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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.

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    Your first error is in your justification for $p_0 \ge 0$, Moreover, that error is repeated at later stages of your argument. – 2017-02-07
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    Maybe it's not an error. Can you defend it? – 2017-02-07
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    The way I would get $p_0 \ge 0$ is by substituting $x=0$ into both the rational function and the power series. – 2017-02-07
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    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. – 2017-02-07
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    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. – 2017-02-07
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    If two power series sum to a power series with nonnegative coefficients, are you claiming that each of the summands must have nonnegative coefficients? – 2017-02-07
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    That was what I was trying to do, yes. But now I see why that is a bad idea. Because $1+3-1\geq 0$ and not each each number being summed is. – 2017-02-07
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    In your very first long division set $x=1$ to get $\dfrac{p_0+p_1}{2}=p_0+p_1$. – 2017-02-07
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    Why am I setting $x=1$? – 2017-02-07
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    Do you know the power series for $\frac{1}{1+x}$? – 2017-02-07
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    @jm.byrnes If the long division was done correctly, the equality should hold for all $x$ (except $-1$), including $1$. But it doesn't hold for $x=1$. I just picked a simple example, $0$ didn't work so I took the next best thing. – 2017-02-07
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    If you replace $\frac{1}{1+x}$ by its power series, the required results will follow more easily -- in particular, no long divisions are needed. – 2017-02-07
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    The power series is $\sum^\infty_{n=0}(-1)^nx^n$ correct? – 2017-02-07
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    Yes, that's correct. – 2017-02-07
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    Ok. I will try using that hint. And add a second attempt to my post here. – 2017-02-07
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    @quasi I added the power series to my proof. But am unsure how to go from here. I see how the $p_0\geq 0$ follows now but not the others. – 2017-02-07
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    You need to _multiply_ those two power series. No need for induction. You can directly write the coefficient of $x^m$ in the product as a summation. – 2017-02-07
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    Once you switch to the multiplicative form, you no longer need the rational function form. – 2017-02-07
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    Even for $x = 0$, you did it the hard way. Just plug $x = 0$ into the product of the two power series. Next try for the coefficient of the linear term of the product. – 2017-02-07
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    @quasi How do you multiply together a finite and infinite series? I am trying to review the product of two series here: http://www2.fiu.edu/~aladrog/ProductPowerSeries.pdf. – 2017-02-07
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    Collect appropriate products yielding each power, low to high. This is the _standard_ way to multiply two power series. – 2017-02-07
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    For example, the constant term of the product is just the product of the two constant terms. The linear term of the product is the sum of two monomial products. Which ones? Next do the quadratic term of the product. Once the pattern is clear, express the degree $m$ term of the product as a sum of monomial products. – 2017-02-08
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    Here is what I have for the product at the moment. $(\sum_{i=0}^np_ix^i)(\sum_{i=0}^\infty (-1)^ix^i)=\sum_{i=0}^\infty[p_0(-1)^i+p_1(-1)^{i-1}+\cdots+p_n(-1)^{i-n}+(-1)^{i-(n+1)}+\cdots+(-1)^0]x^i$ I feel like this isn't right though. Having two variables $n$ and $i$ doesn't seem to work. I know the result of the 1st few is: – 2017-02-08
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    $p_0+(p_1-p_0)*x+(p_2-p_1+p_0)*x^2+(p_3-p_2+p_1-p_0)*x^3+(p_4-p_3+p_2-p_1+p_0)*x^4+(p_5-p_4+p_3-p_2+p_1-p_0)*x^5$. So I see why the inequalities hold. Just trying to show it more rigorously. – 2017-02-08

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