# A fascinating sequence of polynomials

So, I have a sequence of polynomials: $$p_m(z) = 2\sum_{k=1}^{m-1}(1-\tfrac km)z^k = \frac{2z}{1-z}\Big(1-\frac 1m\sum_{k=0}^{m-1}z^k\Big),$$ where the last expression only holds for $$z\neq 1$$. Fascinatingly, when you plot the zeros of $$p_m$$, you see that they lie outside of the unit circle line $$\mathbb T$$ and approach $$\mathbb T$$ for $$m\to\infty$$ so that for large $$m$$ the zeros of $$p_m$$ almost form the unit circle. However, $$z=1$$ is never a zero and the gap around $$z=1$$ is larger than around any other $$z\in\mathbb T$$. This gap becomes smaller and smaller with growing $$m$$.

What I also observed (by plots for many values of $$m$$) is the following: $$\operatorname{Re}p_m(e^{it})$$ is very close to (but always larger than) $$-1$$ on an interval $$[\delta,2\pi-\delta]$$ and approaches $$p_m(1) = m-1$$ rapidly at the boundary of $$[0,2\pi]$$. My question:

For given $$\delta>0$$, I'd like to find an $$m_\delta$$ such that for $$m\ge m_\delta$$ we have $$\operatorname{Re}p_m(e^{it})\le 0$$ for all $$t\in [\delta,2\pi-\delta]$$.

I can write down the trigonometric polynomial $$\operatorname{Re}p_m(e^{it}) = 2\sum_{k=1}^{m-1}(1-\tfrac km)\cos(kt)$$, but I cannot seem to bound this guy. I'm not even able to show that $$\operatorname{Re}p_m(e^{it})\ge -1$$ for all $$t$$. Can anyone help?

Indeed, its real part on the unit circle can be written as, $$\begin{split} \Re p_m(e^{it}) &= \sum_{k=1}^{m-1}\left(1-\tfrac km\right)e^{ikt} + \sum_{k=1}^{m-1}\left(1-\tfrac km\right)e^{-ikt}\\ &=-1+\sum_{k=0}^{m-1}\left(1-\tfrac km\right)e^{ikt} + \sum_{k=1}^{m-1}\left(1-\tfrac km\right)e^{-ikt}\\ &= -1 + \sum_{|k|\leq m-1}\left(1-\tfrac {|k|}m\right)e^{ikt}\\ &= -1+\frac 1 m\left( \frac{\sin\left(\frac{mt}2\right)}{\sin\left(\frac t 2\right)}\right)^2 \end{split}$$ where we have used the known equivalent expressions for the Fejer Kernel $$F_m(t)=\sum_{|k|\leq m-1}\left(1-\tfrac {|k|}m\right)e^{ikt}=\frac 1 m\left( \frac{\sin\left(\frac{mt}2\right)}{\sin\left(\frac t 2\right)}\right)^2$$ The proof of the above is given in the Wikipedia entry for the Fejer kernel.
It immediately follows that $$-1 \leq \Re P_m(e^{it}) \leq m-1$$ The right-hand side is attained at $$t$$ being a multiple of $$2\pi$$. So I don't agree with your conclusion that it should be close to $$-1$$. See the Desmos link.
• Thank you very much for making the connection to the Fejer kernel. "So I don't agree with your conclusion that it should be close to $-1$" - Well, I'm talking about large values of $m$. Now, it is clear to me why this is: $F_m\to\delta$ as $m\to\infty$. So, now my question is: what is the largest $t\in [0,\pi]$ such that $F_m(t)=1$? Dec 29, 2023 at 20:16
Ok, I can now answer my question by myself. Thanks a lot to Stefan Lafon who pointed me to the connection with the Fejér kernel. Let us set $$t_m := \arccos\left(\frac{\sqrt m - 3}{\sqrt m - 1}\right),$$ which for large $$m$$ is a very small number. We shall prove that $$F_m(t)\le 1$$ (and hence $$\operatorname{Re}p_m(e^{it})\le 0$$) for $$t\in [t_m,\pi]$$. To see this, we first observe that \begin{align} F_m(t)\le 1 &\Longleftrightarrow \frac 1m\left(\frac{1-\cos(mt)}{1-\cos t}\right)^2\le 1\\ &\Longleftrightarrow 1-\cos(mt)\le\sqrt m (1-\cos t)\\ &\Longleftrightarrow \sqrt m\cos t - \cos(mt)\le\sqrt m - 1\\ &\Longleftrightarrow \frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1}\le 1. \end{align} Next, we have $$\left|\frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1} - \cos t\right| = \left|\frac{\cos t - \cos(mt)}{\sqrt m - 1}\right|\,\le\,\frac 2{\sqrt m - 1}.$$ Hence, if $$t\ge t_m$$, then $$\cos t\le \frac{\sqrt m - 3}{\sqrt m - 1}$$, and hence $$\frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1}\le \left|\frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1} - \cos t\right| + \cos t\le \frac 2{\sqrt m - 1} + \frac{\sqrt m - 3}{\sqrt m - 1} = 1.$$