Does anyone know of a **closed form** for the function on $\mathbb{N}$ which returns the greatest power of two which divides a given integer?

To be more precise, any positive integer $n\in\mathbb{N}$ can be uniquely expressed as $n=2^pq$ where $p,q\in\mathbb{N}$ and furthermore $q\equiv1\mod2$. I am looking for a closed form of the resulting function $f:\mathbb{N}\to\mathbb{N}$ which is such that $f:n\mapsto p$, as defined e.g. on Wikipedia.

As a starting point, I constructed a summation which does the job: $$f(n)=\sum_{j=1}^{\rho(n)}\left(\prod_{i=1}^{j}\cos\left[\frac{\pi n}{2^i}\right]\right)^2$$ where $\rho(n)=\lfloor\log_2n\rfloor$. Sadly, this expression is not very useful, and I would prefer a closed form expression. Using Morrie's Law, the product can be converted to a limit as follows: $$f(n)=\lim_{\epsilon\to0}t[\pi(n+\epsilon),\rho(n)]$$ where $$t[x,m]=\sum_{j=1}^{m}\left(\frac{2^{-j}\sin[x]\cos[x]}{\sin[2^{-j}x]}\right)^2$$ However, I cannot find a closed form for this summation...

So in summary, I'd be grateful if anyone could give me an expression for $t(x,m)$ which would make my version of $f$ usable, or if anyone could tell me another such $f$.

Thanks!

EDIT: I followed Gerry's answer and derived the following Fourier series for $f$:

$$f(n)=(1+\cos[\pi n])\left(1-2^{-\rho(n)}+\sum_{j=1}^{\rho(n)}\sum_{k=1}^\infty\frac{\sin[2\pi k n 2^{-j}]-\sin[2\pi k (n-1) 2^{-j}]}{k}\right)$$

I will try to further simplify this...