Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

or

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).

share|improve this question
add comment

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

share|improve this answer
    
It seems to me that your argument is more than a heuristic and essentially a complete proof. –  Lucia Aug 24 '13 at 5:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.