Let $k>0$ be a positive integer. Set $n=4k.$ Let $R(t)$ be a polynomial of degree $n-1$
with coefficients in $\lbrace -1,1 \rbrace$.


Consider the discrete average


$$
D(n,R) = \frac{\sum\_{j=0}^{n-1} \vert R(exp(2\pi i j/n)) \vert}{n}
$$


and the average


$$
A(n,R) = \frac{\int\_{0}^{2\pi} \vert R(exp(it) \vert dt}{2\pi}.
$$


When $k=1$ so that $n=4$ we have that for one half of the possible polynomials $R(t)$
$$
D(n,R) \leq A(n,R).
$$


Question: What happens when $k>1.$



Average compared with discrete average for some $\lbrace -1,1 \rbrace$ polynomials
The claims seems to be false. Numerical integration for $k=2, (n=8)$ gives that 
for $152$ out of the $256=2^8$ polynomials ($59.3\%$) you have $D(n,R) \leq A(n,R)$.


I do not see why having n a multiple of 4 should matter. 
The fact that for half of polynomials the inequality holds at $n=4$ seems like a coincident. You get the following numbers for different values of $n$: 
($m$ is the number of polynomials for which $D \leq A$)


 $n$, $2^n$, $m$, $fraction$
----------------------------



```
1    2    2    1.000
2    4    4    1.000
3    8    6    0.750
4   16    8    0.500
5   32   16    0.500
6   64   36    0.562
7  128   66    0.515
8  256  152    0.593

```

