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?


2 Answers 2


Your polynomial is closely related to the Fejer kernel.

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.

  • 1
    $\begingroup$ 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 $F_m(t)=1$? $\endgroup$
    – amsmath
    Dec 29, 2023 at 20:16
  • $\begingroup$ For all those who might come by in 10 years or so: A solution to the problem is given in the answer below (which would not have been possible without the accepted answer by Stefan). $\endgroup$
    – amsmath
    Dec 30, 2023 at 1:39

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 $$ \frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1}\le \left|\frac{\sqrt m\cos t - \cos(mt)}{\sqrt m - 1} - \cos t\right| + \cos t\le \frac 2{\sqrt m - 1} + \frac{\sqrt m - 3}{\sqrt m - 1} = 1. $$


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