5

Consider a 2-dim regular n-gon whose vertices lie on the unit circle. Let $\chi_n$ denote the characteristic function of this polygon and $\widehat{\chi}_n$ its Fourier transform. The special case n = 4 lends itself particularly well to calculation. Namely, without much loss of generality, rotate the square so that its sides are parallel to the coordinate axes. The result is a product interval which leads to an immediate determination of $\widehat{\chi}_4$ as a product of sinc functions.
How does this compare to other values of n ?

Questions: (1) What is the simplest way to evaluate $\widehat{\chi}_3$, the transform of an equilateral triangle?

(2) Have the$\;$ $\widehat{\chi}_n$$\,$ been explicitly worked out for small n ?

(3) Denote by$\,$ $\chi_\infty$ $\,$ the characteristic function of the unit disk and let$\,$ $\widehat{\chi}_\infty$ be its Fourier transform (essentially a Bessel function).$\,$ Are there sharp bounds - using any convenient norm - for the difference $\,$ $\Vert$ $\widehat{\chi}_\infty$ - $\widehat{\chi}_n$$\Vert$ $\,$ ? $\;$ Thanks

user2052
  • 2,427
  • 2
    The Fourier transform of the indicator function of any simple polygon $\Omega$, $\int_\Omega e^{ip\cdot r} dxdy$, is the value of $e^{ip\cdot r}$ at the vertices weighted by some rational function of coordinates of $p$ and $r$. see this answer. – achille hui Sep 15 '17 at 20:53

2 Answers2

1

One way to do this is to set up an integral over a triangle having vertices at the points $(0,1)$, $(-\sqrt{3}/2,-1/2)$, $(\sqrt{3}/2,-1/2)$. The FT may then be written as

$$\int_{-\sqrt{3}/2}^0 dx \, e^{i k_x x} \, \int_{-1/2}^{-\sqrt{3} x+1} dy \, e^{i k_y y} + \int_0^{\sqrt{3}/2} dx \, e^{i k_x x} \, \int_{-1/2}^{\sqrt{3} x+1} dy \, e^{i k_y y}$$

which is messy but straightforward.

Ron Gordon
  • 138,521
  • perhaps it was too optimistic to hope for much simplification (on the other hand, hope never dies) - thanks to both of you for thinking about it – user2052 Sep 28 '17 at 21:23
1

For a $n$-gon $P$ whose boundary in positive orientation is $p_1 \to p_2 \to\ldots \to p_n \to p_1, \ \ p_j = (x_j,y_j)$ and indicator function $\displaystyle\chi(x,y) = 1_{(x,y) \in P}$, then the distributional derivative $\partial_x\chi$ is the distribution $$\partial_x \chi = -\sum_{j=1}^n a_j \delta_{[p_j \to p_{j+1}]}$$ indicating the boundary of $P$, where $a_j = \frac{y_{j+1}-y_j}{\|p_{j+1}-p_j\|}$ is the $\frac{dy}{d\|.\|}$ slope of the edge $[p_j \to p_{j+1}]$ and $\delta_{[p_j \to p_{j+1}]}$ is the distribution defined by $\langle \delta_{[p_j \to p_{j+1}]},\phi \rangle= \int_{p_j}^{p_{j+1}} \phi(x) d\|x\|$.

Thus $$i \omega_x \widehat{\chi}(\omega) = -\sum_{j=1}^n a_j\int_{p_j}^{p_{j+1}} e^{-i (\omega,u)} d \|u\|= -\sum_{i=1}^n a_i \frac{e^{-i (\omega,p_{j+1})}-e^{-i (\omega,p_{j})}}{-i(\omega ,p_{j+1}-p_j)}$$

$$ \widehat{\chi}(\omega) = \iint_P e^{-i (\omega,u)} d^2u= \frac{-1}{ \omega_x}\sum_{i=1}^n \frac{x_{j+1}-x_j}{\|p_{j+1}-p_j\|} \frac{e^{-i (\omega,p_{j+1})}-e^{-i (\omega,p_{j})}}{(\omega ,p_{j+1}-p_j)}$$

(ie. reproving Green's theorem but also making the contribution of the edges clear. And I didn't check correctness on an example)

reuns
  • 77,999