I have a sum ${\phi _1}$ where $w(n)$ is Kaiser window (centered at zero)), and ${T_s}$ is a sampling time.
$${\phi _1} = \sum_{n = - N/2}^{N/2} {w{{(n)}^2}\cos } \left( {4\pi fn{T_s}} \right)$$
If I chose Kaiser (modified Bessel function) window parameter in such a way that ${\phi _1}$=0, is it possible to prove that sums ${\phi _2}$ and ${\phi _3}$ are also equal to zero or very small?
$$ \begin{split} \phi _2 &= \sum_{n = - N/2}^{N/2} {w{{(n)}^2}\cos } \left( {4\pi fn{T_s}} \right){n^2}T_s^2\\ \phi _3 &= \sum_{n = - N/2}^{N/2} {w{{(n)}^2}\cos } \left( {4\pi fn{T_s}} \right){n^4}T_s^4 \end{split} $$
In my numerical tests ${\phi _1}$ is of order $10^{-8}$, and ${\phi _2}$ and ${\phi _3}$ are even smaller, but I don't know how to prove it. I'm note sure is it the property of Kaiser window function, or it can be proved only considering that ${\phi _1}=0$. I also tried to prove this in integral form, but I did not succeed.
Any suggestion is welcome! Thank you in advance!