Consider a function $f:[0,1]\rightarrow \mathbb{C}$ and points $t_0,t_1,\ldots,t_n\in[0,1]$ \begin{equation*} f(t)=\prod_{k=1}^n\frac{(e^{2\pi i t}-e^{2\pi i t_k})}{(e^{2\pi i t_0}-e^{2\pi i t_k})} \end{equation*} I would like to derive conditions on $t_0,t_1,\ldots,t_n$ under which $|f(t)|< 1$ for all $t\in[0,t_0)\cup(t_0,1]$. (note that $f(t_0)=1$ by construction).

I'm looking for conditions like

1) smallest constant $c$ such that

$min_{k,\ell}|t_k-t_\ell|\ge c/n$

suffices, (with the distance meant to be circular that is |0.9-0.1|=0.2).


2) more sophisticated conditions like:

$D_{n+1}(t_0,t_1,\ldots,t_n)$ needs to be small. The discrepancy of a a finite sequence of real numbers $x_1,x_2,\ldots,x_N\in[0,1]$ is defined as \begin{equation*} D_N(x_1,x_2,\ldots,x_N)=\underset{0\le\alpha<\beta\le 1}{sup}\bigg|\frac{A([\alpha,\beta);N)}{N}-(\beta-\alpha)\bigg|, \end{equation*} with $A([\alpha,\beta);N)$ denoting the number of $x_i$ such that $x_i\in[\alpha,\beta)$ (Based on section 2 of Uniform Distribution of Sequences by Kuipers and Niederreiter).


Well I'm not sure if this'll be much help, but I'm doubtful that conditions like those you've described, involving inequalities, will get you what you want. Heuristically at least, I think the condition that $| f(t) | < 1$ for $t \neq t_{0}$ will only be satisfied for $(t_{0}, \ldots, t_{n})$ in an $n$-dimensional subset of $[0, 1]^{n + 1}$, whereas the set of tuples satisfying an inequality typically would still have dimension $n + 1$.

Here's the argument I have in mind. Given $t_{1}, \ldots, t_{n}$, define $$ g(t) = \prod_{k = 1}^{n} | e^{2 \pi i t} - e^{2 \pi i t_{k}} |. $$ The condition on $f$ is then equivalent to $g(t) < g(t_{0})$ for all $t \neq t_{0}$, i.e. $t_{0}$ is the unique absolute maximum of $g$ on the interval $[0, 1]$. So for a given set of points $t_{1}, \ldots, t_{n}$, at most one value of $t_{0}$ will produce a function $f$ as you want. Thus you should be able to parameterize the $(n + 1)$-tuples $(t_{0}, \ldots, t_{n})$ in $[0, 1]^{n + 1}$ producing such functions $f$ by the $n$-tuples $(t_{1}, \ldots, t_{n})$.

  • $\begingroup$ It seems to me that your argument is more than a heuristic and essentially a complete proof. $\endgroup$ – Lucia Aug 24 '13 at 5:06

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