Rudin–Shapiro polynomials

Example

Define polynomials $P_m$ and $Q_m$ by $P_0(z) = Q_0(z) = 1$ and, for $m \ge 0$, $$P_{m+1}(z) = P_m(z) + z^{2^m} Q_m(z), \qquad Q_{m+1}(z) = P_m(z) - z^{2^m} Q_m(z).$$ These are the Rudin–Shapiro polynomials. By induction, $P_m$ and $Q_m$ have degree $2^m - 1$, and all $2^m$ of their coefficients are $+1$ or $-1$. Indeed $P_m$ fills the coefficients of $1, \ldots, z^{2^m - 1}$ and $z^{2^m} Q_m$ fills those of $z^{2^m}, \ldots, z^{2^{m+1} - 1}$. So they are Littlewood polynomials. The first ones are $P_1(z) = 1 + z$, $Q_1(z) = 1 - z$ and $P_2(z) = 1 + z + z^2 - z^3$.

For $\abs{z} = 1$, the parallelogram law $\abs{u + v}^2 + \abs{u - v}^2 = 2\abs{u}^2 + 2\abs{v}^2$ with $\abs{z^{2^m}} = 1$ gives $$\abs{P_{m+1}(z)}^2 + \abs{Q_{m+1}(z)}^2 = 2 \left( \abs{P_m(z)}^2 + \abs{Q_m(z)}^2 \right).$$ Hence $\abs{P_m(z)}^2 + \abs{Q_m(z)}^2 = 2^{m+1}$, and $$\max_{\abs{z} = 1} \abs{P_m(z)} \le \sqrt{2} \cdot \sqrt{2^m}.$$