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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).

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1 Answer 1

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 $(t_{1}, \ldots, t_{n})$.

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