c_2=\frac{a^2}{b^2}-\frac{1}{b}$. The next condition reads $$c_3b+c_2a+c_1\geq 0\Leftrightarrow c_3\geq \frac{1}{b}\left(\frac{a}{b}-\frac{a^3}{b^2}+\frac{a}{b}\right)=\frac{2a}{b^2}-\frac{a^3}{b^3}$$ Again, if $\frac{2a}{b^2}-\frac{a^3}{b^3}\leq 0$ we are done by setting $c_3=c_4=\dots=0$. If not, we set $c_3=\frac{2a}{b^2}-\frac{a^3}{b^3}$. The next condition reads $$c_4b+c_ca+c_2\geq 0\Leftrightarrow c_4\geq \frac{1}{b}\left(\frac{1}{b}-\frac{a^2}{b^2}+\frac{a^4}{b^3}-\frac{2a^2}{b^2}\right)=\frac{a^4}{b^4}-\frac{3a^2}{b^3}+\frac{1}{b^2}$$ This process keeps repeating.

The next step is to find a closed form for this sequence of lower bounds. Therefore we define

$$g(n,m)=\begin{cases} 0 &\text{ if } n>m\\ 0 &\text{ if } n\not\equiv m\mod 2\\ (-1)^{\frac{m+n}{2}}\binom{\frac{m+n}{2}}{n} &\text{ otherwise} \end{cases}$$ Then the conditions read $$c_n\geq \sum_{m=0}^n \frac{a^{m}}{b^{m/2+n/2}}g(n,m)=\frac{1}{b^{n/2}}\sum_{m=0}^n \left(\frac{a}{\sqrt{b}}\right)^mg(n,m)$$ And so it suffices to prove that for all $a,b$ with $a^2-4b<0\Leftrightarrow \frac{a}{\sqrt{b}}<2$, the expression $\sum_{m=0}^n \left(\frac{a}{\sqrt{b}}\right)^mg(n,m)$ will be $\leq 0$ for some $n>0$.

To prove this, we use inspiration from this paper. Write $$H_n(x)=\sum_{m=0}^n (x)^mg(n,m)$$ Then using the substituion $x=2\cosh(z)$ we find that $$H_n(x)=\pm\frac{\sinh((n+1)z)}{\sinh(z)}$$ So $$H_n(x)=0\Leftrightarrow z=\frac{i\pi k}{n+1}, k\in\{1,\dots,n\}\Leftrightarrow x=2\cosh\left(\frac{i\pi k}{n+1}\right)\in(-2,2)$$ From here it is easy to finish the proof: for $n$ big enough we will be able to find $k$ such that $$\frac{a}{\sqrt{b}}\in \left[2\cosh\left(\frac{i\pi k}{n+1}\right),2\cosh\left(\frac{i\pi (k+1)}{n+1}\right)\right]$$ and $H_n(x)$ is non-positive on this interval.

Note that the condition of irreduciblity of $x^2+ax+b$ is nicely used, since it ensures that $\frac{a}{\sqrt{b}}$ is in the range $[-2,2]$ where $H_n(x)$ switches sign

To see what this looks like in the case mentioned by Arthur in the comments: we first create a small table with values for $g(n,n)$:

$$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -3 & 0 & 1\\ 0 & 0 & 0 & 0 & -1 & 0 & 4 & 0 & -3 & 0\\ 0 & 0 & 0 & 1 & 0 & -5 & 0 & 6 & 0 & -1\\ 0 & 0 & -1 & 0 & 6 & 0 & -10 & 0 & 4 & 0\\ 0 & 1 & 0 & -7 & 0 & 15 & 0 & -10 & 0 & 1\\ -1 & 0 & 8 & 0 & -21 & 0 & 20 & 0 & -5 & 0 \end{pmatrix} $$

So the conditions read

$$c_0 \geq 1$$ $$c_1 \geq \frac{1.9}{1}=1.9$$ $$c_2 \geq 1.9^2-1=2.61$$ $$c_3 \geq 2*1.9-1.9^3=3.059$$ $$c_4\geq 3.2021$$ $$c_5\geq 3.02499$$ $$c_6\geq 2.545381$$ $$c_7\geq 1.8112339$$ $$c_8\geq 0.89596341$$ $$c_9\geq -0.108903421$$ This is smaller then $0$, so we can set $c_9=c_{10}=\dots=0$, and find $$T(x)=1+1.9x+2.61x^2+3.059x^3+3.2021x^4+3.02499x^5+2.545381x^6+1.8112339x^7+0.89596341x^8$$ will work, and indeed $$(x^2-1.9x+1)(1+1.9x+2.61x^2+3.059x^3+3.2021x^4+3.02499x^5+2.545381x^6+1.8112339x^7+0.89596341x^8)=1 + 4.440892098500626\dot{}10^{-16} x^3 + 4.440892098500626\dot{}10^{-16} x^6 + 4.440892098500626\dot{}10^{-16} x^7 + 0.108903 x^9 + 0.895963 x^{10}$$ I think it's safe to say that, without rounding errors, we would find $$(x^2-1.9x+1)(1+1.9x+2.61x^2+3.059x^3+3.2021x^4+3.02499x^5+2.545381x^6+1.8112339x^7+0.89596341x^8)=1 + 0.108903 x^9 + 0.895963 x^{10}$$

So it's not suprising that Arthur conjectured that this is a counter example, since it requires a rather arbitrary looking degree $8$ polynomial.

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for the other test case: $F(x)=x^4-x+1$: we factor over $\mathbb{R}$, and find one quadratic that already has positive coefficients, and the quadratic $$G(x)=0.713639-1.45427 x+x^2$$ Using the algorithm, we find $$c_0=1, c_1\geq 2.03782 c_2\geq 2.75145, c_3\geq 2.75144, c_4\geq 1.75142, c_5\geq -0.286423$$ so we are done, and $$(0.713639-1.45427 x+x^2)(1+2.03782x+2.75145x^2+2.75144x^3+1.75142x^4)=1.75142x^6+0.204402x^5+0.713639$$