Do the coefficients of these polynomials alternate in sign

Question

Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=x-1\\
p_4(x)&=x-2\\
p_5(x)&=x^2-3x+1\\
p_6(x)&=x^2-4x+3\\
p_7(x)&=x^3-5x^2+6x-1\\
p_8(x)&=x^3-6x^2+10x-4
\end{align}
It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts?

Answer 1

Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ... \begin{eqnarray*} p_{2i}(x) = \sum_{j=0}^{i-1} \binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\ p_{2i-1}(x) = \sum_{j=0}^{i-1} \binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\ \end{eqnarray*} Proof of the even formula ... \begin{eqnarray*} &p_{2i-1}(x) -p_{2i-2}(x) =\\ & x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \sum_{j=0}^{i-1} \binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\ &= x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2} = p_{2i}(x). \end{eqnarray*} And the answer to the question in the title is $\color{red}{\text{yes}}$.