# Showing inequalities hold within a partial order on $\mathbb{R}[x]$

I am struggling with the following problem.

Consider in $\mathbb{R} [x]$ the following relation:

$$p(x) \succ q(x) \mbox{ if the formal power series } \hspace{3mm} \frac{p(x)-q(x)}{1+x}$$

is a polynomial with nonnegative coefficients. Show that this is a partial order on $\mathbb{R} [x]$ and if $p(x)=p_0 + p_1 x+p_2 x^2 + \cdots + p_n x^n \succ 0$, then the following inequalities hold: \begin{align*} p_{0} &\geq 0 \\ p_{1} - p_0 &\geq 0 \\ p_{2} - p_1 + p_0 &\geq 0 \\ &\hspace{2mm} \vdots \\ p_{n} - p_{n-1}+p_{n-2}-\cdots +(-1)^n p_0 &\geq 0 \end{align*} Conclude that if $p(x) \succ 0$, all its coefficients are nonnegative, but the converse is false.

I had no problem showing that this relation was a partial order or showing that the converse is false. My converse contradiction is $p(x)=2x^3+x^2+1$ because $$\frac{(2x^3+x^2+1)-0}{1+x}=\frac{(1+x)(2x^2-x+1)}{1+x}=2x^2-x+1$$ which is a polynomial with negative coefficients. Proving the inequalities has proven to be a bit of a struggle however... Here is what I have tried so far.

2nd Attempt:

Recall $\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n=1-x+x^2-x^3+\cdots$.

Notice $p_0+p_1x+p_2x^2+\cdots+p_nx^2=\sum_{i=0}^np_ix^i$.

Then $\frac{p_0 + p_1 x+p_2 x^2 + \cdots + p_n x^n - 0}{1+x} = (\sum_{i=0}^np_ix^i)(\sum_{i=0}^\infty (-1)^ix^i)=\sum_{i=0}^\infty[p_0(-1)^i+p_1(-1)^{i-1}+\cdots+p_n(-1)^{i-n}+(-1)^{i-(n+1)}+\cdots+(-1)^0]x^i$.

Suppose $x=1$.

If $x=0$, then $$\sum_{k=1}^\infty[p_0(-1)^k+p_1(-1)^{k-1}+\cdots+p_n(-1)^{k-n}+(-1)^{k-(n+1)}+\cdots+(-1)^0]x^k =0$$

$\frac{p_0 + p_1 x+p_2 x^2 + \cdots + p_n x^n}{1+x}=p_0$. Therefore $p_0\geq 0$ in order to avoid negative coefficients.

Question: How do the other inequalities follow. I know it has been suggest to try using $x=1$, but I don't see how this helps.

1st Attempt:

Consider $$\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_n x^n}{1+x}=\frac{p_0}{1+x}+\frac{p_1 x+p_2 x^2 +\cdots+ p_n x^n}{1+x}.$$ Notice $\frac{p_0}{1+x}\geq 0$ in order to avoid a negative coefficient. Then $p_0\geq 0$.

Consider $$\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_n x^n}{1+x}=\frac{p_0+p_1x}{1+x}+\frac{p_2 x^2 +\cdots+ p_n x^n}{1+x}.$$ By long division, $\frac{p_0+p_1x}{1+x}=(x+1)\left(p_1+\frac{p_0-p_1}{x+1}\right)$. Notice $\frac{p_0-p_1}{x+1}\geq 0$ in order to avoid a negative coefficient. Then $-p_1+p_0\geq 0$.

1st Problem/Question: My signs are reversed here. Is this right? If not, where did I go wrong?

We will prove the rest of the inequalities hold by induction.

Base Case: We want to show true for $n=2$. Consider $$\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_n x^n}{1+x}=\frac{p_0+p_1x+p_2 x^2}{1+x}+\frac{p_3 x^3 +\cdots+ p_n x^n}{1+x}.$$ By long division, $$\frac{p_0+p_1x+p_2 x^2}{1+x}=(x+1)\left(p_2x+(p_1-p_2)+\frac{p_0-p_1+p_2}{x+1}\right).$$ Notice $\frac{p_0-p_1+p_2}{x+1}\geq 0$ in order to avoid a negative coefficient. Then $p_2-p_1+p_0\geq 0$.

2nd Problem/Question: Am I ok to ignore $(p_1-p_2)$ here and keep moving forward?

Induction: Assume true for $n=k$. Then when $\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_kx^k}{1+x}$, we have $$p_{k} - p_{k-1}+p_{k-2}-\cdots +(-1)^k p_0 \geq 0.$$ We want to show this inequality holds true when $n=k+1$. Consider $$\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_k x^k+p_{k+1} x^{k+1}}{1+x}=\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_kx^k}{1+x}+\frac{p_{k+1} x^{k+1}}{1+x}.$$ By the induction hypothesis, we know $p_{k} - p_{k-1}+p_{n-2}-\cdots +(-1)^n p_0 \geq 0$. Therefore we just need to consider $\frac{p_{k+1} x^{k+1}}{1+x}$. Notice $\frac{p_{k+1} x^{k+1}}{1+x}\geq 0$ in order to avoid a negative coefficient. Then $p_{k+1} \geq 0$. Therefore $p_{k+1}+p_{k} - p_{k-1}+p_{n-2}-\cdots +(-1)^n p_0\geq 0$.

3rd Problem/Question: This result is not what I was trying to show. It does not follow the same pattern as in our induction hypothesis. I am struggling to see how I could approach this problem differently.

Any kind of help would be much appreciated. Thank you for taking the time to read and respond if you so choose.

• Your first error is in your justification for $p_0 \ge 0$, Moreover, that error is repeated at later stages of your argument. – quasi Feb 7 '17 at 8:20
• Maybe it's not an error. Can you defend it? – quasi Feb 7 '17 at 8:26
• The way I would get $p_0 \ge 0$ is by substituting $x=0$ into both the rational function and the power series. – quasi Feb 7 '17 at 8:27
• Also, what you're trying to prove by induction is unclear. Isn't the statement to be proved a statement about the polynomial $p(x)$? Also, the statement needs some kind of relation symbol. As it stands, it's not even a statement. – quasi Feb 7 '17 at 8:38
• My logic was that since we can write $\frac{p_0 + p_1 x+p_2 x^2 +\cdots+ p_n x^n}{1+x}$ as $\frac{p_0}{1+x}$+\frac{p_1x}{1+x}+\cdots+\frac{p_nx^n}{1+x} $Then we could look at each$p_i\$ separately if need be. – jm.byrnes Feb 7 '17 at 8:38