A Littlewood polynomial is a polynomial with coefficients from $\{ 1, -1\}$ and the set of Littlewood polynomials with degree $n$ is denoted by $\cal{L}_n$.

I'm interested in a "good" lower bound on the $L^1$-norm of Littlewood polynomials on the unit circle in the complex plane.(which I denote by $\|\cdot \|_1$) To be more precise, consider the polynomial $p(z) = 1+z+\cdots+z^n$, what can be said about $$a_n = \frac{\min_{q\in \cal{L}_n} \| q\|_1}{\| p\|_1}.$$ Does $a_n \to 1$ as $n\to \infty$?

  • $\begingroup$ There seems to be some missing information: is ${\mathcal L}_n$ the set of Littlewood polynomials of degree $n$? $\endgroup$ – Yemon Choi Jun 18 '18 at 15:34
  • $\begingroup$ Computation for $n\leq 9$ suggests strongly that $a_n=1$ for all $n$. $\endgroup$ – Neil Strickland Jun 18 '18 at 16:14
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    $\begingroup$ mathnet.ru/php/… $\endgroup$ – fedja Jun 18 '18 at 16:32
  • $\begingroup$ @fedja: a nice reference! Another interesting case is when considering polynomials whose absolute value of the coefficients are all equal to 1. Do you know is there any similar result in this case? $\endgroup$ – Mahdi Jun 18 '18 at 19:01
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    $\begingroup$ @Mahdi It has been conjectured by Littlewood that the answer is the same but AFAIK it is still open in general. $\endgroup$ – fedja Jun 18 '18 at 21:12

The link in Fedja's comment shows that $a_n=1$ for all $n$. I will just add the following formula: $$ \|p\|_1 = \begin{cases} \frac{2\pi}{n+1} + 4\sum_{k=1}^{n/2}\tan\left(\frac{\pi k}{n+1}\right)/k & \text{ if $n$ is even } \\ 4\sum_{k=1}^{(n+1)/2}\tan\left(\frac{\pi(k-\tfrac{1}{2})}{n+1}\right)/(k-\tfrac{1}{2}) & \text{ if $n$ is odd } \end{cases} $$ or in more machine-friendly form:

b := proc(n)
 if modp(n,2) = 0 then
  4*add(tan(Pi*k/(n+1))/k,k=1..n/2) + 2*Pi/(n+1);

The proof is somewhat intricate, but all the ingredients are basically elementary. This gives a nice smooth graph for $\|p\|_1$ against $n$, as shown below. I haven't found a good asymptotic formula.

enter image description here


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