Moments of Fourier transform of a polygon

Question

I am working on the Fourier transform of a polygon. Based on a publication of Shung Wu Lee I have the following formula for the Fourier transform:

$$S(u,v) = \sum_{n=1}^N e^{i \omega\cdot\gamma_n}\left[\frac{\alpha_n\times\alpha_{n-1}}{(\omega\cdot\alpha_n)(\omega\cdot\alpha_{n-1})}\right] $$

where $\omega$, $\alpha_n$ and $\gamma_n$ are all vectors in 2D space with

$$\alpha_n = \frac{\gamma_{n+1}-\gamma_{n}}{|\gamma_{n+1}-\gamma_{n}|}$$

The vectors $\gamma_n$ are the vertex of the polygon and in the formula the indices are meant modulus N so that $\gamma_0 = \gamma_N$ and $\gamma_{N+1}=\gamma_1$.

If we develop the Taylor series expansion of the exponential and we get the zero-th order term we have:

$$\sum_{n=1}^N \frac{\alpha_n\times\alpha_{n-1}}{(\omega\cdot\alpha_n)(\omega\cdot\alpha_{n-1})} $$

which seems to be zero for any $\omega$ but for the moment I haven't figured out a proof.

Can someone point out a possible proof or otherwise give some hint about how to proceed.

I guess also that the first order term of the expansion is also zero and the second order term should be the oriented area of the polygon.

Answer 1

I think I found the way to proof the value given above is zero for any $\omega$. I don't give here the full details but I proceeded in the following way:

• show it is zero for a triangle
• for induction, it is true for a polygon with n vertex, add a new vertex and show it is still true

To show it is zero for a triangle I expanded everything in terms of $\gamma_n$ and then I noted that:

$$(\gamma_3-\gamma_2)\times(\gamma_2 - \gamma_1) =(\gamma_1-\gamma_3)\times(\gamma_3 - \gamma_2)$$

because both sides are equal to twice the area of the oriented triangle. The same is true for the other permutations of the indices.

To show that it remains true when a triangle is added in a polygon is quite easy if expanded in term of the function:

$$t(\alpha_i,\alpha_j)=\frac{\alpha_i\times\alpha_j}{(\omega\cdot\alpha_i)(\omega\cdot\alpha_j)}$$

It is enough to write the expression and note that the term corresponding to the side in common between the triangle and the polygon vanish because the function $t$ is anti-symmetric.

Otherwise, for the terms arising from the first and second terms of the Taylor expansion of the exponential, I don't know but I suspect they can be shown to be equal to zero and the area of the triangle respectively using a similar approach.