# Math Help - The zeroes of a tricky continuous function

1. ## The zeroes of a tricky continuous function

Here's a tricky one:
I need to disprove that the function [f(x)=sum( sin(2*Pi*(x+1)/n) / (n*sin(Pi*x/n)*sin((x+2)*Pi/n), n=2..trunc(squareroot(x))
=0] in every open interval (k, k+1) for large enough k and forever after. How?

2. Originally Posted by AlexBot
Here's a tricky one:
I need to disprove that the function [f(x)=sum(sin(2*Pi*(x+1)/n)/(n*sin(Pi*x/n)*sin((x+2)*Pi/n), n=2..trunc(sqrt(x))
=0] in every open interval (k, k+1) for large enough k and forever after. How?
$\displaystyle f(x)=\sum_{n=2}^{\text{trunc}(\sqrt{x})}\sin\left( \frac{2\pi(x+1)}{n}\right)\sin\left(\frac{\pi(x+2) }{n}\right)$?

What is the truncation function? The floor function?

3. no, there's a division sign in there and another term. I don't really know how to type using TEX. Sorry. ANy ideas though?

4. yes, the truncated function is the floor function.