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I am attempting to show that there does not exist an N past which every open unit interval (k, k+1) -where k is an integer- contains a zero of the following function:

$h(x)=\sum_{n=2}^{[\sqrt(x)]} \frac{\cot(x\pi/n)}{n}+\frac{\cot((x+2)\pi/n)}{n}$

Where the $[\sqrt(x)]$ is the lowest integer of the square root of $x$. Any thoughts? I had figured that I could consider the interval $(i^2, (i+1)^2)$ in which h(x) is described by the function

$h(x)=h_n(x)=\sum_{n=2}^{i} \frac{\cot(x\pi/n)}{n}+\frac{\cot((x+2)\pi/n)}{n}$

Then try to see on what unit intervals (k, k+1) in here contain a zero of $h_n$ then try to show that $((i+1)^2, (i+2)^2)$ also has such an interval, but I am running out of ideas.

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Just to clarify, you are attempting to show that there exist an infinite number of non-overlapping ranges (N, N+1), where N is greater than, say, 2, such that h(x) has no zero in (N, N+1)? –  Gabriel Benamy Nov 18 '10 at 2:43
8  
You are not the first to run out of ideas when trying to prove the (weak version of the) twin prime conjecture :). Should we close? (For those who do not see it immediately themselves, the only way for the function not to have a jump from $-\infty$ to $+\infty$ at $k$ is either to have both $k$ and $k+2$ prime, or to have $k$ prime and $k+2$ a square of a prime; conversely, if the jump is missing at $k$, then $h$ is decreasing on $(k-1,k+1)$.). –  fedja Nov 18 '10 at 3:18
    
I don't think the tag is appropriate in any case. –  David Roberts Nov 18 '10 at 4:17
    
I'm toying with adding the tag twin-prime, as well as open-problem. If no one objects I will. –  David Roberts Nov 18 '10 at 4:19
    
I'd suggest analytic-number-theory instead. –  Gerry Myerson Nov 18 '10 at 7:50
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