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Abstract
$$\rm Define\quad F\sb{n}: \IR\sp{n} \to \IR\ by\ F\sb{n}(x\sb1,x\sb2,\cdots,x\sb{n}) = \vert\sum\limits\sbsp{j=1}{n} e\sp{ix\sb{j}}\vert\sp2$$and$$\rm f(n) = {\sup\sb{x\sb1 < x\sb2 < \cdots < x\sb{n}}}\ {\inf\sb{\alpha\in\IR}}\ (F\sb{n}(x\sb1\alpha,x\sb2\alpha,\cdots,x\sb{n}\alpha))\sp{1\over2}.$$To evaluate f(n), it suffices to consider only nonnegative integral values for the $\rm x\sb{j},$ j = $\rm 1,2,\cdots,n,$ with $\rm x\sb1$ = 0. A Newman polynomial is defined to be a complex polynomial P(z) = $\rm\sum\sbsp{j = 1}{n}\ z\sp{x\sb{j}}$ with coefficients all zero or one, and constant term equal to one. Consequently, for integral x$\sb{\rm j}$'s, with x$\sb1$ = 0, we have:$$\rm{\inf\sb{\alpha \in \IR}}\ (F\sb{n}(x\sb1 \alpha,x\sb2 \alpha,\cdots,x\sb{n}\alpha))\sp{1\over2} = {\min\sb{\vert z\vert=1}}\ \vert P(z)\vert.$$
In the process of evaluating f(n), we produce a Newman polynomial with n nonzero terms, with the largest minimum modulus on the unit circle.
We specifically calculate f(1), f(2), f(3), and f(4) and determine some computational results for f(5) and f(6).
